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Page No 3.101:

Question 1:

Point A and B are 70 km. a part on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours, but if the travel towards each other, the meet in one hour. Find the speed of the two cars.

Answer:

We have to find the speed of car

Let and be two cars starting from pointsand respectively. Let the speed of car be x km/hr and that of carbe y km/hr.

Case I: When two cars move in the same directions:

Suppose two cars meet at point, Then,

Distance travelled by car

Distance travelled by car

It is given that two cars meet in 7 hours.

Therefore, Distance travelled by car X in hours = km

Distance traveled by car y in 7 hours = km

Clearly

Dividing both sides by common factor 7 we get,

Case II : When two cars move in opposite direction

Suppose two cars meet at point. Then,

Distance travelled by car,

Distance travelled by car.

In this case, two cars meet in 1 hour

Therefore Distance travelled by car X in1 hour km

Distance travelled by car Y in 1 hour km

From the above clearly,

...(ii)

By solving equation (i) and (ii), we get

Substituting in equation (ii) we get

Hence, the speed of car starting from point A is

The speed of car starting from point B is.

Page No 3.101:

Question 2:

A sailor goes 8 km downstream in 40 minutes and returns in 1 hours. Determine the speed of the sailor in still water and the speed of the current.

Answer:

Let the speed of the sailor in still water be x km/hr and the speed of the current be y km/hr

Speed upstream

Speed downstream

Now, Time taken to cover 8km down stream =

Time taken to cover 8km upstream=

But, time taken to cover 8 km downstream in 40 minutes or that is

Dividing both sides by common factor 2 we get

Time taken to cover 8km upstream in1hour

...(ii)

By solving these equation and we get

Substitute in equationwe get

Hence, the speed of sailor is

The speed of current is

Page No 3.101:

Question 3:

The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boat in still water.

Answer:

Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr

Speed upstream =km/hr

Speed down stream =km/hr

Now,

Time taken to cover 30 km upstream = hrs

Time taken to cover 44 km down stream = hrs

But total time of journey is 10 hours

Time taken to cover 40 km upstream=

Time taken to cover 55 km down stream =

In this case total time of journey is given to be 13 hours

Therefore, ...(ii)

Putting = u and in equation and we get

Solving these equations by cross multiplication we get

and

Now,

By solving equation and we get ,

Substituting in equation we get ,

Hence, speed of the boat in still water is

Speed of the stream is



Page No 3.102:

Question 4:

A boat goes 24 km upstream and 28 km downstream in 6 hrs. It goes 30 km upstream and 21 km downstream in 612 hrs. Find the speed of the boat in still water and also speed of the stream.

Answer:

We have to find the speed of the boat in still water and speed of the stream

Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr then

Speed upstream

Sped down stream

Now, Time taken to cover km down stream =

Time taken to cover km upstream =

But, total time of journey is 6 hours

Time taken to cover km upstream =

Time taken to cover km down stream =

In this case total time of journey is given to or

...(ii)

By and in equation (i) and (ii) we get

Solving these equations by cross multiplication we get

and

and

Now,

and

By solving equation and we get,

By substituting in equation we get

Hence, the speed of the stream is

The speed of boat is

Page No 3.102:

Question 5:

A man walks a certain distance with certain speed. If he walks 1/2 km an hour faster, he takes 1 hour less. But, if he walks 1 km an hour slower, he takes 3 more hours. Find the distance covered by the man and his original rate of walking.

Answer:

Let the actual speed of the train be x Km/hr and the actual time taken by y hours. Then,

Distance covered=

If the speed is increased by, then time of journey is reduced by 1 hour i.e., when speed is , time of journey is

Distance covered= km

 ...(ii)

When the speed is reduced by, then the time of journey is increased by i.e., when speed is, time of journey is

Distance covered =

Thus we obtain the following equations

By using elimination method, we have

Putting the value in equation (iii) we get

Putting the value of x and y in equations (i) we get

Distance covered =

=

Hence, the distance is,

The speed of walking is .

 

Page No 3.102:

Question 6:

A person rowing  at the rate of 5 km/h in still water , takes thrice as much time in going 40 km upstream as in going 40 km downstream . Find the speed of the stream .

Answer:

Speed of the boat in still water = 5 km/h
Let the speed of stream = x km/h
∴ Speed of boat upstream = (5 − x) km/h
Speed of boat downstream = (5 + x) km/h
Time taken to row 40 km upstream = 405-x
Time taken to row 40 km downstream = 405+x
According to the given condition, 
405-x=3405+x15-x=35+x5+x=15-3x4x=10x=104=2.5 km/h
Therefore, the speed of the stream is 2.5 km/h.

Page No 3.102:

Question 7:

Ramesh travels 760 km to his home partly by train and partly by car. He takes 8 hours if he travels 160 km. by train and the rest  by car. He takes 12 minutes more if the travels 240 km by train and the rest by car. Find the speed of the train and car respectively.

Answer:

Let the speed of the train be x km/hour that of the car be y km/hr, we have the following cases

Case I: When Ramesh travels 760 Km by train and the rest by car

Time taken by Ramesh to travel 160 Km by train =

Time taken by Ramesh to travel (760-160) =600 Km by car =

Total time taken by Ramesh to cover 760Km = +

It is given that total time taken in 8 hours

Case II: When Ramesh travels 240Km by train and the rest by car

Time taken by Ramesh to travel 240 Km by train =

Time taken by Ramesh to travel (760-240) =520Km by car =

In this case total time of the journey is 8 hours 12 minutes

  ...(ii)

Putting and, , the equations and reduces to

Multiplying equation (iii) by 6 and (iv)by 20 the above system of equation becomes

Subtracting equation from we get

Putting in equation, we get

Now

and

Hence, the speed of the train is ,

The speed of the car is .

Page No 3.102:

Question 8:

A man travels 600 km partly by train and partly by car. If the covers 400 km by train and the rest by car, it takes him 6 hours 30 minutes. But, if the travels 200 km by train and the rest by car, he takes half an hour longer. Find the speed of the train and that of the car.

Answer:

Let the speed of the train be x km/hr that of the car be y km/hr, we have the following cases:

Case I: When a man travels 600Km by train and the rest by car

Time taken by a man to travel 400 Km by train =

Time taken by a man to travel (600-400) =200Km by car =hrs

Total time taken by a man to cover 600Km =

It is given that total time taken in 8 hours

Case II: When a man travels 200Km by train and the rest by car

Time taken by a man to travel 200 Km by train =

Time taken by a man to travel (600-200) = 400 Km by car

In this case, total time of the journey in 6 hours 30 minutes + 30 minutes that is 7 hours,

   ...(ii)

Putting and, , the equations and reduces to

Multiplying equation (iii) by 6 the above system of equation becomes

Substituting equation and, we get

Putting in equation, we get

Now

and

Hence, the speed of the train is,

The speed of the car is.

Page No 3.102:

Question 9:

Places A and B are 80 km apart from each other on a highway. A car starts from A and other from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in opposite directions, they meet in 1 hour and 20 minutes. Find the speeds of the cars.

Answer:

Let x and y be two cars starting from points A and B respectively.

Let the speed of the car X be x km/hr and that of the car Y be y km/hr.

Case I: When two cars move in the same directions:

Suppose two cars meet at point Q, then,

Distance travelled by car X = AQ

Distance travelled by car Y = BQ

It is given that two cars meet in 8 hours.

Distance travelled by car X in 8 hours =km

AQ=

Distance travelled by car Y in 8 hours =km

BQ =

Clearly AQ-BQ = AB

Both sides divided by 8, we get

Case II: When two cars move in opposite direction

Suppose two cars meet at point P, then,

Distance travelled by X car X=AP

Distance travelled by Y car Y=BP

In this case, two cars meet in 1 hour 20 minutes, we can write it as 1 hour or

hours that is hours.

Therefore,

Distance travelled by car y in hours = km

Distance travelled by car y in hours = km

    ...(ii)

By solving (i) and (ii) we get,

By substituting in equation (ii), we get

Hence, speed of car X is , speed of car Y is.

Page No 3.102:

Question 10:

A boat goes 12 km upstream and 40 km downstream in 8 hours. I can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.

Answer:

We have to find the speed of the boat in still water and speed of the stream

Let the speed of the boat in still water be km/hr and the speed of the stream be km/hr then

Speed upstream

Sped down stream

Now, Time taken to cover km upstream =

Time taken to cover km down stream =

But, total time of journey is 8 hours

Time taken to cover km upstream =

Time taken to cover km down stream =

In this case total time of journey is given to

          ...(ii)

By and in equation (i) and (ii) we get

Solving these equations by cross multiplication we get

and

and

Now,

and

By solving equation and we get,

By substituting in equation we get

Hence, the speed of boat in still water is ,

The speed of the stream is.

Page No 3.102:

Question 11:

Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Answer:

Let the speed of the train be x km/hour that of the bus be y km/hr, we have the following cases

Case I: When Roohi travels 300 Km by train and the rest by bus

Time taken by Roohi to travel 60 Km by train =

Time taken by Roohi to travel (300-60) =240 Km by bus =

Total time taken by Roohi to cover 300Km= +

It is given that total time taken in 4 hours

Case II: When Roohi travels 100 km by train and the rest by bus

Time taken by Roohi to travel 100 Km by train =

Time taken by Roohi to travel (300-100) =200Km by bus =

In this case total time of the journey is 4 hours 10 minutes

        ...(i)

Putting and, , the equations and reduces to

Subtracting equation (iv) from (iii)we get

Putting in equation (iii), we get

Now

and

Hence, the speed of the train is,

The speed of the bus is.

Page No 3.102:

Question 12:

Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Answer:

Let the speed of rowing in still water be x km/hr and the speed of the current be y km/hr

Speed upstream

Speed downstream

Now,

Time taken to cover km down stream =

Time taken to cover km upstream =

But, time taken to cover km downstream in

Time taken to cover km upstream in 2 hours

     ...(i)

By solving these equation (i) and (ii) we get

Substitute in equation (i)we get

Hence, the speed of rowing in still water is,

The speed of current is .4 km/ hr

Page No 3.102:

Question 13:

A motor boat can travel 30 km upstream and 28 km downstream in 7 hours . It can travel 21 km upstream and return in 5 hours . Find the speed of the boat in still water   and the speed of the upstream .

Answer:

Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.
Speed of boat upstream = x − y
Speed of boat downstream = xy
It is given that, the boat travels 30 km upstream and 28 km downstream in 7 hours. 
30x-y+28x+y=7                     
Also, the boat travels 21 km upstream and return in 5 hours.
21x-y+21x+y=5                     
Let 1x-y=u and 1x+y=v.
So, the equation becomes 
30u+28v=7                .....(i)21u+21v=5                .....(ii) 
Multiplying (i) by 21 and (ii) by 30, we get
630u+588v=147        ...(iii)630u+630v=150        ...(iv)
Solving (iii) and (iv), we get
v=114 and u=16
But,
1x-y=u and 1x+y=v
So,
 1x-y=16 and 1x+y=114x-y=6 and x+y=14
Solving these two equations, we get
x = 10 and y = 4
So, the speed of boat in still water = 10 km/h and speed of stream = 4 km/h.

Page No 3.102:

Question 14:

Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But if he travels 260 km by train and 240 km by taxi he takes 6 minutes longer. Find the speed of the train and that of the taxi.

Answer:

Let the speed of the train be x km/hour that of the taxi be y km/hr, we have the following cases

Case I: When Abdul travels 300 Km by train and the 200 Km by taxi

Time taken by Abdul to travel 300 Km by train =

Time taken by Abdul to travel 200 Km by taxi =

Total time taken by Abdul to cover 500 Km =

It is given that total time taken in 5 hours 30 minutes

Case II: When Abdul travels 260 Km by train and the 240 km by taxi

Time taken by Abdul to travel 260 Km by train =

Time taken by Abdul to travel 240 Km by taxi =

In this case total time of the journey is 5 hours 36 minutes

Putting and, , the equations and reduces to

Multiplying equation by 6 the above system of equation becomes

Subtracting equation from we get

Putting in equation, we get

Now

and

Hence, the speed of the train is ,

The speed of the taxi is .

Page No 3.102:

Question 15:

A train covered a certain distance at a uniform speed. If the train could have been 10 km/hr. faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/hr; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Answer:

Let the actual speed of the train be and the actual time taken by y hours. Then,

Distance covered=

If the speed is increased by, then time of journey is reduced by 2 hours

when speed is , time of journey is

Distance covered=

When the speed is reduced by, then the time of journey is increased by when speed is, time of journey is

Distance covered=

Thus, we obtain the following system of equations:

By using cross multiplication, we have

Putting the values of x and y in equation (i), we obtain

Distance=

Hence, the length of the journey is .

Page No 3.102:

Question 16:

Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of two cars.

Answer:

Let x and y be two cars starting from points A and B respectively.

Let the speed of the car X be x km/hr and that of the car Y be y km/hr.

Case I: When two cars move in the same directions:

Suppose two cars meet at point Q, then,

Distance travelled by car X=AQ

Distance travelled by car Y=BQ

It is given that two cars meet in 5 hours.

Distance travelled by car X in 5 hours =km AQ=

Distance travelled by car Y in 5 hours =km BQ=

Clearly AQ-BQ = AB

Both sides divided by 5, we get

Case II: When two cars move in opposite direction

Suppose two cars meet at point P, then,

Distance travelled by X car X=AP

Distance travelled by Y car Y=BP

In this case, two cars meet in 1 hour

Therefore, 

Distance travelled by car y in1 hours = km

Distance travelled by car y in 1 hours = km

By solving (i) and (ii) we get,

By substituting in equation (ii), we get

Hence, speed of car X is , speed of car Y is.



Page No 3.103:

Question 17:

While covering a distance of 30 km. Ajeet takes 2 hours more than Amit. If Ajeet doubles his speed, he would take 1 hour less than Amit. Find their speeds of walking.

Answer:

Let the speed of Ajeet and Amit be x Km/hr respectively. Then,

Time taken by Ajeet to cover

Time taken by Amit to cover

By the given conditions, we have

If Ajeet doubles his speed, then speed of Ajeet is

Time taken by Ajeet to cover

Time taken by Amit to cover

According to the given condition, we have

 ...(ii)

Putting and, in equation (i) and (ii), we get

Adding equations (iii) and (iv), we get

Putting in equation (iii), we get

Now,

and

Hence, the speed of Ajeet is

The speed of Amit is

Page No 3.103:

Question 18:

A takes 3 hours more than B to walk a distance of 30 km. But, if A doubles his pace (speed) he is ahead of B by 112 hours. Find the speeds of A and B.

Answer:

Let the speed of A and B be x Km/hr and y Km/hr respectively. Then,

Time taken by A to cover ,

And, Time taken by B to cover .

By the given conditions, we have

If A doubles his pace, then speed of A is

Time taken by A to cover ,

Time taken by B to cover .

According to the given condition, we have

Putting and, in equation (i) and (ii), we get

Adding equations (iii) and (iv), we get,

Putting in equation (iii), we get

Now,

and,

Hence, the A’s speed is,

The B’s speed is.



Page No 3.111:

Question 1:

If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units. Find the area of the rectangle.

Answer:

Let the length and breadth of the rectangle be and units respectively

Then, area of rectangle square units

If length is increased and breadth reduced each by units, then the area is reduced by square units

    x + 2 y - 2 = xy - 28xy - 2x + 2y - 4 = xy - 28-2x + 2y - 4 + 28 = 0-2x + 2y + 24 = 02x - 2y - 24 = 0

Therefore,

Then the length is reduced by unit and breadth is increased by units then the area is increased by square units

     x - 1 y + 2 = xy + 33xy + 2x - y - 2 = xy + 332x - y -2 -33 = 02x - y -35 = 0
 

Therefore, 2x - y - 35 = 0    .....ii

Thus we get the following system of linear equation

2x - 2y - 24 = 02x - y - 35 = 0

By using cross multiplication, we have

and

The length of rectangle is units.

The breadth of rectangle is units.

Area of rectangle =lengthbreadth,

square units

Hence, the area of rectangle is square units

Page No 3.111:

Question 2:

The area of a rectangle remains the same if the length is increased by 7 meters and the breadth is decreased by 3 meters. The area remains unaffected if the length is decreased by 7 meters and breadth in increased by 5 meters. Find the dimensions of the rectangle.

Answer:

Let the length and breadth of the rectangle be and units respectively

Then, area of rectangle =square units

If length is increased by meters and breadth is decreased by meters when the area of a rectangle remains the same

Therefore,

If the length is decreased by meters and breadth is increased by meters when the area remains unaffected, then

Thus we get the following system of linear equation

By using cross-multiplication, we have

and

Hence, the length of rectangle ismeters

The breadth of rectangle is meters

Page No 3.111:

Question 3:

In a rectangle, if the length is increased by 3 meters and breadth is decreased by 4 meters, the area of the rectangle is reduced by 67 square meters. If length is reduced by 1 meter and breadth is increased by 4 meters, the area is increased by 89 Sq. meters. Find the dimensions of the rectangle.

Answer:

Let the length and breadth of the rectangle be and units respectively

Then, area of rectangle = square units

If the length is increased by meters and breath is reduced each by square meters the area is reduced by square units

Therefore,

Then the length is reduced by meter and breadth is increased by meter then the area is increased by square units

Therefore,

Thus, we get the following system of linear equation

By using cross multiplication we have

and

Hence, the length of rectangle ismeter,

The breath of rectangle is meter.

Page No 3.111:

Question 4:

The incomes of X and Y are in the ratio of 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves Rs 1250, find their incomes.

Answer:

Let the income of be Rs and the income of be Rs.further let the expenditure of be and the expenditure of be respectively then,

Saving of =

Saving of =

Solving equation and by cross- multiplication, we have

The monthly income of =

The monthly income of

Hence the monthly income of is Rs

The monthly income of is Rs

Page No 3.111:

Question 5:

A and B each has some money. If A gives Rs 30 to B, then B will have twice the money left with A. But, if B gives Rs 10 to A, then A will have thrice as much as is left with B. How much money does each have?

Answer:

Let the money with A be Rs x and the money with B be Rs y.

If A gives Rs 30 to B, Then B will have twice the money left with A, According to the condition we have,

If B gives Rs 10 to A, then A will have thrice as much as is left with B,

By multiplying equation with 2 we get,

By subtracting from we get,

By substituting in equation we get

Hence the money with A be and the money with B be

Page No 3.111:

Question 6:

ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)°, ∠B = (3y − 5)°, ∠C = (−4x)° and ∠D = (7x + 5)°. Find the four angles.

Answer:

We know that the sum of the opposite angles of cyclic quadrilateral is .in the cyclic quadrilateral angles and and angles and pairs of opposite angles

Therefore

and

By substituting and we get

Divide both sides of equation by 4 we get

By substituting and we get

By multiplying equation by 3 we get

By subtracting equation from we get

By substituting in equation we get

The angles of a cyclic quadrilateral are


Hence the angles of quadrilateral are

Page No 3.111:

Question 7:

2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men and 4 boys. How long would it take one man and one boy to do it?

Answer:

A man can alone finish the work in days and one boy alone can finish it in days then

One mans one days work =

One boys one days work=

2men one day work=

7boys one day work=

Since 2 men and 7 boys can finish the work in 4 days

Again 4 men and 4 boys can finish the work in 3 days

Putting and in equation and we get

By using cross multiplication we have

Now,

and

Hence, one man alone can finish the work in and one boy alone can finish the work in .

Page No 3.111:

Question 8:

In a ∆ABC, ∠A = x°, ∠B = (3x − 2)°, ∠C = y°. Also, ∠C − ∠B = 9°. Find the three angles.

Answer:

Let , and

C-B=9° C=9°+BC=9°+3x°-2°C=7°+3x°

Substitute in above equation we get ,

y°=7°+3x°

A+B+C=180°x°+ (3x°-2°)+(7°+3x°) = 180°7x°+5°=180°7x°=180°-5°=175°x=1757=25 A = x = 25B = (3x-2) = 3(25)-2=73C = (7°+3x°) = 7+3(25)=82A=25, B=73, C=82Hence, the answer.

Page No 3.111:

Question 9:

In a cyclic quadrilateral ABCD, ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°, ∠D = (4x − 5)°. Find the four angles.

Answer:

We know that the sum of the opposite angles of cyclic quadrilateral is .in the cyclic quadrilateral, angles and and angles and pairs of opposite angles

Therefore and

Taking

By substituting and we get

Taking

By substituting and we get

By multiplying equation by we get

By subtracting equation from we get

By substituting in equation we get

The angles of a cyclic quadrilateral are

Hence, the angles of cyclic quadrilateral ABCD are .

Page No 3.111:

Question 10:

Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awanded for each correct answer and 2 marks been deducted for each incorrect answer, the Yash would have scored 50 marks. How many question were there in the test?

Answer:

Let take right answer will beand wrong answer will be .

Hence total number of questions will be

If yash scored marks in atleast getting marks for each right answer and losing mark for each wrong answer then

If 4 marks awarded for each right answer and 2 marks deduced for each wrong answer the he scored 50 marks

By multiplying equation by 2 we get

By subtracting from we get

Putting in equation we have

Total number question will be

Hence, the total number of question is .

Page No 3.111:

Question 11:

In a ∆ABC, ∠A = x°, ∠B = 3x° and ∠C = y°. If 3y − 5x = 30, prove that the triangle is right angled.

Answer:

We have to prove that the triangle is right

Given and

Sum of three angles in triangle are

By solving with we get,

Multiplying equation by 3 we get

Subtracting equation from

Substituting in equation we get

Angles and are

A right angled triangle is a triangle in which one side should has a right angle that is in it.

Hence, The triangle ABC is right angled

Page No 3.111:

Question 12:

The car hire charges in a city comprise of a fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is Rs 89 and the journey of 20 km, the charge paid is Rs 145. What will a person have to pay for travelling a distance of 30 km?

Answer:

Let the fixed charges of car be per km and the running charges be km/hr

According to the given condition we have

Putting in equation we get

Therefore, Total charges for travelling distance of km

= Rs

Hence, A person have to pay for travelling a distance of km.



Page No 3.112:

Question 13:

A part of monthly hostel charges in a college are fixed and the remaining depend on the number of days one has taken food in the mess. When a student A takes foods for 20 days, he has to pay Rs 1000 as hostel charges whereas a students B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charge and the cost of food per day.

Answer:

Let the fixed charges of hostel be and the cost of food charges be per day

According to the given condition we have,

Subtracting equation from equation we get

Putting in equation we get

   

Hence, the fixed charges of hostel is .

The cost of food per day is .

Page No 3.112:

Question 14:

Half the perimeter of a garden, whose length is 4 more than its width is 36 m. Find the dimension of the garden.

Answer:

Let perimeter of rectangular garden will be .if half the perimeter of a garden will be

When the length is four more than its width then

Substituting in equation we get

Putting in equation we get

Hence, the dimensions of rectangular garden are and

Page No 3.112:

Question 15:

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Answer:

We know that the sum of supplementary angles will be.

Let the longer supplementary angles will be.

Then,

If larger of supplementary angles exceeds the smaller by degree, According to the given condition. We have,

Substitute in equation, we get,

Put equation, we get,

Hence, the larger supplementary angle is ,

The smaller supplementary angle is.

Page No 3.112:

Question 16:

2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroidery, and that taken by 1 man alone.

Answer:

1 women alone can finish the work in days and 1 man alone can finish it in days .then

One woman one day work=

One man one days work =

2 women’s one days work=

5 man’s one days work =

Since 2 women and 5 men can finish the work in 4 days

3 women and 6 men can finish the work in 3 days

Putting and in equation and we get

By using cross multiplication we have

Now ,

Hence, the time taken by 1 woman alone to finish the embroidery is,

The time taken by 1 man alone to finish the embroidery is .

Page No 3.112:

Question 17:

Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes Rs 50 and Rs 100 she received.

Answer:

Let be the notes of and notes will be

If Meena ask for and notes only, then the equation will be,

Divide both sides by then we get,

If Meena got 25 notes in all then the equation will be,

By subtracting the equation from we get,

Substituting in equation, we get

Therefore and

Hence, Meena has notes of and notes of

Page No 3.112:

Question 18:

There are two examination rooms A and B. If 10 candidates are sent from A to B, the number of students in each room is same. If 20 candidates are sent from B to A, the number of students in A is double the number of students in B. Find the number of students in each room.

Answer:

Let us take the A examination room will be x and the B examination room will be y

If 10 candidates are sent from A to B, the number of students in each room is same. According to the above condition equation will be

If 20 candidates are sent from B to A, the number of students in A is double the number of students in B, then equation will be,

By subtracting the equationfromwe get,

Substituting in equation, we get

Hence candidates are in A examination Room,

candidates are in B examination Room.

Page No 3.112:

Question 19:

A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Ahmedabad costs Rs 216 and one full and one half reserved first class tickets cost Rs 327. What is the basic first class full fare and what is the reservation charge?

Answer:

Let take first class full of fare is Rs and reservation charge is Rs per ticket

Then half of the ticket as on full ticket =

According to the given condition we have

Multiplying equation by 2 we have

Subtracting from we get

Putting in equation we get

Hence, the basic first class full fare is

The reservation charge is .

Page No 3.112:

Question 20:

A wizard having powers of mystic in candations and magical medicines seeing a cock, fight going on, spoke privately to both the owners of cocks. To one he said; if your bird wins, than you give me your stake-money, but if you do not win, I shall give you two third of that'. Going to the other, he promised in the same way to give three fourths. From both of them his gain would be only 12 gold coins. Find the the stake of money each of the cock-owners have.

Answer:

Let the strike money of first cock-owner be and of second cock-owner be respectively. Then we have,

For second cock-owner according to given condition we have,

By subtracting from, we have,

Putting in equation we get,

Hence the stake of money first cock-owner is and of second cock-owner is respectively.

Page No 3.112:

Question 21:

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row there would be 2 rows more. Find the number of student in the class.

Answer:

Let the number of students be and the number of row be .then,

Number of students in each row

Where three students is extra in each row, there are one row less that is when each row has students the number of rows is

Total number of students =no. of rowsno. of students in each row

If three students are less in each row then there are rows more that is when each row has

Therefore, total number of students=Number of rowsNumber of students in each row

Putting in and equation we get

Adding and equation we get

Putting in equation we get

Hence, the number of students in the class is.

Page No 3.112:

Question 22:

One says. "give me hundred, friend! I shall then become twice as rich as you" The other replies, "If you give me ten, I shall be six times as rich as you". Tell me what is the amount of their respective capital?

Answer:

21. Let the money with first person be and the money with the second person be. Then,

If first person gives to second person then the second person will become twice as rich as first person, According to the given condition, we have,

if second person gives to first person then the first person will becomes six times as rich as second person, According to given condition, we have,

Multiplying equation by we get,

By subtracting from, we get

Putting in equation, we get,

Hence, first person’s capital will be ,

Second person’s capital will be .

Page No 3.112:

Question 23:

A shopkeeper sells  a saree at 8% profit and a sweater at 10% discount, thereby getting a sum of ₹1008. If she had sold the saree at 10% profit and the sweater at 8% discount , she would have got ₹ 1028 . Find the cost price of the saree and the list price (price before discount) of the sweater.

Answer:

Let the CP of saree be ₹x and the list price of sweater be ₹y
Case I: When saree is sold at 8% profit and sweater at 10% discount.
SP = CP + Profit
⇒ SP of saree = x 8100x = 108100x
SP of sweater = List price − Discount
⇒ SP of sweater = y-10100y = 90100y
Total sum received by the shopkeeper = ₹1008
⇒ 108100x90100y = 1008
⇒ 108x + 90y = 100800            
⇒ 6x + 5y = 5600            .....(i)
Case II: When saree is sold at 10% profit and sweater at 8% discount.
⇒ SP of saree = x10100x = 110100x
SP of sweater = List price − Discount
⇒ SP of sweater = y-8100y = 92100y
Total sum received by the shopkeeper = ₹1028
⇒ 110100x + 92100y = 1028
⇒ 110x + 92y = 102800              .....(ii)
Multiplying (i) by 110 and (ii) by 6, we get
660x + 550y = 616000                .....(iii)
660x + 552y = 616800                .....(iv)
Subtracting (iii) from (iv), we get
2y = 800
y = 400
Putting y = 400 in (i), we get
6x + 2000 = 5600
⇒ 6x = 5600 − 2000 = 3600
x = 600
Hence, CP of saree = ₹600 and list price of sweater = ₹400.

Page No 3.112:

Question 24:

In a competitive examination , one mark is awarded for each correct answer while 1/2 mark is deducted for everey wrong answer . Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly.

Answer:

Let the number of correct answers be x and the number of wrong answers be y.
Total questions Jayanthi answered = 120
So, x + y = 120          .....(i)
Now, marks obtained for answering correctly = 1 × x = x
Marks deducted for answering incorrectly = 12×y = y2
Total marks obtained = x-y2=90             .....(ii)
Subtracting (ii) from (i), we get
y+y2=303y2=30y=20
Putting y = 20 in (i), we get
x + 20 = 120
x = 100
Thus, Jayanti answered 100 questions correctly.



Page No 3.113:

Question 25:

A shopkeeper gives books on rent for reading . She takes a fixed charge for the first two days, and an additional charge for each day thereafter . Latika paid ₹22 for a book kept for 6 days, while  Anand paid ₹16 for the book kept for four days . Find the fixed charges and charge for each extraday.

Answer:

Let the fixed charge for first two days be ₹x and the additional charge for each day extra be ₹y.
It is given that Latika kept the book for 6 days and paid ₹22.
So,
Fixed charge for the first 2 days + Additional charge for 4 days = ₹22
x + 4y = 22      .....(i)
Anand kept the book for 4 days and paid ₹16. So,
Fixed charge for the first 2 days + Additional charge for 2 days = ₹16
x + 2y = 16      .....(ii)
Subtracting (ii) from (i), we get
2y = 6
y = 3
Putting y = 3 in (i), we get
x + 12 = 22
x = 10
Thus, the fixed charge is ₹10 and the additional charge for each extraday is ₹3.

Page No 3.113:

Question 1:

The value of k for which the system of equations has a unique solution, is

kx y = 2
6x − 2y = 3

(a) =3
(b) ≠3
(c) ≠0
(d) =0

Answer:

The given system of equations are

for unique solution

Here

By cross multiply we get

Hence, the correct choice is.

Page No 3.113:

Question 2:

The value of k for which the system, of equations has infinite number of solutions, is

2x + 3y = 5
4x + ky = 10

(a) 1
(b) 3
(c) 6
(d) 0

Answer:

The given system of equations are

For the equations to have infinite number of solutions,

Here,

Therefore

By cross multiplication of we get,

And

Therefore the value of k is 6

Hence, the correct choice is .

Page No 3.113:

Question 3:

The value of k for which the system of equations x + 2y − 3 = 0 and 5x + ky + 7 = 0 has no solution, is

(a) 10
(b) 6
(c) 3
(d) 1

Answer:

The given system of equations are

For the equations to have no solutions,

If we take

Therefore the value of k is10.

Hence, correct choice is.



Page No 3.114:

Question 4:

The value of k for which the system of equations 3x + 5y = 0 and kx + 10y = 0 has non-zero solution, is

(a) 0
(b) 2
(c) 6
(d) 8

Answer:

The given system of equations are,

Here,

By cross multiply we get

Therefore the value of k is 6,

Hence, the correct choice is.

Page No 3.114:

Question 5:

If the system of equations has infinitely many solutions, then

2x + 3y = 7
(a + b)x + (2ab)y = 21

(a) a = 1, b = 5
(b) a = 5, b = 1
(c) a = −1, b = 5
(d) a = 5, b = −1

Answer:

The given systems of equations are

For the equations to have infinite number of solutions,

Here ,

Let us take

By cross multiplication we get,

Now take

By cross multiplication we get,

Substitute in the above equation

Substitute in equation we get,

Therefore and.

Hence, the correct choice is.

Page No 3.114:

Question 6:

If the system of equations is inconsistent, then k =

3x+y=12k-1x+k-1y=2k+1

(a) 1
(b) 0
(c) −1
(d) 2

Answer:

The given system of equations is inconsistent,

If the system of equations is in consistent, we have

Therefore, the value of k is2.

Hence, the correct choice is .

Page No 3.114:

Question 7:

If ambl, then the system of equations

ax+by=clx+my=n

(a) has a unique solution
(b) has no solution
(c) has infinitely many solutions
(d) may or may not have a solution

Answer:

Given the system of equations has

We know that intersecting lines have unique solution

Here

Therefore intersecting lines, have unique solution

Hence, the correct choice is

Page No 3.114:

Question 8:

If the system of equations has infinitely many solutions, then

2x+3y=72ax+a+by=28

(a) a = 2b
(b) b = 2a
(c) a + 2b = 0
(d) 2a + b = 0

Answer:

Given the system of equations are

For the equations to have infinite number of solutions,

By cross multiplication we have

Divide both sides by 2. we get

Hence, the correct choice is .

Page No 3.114:

Question 9:

The value of k for which the system of equations has no solution is

x+2y=53x+ky+15=0

(a) 6
(b) −6
(c) 3/2
(d) None of these

Answer:

The given system of equation is

If then the equation have no solution.

By cross multiply we get

Hence, the correct choice is.

Page No 3.114:

Question 10:

If 2x − 3y = 7 and (a + b)x − (a + b − 3)y = 4a + b represent coincident lines, then a and b satisfy the equation

(a) a + 5b = 0
(b) 5a + b = 0
(c) a − 5b = 0
(d) 5ab = 0

Answer:

The given system of equations are

For coincident lines , infinite number of solution

a1a2=b1b2=c1c2 2a+b=-3-a+b-3=74a+b 2a+b=3a+b-3=74a+b 2a+b-3=3a+b 2a+2b-6=3a+3b 2a+2b-3a-3b=6 -a-b=6 a+b=-6 ---(i)34a+b=7a+b-312a+3b=7a+7b-215a-4b=-21 ---(ii)multiply equation (i) by 5, we get 5a+5b=-30 ---(iii)subtract (ii) from (iii),5a+5b-5a-4b=-30+215a+5b-5a+4b=-99b=-9b=-1substitute b=-1 in equation (1)a+-1=-6a=-6+1 = -5


Option A.:

a+5b=0-5+5-1 = -5-5 = -10 0

Option B:

5a+b=05-5+-1 = -25-1 = -260

Option.C:

a - b = 0

-5 - (-1) = -4 0


None of the option satisfies the values.

 

Page No 3.114:

Question 11:

If a pair of linear equations in two variables is consistent, then the lines represented by two equations are

(a) intersecting
(b) parallel
(c) always coincident
(d) intersecting or coincident

Answer:

If a pair of linear equations in two variables is consistent, then its solution exists.

∴The lines represented by the equations are either intersecting or coincident.

Hence, correct choice is.

 

Page No 3.114:

Question 12:

The area of the triangle formed by the line xa+yb=1 with the coordinate axes is

(a) ab
(b) 2ab
(c) 12ab
(d) 14ab

Answer:

Given the area of the triangle formed by the line

If in the equation either A and B approaches infinity, The line become parallel to either x axis or y axis respectively,

Therefore

Area of triangle

Hence, the correct choice is .

Page No 3.114:

Question 13:

The area of the triangle formed by the lines y = x, x = 6 and y = 0 is

(a) 36 sq. units
(b) 18 sq. units
(c) 9 sq. units
(d) 72 sq. units

Answer:

Given and

We have plotting points as when

Therefore, area of

Area of triangle is square units

Hence, the correct choice is .

Page No 3.114:

Question 14:

If the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely many solutions, then k =

(a) 1
(b) ½
(c) 3
(d) 6

Answer:

The given system of equations

For the equations to have infinite number of solutions

If we take

And

Therefore, the value of k is 6.

Hence, the correct choice is .



Page No 3.115:

Question 15:

If the system of equations kx − 5y = 2, 6x + 2y = 7 has no solution, then k =

(a) −10
(b) −5
(c) −6
(d) −15

Answer:

The given systems of equations are

If

Here

Hence, the correct choice is .

Page No 3.115:

Question 16:

The area of the triangle formed by the lines x = 3, y = 4 and x = y is

(a) ½ sq. unit
(b) 1 sq. unit
(c) 2 sq. unit
(d) None of these

Answer:

Given and

We have plotting points as when

 

Therefore, area of

Area of triangle is square units

Hence, the correct choice is

Page No 3.115:

Question 17:

The area of the triangle formed by the lines 2x + 3y = 12, xy − 1 = 0 and x = 0 (as shown in Fig. 3.23), is


(a) 7 sq. units
(b) 7.5 sq. units
(c) 6.5 sq. units
(d) 6 sq. units

Answer:

Given and

If We have plotting points as

Therefore, area of

Area of triangle is square units

Hence, the correct choice is

Page No 3.115:

Question 18:

The sum of the digits of a two digit number is 9 . If 27 is added to it , the digits of the number get reversed . th number is 
(a) 25     (b) 72        (c)  63     (d)    36

Answer:

Let the digits at the tens and the ones place be x and y, respectively. So, the two digit number is 10x + y.
Now,
x + y = 9         .....(i)
Also,
10x + y + 27 = 10y + x
⇒ 9− 9y = −27
xy = −3            .....(ii)
Adding (i) and (ii), we get
2x = 6
x = 3
Putting x = 3 in (i), we get
3 + y = 9
y = 6 
Thus, the required number is 10 × 3 + 6 = 36.

Hence, the correct answer is option (d).

Page No 3.115:

Question 19:

If x = a, y = b is the solution of the systems of equations x - y = 2  and x + y = 4 , then the values of a and b are, respectively

(a) 3 and 1           (b) 3 and 5      (c) 5 and 3        (d)  - 1 and -3

Answer:

The given equations are 
x-y=2       .....1x+y=4       .....2
Adding (1) and (2), we get
2x = 6
x = 3
Putting x = 3 in (1), we get
3 + y = 4
y = 1
So, x = a = 3 and y = b = 1.
Thus, the values of a and b are 3 and 1, respectively.

Hence, the correct answer is option (a).

Page No 3.115:

Question 20:

For what value k , do the equations 3x - y + 8 = 0  and 6x - ky + 16 = 0 reperesent coincident lines ?
(a) 12      (b)  -12     (c)  2     (d)  - 2

Answer:

The given system of equations is
3x-y+8=06x-ky+16=0
We know that the lines 
a1x+b1y+c1=0a2x+b2y+c2=0
are coincident iff
a1a2=b1b2=c1c2
36=-1-k=81612=1k=12k=2
Thus, the value of k = 2.

Hence, the correct answer is option (c).



Page No 3.116:

Question 21:

Aruna has only ₹1 and ₹2 coins with her . If the total number of coins that she has is 50 and the amount of money with her is ₹75 , then the number of ₹1 and ₹2 coins are , respectively
(a) 35 and 15   (b) 35 and 20    (c) 15 and 35         (d) 25 and 25 

Answer:

Let the number of ₹1 coins be x and that of ₹2 coins be y.
Now,
Total number of coins = 50
So, x + y = 50            .....(i)
Also,
₹1 × x + ₹2 × y = ₹75
x + 2y = 75            .....(ii)
Subtracting (i) from (ii), we get
y = 25
Putting y = 25 in (i), we get
x + 25 = 50
x = 25
So, the number of ₹1 coins and ₹2 coins are 25 and 25, respectively. 

Hence, the correct answer is option (d).

Disclaimer: The answer given in the book does not match with the one obtained.

Page No 3.116:

Question 1:

The pair of equations y = 0 and y = –7 has  _________ solution.

Answer:

The equation y = 0 is the x-axis and y = –7 is the line parallel to x-axis.

Therefore, both the lines are parallel lines.

Hence, the pair of equations y = 0 and y = –7 has no solution.

Page No 3.116:

Question 2:

The pair of equations x = a and y = b has solution ________.

Answer:

The equation x = a is the line parallel to y-axis and y = b is the line parallel to x-axis.

​Therefore, both the lines intersect each other.

The point of intersection is x = a and y = b i.e. (a, b).

Hence, the pair of equations  x = a and y = b has solution (a, b).

Page No 3.116:

Question 3:

If the pair of equations 3x + 2ky = 2 and 2x + 5y + 1 = 0 has no solution, then k = _________.

Answer:

If the system of equations a1x+b1y+c1=0 and a2x+b2y+c2=0 has no solution, then a1a2=b1b2c1c2.

Since, the system of equations 3x + 2ky = 2 and 2+ 5y + 1 = 0 has no solution
Therefore,
32=2k5-2132=2k5 and 2k5-2115=4k and 2k-10k=154 and k-5

Hence, k = 154.


 

Page No 3.116:

Question 4:

If a pair of linear equations is consistent, then the lines representing them are either _______ or ________.

Answer:

If a pair of linear equations is consistent, then two case arises:

Case 1: When the lines are coincident, they have infinitely many solutions.

Case 2: When the lines are intersecting, they have a unique solution.

Hence, If a pair of linear equations is consistent, then the lines representing them are either coincident or intersecting.

 

Page No 3.116:

Question 5:

There is (are) ________ value(s) of c for which the pair of equations cx – y = 2 and 6x – 2y = 3 have infinitely many solutions.

Answer:

If the system of equations a1x+b1y+c1=0 and a2x+b2y+c2=0 has infinitely many solutions, then a1a2=b1b2=c1c2.

Since, the system of equations cx – y = 2 and 6x – 2y = 3 has infinitely many solutions
Therefore,
c6=-1-2=-2-3c6=12=23But, 1223

Therefore, for no value of the pair of equations cx – y = 2 and 6x – 2y = 3 have infinitely many solutions.

Hence, there is (are)  no  value(s) of c for which the pair of equations cx – y = 2 and 6x – 2y = 3 have infinitely many solutions.

 

Page No 3.116:

Question 6:

If one equation of a pair of dependent linear equations is 5x – 7y + 2 = 0, then the second equation is given by _______.

Answer:

To get the dependent linear equation, multiply the given equation by any non-zero integer.

If we multiply the given equation by any non-zero integer a, we get
5ax – 7ay + 2a = 0

Hence, the second equation is given by 5ax – 7ay + 2a = 0, where a is any non-zero integer.

Page No 3.116:

Question 7:

If a pair of linear equations is consistent with a unique solution, then the lines representing them are _______.

Answer:

If a pair of linear equations is consistent, then two case arises:

Case 1: When the lines are coincident, they have infinitely many solutions.

Case 2: When the lines are intersecting, they have a unique solution.

Hence, if a pair of linear equations is consistent with a unique solution, then the lines representing them are intersecting.

Page No 3.116:

Question 8:

The pair of equations λx + 3y = 7, 2x + 6y = 14 will have infinitely many solutions for λ = ________.

Answer:

If the system of equations a1x+b1y+c1=0 and a2x+b2y+c2=0 has infinitely many solutions, then a1a2=b1b2=c1c2.

Since, the system of equations λx + 3y = 7, 2x + 6y = 14 has infinitely many solutions
Therefore,
λ2=36=-7-14λ2=12=12λ2=12λ=1

Hence, the pair of equations λx + 3y = 7, 2x + 6y = 14 will have infinitely many solutions for λ =  1 .

Page No 3.116:

Question 9:

If the pair of equations 2x + 3y – 5 = 0 and px – 6y – 8 = 0 has a unique solution for all real values of p except _______.

Answer:

If the system of equations a1x+b1y+c1=0 and a2x+b2y+c2=0 has a unique solution, then a1a2b1b2.

Since, the system of equations 2x + 3y – 5 = 0 and px – 6y – 8 = 0 has a unique solution
Therefore,
2p3-62p-124-pp-4

Hence, if the pair of equations 2x + 3y – 5 = 0 and px – 6y – 8 = 0 has a unique solution for all real values of p except –4.

Page No 3.116:

Question 10:

If the lines represented by the equations 3x – y – 5 = 0 and 6x2yp = 0 are parallel, then p is equal to ________.

Answer:

If the system of equations a1x+b1y+c1=0 and a2x+b2y+c2=0 has no solution, i.e., they are parallel, then a1a2=b1b2c1c2.

Since, the system of equations  3x – y – 5 = 0 and 6x – 2y – p = 0 are parallel
Therefore,
36=-1-2-5-p12=12 5p12 5pp10

Hence, if the lines represented by the equations 3x – y – 5 = 0 and 6x – 2y – p = 0 are parallel, then p is equal to all real values except 10.
 

Page No 3.116:

Question 11:

If x = a, y = b is the solution of the pair of equations x – y = 2 and x + y = 4, then a = _______ and b = _______.

Answer:

Since, x = ay = b is the solution of the pair of equations x – y = 2 and x + y = 4
Therefore, x = ay = b satisfies the equations x – y = 2 and x + y = 4.

a-b=2    ...1a+b=4    ...(2)Solving 1 and 2, we geta=3 and b=1

Hence, a = 3 and = 1.

Page No 3.116:

Question 12:

If the pair of equations ax + 2y = 7 and 3x + by = 16 represent parallel lines, then ab = ________.

Answer:

If the system of equations a1x+b1y+c1=0 and a2x+b2y+c2=0 has no solution, i.e., they are parallel, then a1a2=b1b2c1c2.

Since, the system of equations ax + 2y = 7 and 3x + by = 16 represent parallel lines
Therefore,
a3=2b-7-16a3=2b716a3=2bab=6

Hence, If the pair of equations ax + 2y = 7 and 3x + by = 16 represent parallel lines, then ab = 6.

Page No 3.116:

Question 13:

The line 4x + 3y – 12 = 0 cuts the coordinate axes at A and B. The area of ΔOAB is _________.

Answer:

The given equation is 4x + 3y – 12 = 0    ...(i)

Solving equation (i), we get

4x+3y-12=03y=12-4xy=12-4x3

When x = 0, y = 4
When x = 3, y = 0
 

x 0 3
y 4 0

On plotting these points on a graph, we get



We can see that OA = 3 units and OB = 4 units

Area of ΔOAB12×OA×OB
                         = 12×3×4
                         = 6

Hence, the area of ΔOAB is 6 square units.

Page No 3.116:

Question 14:

If 2x+y = 2x–y = 8, then x = _______ and y = _______.

Answer:

Given: 2x+y = 2x–y = 8

Now,
2x+y=82x+y=232x+y=232x+y=32    ...(1)2x-y=82x-y=232x-y=232x-y=32    ...(2)Solving (1) and (2), we getx=32 and y=0


Hence, x = 32 and y = 0.

Page No 3.116:

Question 15:

The system of equations ax + 3y = 1, –12x + ay = 2 has ________ for all real values of a.

Answer:

The given system of equations are:
ax + 3y = 1     ...(i)
–12x + ay = 2     ...(ii)

Now, 
a1a2=a-12b1b2=3ac1c2=-1-2=12If a1a2=b1b2a-12=3aa2=-36Which is not possible for any value of a.Thus, a1a2b1b2 for all real values of a.

Hence, the system of equations ax + 3y = 1, –12x + ay = 2 has a unique solution for all real values of a.



 



Page No 3.117:

Question 1:

Write the value of k for which the system of equations x + y − 4 = 0 and 2x + ky − 3 = 0 has no solution.

Answer:

The given system of equations is

.

For the equations to have no solutions

By cross multiplication we get,

Hence, the value of k is when system equations has no solution.

Page No 3.117:

Question 2:

Write the value of k for which the system of equations has infinitely many solutions.

2x-y=56x+ky=15

Answer:

The given systems of equations are

For the equations to have infinite number of solutions,

By cross Multiplication we get,

Hence the value of k is when equations has infinitely many solutions.

Page No 3.117:

Question 3:

Write the value of k for which the system of equations 3x − 2y = 0 and kx + 5y = 0 has infinitely may solutions.

Answer:

The given equations are

For the equations to have infinite number of solutions,

Therefore,

By cross multiplication we have

Hence, the value of k for the system of equation and is.

Page No 3.117:

Question 4:

Write the value of k for which the system of equations x + ky = 0, 2xy = 0 has unique solution.

Answer:

The given equations are

x+ky=02x-y=0a1=1, a2=2, b1=k, b2=-1a1a2=12,b1b2=k-1
For unique solution

For all real values of k, except the equations have unique solutions.

Page No 3.117:

Question 5:

Write the set of values of a and b for which the following system of equations has infinitely many solutions.

2x+3y=72ax+a+by=28

Answer:

The given equations are

For the equations to have infinite number of solutions,

Therefore

Let us take

By dividing both the sides by 7 we get,

By multiplying equations by 2 we get

Substituting from we get

Subtracting in equation we have

Hence, the value of when system of equations has infinity many solutions.

Page No 3.117:

Question 6:

For what value of k, the following pair of linear equation has infinitely many solutions?

10x+5y-k-5=020x+10y-k=0

Answer:

The given equations are

For the equations to have infinite number of solutions

Let us take

Hence, the value of when the pair of linear equations has infinitely many solutions.

Page No 3.117:

Question 7:

Write the number of solution of the following pair of linear equations:

x + 2y − 8 = 0
2x + 4y = 16

Answer:

The given equations are

Every solution of the second equation is also a solution of the first equation.

Hence, there are, the system equation is consistent.

Page No 3.117:

Question 8:

Write the number of solutions of the following pair of linear equations:

x + 3y − 4 = 0
2x + 6y = 7

Answer:

The given linear pair of equations are

If then

Hence, the number of solutions of the pair of linear equation is.

Therefore, the equations have no solution.



Page No 3.12:

Question 1:

Akhila went to a fair in her village. She wanted to enjoy rides in the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it.) The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game of Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically and graphically.

Answer:

Let no. of ride is   and no. of Hoopla is .He paid Rs 20 for ride and for Hoopla.

The cost of ride is Rs and cost of Hoopla is Rs.then

The number of Hoopla is the half number of ride, then

Hence algebraic equations are and

Now, we draw the graph for algebraic equations.

Page No 3.12:

Question 2:

Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be". Is not this interesting? Represent this situation algebraically and graphically.

Answer:

Let age of Aftab is  years and age of his daughter is years. Years ago his age wastimes older as her daughter was. Then

Three years from now, he will be three times older as his daughter will be, then

 

Hence the algebraic representation are and

Page No 3.12:

Question 3:

The path of a train A is given by the equation 3x + 4y − 12 = 0 and the path of another train B is given by the equation 6x + 8y − 48 = 0. Represent this situation graphically.

Answer:

The given equation are   and.

In order to represent the above pair of linear equation graphically, we need

Two points on the line representing each equation. That is, we find two solutions

of each equation as given below:

We have,

Putting we get

Putting we get

Thus, two solution of equation are

We have

Putting we get

Putting we get

Thus, two solution of equation are

8
6

Now we plot the pointand and draw a line passing through

These two points to get the graph o the line represented by equation

We also plot the points and and draw a line passing through

These two points to get the graph O the line represented by equation

We observe that the line parallel and they do not intersect anywhere.

Page No 3.12:

Question 4:

Gloria is walking along the path joining (−2, 3) and (2, −2), while Suresh is walking along the path joining (0, 5) and (4, 0). Represent this situation graphically.

Answer:

Gloria is walking the path joining and

Suresh is walking the path joining and

 

0 4
5 0

The graphical representations are

Page No 3.12:

Question 5:

On comparing the ratios a1a2, b1b2 and c1c2, and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide :

(i) 5x − 4y + 8 = 0
7x + 6y − 9 = 0

(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0

(iii) 6x − 3y + 10 = 0
2xy + 9 = 0

Answer:

(i) Given equation are: 5x + 4y + 8 = 0 

7x + 6y − 9 = 0

We have And

Thus the pair of linear equation is intersecting.

(ii) Given equation are: 9x + 3y + 12 = 0

18x + 6y + 24 = 0

We have

Thus the pair of linear is coincident lines.

(iii) Given equation are:

We have

Thus the pair of line is parallel lines.

Page No 3.12:

Question 6:

Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :

(i) intersecting lines
(ii) parallel lines
(iii) coincident lines

Answer:

(i) Given the linear equation are:

We know that intersecting condition:

Where

Hence the equation of other line is

(ii) We know that parallel line condition is:

Where

Hence the equation is

(iii) We know that coincident line condition is:

Where

Hence the equation is

Page No 3.12:

Question 7:

The cost of 2kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2kg of grapes is Rs. 300 Represent th situation algebraically and geometrically.

Answer:

Let the cost of 1 kg of apples be Rs x.

And, cost of 1 kg of grapes = Rs y

According to the question, the algebraic representation is

For ,

The solution table is

x 50 60 70
y 60 40 20

 

For 4x + 2y = 300,

The solution table is

x 70 80 75
y 10 –10 0

The graphical representation is as follows.



Page No 3.29:

Question 1:

Solve the following systems of equations graphically:

x + y = 3
2x + 5y = 12

Answer:

The given equations are:

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

0 3
3 0

Draw the graph by plotting the two points from table.

Graph of the equation:

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

0 6
12/5 0

Draw the graph by plotting the two points from the table.

The two lines intersect at point P.

Hence, and is the solution.

Page No 3.29:

Question 2:

Solve the following systems of equations graphically:

x − 2y = 5
2x + 3y = 10

Answer:

The given equations are 

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

0 5
0

Draw the graph by plotting the two points from table.

Graph the equation

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

0 5
10/3 0

Draw the graph by plotting the two points from table.

The two lines intersects at point B

Hence is the solution

Page No 3.29:

Question 3:

Solve the following systems of equations graphically:

3x + y + 1 = 0
2x − 3y + 8 = 0

Answer:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

0 –1/3
–1 0

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we ge

Putting in equationwe get

Use the following table to draw the graph.

0 –4
8/3 0

Draw the graph by plotting the two points from table.

The two lines intersect at points P.

Hence is the solution.

Page No 3.29:

Question 4:

Solve the following systems of equations graphically:

2x + y − 3 = 0
2x − 3y − 7 = 0

Answer:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph. 

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

Putting in equationwe get

Use the following table to draw the graph.

0 7/2
–7/3 0

Draw the graph by plotting the two points from table.

The two lines intersect at points P.

Hence is the solution.

Page No 3.29:

Question 5:

Solve the following systems of equations graphically:
x – y + 1 = 0
3x + 2y – 12 = 0

Answer:

The given equations are:

x – y + 1 = 0    ...(i)
3x + 2y – 12 = 0     ...(ii)

Solving equation (i), we get

x-y+1=0y=x+1

When x = 0, y = 1
When x = 1, y = 2
 

x 0 1
y 1 2

Solving equation (ii), we get

3x+2y-12=02y=12-3xy=12-3x2

When x = 0, y = 6
When x = 4, y = 0
 
x 0 4
y 6 0

On plotting these points on a graph, we get



Hence, point A(2, 3) is the point of intersection.

Page No 3.29:

Question 6:

Solve the following systems of equations graphically:

x − 2y = 6
3x − 6y = 0

Answer:

The given equations are:

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Plotting the two points equation (i) can be drawn.

Graph of the equation….

Putting in equation, we get:

Putting x=2 in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

We see that the two lines are parallel, so they won’t intersect

Hence there is no solution

Page No 3.29:

Question 7:

Solve the following systems of equations graphically:

x + y = 4
2x − 3y = 3

Answer:

The given equations are

Putting in equation we get:

Putting in equation we get

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at points P.

Hence is the solution

Page No 3.29:

Question 8:

Solve the following systems of equations graphically:

2x + 3y = 4
xy + 3 = 0

Answer:

The given equations are:

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table 

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table

The two lines intersect at points P.

Hence, and is the solution.

Page No 3.29:

Question 9:

Solve the following systems of equations graphically:

2x − 3y + 13 = 0
3x − 2y + 12 = 0

Answer:

The given equations are:

Putting in equation we get:

Putting in equationwe get

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equationwe get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at points P.

Hence, and is the solution.

Page No 3.29:

Question 10:

Solve the following systems of equations graphically:

2x + 3y + 5 = 0
3x − 2y − 12 = 0

Answer:

The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equationwe get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersects at points P

Hence, and is the solution.

Page No 3.29:

Question 11:

Show graphically that each one of the following systems of equations has infinitely many solutions:

2x + 3y = 6
4x + 6y = 12

Answer:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table. 

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Thus the graph of the two equations coincide 

Consequently, every solution of one equation is a solution of the other.

Hence the equations have infinitely many solutions.

Page No 3.29:

Question 12:

Show graphically that each one of the following systems of equations has infinitely many solutions:

x − 2y = 5
3x − 6y = 15

Answer:

The given equations are

Putting in equation we get:

Putting in equationswe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table

Graph of the equation….

Putting in equations we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Thus the graph of the two equations coincide 

Consequently, every solution of one equation is a solution of the other.

Hence the equations have infinitely many solutions.

Page No 3.29:

Question 13:

Show graphically that each one of the following systems of equations has infinitely many solutions:

3x + y = 8
6x + 2y = 16

Answer:

The given equations are

Putting in equation we get:

Putting in equationswe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table. 

Graph of the equation….

Putting in equations we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Thus the graph of the two equations coincide 

Consequently, every solution of one equation is a solution of the other.

Hence the equations have infinitely many solutions.

Page No 3.29:

Question 14:

Show graphically that each one of the following systems of equations has infinitely many solutions:

x − 2y + 11 = 0
3x − 6y + 33 = 0

Answer:

The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

Putting in equationwe get

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Thus the graph of the two equations are coincide 

Consequently, every solution of one equation is a solution of the other.

Hence the equations have infinitely many solutions.

Page No 3.29:

Question 15:

Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution) :

3x − 5y = 20
6x − 10y = −40

Answer:

The given equations are

Putting in equation we get:

Putting in equationwe get

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Here we see that the two lines are parallel 

Hence the given system of equations has no solution.

Page No 3.29:

Question 16:

Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution) :

x − 2y = 6
3x − 6y = 0

Answer:

The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

The graph of (i) can be obtained by plotting the two points .

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.


Draw the graph by plotting the two points from table.

Here the two lines are parallel and so there is no point in common

Hence the given system of equations has no solution.

Page No 3.29:

Question 17:

Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution) :

2yx = 9
6y − 3x = 21

Answer:

The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Here two lines are parallel and so don’t have common points

Hence the given system of equations has no solution.

Page No 3.29:

Question 18:

Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution) :

3x − 4y − 1 = 0
2x-83y+5=0

Answer:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

The graph of (i) can be obtained by plotting the two points.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Here, the two lines are parallel.

Hence the given system of equations is inconsistent.

Page No 3.29:

Question 19:

Determine graphically the vertices of the triangle, the equations of whose sides are given below :

(i) 2yx = 8, 5yx = 14 and y − 2x = 1
(ii) y = x, y = 0 and 3x + 3y = 10

Answer:

(i) Draw the 3 lines as given by equations 

By taking x=1 = 1 cm on x−axis

And y =1=1cm on y−axis

Clearly from graph points of intersection three lines are

(−4,2) , (1,3), (2,5) 

(ii) Draw the 3 lines as given by equations 

By taking x=1 = 1 cm on x−axis

And y =1=1cm on y−axis

From graph point of intersection are (0,0) (10/3,0) (5/3,5/3)

Page No 3.29:

Question 20:

Determine, graphically whether the system of equations x − 2y = 2, 4x − 2y = 5 is consistent or in-consistent.

Answer:

The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

0 5/4
–5/2 0

Draw the graph by plotting the two points from table.

It has unique solution.

Hence the system of equations is consistent

Page No 3.29:

Question 21:

Determine, by drawing graphs, whether the following system of linear equations has a unique solution or not :

(i) 2x − 3y = 6, x + y = 1
(ii) 2y = 4x − 6, 2x = y + 3

Answer:

(i) The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

 

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at point .

Hence the equations have unique solution.

(ii) The equations of graphs is

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

The graph of (i) can be obtained by plotting the two points .

Graph of the equation

Putting in equation we get.

Putting in equation we get.

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines are coincident.

Hence the equations have infinitely much solution.

Hence the system is consistent

Page No 3.29:

Question 22:

Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.

(i) 2x − 5y + 4 = 0,
2x + y − 8 = 0

(ii) 3x + 2y = 12,
5x − 2y = 4

(iii) 2x + y − 11 = 0,
xy − 1 = 0

(iv) x + 2y − 7 = 0,
2xy − 4 = 0

(v) 3x + y − 5 = 0,
2xy − 5 = 0

(vi) 2xy − 5 = 0,
xy − 3 = 0

Answer:

(i) The given equations are 

The two points satisfying (i) can be listed in a table as,

2 8
0 4

The two points satisfying (ii) can be listed in a table as,

4 2
0 4

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 3, y = 2.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.

(ii) The given equations are 

The two points satisfying (i) can be listed in a table as,

4 6
0 –3

The two points satisfying (ii) can be listed in a table as,

3 2
5.5 3

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 2, y = 3.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.

(iii) The given equations are 

The two points satisfying (i) can be listed in a table as,

3 1
5 9

The two points satisfying (ii) can be listed in a table as,

1 5
0 4

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 4, y = 3.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.

(iv) The given equations are 

The two points satisfying (i) can be listed in a table as,

5 7
1 0

The two points satisfying (ii) can be listed in a table as,

2 1
0 –2

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 3, y = 2.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.(v) The given equations are 

The two points satisfying (i) can be listed in a table as,

1 3
2 –4

The two points satisfying (ii) can be listed in a table as,

1 4
–3 3

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 2, y = −1.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.

(vi) The given equations are 

The two points satisfying (i) can be listed in a table as,

1 3
–3 1

The two points satisfying (ii) can be listed in a table as,

1 5
–2 2

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 2, y = −1.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.



Page No 3.30:

Question 23:

Solve the following system of linear equation graphically and shade the region between the two lines and x-axis:

(i) 2x + 3y = 12,
xy = 1

(ii) 3x + 2y − 4 = 0,
2x − 3y − 7 = 0

(iii) 3x + 2y − 11 = 0
2x − 3y + 10 = 0

Answer:

The given equations are:

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table. 

The two lines intersect at. The region enclosed by the lines represented by the given equations and x−axis are shown in the above figure

Hence, and is the solution.

(ii) The given equations are:

Putting in equation we get:

Putting in equationwe get:

 

Use the following table to draw the graph.

0
2 0

The graph of (i) can be obtained by plotting the two points.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table. 

The two lines intersect at. The area enclosed by the lines represented by the given equations and the coordinates x−axis and shaded the area in graph.

Hence, and is the solution.

(iii) The given equations are:

Putting in equation we get:

Putting in equationwe get:

 

Use the following table to draw the graph.

Draw the graph by plotting the two points from table

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table. 

The two lines intersect at. The area enclosed by the lines represented by the given equations and the coordinates x−axis and shaded the area in graph.

Hence, and is the solution.

Page No 3.30:

Question 24:

Draw the graphs of the following equations on the same graph paper.

2x + 3y = 12,
xy = 1

Find the coordinates of the vertices of the triangle formed by the two straight lines and the y-axis.

Answer:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

 

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Draw the graph by plotting the two points from table.

The intersection point is P(3, 2)

Three points of the triangle are

Hence the value of and

Page No 3.30:

Question 25:

Draw the graphs of xy + 1 = 0 and 3x + 2y − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and x-axis and shade the triangular area. Calculate the area bounded by these lines and x-axis.

Answer:

The given equations are

Putting in equation we get:

Putting in equation (i) we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at.

Now, Required area = Area of shaded region

Required area = Area of PBD

Required area =

Required area =

Required area =

Hence the area =

Page No 3.30:

Question 26:

Solve graphically the system of linear equations:

4x − 3y + 4 = 0
4x + 3y − 20 = 0

Find the area bounded by these lines and x-axis.

Answer:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

The graph of (i) can be obtained by plotting the points (0, 4/3), (−1, 0).

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at.

Hence is the solution of the given equations.

Now, 

Required area = Area of PBD

Required area =

Required area =

Required area =

Hence, the area =

Page No 3.30:

Question 27:

Solve the following system of linear equations graphically :

3x + y − 11 = 0, x y − 1 = 0.
Shade the region bounded by these lines and y-axis. Also, find the area of the region bounded by the these lines and y-axis.

Answer:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at.

Hence is the solution of the given equations

The area enclosed by the lines represented by the given equations and the y−axis is shaded region in the figure 

Now, Required area = Area of shaded region

Required area = Area of PAC

Required area =

Required area =

Required area =

Hence the required area is sq. unit

Page No 3.30:

Question 28:

Solve graphically each of the following systems of linear equations. Also, find the coordinates of the points where the lines meet the axis of x in each system.

(i) 2x + y = 6
x − 2y = −2

(ii) 2xy = 2
4xy = 8

(iii) x + 2y = 5
2x − 3y = −4

(iv) 2x + 3y = 8
x − 2y = −3

Answer:

(i)

The given equations are 

The two points satisfying (i) can be listed in a table as,

4 0
−2 6

The two points satisfying (ii) can be listed in a table as,

4 6
3 4

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 2, y = 2.

Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.

(ii) The given equations are 

The two points satisfying (i) can be listed in a table as,

0 2
2 2

The two points satisfying (ii) can be listed in a table as,

1 3
–4 4

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 3, y = 4.

Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.

(iii) The given equations are 

The two points satisfying (i) can be listed in a table as,

0 1
2.5 3

The two points satisfying (ii) can be listed in a table as,

4 –5
4 –2

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 1, y = 2.

Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.

Solution is missing

(iv) The given equations are 

The two points satisfying (i) can be listed in a table as,

–2 7
4 –2

The two points satisfying (ii) can be listed in a table as,

–1 3
1 3

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 1, y = 2.

Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.

Page No 3.30:

Question 29:

Draw the graphs of the following equations:

2x − 3y + 6 = 0
2x + 3y − 18 = 0
y − 2 = 0

Find the vertices of the triangle so obtained. Also, find the area of the triangle.

Answer:

The given equations are 

The two points satisfying (i) can be listed in a table as,

3 6
0 6

The two points satisfying (ii) can be listed in a table as,

0 9
6 0

The two points satisfying (iii) can be listed in a table as,

1 8
2 2

Now, graph of equations (i), (ii) and (iii) can be drawn as,

It is seen that the coordinates of the vertices of the obtained triangle are 

Area of ΔABC =

Page No 3.30:

Question 30:

Solve the following system of equations graphically.

2x − 3y + 6 = 0
2x + 3y − 18 = 0

Also, find the area of the region bounded by these two lines and y-axis.

Answer:

The given equations are: 

Putting in equation (i) we get:

Putting in equation (i) we get:

Use the following table to draw the graph

Draw the graph by plotting the two points from table.

Putting in equationwe get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at.

Hence is the solution of the given equations.

The area enclosed by the lines represented by the given equations and the y−axis 

Now, 

Required area = Area of PCA

Required area =

Required area =

Required area =

Hence the required area is

Page No 3.30:

Question 31:

Solve the following system of linear equations graphically.

4x − 5y − 20 = 0
3x + 5y − 15 = 0

Determine the vertices of the triangle formed by the lines representing the above equation and the y-axis.

Answer:

The given equations are: 

Putting in equation we get:

Putting in equation (i) we get:

Use the following table to draw the graph.

x

0

5

y

0

Draw the graph by plotting the two points from table

Putting in equation (ii) we get:

Putting in equation (ii) we get:

Use the following table to draw the graph.

x

0

5

y

3

0

Draw the graph by plotting the two points from table.

The three vertices of the triangle are

Hence the solution of the equation is and



Page No 3.31:

Question 32:

Draw the graphs of the equations 5xy = 5 and 3xy = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and y-axis. Calculate the area of the triangle so formed.

Answer:

The given equations are: 

Putting in equation (i) we get:

Putting in equation (i) we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

–3

Draw the graph by plotting the two points from table.

Hence the vertices of the required triangle are.

Now, 

Required area = Area of PCA

Required area =

Required area =

Hence the required area is

Page No 3.31:

Question 33:

Form the pair of linear equations in the following problems, and find their solution graphically:

(i) 10 students of class X took part in Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together  cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and a pen.

(iii) Champa went to a 'sale' to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, "The number of skirts is two less than twice the number of pants purchased. Also the number of skirts is four less than four times the number of pants purchased." Help her friends to find how many pants and skirts Champa bought.

Answer:

(i) Let the number of girls be x and the number of boys be y.

According to the question, the algebraic representation is

x + y = 10

xy = 4

For x + y = 10,

x = 10 − y 

x 5 4 6
y 5 6 4

For xy = 4,

x = 4 + y

x 5 4 3
y 1 0 1

Hence, the graphic representation is as follows.

 

From the figure, it can be observed that these lines intersect each other at point (7, 3).

Therefore, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.

According to the question, the algebraic representation is

5x + 7y = 50

7x + 5y = 46

For 5x + 7y = 50,

x 3 10 4
y 5 0 10

7x + 5y = 46

x 8 3 2
y 2 5 12

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point (3, 5).

Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.

(iii) Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are :

y = 2x − 2 … (i)

y = 4x − 4 … (ii)

The graphs of the equations (i) and (ii) can be drawn by finding two solutions for each of the equations. They are given in the following table.

x 2 0
y = 2x 2 2 2

Hence, the graphic representation is as follows.

The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.

Page No 3.31:

Question 34:

Solve the following system of equations graphically:
Shade the region between the lines and the y-axis

(i) 3x − 4y = 7
5x + 2y = 3

(ii) 4xy = 4
3x + 2y = 14

Answer:

The given equations are:

Putting in equation (i) we get:

Putting in equation (i) we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation (ii) we get:

Putting in equation (ii) we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at points of y−axis.

Hence, and is the Solution.

(ii) The equations are:

Putting in equation (1) we get:

Putting in equation (1) we get:

Use the following table to draw the graph:

Draw the graph by plotting the two points from table.

Putting in equation (2) we get:

Putting in equation (2) we get:

Use the following table to draw the graph.

x

Draw the graph by plotting the two points from table.

Two lines intersect at points of y−axis.

Hence and is the solution.

Page No 3.31:

Question 35:

Represent the following pair of equations graphically and write the coordinates of points where the lines intersects y-axis.

x + 3y = 6
2x − 3y = 12

Answer:

The given equations are

Putting in equation (i) we get:

Putting in equationwe get:

Use the following table to draw the graph.

x

The graph of (i) can be obtained by plotting the two points.

Putting in equation (ii) we get:

Putting in equation (ii) we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of lines represented by the equations meet y−axis at respectively.

Page No 3.31:

Question 36:

Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is

(i) intersecting lines
(ii) Parallel lines
(iii) coincident lines

Answer:

(i) For intersecting lines,

Equation of another intersecting line to the given line is−

Since, condition for intersecting lines and unique solution is−

(ii) For parallel lines,

Equation of another parallel line to the given line is−

Since, condition for parallel lines and no solution is−

(iii) For co−incident lines,

Equation of another coincident line to the given line is−

Since, condition for coincident lines and infinite solution is−

Page No 3.31:

Question 37:

Determine graphically the coordinates of the vertices of a triangle, the equations of whose sides are :

(i) y = x, y = 2x and y + x = 6
(ii) y = x, 3y = x, x + y = 8

Answer:

(i) The given equations are 

The two points satisfying (i) can be listed in a table as,

0 1
0 1

The two points satisfying (ii) can be listed in a table as,

1 3
2 6

The two points satisfying (iii) can be listed in a table as,

0 6
6 6

Now, graph of equations (i), (ii) and (iii) can be drawn as,

It is seen that the coordinates of the vertices of the obtained triangle are

(ii) The given equations are 

The two points satisfying (i) can be listed in a table as,

0 2
0 2

The two points satisfying (ii) can be listed in a table as,

3 –3
1 –1

The two points satisfying (iii) can be listed in a table as,

3 5
5 3

Now, graph of equations (i), (ii) and (iii) can be drawn as,

It is seen that the coordinates of the obtained triangle are

Page No 3.31:

Question 38:

Graphically , solve the following pair of equations:

2x + y = 62x - y + 2 =0

Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.

Answer:

The given linear equations are:
2x+y=6             .....i2x-y+2=0       .....ii
For (i), we have

x 0 3
y 6 0

For (ii), we have
x 0 −1
y 2 0

Thus, we plot the graph for these two equations and mark the point where these two lines intersect.


From the graph we see that the two lines intersect at point E(1, 4). 
Now, the area of triangle CEB is
A1=12×4×4=8 square unit
The area of triangle AED is
A2=12×4×1=2 square unit 
So, the ratio of the areas of the two triangles will be
A1A2=82=41
Thus, the required ratio is 4 : 1.

Page No 3.31:

Question 39:

Determine, graphically, the vertices of the triangle formed by the lines y = x, 3y = x , x + y = 8.

Answer:

The given lines are:
y=x          .....i3y=x        .....iix+y=8    .....iii
From (i), we have

x 1 2
y 1 2

For (ii), we have
x 0 3
y 0 1

For (iii), we have
x 0 8
y 8 0

Thus, we plot the graph for these three equations and mark the point where these two lines intersect.


From the graph we find that the vertices of the triangle thus formed are H(4, 4), I(6, 2) and D(0, 0).

Page No 3.31:

Question 40:

Draw the graph of the equations  = 3 , x = 5  and  2x − y − 4 = 0 . Also, find the area of the quadrilateral formed by the lines and the x-axis. 

Answer:

The given equations are 
x = 3                              .....(i)
x = 5                              .....(ii)
2x − y − 4 = 0                .....(iii)
For (iii), we have

x 0 2
y −4 0

We plot the lines on the graph as follows:

The vertices of the quadrilateral thus formed are A(5, 0), B(5, 6), C(3, 2) and D(3, 0). 
∴ Area of the quadrilateral DABC
= Area of a trapezium
12ha+b
=12×2×2+6=8 square units
Thus, the area of the quadrilateral formed by the given lines and the x-axis is 8 square units.
 

Page No 3.31:

Question 41:

Draw the graph of the lines x = -2 and y = 3 . Write the vertices of the figure  formed by these lines , the x-axis  and the y-axis . Also , find the area of the figure.

Answer:

The given lines are 
x=-2              .....iy=3                  .....ii
The graph thus obtained will be as follows:

The figure thus obtained is a rectangle ABCD. The vertices of the rectangle are A(−2, 3), B(0, 3), C(0, 0) and D(−2, 0).
Length of the rectangle = AD = BC = 3 units
Breadth of the rectangle = CD = AB =  2 units
∴ Area of rectangle ABCD
=3×2=6 square units



Page No 3.32:

Question 42:

Draw the graphs of the pair of linear equations x - y + 2 = 0 and 4x - y - 4 = 0 . Calculate the area of the triangle formed by the lines so drawn and the x-axis .

Answer:

The given linear equations are 
x − y + 2 = 0                   .....(i)
4x − y − 4 = 0                 .....(ii)
For (i), we have

x 0 −2
y 2 0

For (ii), we have
x 0 1
y −4 0

The graph of the lines represented by the given equations is shown below:

The triangle thus obtained has the vertices A(2, 4), B(−2, 0) and D(1, 0).
∴ Area of triangle ADB
=12×3×4=6 square units
Thus, the area of the triangle formed by the given lines and the x-axis is 6 square units.



Page No 3.44:

Question 1:

Solve the following systems of equations:

11x+15y+23=07x-2y-20=0

Answer:

The given equations are:

Multiply equation by 2 and equation by 15, and add both equations we get

Put the value of in equationwe get

Hence the value of and

Page No 3.44:

Question 2:

Solve the following systems of equations:

3x − 7y + 10 = 0
y − 2x − 3 = 0

Answer:

The given equations are:

Multiply equation by 2 and equation by, and add both equations we get

Put the value of in equationwe get

Hence the value of and

Page No 3.44:

Question 3:

Solve the following systems of equations:

0.4x + 0.3y = 1.7
0.7x − 0.2y = 0.8

Answer:

The given equations are:

Multiply equation by 2 and equation by, and add both equations we get

Put the value of in equationwe get

Hence the value of and

Page No 3.44:

Question 4:

Solve the following systems of equations:

x2+y=0.87x+y2=10

Answer:

The given equations are:

Subtract (ii) from (i) we get

Put the value of in equation we get

Hence the value of and .

Page No 3.44:

Question 5:

Solve the following systems of equations:

7(y + 3) − 2(x + 2) = 14
4(y − 2) + 3(x − 3) = 2

Answer:

The given equations are:

Multiply equation by and equation by and add both equations we get

Put the value of in equationwe get

Hence the value of and

Page No 3.44:

Question 6:

Solve the following systems of equations:

x7+y3=5x2-y9=6

Answer:

The given equations are:

Multiply equation by and add both equations we get

Put the value of in equation we get

Hence the value of and .

Page No 3.44:

Question 7:

Solve the following systems of equations:

x3+y4=115x6-y3=-7

Answer:

The given equations are:

Multiply equation by and equation by and add both equations we get

Put the value of in equation we get

Hence the value of and .

Page No 3.44:

Question 8:

Solve the following systems of equations:

4x+3y=86x-4y=-5

Answer:

The given equations are:

Multiply equation by and equation by and add both equations we get

Put the value of in equation we get

Hence the value of and .

Page No 3.44:

Question 9:

Solve the following systems of equations:

x+y2=4x3+2y=5

Answer:

The given equations are:

Multiply equation by and subtract equations, we get

Put the value of in equation, we get

Hence the value of x and y are and

Page No 3.44:

Question 10:

Solve the following systems of equations:

x+2y=322x+y=32

Answer:

The given equations are:

Multiply equation by and subtract equation (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

Page No 3.44:

Question 11:

Solve the following systems of equations:

2x-3y=03x-8y=0

Answer:

The given equations are:

Multiply equation by and equation by and subtract equation (ii) from (i), we get

Put the value of in equation, we get

Hence the value of and

Page No 3.44:

Question 12:

Solve the following systems of equations:

3x-y+711+2=102y+x+117=10

Answer:

The given equations are:

3x - y + 711 + 2 = 103x - y + 711 = 833x - y - 7 11 = 833x - y - 7 = 8833x - y = 95    ........1 


    2y + x + 117 = 1014y + x + 117 = 1014y + x + 11 = 7014y + x = 59    x + 14y = 59     .........2

Multiply equation (1) by , we get

462x - 14y = 1330     ......3

adding (2) and (3), we get

  x + 14y + 462x - 14y = 59 + 1330463x = 1389x = 3


Substituting the value of x in 2, we get3 + 14y = 5914y = 59 - 314y = 56y = 4

Hence the value of x and y are and

Page No 3.44:

Question 13:

Solve the following systems of equations:

2x-3y=93x+7y=2, y0

Answer:

The given equations are:

Multiply equation by and by and subtract equation (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

Page No 3.44:

Question 14:

Solve the following systems of equations:

0.5x + 0.7y = 0.74
0.3x + 0.5y = 0.5

Answer:

The given equations are:

Multiply equation by and by and subtract equation (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and



Page No 3.45:

Question 15:

Solve the following systems of equations:

17x+16y=312x-13y=5

Answer:

The given equations are:

Multiply equation by and add both equations we get

Put the value of in equation, we get

Hence the value of and

Page No 3.45:

Question 16:

Solve the following systems of equations:

12x+13y=213x+12y=136

Answer:

The given equations are:

Multiply equation by and by and subtract equation (ii) from (i) we get

Put the value of in equation, we get
12×12+13y=213y=2-113y=1y=13

Hence the value of and y=13

Page No 3.45:

Question 17:

Solve the following systems of equations:

15u+2ν=171u+1ν=365

Answer:

The given equations are:

Multiply equation by and subtract (ii) from (i), we get

Put the value of in equation, we get

Hence the value of and .

Page No 3.45:

Question 18:

Solve the following systems of equations:

3x-1y=-92x+3y=5

Answer:

The given equations are:

Multiply equation by and add both equations, we get

Put the value of in equation, we get

Hence the value of and

Page No 3.45:

Question 19:

Solve the following systems of equations:

2x+5y=160x+40y=19, x0, y0

Answer:

The given equations are:

Multiply equation by and subtract (ii) from equation (i), we get

Put the value of in equation, we get

Hence the value of and .

Page No 3.45:

Question 20:

Solve the following systems of equations:

15x+16y=1213x-37y=8, x0, y0

Answer:

The given equations are:

Multiply equation by and equation by, add both equations, we get

Put the value of in equation, we get

Hence the value of and

Page No 3.45:

Question 21:

Solve the following systems of equations:

4x+3y=143x-4y=23

Answer:

The given equations are:

Multiply equation by and equation by, add both equations, we get

Put the value of in equation, we get

Hence the value of and .

Page No 3.45:

Question 22:

Solve the following systems of equations:

4x+5y=73x+4y=5

Answer:

The given equations are:

Multiply equation by and equation by and subtract (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

Page No 3.45:

Question 23:

Solve the following systems of equations:

2x+3y=135x-4y=-2

Answer:

The given equations are:

Multiply equation by and equation by 3 and add both equations we get

Put the value of in equation, we get

Hence the value of and .

Page No 3.45:

Question 24:

Solve the following systems of equations:

2x+3y=24x-9y=-1

Answer:

The given equations are:

Multiply equation by and add both equations we get

Put the value of in equation, we get

Hence the value of and

Page No 3.45:

Question 25:

Solve the following systems of equations:

x+yxy=2x-yxy=6

Answer:

The given equations are:

Adding both equations, we get

Put the value of in equation, we get

Hence the value of and

Page No 3.45:

Question 26:

Solve the following systems of equations:

2x+3y=9xy4x+9y=21xy, x0, y0

Answer:

The given equations are:

Multiply equation by and subtract (ii) from (i), we get

Put the value of in equation, we get

Hence the value of and

Page No 3.45:

Question 27:

Solve the following systems of equations:

6x+y=7x-y+312x+y=13x-y'

where x + y ≠ 0 and xy ≠ 0

Answer:

The given equations are:

Let and then equations are

Multiply equation by and subtract (ii) from (i), we get

Put the value of in equation, we get

Then 

Add both equations, we get

Put the value of in second equation, we get

Hence the value of and.

Page No 3.45:

Question 28:

Solve the following systems of equations:

xyx+y=65xyy-x=6,

where x + y ≠ 0, yx ≠ 0

Answer:

The given equations are:

Add both equations, we get

Put the value of in equation, we get

Hence the value of and

Page No 3.45:

Question 29:

Solve the following systems of equations:

22x+y+15x-y=555x+y+45x-y=14

Answer:

The given equations are:

Let and then equations are

Multiply equation by and subtracting (ii) from (i), we get

Put the value of in equation, we get

Then 

Add both equations, we get

Put the value of in second equation, we get

Hence the value of and .

Page No 3.45:

Question 30:

Solve the following systems of equations:

5x+y-2x-y=-115x+y+7x-y=10

Answer:

The given equations are:

Let and then equations are

Multiply equation by and equation by and add both equations, we get

Put the value of in equation, we get

Then

Add both equations, we get

Put the value of in first equation, we get

Hence the value of and .

Page No 3.45:

Question 31:

Solve the following systems of equations:

3x+y+2x-y=29x+y-4x-y=1

Answer:

The given equations are:

Let and then equations are

Multiply equation by and add both equations, we get

Put the value of in equation, we get

Then

Add both equations, we get

Put the value of in first equation, we get

Hence the value of and .

Page No 3.45:

Question 32:

Solve the following systems of equations:

12x+2y+533x-2y=-3254x+2y-353x-2y=6160

Answer:

The given equations are:

Let and then equations are

Multiply equation by and equation by add both equations, we get

Put the value of in equation, we get

Then 

Add both equations, we get

Put the value of in equation we get

Hence the value of and .



Page No 3.46:

Question 33:

Solve the following systems of equations:

5x+1-2y-1=1210x+1+2y-1=52

where x ≠ −1 and y ≠ 1

Answer:

The given equations are:

Let and then equations are

Add both equations, we get

Put the value of in equation, we get

Then 

Hence the value of and .

Page No 3.46:

Question 34:

Solve the following systems of equations:

x + y = 5xy
3x + 2y = 13xy,

x ≠ 0, y ≠ 0

Answer:

The given equations are:

Multiply equation by and subtract (ii) from (i), we get

Put the value of in equation, we get

Hence the value of and

Page No 3.46:

Question 35:

Solve the following systems of equations:

x+y=2xyx-yxy=6

x ≠ 0, y ≠ 0

Answer:

The given equations are:

Add both equations we get

Put the value of in equation, we get

Hence the value of and .

Page No 3.46:

Question 36:

Solve the following systems of equations:

2(3u − ν) = 5uν
2(u + 3ν) = 5uν

Answer:

The given equations are:

Multiply equation by and add both equations, we get

Put the value of in equation, we get

Hence the value of and .

Page No 3.46:

Question 37:

Solve the following systems of equations:

23x+2y+33x-2y=17553x+2y+13x-2y=2

Answer:

The given equations are:

Let and then equations are

Multiply equation by and subtract (ii) from (i), we get

Put the value of in equation, we get

Then 

Add both equations, we get

Put the value of in equation we get

Hence the value of and .

Page No 3.46:

Question 38:

Solve the following systems of equations:

44x+y+30x-y=1055x+y+40x-y=13

Answer:

The given equations are:

Let and then equations are

Multiply equation by and equation by add both equations, we get

Put the value of in equation, we get

Then 

Add both equations, we get

Put the value of in equation we get

Hence the value of and

Page No 3.46:

Question 39:

Solve the following systems of equations:

5x-1+1y-2=26x-1-3y-2=1

Answer:

The given equations are:

Let and then equations are 

Multiply equation by and add both equations, we get

Put the value of in equation, we get 

Then 

Hence the value of and .

Page No 3.46:

Question 40:

Solve the following systems of equations:

10x+y+2x-y=415x+y-9x-y=-2

Answer:

The given equations are:

Let and then equations are

Multiply equation by and equation by and add both equations, we get

Put the value of in equation, we get

Then

Add both equations, we get

Put the value of in equation we get

Hence the value of and .

Page No 3.46:

Question 41:

Solve the following systems of equations:

13x+y+13x-y=34123x+y-123x-y=-18

Answer:

The given equations are:

Let and then equations are

Multiply equation by and add both equations, we get

Put the value of in equation, we get

Then 

Add both equations, we get

Put the value of in equation we get

Hence the value of and

Page No 3.46:

Question 42:

Solve the following systems of equations:

7x-2yxy=58x+7yxy=15

Answer:

The given equations are:

Multiply equation by and equation by, add both equations we get

Put the value of in equation, we get

Hence the value of and

Page No 3.46:

Question 43:

Solve the following systems of equations:

152x − 378y = −74
−378x + 152y = −604

Answer:

The given equations are:

Multiply equation by and equation by and add both equations we get

Put the value of in equation, we get

Hence the value of and

Page No 3.46:

Question 44:

Solve the following systems of equations:

99x + 101y = 499
101x + 99y = 501

Answer:

The given equations are:

Multiply equation by and equation by, and subtract (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

Page No 3.46:

Question 45:

Solve the following systems of equations:

23x − 29y = 98
29x − 23y = 110

Answer:

The given equations are:

Multiply equation by and equation by and subtract (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

Page No 3.46:

Question 46:

Solve the following systems of equations:

xy + z = 4
x − 2y − 2z = 9
2x + y + 3z = 1

Answer:

The given equations are:

First of all we find the value of

Put the value of in equation, we get

Put the value of and in equation in we get

Multiply equation by and add equations and, we get

Put the value of in equation, we get

Put the value of and in equation we get

Hence the value of, and.

Page No 3.46:

Question 47:

Solve the following systems of equations:

xy + z = 4
x + y + z = 2
2x + y − 3z = 0

Answer:

The given equations are:

First of all we find the value of

Put the value of in equation, we get

Put the value of and in equation in we get

Put the value of and in equation, we get

Hence the value of, and.

Page No 3.46:

Question 48:

21x + 47y = 110
47x + 21y = 162

Answer:

21x + 47y = 110            .....(i)
47x + 21y = 162            .....(ii)
Adding(i) and (ii), we get
68x + 68y = 272
x+y=4                    .....(iii)
Subtracting (i) from (ii), we get
26x-26y=52x-y=2            .....(iv)               
Adding (iii) and (iv), we get
2x=6x=3
Putting x = 3 in (iv), we get
3 − y = 2
y = 1




 

Page No 3.46:

Question 49:

If (+ 1) is a factor of 2x3+ax2+2bx+1, then find the values of a and b given that 2a − 3b = 4.  

Answer:

Since (x + 1) is a factor of 2x3+ax2+2bx+1, so
2-13+a-12+2b-1+1=0-2+a-2b+1=0a-2b-1=0a-2b=1                         .....i
Also, we are given
2a − 3b = 4          .....(ii)
From (i) and (ii) we get
a=1+2b      .....iii   Substituting the value of a in ii, we get  21+2b-3b=42+4b-3b=4b=2
Putting b = 2 in (iii), we get
a = 1 + 2 × 2 = 5
Thus, the value of a = 5 and b = 2.

Page No 3.46:

Question 50:

Find the solution of the pair of equations x10+y5-1=0     and  x8+y6=15    . Hence , find  λ , if y = λx + 5.

Answer:

The given equations are 
x10+y5-1=0x8+y6=15
x10+y5=1x+2y=10          .....ix8+y6=153x+4y=360          .....ii
Multiplying (i) by 2, we get
2x+4y=20         .....iii
Subtracting (ii) from (iii), we get
x = 340
Putting x = 340 in (i), we get
340 + 2y = 10
⇒ 2y = 10 − 340 = −330
y = −165
Now, in order to find the value of λ, we simply put the value of x and y in the equation y=λx+5.
-165=λ340+5λ=-170340λ=-12
Thus, the value of λ=-12.



Page No 3.47:

Question 51:

Find the values of x and y in the following rectangle.

Answer:


ABCD is the given rectangle. So, AB = CD and AD = BC.
Thus,
x+3y=13     .....i3x+y=7       .....ii
Adding (i) and (ii), we get
4x + 4y = 20    
x+y=5    .....iii
Subtracting (i) from (ii), we get
2− 2y = −6
x-y=-3    .....iv
Adding (iii) and (iv), we get
2x = 2
x=1
Putting x = 1 in (iii), we get
1 + y = 5
y = 4
Thus, x = 1 and y = 4.

Page No 3.47:

Question 52:

Write an equation of a line passing through the point representing solution of the pair of linear equations x + y =2 and 2x - y = 1 . How many such lines can we find ?

Answer:

The given equations are
x+y=2         .....i2x-y=1       .....ii
Adding (i) and (ii), we get
3x = 3
x = 1
Putting x = 1 in (i), we get
1 + y = 2
y = 1
Thus, the solution of the given equations is (1, 1). 
We know that, infinitely many straight lines pass through a single point.
So, the equation of one such line can be 3x + 2y = 5 or 2x + 3y = 5.

Page No 3.47:

Question 53:

Write a pair of linear equations which has the unique solution x = -1 , y = 3 .How many such  pairs can you write ?

Answer:

The unique solution is given as x = −1 and y = 3.
The one pair of linear equations having x = −1 and y = 3 as unique solution can be
12x + 5y = 3
2x + y = 1
Similarly, infinitely many pairs of linear equations are possible. 



Page No 3.57:

Question 1:

Solve each of the following systems of equations by the method of cross-multiplication :

x + 2y + 1 = 0
2x − 3y − 12 = 0

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

By cross multiplication method we get

and

Hence we get the value of and

Page No 3.57:

Question 2:

Solve each of the following systems of equations by the method of cross-multiplication :

3x + 2y + 25 = 0
2x + y + 10 = 0

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

By cross multiplication method we get

Also

Hence we get the value of and

Page No 3.57:

Question 3:

Solve each of the following systems of equations by the method of cross-multiplication :

2x + y = 35
3x + 4y = 65

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Also

Hence we get the value of and

Page No 3.57:

Question 4:

Solve each of the following systems of equations by the method of cross-multiplication :

2xy = 6
xy = 2

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Hence we get the value of and

Page No 3.57:

Question 5:

Solve each of the following systems of equations by the method of cross-multiplication :

x+yxy=2, x-yxy=6

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Let

By cross multiplication method we get

So

We know that

Hence we get the value of and

Page No 3.57:

Question 6:

Solve each of the following systems of equations by the method of cross-multiplication :

ax + by = ab
bxay = a + b

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

T

Thereforeand

Hence we get the value of and

Page No 3.57:

Question 7:

Solve each of the following systems of equations by the method of cross-multiplication :

x + ay = b
axby = c

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

, and  

Hence we get the value of and

Page No 3.57:

Question 8:

Solve each of the following systems of equations by the method of cross-multiplication :

ax + by = a2
bx + ay = b2

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And

Hence we get the value of and

Page No 3.57:

Question 9:

Solve each of the following systems of equations by the method of cross-multiplication :

5x+y-2x-y=-115x+y+7x-y10,

where x ≠ 0 and y ≠ 0

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Rewriting the equation again

By cross multiplication method we get

And

We know that 

Adding equation (3) and (4)

Substituting value of x in equation (3) we get

Hence we get the value ofand

Page No 3.57:

Question 10:

Solve each of the following systems of equations by the method of cross-multiplication :

2x+3y=135x-4y=-2

where x ≠ 0 and y ≠ 0

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Rewriting the equation again

By cross multiplication method we get from eq. (1) and eq. (2)

And

We know that

Hence we get the value of and

Page No 3.57:

Question 11:

Solve each of the following systems of equations by the method of cross-multiplication :

57x+y+6x-y=538x+y+21x-y=9

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Now rewriting the given equation as

By cross multiplication method we get

Consider the following for u

Consider the following for v

We know that

Now adding eq. (3) and (4) we get

And after substituting the value of x in eq. (4) we get

Hence we get the value of and

Page No 3.57:

Question 12:

Solve each of the following systems of equations by the method of cross-multiplication :

xa+yb=2ax-by=a2-b2

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

So for x we have

And 

 

Hence we get the value of and

Page No 3.57:

Question 13:

Solve each of the following systems of equations by the method of cross-multiplication :

xa+yb=a+bxa2+yb2=2

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And 

Hence we get the value of and



Page No 3.58:

Question 14:

Solve each of the following systems of equations by the method of cross-multiplication :

xa=ybax+by=a2+b2

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And 

Hence we get the value ofand

Page No 3.58:

Question 15:

Solve each of the following systems of equations by the method of cross-multiplication :

2ax + 3by = a + 2b
3ax + 2by = 2a + b

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Now consider

And

Hence we get the value of and

Page No 3.58:

Question 16:

Solve each of the following systems of equations by the method of cross-multiplication :

5ax + 6by = 28
3ax + 4by = 18

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following to calculate x

And 

Hence we get the value of and

Page No 3.58:

Question 17:

Solve each of the following systems of equations by the method of cross-multiplication :

(a + 2b)x + (2ab)y = 2
(a − 2b)x + (2a + b)y = 3

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And

Hence we get the value of and

Page No 3.58:

Question 18:

Solve each of the following systems of equations by the method of cross-multiplication :

xa-b+aba-b=ya+b-aba+bx+y=2a2

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following for x

And 

Hence we get the value of and

Page No 3.58:

Question 19:

Solve each of the following systems of equations by the method of cross-multiplication :

bx+cy=a+bax1a-b-1a+b+cy1b-a-1b+a=2aa+b

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following for x

Now for y

Hence we get the value of and

Page No 3.58:

Question 20:

Solve each of the following systems of equations by the method of cross-multiplication :

a-bx+a+by=2a2-2b2a+b x+y=4ab

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following for x

Now consider the following for y 

Hence we get the value of and

Page No 3.58:

Question 21:

Solve each of the following systems of equations by the method of cross-multiplication :

a2x+b2y=c2b2x+a2y=d2

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following for x

Now consider the following for y

Hence we get the value of and

Page No 3.58:

Question 22:

Solve each of the following systems of equations by the method of cross-multiplication :

ax+by=a+b23x+5y=4

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And

Hence we get the value of and

Page No 3.58:

Question 23:

Solve each of the following systems of equations by the method of cross-multiplication :

2ax-by+a+4b=02bx+ay+b-4a=0

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

After rewriting equations

By cross multiplication method we get

For y consider the following

Hence we get the value of and

Page No 3.58:

Question 24:

Solve each of the following systems of equations by the method of cross-multiplication :

6ax+by=3a+2b6bx-ay=3b-2a

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation, after rewriting equations

By cross multiplication method we get 

Consider the following for y 

Hence we get the value of and

Page No 3.58:

Question 25:

Solve each of the following systems of equations by the method of cross-multiplication :

a2x-b2y=0a2bx+b2ay=a+b, x, y0

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Let

Rewriting equations 

Now, by cross multiplication method we get

For u consider the following

For y consider

We know that

Now 

Hence we get the value of and

Page No 3.58:

Question 26:

Solve each of the following systems of equations by the method of cross-multiplication :

mx-ny=m2+n2x+y=2m

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Now for y

Hence we get the value of and

Page No 3.58:

Question 27:

Solve each of the following systems of equations by the method of cross-multiplication :

axb-bya=a+bax-by=2ab

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

For y

Hence we get the value of and

Page No 3.58:

Question 28:

Solve each of the following systems of equations by the method of cross-multiplication :

bax+aby=a2+b2x+y=2ab

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get 

Hence we get the value of



Page No 3.73:

Question 1:

In each of the following systems of equations determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

x − 3y = 3
3x − 9y = 2

Answer:

GIVEN: 

To find: To determine whether the system has a unique solution, no solution or infinitely many solutions 

We know that the system of equations

For unique solution

For no solution

For infinitely many solution 

Here,

Since which means hence the system of equation has no solution.

Hence the system of equation has no solution

Page No 3.73:

Question 2:

In each of the following systems of equations determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it

2x + y = 5
4x + 2y = 10

Answer:

GIVEN: 

To find: To determine whether the system has a unique solution, no solution or infinitely many solutions 

We know that the system of equations

For unique solution

For no solution

For infinitely many solution 

Here,

Since which means hence the system of equation has infinitely many solution.

Hence the system of equation has infinitely many solutions

Page No 3.73:

Question 3:

In each of the following systems of equations determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it

3x − 5y = 20
6x − 10y = 40

Answer:

GIVEN: 

To find: To determine whether the system has a unique solution, no solution or infinitely many solutions 

We know that the system of equations

For unique solution

For no solution

For infinitely many solution 

Here,

Since which means hence the system of equation has infinitely many solution.

Hence the system of equation has infinitely many solutions

Page No 3.73:

Question 4:

In each of the following systems of equations determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it

x − 2y = 8
5x − 10y = 10

Answer:

GIVEN: 

To find: To determine whether the system has a unique solution, no solution or infinitely many solutions 

We know that the system of equations

For unique solution

For no solution

For infinitely many solution 

Here,

Since which means hence the system of equation has no solution.

Hence the system of equation has no solution

Page No 3.73:

Question 5:

 Find the value of k for which the following system of equations has a unique solution:

kx + 2y = 5
3x + y = 1

Answer:

GIVEN: 

To find: To determine to value of k for which the system has a unique solution.

We know that the system of equations

For unique solution

Here,

Hence for the system of equation has unique solution.

Page No 3.73:

Question 6:

Find the value of k for which the following system of equations has a unique solution:

4x+ky+8=02x+2y+2=0

Answer:

GIVEN: 

To find: To determine to value of k for which the system has a unique solution.

We know that the system of equations

For unique solution

Here,

Hence for the system of equation has unique solution

Page No 3.73:

Question 7:

Find the value of k for which the following system of equations has a unique solution:

4x-5y=k2x-3y=12

Answer:

GIVEN: 

To find: To determine to value of k for which the system has a unique solution.

We know that the system of equations

For unique solution

Here,

Hence already for the system of equation to have unique solution but the value of k should be a real number

Hence for the system of equation has unique solution.

Page No 3.73:

Question 8:

Find the value of k for which the following system of equations has a unique solution:

x+2y=35x+ky+7=0

Answer:

GIVEN: 

To find: To determine to value of k for which the system has a unique solution.

We know that the system of equations

For unique solution

Here,

Hence for the system of equation has unique solution

Page No 3.73:

Question 9:

Find the value of k for which each of the following system of equations have infinitely many solutions :

2x + 3y − 5 = 0
6x + ky − 15 = 0

Answer:

GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here,

Hence for the system of equation have infinitely many solutions

Page No 3.73:

Question 10:

Find the value of k for which each of the following system of equations have infinitely many solutions :

4x+5y=3kx+15y=9

Answer:

GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here,

Hence for the system of equation have infinitely many solutions.

Page No 3.73:

Question 11:

Find the value of k for which each of the following system of equations have infinitely many solutions :

kx-2y+6=04x-3y+9=0

Answer:

GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here,

Hence for the system of equation have infinitely many solutions.

Page No 3.73:

Question 12:

Find the value of k for which each of the following system of equations have infinitely many solutions :

8x+5y=9kx+10y=18

Answer:

GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here,

Hence for the system of equation have infinitely many solutions

Page No 3.73:

Question 13:

Find the value of k for which each of the following system of equations have infinitely many solutions :

2x-3y=7k+2x-2k+1y=32k-1

Answer:

GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here

Consider the following

Now consider the following

Hence for the system of equation have infinitely many solutions.

Page No 3.73:

Question 14:

Find the value of k for which each of the following system of equations have infinitely many solutions :

2x+3y=2k+2x+2k+1y=2k-1

Answer:

GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here,

Consider the following for k

Now consider the following

Hence for the system of equation have infinitely many solutions.

Page No 3.73:

Question 15:

Find the value of k for which each of the following system of equations have infinitely many solutions :

x+k+1y=4k+1x+9y=5k+2

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here,

Hence for the system of equation have infinitely many solutions.

Page No 3.73:

Question 16:

Find the value of k for which each of the following system of equations have infinitely many solutions :

kx+3y=2k+12k+1x+9y=7k+1

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution

Here,

Consider the following relation to find k

Now consider the following

Hence for the system of equation have infinitely many solutions

Page No 3.73:

Question 17:

Find the value of k for which each of the following system of equations have infinitely many solutions :

2x+k-2y=k6x+2k-1y=2k+5

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here,

Consider the following relation to find k

Now again consider the following

Hence for the system of equation have infinitely many solutions

Page No 3.73:

Question 18:

Find the value of k for which each of the following system of equations have infinitely many solutions :

2x+3y=7k+1x+2k-1y=4k+1

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here

Now again consider the following to find k

Hence for the system of equation have infinitely many solutions

Page No 3.73:

Question 19:

Find the value of k for which each of the following system of equations have infinitely many solutions :

2x+3y=kk-1x+k+2y=3k

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here,

Consider the following to find out k

Now again consider the following relation 

So the common solution is 7

Hence for the system of equation have infinitely many solutions

Page No 3.73:

Question 20:

Find the value of k for which each of the following system of equations have no solution :

kx-5y=26x+2y=7

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has no solution 

We know that the system of equations

For no solution

Here,

Hence for the system of equation have infinitely many solutions.

Page No 3.73:

Question 21:

Find the value of k for which each of the following system of equations have no solution :

x+2y=02x+ky=5

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has no solution 

We know that the system of equations

For no solution

Here,

Hence for the system of equation has no solution



Page No 3.74:

Question 22:

Find the value of k for which each of the following system of equations have no solution :

3x − 4y + 7 = 0
kx + 3y − 5 = 0

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has no solution 

We know that the system of equations

For no solution

Here,

Hence for the system of equation has no solution.

Page No 3.74:

Question 23:

Find the value of k for which each of the following system of equations have no solution :

2x-ky+3=03x+2y-1=0

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has no solution 

We know that the system of equations

For no solution

Here,

Hence for the system of equation has no solution.

Page No 3.74:

Question 24:

Find the value of k for which each of the following system of equations have no solution :

2x + ky = 11
5x − 7y = 5

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has no solution 

We know that the system of equations

For no solution

Here,

Hence for the system of equation has no solution.

Page No 3.74:

Question 25:

Find the value of k for which each of the following system of equations have no solution :

cx + 2y = 3
12x + cy = 6

Answer:

GIVEN:

To find: To determine for what value of c the system of equation has no solution 

We know that the system of equations

For no solution

Here,

Hence for the system of equation has no solution.

Page No 3.74:

Question 26:

For what value of k the following system of equations will be inconsistent?

4x+6y=112x+ky=7

Answer:

GIVEN:

To find: To determine for what value of k the system of equation will be inconsistent 

We know that the system of equations

For the system of equation to be inconsistent

Here,

Hence for the system of equation will be inconsistent.

Page No 3.74:

Question 27:

For what value of α, the system of equations will have no solution?

αx + 3y = α − 3
12x + αy = α

Answer:

GIVEN:
αx+3y=α-312x+αy=α
To find: To determine for what value of k the system of equation has no solution 

We know that the system of equations

For no solution

Here,

α12=3αα-3α
Consider the following for α

α12=3αα2=12×3α2=36α=±6

Now consider the following

3αα-3α3αα(α-3)3αα2-3α6αα2α6
Hence the common value of α is − 6

Hence for α = -6 the system of equation has no solution

Page No 3.74:

Question 28:

Find the value of k for which the system has (i) a unique solution, and (ii) no solution.

kx + 2y = 5
3x + y = 1

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has

(1) Unique solution 

(2) No solution

We know that the system of equations

(1) For Unique solution

Here,

Hence for the system of equation has unique solution.

(2) For no solution

Hence for the system of equation has no solution

Page No 3.74:

Question 29:

(i) Prove that there is a value of c(≠ 0) for which the system
6x + 3y = c − 3
12x + cy = c has infinitely many solutions. Find this value.

(ii) Find c if the system  of equations cx + 3y + 3 – c = 0, 12x + cy c = 0 has infinitely many solutions?
 

Answer:

(i) 

To find: To determine for what value of c the system of equation has infinitely many solution 

We know that the system of equations

For infinitely many solution

Here

Consider the following

Now consider the following for c

But it is given that c  0. Hence = 6

Hence for  the system of equation have infinitely many solutions.

(ii) If the system of equations a1x+b1y+c1=0 and a2x+b2y+c2=0 has infinitely many solutions, then a1a2=b1b2=c1c2.

Since, the system of equations cx + 3y + 3 – = 0, 12cy – c = 0 has infinitely many solutions
Therefore,
c12=3c=3-c-cc12=3c=3-c-cc12=3cc×c=3×12c2=36c=±6For c=6,c12=3c=3-c-c612=36=3-6-6=12Hence, it holds for c=6.For c=-6,c12=3c=3-c-c-612=3-6=3--66-12Hence, it does not holds for c=-6.

Hence, the value of c is 6.



 

Page No 3.74:

Question 30:

Find the values of k for which the system will have (i) a unique solution, and (ii) no solution. Is there a value of k for which the system has infinitely many solutions?

2x + ky = 1
3x − 5y = 7

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has

(1) Unique solution 

(2) No solution

(3) Infinitely many solution

We know that the system of equations

(1) For Unique solution

Here,

Hence for the system of equation has unique solution

(2) For no solution

Here,

Hence for the system of equation has no solution

(3) For infinitely many solution

Here,

But since here

Hence the system does not have infinitely many solutions.

Page No 3.74:

Question 31:

For what value of k, the following system of equations will represent the coincident lines?

x + 2y + 7 = 0
2x + ky + 14 = 0

Answer:

GIVEN: 

To find: To determine for what value of k the system of equation will represents coincident lines

We know that the system of equations

For the system of equation to represent coincident lines we have the following relation

Here,

Hence for the system of equation represents coincident lines

Page No 3.74:

Question 32:

Obtain the condition for the following system of linear equations to have a unique solution

ax + by = c
lx + my = n

Answer:

GIVEN: 

To find: To determine the condition for the system of equation to have a unique equation

We know that the system of equations

For unique solution

Here

Hence for the system of equation have unique solution.

Page No 3.74:

Question 33:

Determine the values of a and b so that the following system of linear equations have infinitely many solutions :

(2a − 1) x + 3y − 5 = 0
3x + (b − 1)y − 2 = 0

Answer:

GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here

Again consider

Hence for and the system of equation has infinitely many solution.



Page No 3.75:

Question 34:

Find the values of a and b for which the following system of linear equations has infinite number of solutions :

2x-3y=7a+bx-a+b-3y=4a+b

Answer:

GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here

Consider the following

5a − 4b + 21 = 0……(1)

Again

a + b + 6 = 0…… (2)

Multiplying eq. (2) by 4 and adding eq. (1)

Putting the value of a in eq. (2)

Hence for and the system of equation has infinitely many solution.

Page No 3.75:

Question 35:

Find the values of p and q for which the following system of linear equations has infinite number of solutions:

2x+3y=9p+qx+2p-qy=3p+q+1

Answer:

GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here

…… (1) 

Again consider

…… (2)

Multiplying eq. (2) by 3 and subtracting from eq. (1)

Putting the value of q in eq. (2)

Hence for and the system of equation has infinitely many solution.

Page No 3.75:

Question 36:

Find the values of a and b for which the following system of equations has infinitely many solutions:

i 2a-1x-3y=5     3x+b-2y=3(ii) 2x-2a+5y=5     2b+1x-9y=15(iii) a-1x+3y=2       6x+1-2by=6(iv) 3x+4y=12      a+bx+2a-by=5a-1v 2x+3y=7     a-bx+a+by=3a+b-2vi 2x+3y-7=0      a-1x+a+1y=3a-1vii 2x+3y=7       a-1x+a+2y=3a
viii x+2y=1a-bx+a+by=a+b-2 
ix 2x+3y=72ax+ay=28-by

Answer:

(i) GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here

Consider

Again consider

Hence forand the system of equation has infinitely many solution.

(ii) GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here

Consider the following

Again consider

Hence for and the system of equation has infinitely many solution. 

(iii) GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here

Consider the following

Again consider

Hence forand the system of equation has infinitely many solution.

(iv) GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here

Consider the following

…… (1)

Again consider

…… (2)

Multiplying eq. (2) by 2 and subtracting eq. (1) from eq. 2

Substituting the value of ‘a’ in eq. (2) we get

Hence forand the system of equation has infinitely many solution.

(v) GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here

Consider the following

…… (1)

Again consider the following

…… (2)

Multiplying eq. (2) by 2 and subtracting eq. (1) from eq. (2)

Substituting the value of b in eq. (2) we get

Hence forand the system of equation has infinitely many solution.

(vi) GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

Rewrite the given equations

We know that the system of equations

For infinitely many solution 

Here

Consider the following

Hence for the system of equation have infinitely many solutions.

(vii) GIVEN :

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here

Consider the following

Hence for the system of equation have infinitely many solutions.
(viii) Given:
x+2y=1a-bx+a+by=a+b-2 
We know that the system of equations

has infinitely many solutions if

So,
 1a-b=2a+b=1a+b-21a-b=2a+b and 2a+b=1a+b-2a+b=2a-2b and 2a+2b-4=a+ba=3b and a+b=4a-3b=0 and a+b=4
Solving these two equations, we get
-4b=-4b=1
Putting b = 1 in a + b = 4, we get
a = 3

(ix) Given:
2x+3y=72ax+ay=28-by
2ax+a+by=28
We know that the system of equations

has infinitely many solutions if

22a=3a+b=7281a=3a+b=141a=3a+b and 3a+b=14
Now,
1a=3a+b
⇒ a + b = 3a 
b = 2a              .....(1)
Also, 3a+b=14
a + b = 12     .....(2)
Solving (1) and (2), we get
a = 4 and b = 8

Page No 3.75:

Question 37:

For which value(s) of λ , do the pair of linear equations    λx + y = λ2 and   x + λy = 1   have 
(i) no solution ?      (ii) infinitely many solutions ?

(iii)  a unique solution  ?

Answer:

The given linear equations are 
λx+y=λ2x+λy=1
(i) We know that the system of equations


will have no solution if

So, 
λ1=1λλ21λ2=1λ=±1
Also, 
1λλ21λ31λ1
Thus, the given system of equations has no solutions when λ=-1.
(ii) We know that the system of equations


will have infinitely many solutions if


λ1=1λ=λ21λ2=1λ=±1
Also, 
1λ=λ21λ3=1λ=1
Thus, the given system of equations has infinitely many solutions when λ=1.
(iii) We know that the system of equations

will have a unique solution if

λ11λλ21λ±1
Thus, the given system of equations has a unique solution for all real values of λ except ±1.



Page No 3.78:

Question 1:

5 pens and 6 pencils together cost Rs 9 and 3 pens and 2 pencils cost Rs 5. Find the cost of 1 pen and 1 pencil.

Answer:

Given: 

(i) 5 pens and 6 pencils together cost of Rs. 9.

(ii) 3 pens and 2 pencils cost Rs. 5.

To Find: Cost of 1 pen and 1 pencil.

Let 

(i) The cost of 1 pen = Rs x.

(ii) The cost of 1 pencil = Rs y.

According to question

Thus we get the following system of linear equation

By using cross multiplication we have

Cost of one pen =

Cost of one pencil =

Page No 3.78:

Question 2:

7 audio cassettes and 3 video cassettes cost Rs 1110, while 5 audio cassettes and 4 video cassettes cost Rs 1350. Find the cost of an audio cassette and a video cassette.

Answer:

Given: 

(i) 7 Audio cassettes and 3 Video cassettes cost is 1110.

(ii) 5 Audio cassettes and 4 Video cassettes cost Rs. 1350.

To Find: Cost of 1 audio cassette and 1 video cassettes.

Let (i) the cost of 1 audio cassette = Rs. x.

(ii) the cost of 1 video cassette = Rs. y.

According to the given conditions, we have

Thus, we get the following system of linear equation,

By using cross multiplication, we have

Hence cost of 1 audio cassette =

Hence cost of 1 video cassette =

Page No 3.78:

Question 3:

Reena has pend and pencils which together are 40 in number. If she has 5 more pencils and 5 less pens, the number of pencils would become 4 times the number of pens. Find the original number of pens and pencils.

Answer:

Given:

(i) Total numbers of pens and pencils = 40.

(ii) If she has 5 more pencil and 5 less pens, the number of pencils would be 4 times the number of pen.

To find: Original number of pens and pencils.

Suppose original number of pencil = x

And original number of pen = y

According the given conditions, we have,

Thus we got the following system of linear equations

Substituting the value of y from equation 1 in equation 2 we get

Substituting the value of y in equation 1 we get

Hence we got the result number of pencils is and number of pens are

Page No 3.78:

Question 4:

4 tables and 3 chairs, together, cost Rs 2,250 and 3 tables and 4 chairs cost Rs 1950. Find the cost of 2 chairs and 1 table.

Answer:

Given:

(i) Cost of 4 tables and 3 chairs = Rs 2250.

(ii) Cost of 3 tables and 4 chairs = Rs 1950.

To find: The cost of 2 chairs and 1 table.

Suppose, the cost of 1 table = Rs x.

The cost of 1 chair = Rs y.

According to the given conditions,

4x + 3y = 2250,

4x + 3y − 2200 = 0 …… (1)

3x + 4y = 1950,

3x + 4y − 1950 = 0 …… (2)

Solving eq. (1) and Eq. (2) by cross multiplication

∴ cost of 1 chairs = Rs. 150.

Hence total cost of 2 chairs and 1 table =



Page No 3.79:

Question 5:

3 bags and 4 pens together cost Rs 257 whereas 4 bags and 3 pens together cost Rs 324. Find the total cost of 1 bag and 10 pens.

Answer:

Given:

(i) Cost of 3 bags and 4 pens = Rs. 257.

(ii) Cost of 4 bags and 3 pens = Rs. 324.

To Find: Cost of 1 bag and 10 pens.

Suppose, the cost of 1 bag = Rs. x.

and the cost 1 pen = Rs. y.

According to the given conditions, we have

3x + 4y = 257,

3x + 4y − 257 = 0 …… (1)

4x + 3y = 324

4x +3y − 324 = 0 …… (2)

Solving equation 1 and 2 by cross multiplication

Total cost of 1 bag and 10 pens =

Hence total cost of 1 bag and 10 pens =

Page No 3.79:

Question 6:

5 books and 7 pens together cost Rs 79 whereas 7 books and 5 pens together cost Rs 77. Find the total cost of 1 book and 2 pens.

Answer:

Given:

(i) Cost of 5 books and 7 pens = Rs. 79.

(ii) Cost of 7 books and 5 pens = Rs. 77.

To find: Cost of 1 book and 2 pens.

Suppose the cost of 1 book = Rs x.

and the cost of 1 pen = Rs y.

According to the given conditions, we have

5x + 7y = 79

5x + 7y − 79 = 0 …… (1)

7x + 5y = 77,

5x + 7y − 77 = 0 …… (2)

Thus we get the following system of linear equation,

Hence, the cost of 1 book = Rs 6

and the cost of 1 pen = Rs 7.

Therefore the cost of 2 pen = Rs 14.

Total cost of 1 book and 2 pens = 14 + 6 = 20

Total cost of 1 book and 2 pens =

Hence total cost of 1 book and 2 pens =

Page No 3.79:

Question 7:

Jamila sold a table and a chair for ₹ 1050 , thereby making a profit of 10% on the table and 25% on the chair. If she  had taken a profit of 25% on the table and 10% on the chair she would have got   ₹ 1065. Find the cost price of each .

Answer:

Let the CP of the table be Rs x and that of the chair be Rs y
Case I: Profit on table = 10100x
We know SP − CP = Profit
⇒ SP = Profit + CP
⇒ SP = 10100x + x = 110100x=1110x
Profit on chair = 25100y
SP = 25100y + y125100y
Total SP = 110100x+125100y=1050 
110x + 125y = 105000                        .....(i)
Case II: Profit on table = 25100x
⇒ SP = Profit + CP
⇒ SP = 25100x + x = 125100x
Profit on chair = 10100y
SP = 10100y + y110100y
Total SP = 125100x110100y = 1065
125x + 110y = 106500                       .....(ii)
From (i) and (ii), we have
235(x + y) = 211500
x + y = 900                                    .....(iii)
Subtracting (i) from (ii), we have
15(x − y) = 1500
x − y = 100                                    .....(iv)
Solving (iii) and (iv), we get
x = 500 and y = 400
Thus, CP of table = Rs 500 and CP of chair = Rs 400.

Page No 3.79:

Question 8:

Susan invested certain amount of money in two schemes A and B , which offer interest at the rate of 8% per annum and 9% per annum , respectively . She received ₹1860 as annual interest .However, had she interchanged the amount of investment in the two schemes , she would have received  ₹20 more as annual interest . How much money did she invest in each scheme ?

Answer:

Let the money invested in Scheme A be Rs x and that in Scheme B be Rs y.
I=PRT100
So, Interest in scheme A = 8x100
Interest in scheme B = 9y100
Total annual interest = 8x100+9y100 = 1860
⇒ 8x + 9y = 186000                          .....(i)
After interchaning the amounts in the two schemes, the new total annual interest = 9x100+8y100.
Now,
9x100+8y100 = 1860 + 20 = 1880
⇒ 9x + 8y = 188000                          .....(ii)
Adding (i) and (ii), we get
17x + 17y = 374000
x + y = 22000                               .....(iii)
Subtracting (i) from (ii), we get
xy = 2000                                     .....(iv)
Adding (iii) and (iv), we get
2x = 24000
x = 12000
Puting x = 12000 in (iii), we get
12000 + y = 22000
y = 10000
So, the money invested in scheme A = Rs 12,000 and in scheme B = Rs 10,000.

Page No 3.79:

Question 9:

The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, he buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Answer:

Given: (i) 7 bats and 6balls cost is Rs3800

(ii) 3 bats and 5balls cost is Rs1750

To find: Cost of 1 bat and 1 ball

Let (i) the cost of 1 bat = Rs. x.

(ii) the cost of 1 ball = Rs. y.

According to the given conditions, we have

Thus, we get the following system of linear equation,

7x + 6y − 3800 = 0 …… (1)

3x + 5y − 1750 = 0 …… (2)

By using cross multiplication, we have

Hence cost of 1 bat =

Hence cost of 1 ball =

Page No 3.79:

Question 10:

A lending library has a fixed charge for the first three days and additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer:

To find:

(1) the fixed charge

(2) The charge for each day

Let the fixed charge be Rs x

And the extra charge per day be Rs y.

According to the given conditions,

Subtracting equation 1 and 2 we get

Substituting the value of y in equation 1 we get

Hence the fixed charge is and the charge of each day

Page No 3.79:

Question 11:

The cost of 4 pens and 4 pencil boxes is  ₹100 . Three times the cost of a pen is  ₹15 more than the cost of a pencil box . Form the pair of linear equations for the above situation . Find the cost of a pen and a pencil box.

Answer:

Let the cost of 1 pen be ₹x and that of 1 pencil box be ₹y.
Now,
Cost of 4 pens + Cost of 4 pencil boxes = ₹100                (Given)
⇒ 4x + 4y = 100
x + y = 25                  .....(i)
Also,
3 × Cost of  a pen = Cost of a pencil box + ₹15
3x = y + 15
⇒ 3x − y = 15               .....(ii)
Adding (i) and (ii), we get
4x = 40
x = 10
Putting x = 10 in (i), we get
10 + y = 25
y = 15
Thus, the cost of a pen = ₹10 and that of a pencil box = ₹15.

Page No 3.79:

Question 12:

One says, "Give me a hundred, friend! I shall then become twice as rich as you." The other replies, "If you give me ten, I shall be six times as rich as you." Tell me what is the amount of their respective capital?

Answer:

To find:

(1) Total amount of A.

(2) Total amount of B.

Suppose A has Rs x and B has Rs y

According to the given conditions,

x + 100 = 2(y − 100)
x + 100 = 2− 200
− 2y = −300                  ....(1)
and

y + 10 = 6(x − 10)
y + 10 = 6x − 60
6x − y = 70                      ....(2)

Multiplying equation (2) by 2 we get

12x − 2y = 140               ....(3)

Subtracting (1) from (3),  we get

11x = 440
 x = 40

Substituting the value of x in equation (1), we get

40 − 2y = −300
−2y = −340
    y = 170

Hence A has and B has

Page No 3.79:

Question 13:

A and B each have a certain number of mangoes. A says to B, "if you give 30 of your mangoes, I will have twice as many as left with you." B replies, "if you give me 10, I will have thrice as many as left with you." How many mangoes does each have?

Answer:

To find:

(1) Total mangoes of A.

(2) Total mangoes of B.

Suppose A has x mangoes and B has y mangoes,

According to the given conditions,

3x + 6y + 270 = 0 …… (3) and

Now adding eq.2 and eq.3

5y = 310

y =

Hence A has 34 mangoes and B has 62 mangoes.

Page No 3.79:

Question 14:

Vijay had some bananas , and he divided them into two lots A and B . He sold first lot at the rate of ₹2 for 3 bananas and the second lot at the rate of  ₹1 per banana  and got a total of  ₹400 . If he had sold the first lot at the rate of  ₹1 per banana and the second lot at the rate  of  ₹4 per five bananas, his total collection would have been  ₹460 . Find the total number of bananas he had .

Answer:

Let the bananas in lot A be x and that in lot B be y
Vijay sold 3 bananas for Rs 2 in lot A.
So, the cost of x bananas in lot A = 23x
Now,
Cost of x bananas in lot A + Cost of y bananas in lot B = ₹400
23x+y=400
⇒ 2x + 3y = 1200        .....(i)
Now, if he sells the first lot at the rate of Rs 1 per banana and second for Rs 4 for 5 bananas, then
x+45y = ₹460
⇒ 5x + 4y = 2300        .....(ii)
Solving (i) and (ii), we get
x = 300 and y = 200
So, the total number of bananas = x + y = 300 + 200 = 500.

Page No 3.79:

Question 15:

On selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains Rs 2000. But if he sells the T.V. at 10% gain the fridge at 5% loss. He gains Rs 1500 on the transaction. Find the actual prices of T.V. and fridge.

Answer:

Given:

(i) On selling of a T.V. at 5% gain and a fridge at 10% gain, shopkeeper gain Rs.2000.

(ii) Selling T.V. at 10% gain and fridge at 5% loss. He gains Rs. 1500.

To find: Actual price of T.V. and fridge.

According to the question:

Hence we got the pair of equations

1x + 2y − 40000 = 0 …… (1)

2x − 1y − 30000 = 0 …… (2)

Solving the equation by cross multiplication method;

Cost of T.V. =

Cost of fridge =

Hence the cost of T.V. is and that of fridge is .



Page No 3.85:

Question 1:

The sum of two numbers is 8. If their sum is four times their difference, find the numbers.

Answer:

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The sum of the two numbers is 8. Thus, we have

The sum of the two numbers is four times their difference. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Multiplying the first equation by 5 and then adding with the second equation, we have

Substituting the value of x in the first equation, we have

Hence, the numbers are 5 and 3.

Page No 3.85:

Question 2:

The sum of digits of a two digit number is 13. If the number is subtracted from the one obtained by interchanging the digits, the result is 45. What is the number?

Answer:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The sum of the digits of the number is 13. Thus, we have

After interchanging the digits, the number becomes.

The difference between the number obtained by interchanging the digits and the original number is 45. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.



Page No 3.86:

Question 3:

A number consist of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.

Answer:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The sum of the digits of the number is 5. Thus, we have

After interchanging the digits, the number becomes.

The number obtained by interchanging the digits is greater by 9 from the original number. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

Page No 3.86:

Question 4:

The sum of digits of a two number is 15. The number obtained by reversing the order of digits of the given number exceeds the given number by 9. Find the given number.

Answer:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The sum of the digits of the number is 15. Thus, we have

After interchanging the digits, the number becomes.

The number obtained by interchanging the digits is exceeding by 9 from the original number. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

Page No 3.86:

Question 5:

The sum of a two-digit number and the number formed by reversing the order of digit is 66. If the two digits differ by 2, find the number. How many such numbers are there?

Answer:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The two digits of the number are differing by 2. Thus, we have

After interchanging the digits, the number becomes.

The sum of the numbers obtained by interchanging the digits and the original number is 66. Thus, we have

So, we have two systems of simultaneous equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

(i) First, we solve the system

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

(ii) Now, we solve the system

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

There are two such numbers.

Page No 3.86:

Question 6:

The sum of two numbers is 1000 and the difference between their squares is 256000. Find the numbers.

Answer:

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The sum of the two numbers is 1000. Thus, we have

The difference between the squares of the two numbers is 256000. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the numbers are 628 and 372.

Page No 3.86:

Question 7:

The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.

Answer:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The two digits of the number are differing by 3. Thus, we have

After interchanging the digits, the number becomes.

The sum of the numbers obtained by interchanging the digits and the original number is 99. Thus, we have

So, we have two systems of simultaneous equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

(i) First, we solve the system

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

(ii) Now, we solve the system

Adding the two equations, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

Note that there are two such numbers.

Page No 3.86:

Question 8:

A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Answer:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The number is 4 times the sum of the two digits. Thus, we have

After interchanging the digits, the number becomes.

If 18 is added to the number, the digits are reversed. Thus, we have

So, we have the systems of equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Subtracting the first equation from the second, we have

Substituting the value of y in the first equation, we have

Hence, the number is.

Page No 3.86:

Question 9:

A two-digit number is 3 more than 4 times the sum of its digits. If 8 is added to the number, the digits are reversed. Find the number.

Answer:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The number is 3 more than 4 times the sum of the two digits. Thus, we have

After interchanging the digits, the number becomes.

If 18 is added to the number, the digits are reversed. Thus, we have

So, we have the systems of equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Subtracting the first equation from the second, we have

Substituting the value of y in the first equation, we have

Hence, the number is.

Page No 3.86:

Question 10:

A two-digit number is 4 more than 6 times the sum of its digits. If 18 is subtracted from the number, the digits are reversed. Find the number.

Answer:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The number is 4 more than 6 times the sum of the two digits. Thus, we have

After interchanging the digits, the number becomes.

If 18 is subtracted from the number, the digits are reversed. Thus, we have

So, we have the systems of equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Multiplying the second equation by 5 and then subtracting from the first, we have

Substituting the value of y in the second equation, we have

Hence, the number is.

Page No 3.86:

Question 11:

A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number.

Answer:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The number is 4 times the sum of the two digits. Thus, we have

After interchanging the digits, the number becomes.

The number is twice the product of the digits. Thus, we have

So, we have the systems of equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Substituting in the second equation, we get

Or

Substituting the value of y in the first equation, we have

Hence, the number is.

Note that the first pair of solution does not give a two digit number.

Page No 3.86:

Question 12:

A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number.

Answer:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The product of the two digits of the number is 20. Thus, we have

After interchanging the digits, the number becomes.

If 9 is added to the number, the digits interchange their places. Thus, we have

So, we have the systems of equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Substituting from the second equation to the first equation, we get

Or

Substituting the value of y in the second equation, we have

Hence, the number is.

Note that in the first pair of solution the values of x and y are both negative. But, the digits of the number can’t be negative. So, we must remove this pair.

Page No 3.86:

Question 13:

The difference between two numbers is 26 and one number is three times the other. Find them.

Answer:

Let the numbers are x and y. One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y.

The difference between the two numbers is 26. Thus, we have

One of the two numbers is three times the other number. Here, we are assuming that x is greater than or equal to y. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Substitutingfrom the second equation in the first equation, we get

Substituting the value of y in the first equation, we have

Hence, the numbers are 39 and 13.

Page No 3.86:

Question 14:

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Answer:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The sum of the two digits of the number is 9. Thus, we have

After interchanging the digits, the number becomes.

Also, 9 times the number is equal to twice the number obtained by reversing the order of the digits. Thus, we have

So, we have the systems of equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Substituting from the second equation to the first equation, we get

Substituting the value of y in the second equation, we have

Hence, the number is.

Page No 3.86:

Question 15:

Seven times a two-digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is 3. Find the number.

Answer:

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is.

The difference between the two digits of the number is 3. Thus, we have

After interchanging the digits, the number becomes.

Seven times the number is equal to four times the number obtained by reversing the order of the digits. Thus, we have

So, we have two systems of simultaneous equations

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

(i) First, we solve the system

Multiplying the first equation by 2 and then subtracting from the second equation, we have

Substituting the value of x in the first equation, we have

Hence, the number is.

(ii) Now, we solve the system

Multiplying the first equation by 2 and then subtracting from the second equation, we have

Substituting the value of x in the first equation, we have

But, the digits of the number can’t be negative. Hence, the second case must be removed.

Page No 3.86:

Question 16:

Two numbers are in the ratio 5 : 6 . If 8 is subtracted from each of the numbers , the ratio becomes 4 : 5 . Find the numbers .

Answer:

Let the two numbers be x and y.
So, 
xy=566x=5y6x-5y=0       .....i 
Now, when 8 is subtracted from each of the numbers, then
x-8y-8=455x-40=4y-325x-4y=8           .....ii
Multiplying (i) with 4 and (ii) with 5, we get
24x − 20y = 0          .....(iii)
25x − 20y = 40        .....(iv)
Subtracting (iii) from (iv), we get
x = 40
Putting x =  40 in (i), we get
240 − 5y = 0
y = 48
Thus, the two numbers are 40 and 48.

Page No 3.86:

Question 17:

A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3 . Find the number .

Answer:

Let the digits of the two digit number be x and y. So, the two digit number will be 10x + y
Now, according to the given condition the number is obtained in two ways.
Case I: 8(x + y) − 5 = 10x + y
⇒ 8x + 8y − 5 = 10x + y
⇒ 2x −7y = −5               .....(1)
Case II: 16(x − y) + 3 = 10x + y
⇒ 16x − 16y + 3 = 10x + y
⇒ 6x − 17y = −3            .....(2)
Multiplying (1) by 3, we get
6x − 21y = −15              .....(3)
Subtracting (2) from (3), we get
− 4y = −12
y = 3
Putting y = 3 in (1), we get
2x − 21 = −5
⇒ 2x = 16
x = 8
So, the required number is 10x + y =  10 × 8 + 3 = 83.



Page No 3.88:

Question 1:

The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.

Answer:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

The numerator of the fraction is 4 less the denominator. Thus, we have

If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is 8 times the numerator. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

Subtracting the second equation from the first equation, we get

Substituting the value of x in the first equation, we have

Hence, the fraction is.



Page No 3.89:

Question 2:

A fraction becomes 9/11 if 2 is added to both numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Answer:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

If 2 is added to both numerator and the denominator, the fraction becomes. Thus, we have

If 3 is added to both numerator and the denominator, the fraction becomes. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

Page No 3.89:

Question 3:

A fraction becomes 1/3 if 1 is subtracted from both its numerator and denominator. It 1 is added to both the numerator and denominator, it becomes 1/2. Find the fraction.

Answer:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

If 1 is subtracted from both numerator and the denominator, the fraction becomes. Thus, we have

If 1 is added to both numerator and the denominator, the fraction becomes. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

Page No 3.89:

Question 4:

If we add 1 to the numerator and subtract 1 from the denominator, a fraction becomes 1. It also becomes 1/2 if we only add 1 to the denominator. What is the fraction?

Answer:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

If 1 is added to the numerator and 1 is subtracted from the denominator, the fraction becomes. Thus, we have

If 1 is added to the denominator, the fraction becomes. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

Page No 3.89:

Question 5:

If the numerator of a fraction is multiplied by 2 and the denominator is reduced by 5 the fraction becomes 6/5. And, if the denominator is doubled and the numerator is increased by 8, the fraction becomes 2/5. find the fraction.

Answer:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

If the numerator is multiplied by 2 and the denominator is reduced by 5, the fraction becomes. Thus, we have

If the denominator is doubled and the numerator is increased by 8, the fraction becomes. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

Page No 3.89:

Question 6:

When 3 is added to the denominator and 2 is subtracted from the numerator a fraction becomes 1/4. And when 6 is added to numerator and the denominator is multiplied by 3, it becomes 2/3. Find the fraction.

Answer:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

If 3 is added to the denominator and 2 is subtracted from the numerator, the fraction becomes. Thus, we have

If 6 is added to the numerator and the denominator is multiplied by 3, the fraction becomes. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

Page No 3.89:

Question 7:

The sum of a numerator and denominator of a fraction is 18. If the denominator is increased by 2, the fraction reduces to 1/3. Find the fraction.

Answer:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

The sum of the numerator and the denominator of the fraction is. Thus, we have

If the denominator is increased by 2, the fraction reduces to. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

Page No 3.89:

Question 8:

If 2 is added to the numerator of a fraction, it reduces to 1/2 and if 1 is subtracted from the denominator, it reduces to 1/3. Find the fraction.

Answer:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

If 2 is added to the numerator of the fraction, it reduces to. Thus, we have

If 1 is subtracted from the denominator, the fraction reduces to. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

 

Page No 3.89:

Question 9:

A fraction becomes 1/3 when 2 is subtracted from the numerator and it becomes 1/2 when 1 is subtracted from the denominator. Find the fraction.

Answer:

Let the numerator of the fraction be x and the denominator be y.

According to the question,

x-2y=13   ...(i) and xy-1=12     ...(ii)Solving equation i, we get3x-2=y3x-6=y      ...iiiSubstituting the value of y in equation ii, we getx3x-6-1=122x=3x-6-12x=3x-72x-3x=-7-x=-7x=7      ...(iv)From (iii) and (iv), we getx=7 and y=15Hence, the fraction is 715.

Page No 3.89:

Question 10:

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.

Answer:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

The sum of the numerator and denominator of the fraction is 4 more than twice the numerator. Thus, we have

If the numerator and denominator are increased by 3, they are in the ratio 2:3. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

Page No 3.89:

Question 11:

The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Determine the fraction.

Answer:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

The sum of the numerator and denominator of the fraction is 3 less than twice the denominator. Thus, we have

If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.

Page No 3.89:

Question 12:

The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.

Answer:

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is

The sum of the numerator and denominator of the fraction is 12. Thus, we have

If the denominator is increased by 3, the fraction becomes. Thus, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the fraction is.



Page No 3.92:

Question 1:

A father is three times as old as his son. After twelve years, his age will be twice as that of his son then. Find the their present ages.

Answer:

Let the present age of father be x years and the present age of son be y years.

Father is three times as old as his son. Thus, we have

After 12 years, father’s age will beyears and son’s age will beyears. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of father isyears and the present age of son isyears.

Page No 3.92:

Question 2:

Ten years later, A will be twice as old as B and five years ago, A was three times as old as B. What are the present ages of A and B?

Answer:

Let the present age of A be x years and the present age of B be y years.

After 10 years, A’s age will beyears and B’s age will beyears. Thus using the given information, we have

Before 5 years, the age of A wasyears and the age of B wasyears. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of A isyears and the present age of B isyears.

Page No 3.92:

Question 3:

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Answer:

Let the present age of Nuri be x years and the present age of Sonu be y years.

After 10 years, Nuri’s age will be(x + 10) years and the age of Sonu will be(y + 10) years. Thus using the given information, we have

Before 5 years, the age of Nuri was(x – 5)years and the age of Sonu was(y – 5)years. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of Nuri isyears and the present age of Sonu isyears.

Page No 3.92:

Question 4:

Six years hence a man's age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.

Answer:

Let the present age of the man be x years and the present age of his son be y years.

After 6 years, the man’s age will beyears and son’s age will beyears. Thus using the given information, we have

Before 3 years, the age of the man wasyears and the age of son’s wasyears. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of the man isyears and the present age of son isyears.

Page No 3.92:

Question 5:

Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then. Find their present ages.

Answer:

Let the present age of father be x years and the present age of his son be y years.

After 10 years, father’s age will beyears and son’s age will beyears. Thus using the given information, we have

Before 10 years, the age of father wasyears and the age of son wasyears. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of father isyears and the present age of son isyears.

Page No 3.92:

Question 6:

The present age of a father is three years more than three times the age of the son. Three years hence father's age will be 10 years more than twice the age of the son. Determine their present ages.

Answer:

Let the present age of father be x years and the present age of his son be y years.

The present age of father is three years more than three times the age of the son. Thus, we have

After 3 years, father’s age will beyears and son’s age will beyears. 

Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of father isyears and the present age of son isyears.

Page No 3.92:

Question 7:

A father is three times as old as his son. In 12 years time, he will be twice as old as his son. Find the present ages of father and the son.

Answer:

Let the present age of father be x years and the present age of his son be y years.

The present age of father is three times the age of the son. Thus, we have

After 12 years, father’s age will beyears and son’s age will beyears. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of father isyears and the present age of son isyears.

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Question 8:

Father's age is three times the sum of age of his two children. After 5 years his age will be twice the sum of ages of two children. Find the age of father.

Answer:

Let the present age of father be x years and the present ages of his two children’s be y and z years.

The present age of father is three times the sum of the ages of the two children’s. Thus, we have

After 5 years, father’s age will beyears and the children’s age will beandyears. Thus using the given information, we have

So, we have two equations

Here x, y and z are unknowns. We have to find the value of x.

Substituting the value offrom the first equation in the second equation, we have

By using cross-multiplication, we have

Hence, the present age of father isyears.

Page No 3.92:

Question 9:

Two years ago, a father was five times as old as his son. Two year later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.

Answer:

Let the present age of father be x years and the present age of his son be y years.

After 2 years, father’s age will beyears and the age of son will beyears. Thus using the given information, we have

Before 2 years, the age of father wasyears and the age of son wasyears. Thus using the given information, we have

So, we have two equations

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

Hence, the present age of father isyears and the present age of son isyears.

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Question 10:

A is elder to B by 2 years. A's father F is twice as old as A and B is twice as old as his sister S. If the ages of the father and sister differ by 40 years, find the age of A.

Answer:

Let the present ages of A, B, F and S be x, y, z and t years respectively.

A is elder to B by 2 years. Thus, we have

F is twice as old as A. Thus, we have

B is twice as old as S. Thus, we have

The ages of F and S is differing by 40 years. Thus, we have

So, we have four equations

……(1)

……(2)

……(3)

……(4)

Here x, y, z and t are unknowns. We have to find the value of x.

By using the third equation, the first equation becomes

From the fourth equation, we have

Hence, we have

Using the second equation, we have

Hence, the age of A isyears.

Page No 3.92:

Question 11:

The ages of two friends Ani and Biju differ by 3 years. Ani's father Dharma is twice as old as Ani and Biju as twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Answer:

Let the present ages of Ani, Biju, Dharam and Cathy be x, y, z and t years respectively.

The ages of Ani and Biju differ by 3 years. Thus, we have

Dharam is twice as old as Ani. Thus, we have

Biju is twice as old as Cathy. Thus, we have

The ages of Cathy and Dharam differ by 30 years. Clearly, Dharam is older than Cathy. Thus, we have

So, we have two systems of simultaneous equations

(i)

(ii)

Here x, y, z and t are unknowns. We have to find the value of x and y.

(i) By using the third equation, the first equation becomes

From the fourth equation, we have

Hence, we have

Using the second equation, we have

From the first equation, we have

Hence, the age of Ani isyears and the age of Biju isyears.

(ii) By using the third equation, the first equation becomes

From the fourth equation, we have

Hence, we have

Using the second equation, we have

From the first equation, we have

Hence, the age of Ani isyears and the age of Biju isyears.

Note that there are two possibilities.

Page No 3.92:

Question 12:

Two years ago, Salim was thrice as old as his daughter and six years later , he will be four years older than twice her age . How old are they now ?

Answer:

Let the present ages of Salim be x years and that of her daughter be y years.
Two years ago, the age of Salim was (x − 2) years and that of her daughter was (y − 2).
It is given that Salim was thrice as old as her daughter two years ago. So,
x − 2 = 3(y − 2)
x − 2 = 3y − 6
x − 3y = −4                   .....(i)
Six years later, the age of Salim will be (x + 6) and that of her daughter will be (y + 6).
x + 6 = 2(y + 6) + 4
x − 2y = 10                    .....(ii)
Subtracting (ii) from (i), we get
y = −14
y = 14
Putting y = 14 in (ii), we get
x − 28 = 10
x = 38
Hence, the present age of Salim is 38 years and that of her daughter is 14 years.

Page No 3.92:

Question 13:

The age of the father is twice the sum of the ages of his two children . After 20 years , his age will be eual to the sum of the ages of his children . Find the age of the father.

Answer:

Let the present age of the father be x years and the sum of the present ages of his two children be y years. 
Now according to the given conditions,
Case I: x = 2y
− 2y = 0           .....(i)
Case II: After 20 years, the age of the father will be (x + 20) years and the sum of the ages of the two children will be y + 20 + 20 = (y + 40) years.
So, x + 20 = y + 40
x − y = 20           .....(ii)
Subtracting (ii) from (i), we get
y = −20
y = 20
Putting y = 20 in (i), we get
x − 40 = 0
x = 40
Hence, the present age of the father is 40 years.



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