Page No 175:
Answer:
(ix)
This is of the form ax2 + bx + c = 0.
Hence, the given equation is a quadratic equation.
(x)
This is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
(xi)
This is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
Page No 175:
Answer:
Page No 175:
Answer:
(i)
So, the equation becomes
On factorising we get;
Hence, the other root is 3.
(ii)
Page No 175:
Answer:
LHS;
Consider the quadratic equation;
Put in the given equation.
Hence, is a solution to the given quadratic equation.
Page No 175:
Answer:
(2x − 3)(3x + 1) = 0
⇒ 2x − 3 = 0 or 3x + 1 = 0
⇒ 2x = 3 or 3x = −1
⇒ x = or x =
Hence, the roots of the given equation are and .
Page No 175:
Answer:
4x2 + 5x = 0
⇒ x(4x + 5) = 0
⇒ x = 0 or 4x + 5 = 0
⇒ x = 0 or x =
Hence, the roots of the given equation are 0 and .
Page No 175:
Answer:
Page No 175:
Answer:
We write, as
Hence, the roots of the given equation are and .
Page No 175:
Answer:
We write, as
Hence, the roots of the given equation are −1 and −5.
Page No 175:
Answer:
We write, as
Hence, the roots of the given equation are and .
Page No 175:
Answer:
Page No 175:
Answer:
Page No 176:
Answer:
Page No 176:
Answer:
Page No 176:
Answer:
We write, as
Hence, the roots of the given equation are and .
Page No 176:
Answer:
Page No 176:
Answer:
Page No 176:
Answer:
Page No 176:
Answer:
Page No 176:
Answer:
We write, as
Hence, the roots of the given equation are and .
Page No 176:
Answer:
Consider
Factorising by splitting the middle term;
Hence, the roots of the given equation are and .
Page No 176:
Answer:
Page No 176:
Answer:
Page No 176:
Answer:
We write, as
Hence, the roots of the given equation are and .
Page No 176:
Answer:
Page No 176:
Answer:
We write, as
Hence, is the repreated root of the given equation.
Page No 176:
Answer:
We write, as
Hence, the roots of the given equation are and .
Page No 176:
Answer:
We write, as
Hence, the roots of the given equation are and .
Page No 176:
Answer:
Hence, 1 and are the roots of the given equation.
Page No 176:
Answer:
We write, as
Hence, the roots of the given equation are and .
Page No 176:
Answer:
We write, as
Hence, the roots of the given equation are and .
Page No 176:
Answer:
We write, 13x = 5x + 8x as
Hence, and are the roots of the given equation.
Page No 176:
Answer:
Page No 176:
Answer:
Page No 176:
Answer:
We write, as
Hence, is the repreated root of the given equation.
Page No 176:
Answer:
We write, as
Hence, is the repeated root of the given equation.
Page No 176:
Answer:
Page No 176:
Answer:
Page No 176:
Answer:
We write, as
Hence, and are the roots of the given equation.
Page No 176:
Answer:
We write, as
Hence, and are the roots of the given equation.
Page No 176:
Answer:
We write, as
Hence, and are the roots of the given equation.
Page No 176:
Answer:
We write, as
Hence, and are the roots of the given equation.
Page No 176:
Answer:
We write, as
Hence, and are the roots of the given equation.
Page No 176:
Answer:
We write, as
Hence, and are the roots of the given equation.
Page No 176:
Answer:
We write, as
Hence, and are the roots of the given equation.
Page No 176:
Answer:
Page No 176:
Answer:
We write, as
Hence, and are the roots of the given equation.
Page No 176:
Answer:
Page No 176:
Answer:
Page No 176:
Answer:
Page No 176:
Answer:
We write, as
Hence, and are the roots of the given equation.
Page No 176:
Answer:
Hence, −4 and 4 are the roots of the given equation.
Page No 176:
Answer:
Hence, −2 and 1 are the roots of the given equation.
Page No 176:
Answer:
Hence, 1 and 3 are the roots of the given equation.
Page No 177:
Answer:
(i)
Hence, −6 and 2 are the roots of the given equation.
(ii)
Page No 177:
Answer:
Hence, and are the roots of the given equation.
Page No 177:
Answer:
Page No 177:
Answer:
Hence, 6 and are the roots of the given equation.
Page No 177:
Answer:
(i)
Hence, and are the roots of the given equation.
(ii)
Page No 177:
Answer:
Hence, and are the roots of the given equation.
Page No 177:
Answer:
Hence, 8 and are the roots of the given equation.
Page No 177:
Answer:
Hence, 5 and are the roots of the given equation.
Page No 177:
Answer:
Page No 177:
Answer:
(i)
Hence, 2 and are the roots of the given equation.
(ii)
Page No 177:
Answer:
Hence, 0 and −7 are the roots of the given equation.
Page No 177:
Answer:
Hence, 0 and 1 are the roots of the given equation.
Page No 177:
Answer:
Page No 177:
Answer:
Page No 177:
Answer:
Page No 177:
Answer:
Page No 177:
Answer:
Page No 177:
Answer:
Page No 177:
Answer:
Page No 185:
Answer:
Hence, and are the roots of the given equation.
Page No 185:
Answer:
Hence, and are the roots of the given equation.
Page No 185:
Answer:
Hence, and are the roots of the given equation.
Page No 185:
Answer:
Hence, is the repeated root of the given equation.
Page No 185:
Answer:
Hence, and are the roots of the given equation.
Page No 185:
Answer:
Hence, 1 and are the roots of the given equation.
Page No 185:
Answer:
Hence, and are the roots of the given equation.
Page No 185:
Answer:
Hence, and −1 are the roots of the given equation.
Page No 185:
Answer:
Hence, 1 and are the roots of the given equation.
Page No 185:
Answer:
Hence, and are the roots of the given equation.
Page No 185:
Answer:
Hence, 2 and are the roots of the given equation.
Page No 185:
Answer:
Hence, and are the roots of the given equation.
Page No 185:
Answer:
Hence, and 1 are the roots of the given equation.
Page No 185:
Answer:
Hence, and are the roots of the given equation.
Page No 185:
Answer:
Hence, and are the roots of the given equation.
Page No 185:
Answer:
But, cannot be negative for any real value of x.
So, there is no real value of x satisfying the given equation.
Hence, the given equation has no real roots.
Page No 191:
Answer:
​
​
(v)
Comparing it with
, we get
a = 2,
b = −3 and
c = 1
∴ Discriminant,
D =
​
Page No 191:
Answer:
Page No 191:
Answer:
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
a = 2, b = 1 and c = −4
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
Page No 191:
Answer:
Page No 191:
Answer:
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
a = 2, b = and c = 1
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, is the repeated root of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
a = , b = 7 and c =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
a = , b = and c =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
a = 2, b = and c =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
a = , b = 5 and c =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
a = 3, b = and c = 2
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, is the repeated root of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
a = , b = and c =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
a = 1, b = 1 and c = 2
∴ Discriminant, D =
Hence, the given equation has no real roots (or real roots does not exist).
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
A = 2, B = a and C =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
a = 1, b = and c =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and 1 are the roots of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
a = 2, b = and c = 6
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
a = 3, b = −2 and c = 2
∴ Discriminant, D =
Hence, the given equation has no real roots (or real roots does not exist).
Page No 191:
Answer:
The given equation is
This equation is of the form , where a = 1, b = −3 and c = 1.
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is
This equation is of the form , where a = 3, b = −6 and c = 2.
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is
This equation is of the form , where a = 1, b = −3 and c = −1.
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is
This equation is of the form , where a = , b = 2mn and c = .
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
A = 36, B = and C =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
A = 1, B = and C =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
A = 1, B = 6 and C =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
A = 1, B = 5 and C =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
A = 1, B = −4a and C =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 191:
Answer:
The given equation is .
Comparing it with , we get
A = 4, B = −4a2 and C =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 192:
Answer:
The given equation is .
Comparing it with , we get
A = 4, B = 4b and C =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 192:
Answer:
The given equation is .
Comparing it with , we get
A = 1, B = and C =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 192:
Answer:
Page No 192:
Answer:
The given equation is .
Comparing it with , we get
A = , B = and C =
∴ Discriminant, D =
So, the given equation has real roots.
Now,
Hence, and are the roots of the given equation.
Page No 192:
Answer:
Page No 199:
Answer:
(i) The given equation is .
This is of the form , where a = 2, b = −8 and c = 5.
∴ Discriminant, D =
Hence, the given equation has real and unequal roots.
(ii) The given equation is .
This is of the form , where a = 3, b = and c = 2.
∴ Discriminant, D =
Hence, the given equation has real and equal roots.
(iii) The given equation is .
This is of the form , where a = 5, b = −4 and c = 1.
∴ Discriminant, D =
Hence, the given equation has no real roots.
(iv) The given equation is
This is of the form , where a = 5, b = −10 and c = 6.
∴ Discriminant, D =
Hence, the given equation has no real roots.
(v) The given equation is .
This is of the form , where a = 12, b = and c = 5.
∴ Discriminant, D =
Hence, the given equation has real and equal roots.
(vi) The given equation is .
This is of the form , where a = 1, b = −1 and c = 2.
∴ Discriminant, D =
Hence, the given equation has no real roots.
Page No 199:
Answer:
The given equation is .
Hence, the given equation has no real roots.
Page No 199:
Answer:
Page No 199:
Answer:
Page No 199:
Answer:
The given equation is
This is of the form , where a = k, b = and c = 10.
The given equation will have real and equal roots if D = 0.
But, for k = 0, we get 10 = 0, which is not true.
Hence, 2 is the required value of k.
Page No 199:
Answer:
The given equation is .
This is of the form , where a = 4, b = p and c = 3.
The given equation will have real and equal roots if D = 0.
Hence, and are the required values of p.
Page No 200:
Answer:
The given equation is .
This is of the form , where a = 9, b = −3k and c = k.
The given equation will have real and equal roots if D = 0.
But, k ≠ 0 (Given)
Hence, the required value of k is 4.
Page No 200:
Answer:
(i)
The given equation is .
This is of the form , where a = 3k +1, b = 2(k + 1) and c = 1.
The given equation will have real and equal roots if D = 0.
Hence, 0 and 1 are the required values of k.
(ii)
Page No 200:
Answer:
The given equation is .
This is of the form , where a = 2p +1, b = −(7p + 2) and c = 7p − 3.
The given equation will have real and equal roots if D = 0.
Hence, 4 and are the required values of p.
Page No 200:
Answer:
The given equation is .
This is of the form , where a = p +1, b = −6(p + 1) and c = 3(p + 9).
The given equation will have real and equal roots if D = 0.
But, (Given)
Thus, the value of p is 3.
Putting p = 3, the given equation becomes .
Hence, 3 is the repeated root of this equation.
Page No 200:
Answer:
It is given that −5 is a root of the quadratic equation .
The roots of the equation = 0 are equal.
Thus, the value of k is .
Page No 200:
Answer:
It is given that 3 is a root of the quadratic equation .
The roots of the equation are equal.
Hence, the value of p is 12.
Page No 200:
Answer:
It is given that −4 is a root of the quadratic equation .
The equation has equal roots.
Hence, the required value of k is 2 or .
Page No 200:
Answer:
Page No 200:
Answer:
Page No 200:
Answer:
Page No 200:
Answer:
Page No 200:
Answer:
Page No 200:
Answer:
(i) The given equation is .
The given equation has real and distinct roots if D > 0.
(ii) The given equation is .
The given equation has real and distinct roots if D > 0.
(iii) The given equation is .
The given equation has real and distinct roots if D > 0.
(iv) The given equation is .
The given equation has real and distinct roots if D > 0.
Page No 200:
Answer:
The given equation is .
Since a and b are real and a ≠ b, so and .
∴ .....(1) (Product of two positive numbers is always positive)
Also, .....(2) (Product of two positive numbers is always positive)
Adding (1) and (2), we get
(Sum of two positive numbers is always positive)
Hence, the roots of the given equation are real and unequal.
Page No 201:
Answer:
(i)
It is given that the roots of the equation are equal.
Hence Proved.
(ii)
Consider the equation
On comparing the above equation with the general equation , we get
We know that
Since, the value of D is negative, the given equation has no real roots.
Page No 201:
Answer:
It is given that the roots of the equation are real.
Also, the roots of the equation are real.
The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if
Page No 221:
Answer:
Let the required natural number be x.
According to the given condition,
∴ x = 12 (x cannot be negative)
Hence, the required natural number is 12.
Page No 221:
Answer:
Let the required natural number be x.
According to the given condition,
Putting or x = y2, we get
∴ y = 11 (y cannot be negative)
Now,
Hence, the required natural number is 121.
Page No 221:
Answer:
Let the required numbers be x and (28 − x).
According to the given condition,
When x = 12,
28 − x = 28 − 12 = 16
When x = 16,
28 − x = 28 − 16 = 12
Hence, the required numbers are 12 and 16.
Page No 221:
Answer:
Let the required two consecutive positive integers be x and (x + 1).
According to the given condition,
∴ x = 13 (x is a positive integer)
When x = 13,
x + 1 = 13 + 1 = 14
Hence, the required positive integers are 13 and 14.
Page No 221:
Answer:
Let the two consecutive positive odd numbers be x and (x + 2).
According to the given condition,
∴ x = 15 (x is a positive odd number)
When x = 15,
x + 2 = 15 + 2 = 17
Hence, the required numbers are 15 and 17.
Page No 221:
Answer:
Let the two consecutive positive even numbers be x and (x + 2).
According to the given condition,
∴ x = 14 (x is a positive even number)
When x = 14,
x + 2 = 14 + 2 = 16
Hence, the required numbers are 14 and 16.
Page No 221:
Answer:
Let the two consecutive positive integers be x and (x + 1).
According to the given condition,
∴ x = 17 (x is a positive integer)
When x = 17,
x + 1 = 17 + 1 = 18
Hence, the required integers are 17 and 18.
Page No 221:
Answer:
Page No 221:
Answer:
Let the required consecutive multiples of 3 be 3x and 3(x + 1).
According to the given condition,
∴ x = 8 (Neglecting the negative value)
When x = 8,
3x = 3 × 8 = 24
3(x + 1) = 3 × (8 + 1) = 3 × 9 = 27
Hence, the required multiples are 24 and 27.
Page No 221:
Answer:
Let the two consecutive positive odd integers be x and (x + 2).
According to the given condition,
∴ x = 21 (x is a positive odd integer)
When x = 21,
x + 2 = 21 + 2 = 23
Hence, the required integers are 21 and 23.
Page No 221:
Answer:
Let the two consecutive positive even integers be x and (x + 2).
According to the given condition,
∴ x = 16 (x is a positive even integer)
When x = 16,
x + 2 = 16 + 2 = 18
Hence, the required integers are 16 and 18.
Page No 222:
Answer:
Let the required natural numbers be x and (9 − x).
According to the given condition,
When x = 3,
9 − x = 9 − 3 = 6
When x = 6,
9 − x = 9 − 6 = 3
Hence, the required natural numbers are 3 and 6.
Page No 222:
Answer:
Let the required natural numbers be x and (15 − x).
According to the given condition,
When x = 5,
15 − x = 15 − 5 = 10
When x = 10,
15 − x = 15 − 10 = 5
Hence, the required natural numbers are 5 and 10.
Page No 222:
Answer:
Let the required natural numbers be x and (x + 3).
Now, x < x + 3
According to the given condition,
∴ x = 4 (−7 is not a natural number)
When x = 4,
x + 3 = 4 + 3 = 7
Hence, the required natural numbers are 4 and 7.
Page No 222:
Answer:
Let the required natural numbers be x and (x + 5).
Now, x < x + 5
According to the given condition,
∴ x = 2 (−7 is not a natural number)
When x = 2,
x + 5 = 2 + 5 = 7
Hence, the required natural numbers are 2 and 7.
Page No 222:
Answer:
Let the required consecutive multiples of 7 be 7x and 7(x + 1).
According to the given condition,
∴ x = 3 (Neglecting the negative value)
When x = 3,
7x = 7 × 3 = 21
7(x + 1) = 7(3 + 1) = 7 × 4 = 28
Hence, the required multiples are 21 and 28.
Page No 222:
Answer:
Let the natural number be x.
According to the given condition,
∴ x = 8 (x is a natural number)
Hence, the required number is 8.
Page No 222:
Answer:
Let the two parts be x and (57 − x).
According to the given condition,
When x = 40,
57 − x = 57 − 40 = 17
When x = 17,
57 − x = 57 − 17 = 40
Hence, the required parts are 17 and 40.
Page No 222:
Answer:
Let the two parts be x and (27 − x).
According to the given condition,
When x = 12,
27 − x = 27 − 12 = 15
When x = 15,
27 − x = 27 − 15 = 12
Hence, the required parts are 12 and 15.
Page No 222:
Answer:
Page No 222:
Answer:
Page No 222:
Answer:
Page No 222:
Answer:
Let the three consecutive positive integers be x, x + 1 and x + 2.
According to the given condition,
∴ x = 4 (x is a positive integer)
When x = 4,
x + 1 = 4 + 1 = 5
x + 2 = 4 + 2 = 6
Hence, the required integers are 4, 5 and 6.
Page No 222:
Answer:
Page No 222:
Answer:
​​
Page No 222:
Answer:
Page No 222:
Answer:
Let the denominator of the required fraction be x.
Numerator of the required fraction = x − 3
∴ Original fraction =
If 1 is added to the denominator, then the new fraction obtained is .
According to the given condition,
When x = 5,
When x = 9,
(This fraction is neglected because this does not satisfies the given condition.)
Hence, the required fraction is .
Page No 223:
Answer:
Let the required number be x.
According to the given condition,
Hence, the required number is or .
Page No 223:
Answer:
Page No 223:
Answer:
Page No 223:
Answer:
Page No 223:
Answer:
Let x be the number of students who planned a picnic.
∴ Original cost of food for each member = â‚ą
Five students failed to attend the picnic. So, (x − 5) students attended the picnic.
∴ New cost of food for each member = â‚ą
According to the given condition,
â‚ą − â‚ą = â‚ą20
∴ x = 25 (Number of students cannot be negative)
Number of students who attended the picnic = x − 5 = 25 − 5 = 20
Amount paid by each student for the food = â‚ą = â‚ą = â‚ą100
Page No 223:
Answer:
Let the original price of the book be â‚ąx.
∴ Number of books bought at original price for â‚ą600 =
If the price of a book is reduced by â‚ą5, then the new price of the book is â‚ą(x − 5).
∴ Number of books bought at reduced price for â‚ą600 =
According to the given condition,
∴ x = 30 (Price cannot be negative)
Hence, the original price of the book is â‚ą30.
Page No 223:
Answer:
Let the original duration of the tour be x days.
∴ Original daily expenses = â‚ą
If he extends his tour by 4 days, then his new daily expenses = â‚ą
According to the given condition,
â‚ą − â‚ą = â‚ą90
∴ x = 20 (Number of days cannot be negative)
Hence, the original duration of the tour is 20 days.
Page No 223:
Answer:
Let the marks obtained by P in mathematics and science be x and (28 − x), respectively.
According to the given condition,
When x = 12,
28 − x = 28 − 12 = 16
When x = 9,
28 − x = 28 − 9 = 19
Hence, he obtained 12 marks in mathematics and 16 marks in science or 9 marks in mathematics and 19 marks in science.
Page No 223:
Answer:
Let number of pens bought = x
Let the price per pen = Rs y
It is given that the price of x pens is Rs 180
Also, given that if 3 more pens are purchased for the same amount, the price per pen gets reduced by Rs 3
Put the value of x in (ii) we get
We ignore the negative value.
Therefore, number of pens is 12.
Page No 223:
Answer:
Let the cost price of the article be â‚ąx.
∴ Gain percent = x%
According to the given condition,
â‚ąx + â‚ą = â‚ą75 (Cost price + Gain = Selling price)
∴ x = 50 (Cost price cannot be negative)
Hence, the cost price of the article is â‚ą50.
Page No 223:
Answer:
(i)
Let the present age of the son be x years.
∴ Present age of the man = x2 years
One year ago,
Age of the son = (x − 1) years
Age of the man = (x2 − 1) years
According to the given condition,
Age of the man = 8 × Age of the son
∴ x = 7 (Man's age cannot be 1 year)
Present age of the son = 7 years
Present age of the man = 72 years = 49 years
(ii)
Let the age of man be m and the age of son be s
It is given that man is times as old as his son.
Also given that
Put value of (i) in (ii), we get
Ignore the negative value
So, the age of son = s = 10 years
Also, from (i) we have
So, age of man = 35 years
Age of son = 10 years
Page No 224:
Answer:
Let the present age of Meena be x years.
Meena's age 3 years ago = (x − 3) years
Meena's age 5 years hence = (x + 5) years
According to the given condition,
∴ x = 7 (Age cannot be negative)
Hence, the present age of Meena is 7 years.
Page No 224:
Answer:
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Answer:
Page No 224:
Answer:
Let son's age 2 years ago be x years. Then,
Man's age 2 years ago = 3x2 years
∴ Son's present age = (x + 2) years
Man's present age = (3x2 + 2) years
In three years time,
Son's age = (x + 2 + 3) years = (x + 5) years
Man's age = (3x2 + 2 + 3) years = (3x2 + 5) years
According to the given condition,
Man's age = 4 × Son's age
∴ 3x2 + 5 = 4(x + 5)
∴ x = 3 (Age cannot be negative)
Son's present age = (x + 2) years = (3 + 2) years = 5 years
Man's present age = (3x2 + 2) years = (3 × 9 + 2) years = 29 years
Page No 224:
Answer:
Let the first speed of the truck be x km/h.
∴ Time taken to cover 150 km = h
New speed of the truck = (x + 20) km/h
∴ Time taken to cover 200 km = h
According to the given condition,
Time taken to cover 150 km + Time taken to cover 200 km = 5 h
∴ x = 60 (Speed cannot be negative)
Hence, the first speed of the truck is 60 km/h.
Page No 224:
Answer:
Let the original speed of the plane be x km/h.
∴ Actual speed of the plane = (x + 100) km/h
Distance of the journey = 1500 km
Time taken to reach the destination at original speed = h
Time taken to reach the destination at actual speed = h
According to the given condition,
Time taken to reach the destination at original speed = Time taken to reach the destination at actual speed + 30 min
∴ x = 500 (Speed cannot be negative)
Hence, the original speed of the plane is 500 km/h.
Yes, we appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time. This reflects the caring nature of the pilot and his dedication to the work.
Page No 224:
Answer:
Let the usual speed of the train be x km/h.
∴ Reduced speed of the train = (x − 8) km/h
Total distance to be covered = 480 km
Time taken by the train to cover the distance at usual speed = h
Time taken by the train to cover the distance at reduced speed = h
According to the given condition,
Time taken by the train to cover the distance at reduced speed = Time taken by the train to cover the distance at usual speed + 3 h
∴ x = 40 (Speed cannot be negative)
Hence, the usual speed of the train is 40 km/h.
Page No 224:
Answer:
Let the first speed of the train be x km/h.
∴ Time taken to cover 54 km = h
New speed of the train = (x + 6) km/h
∴ Time taken to cover 63 km = h
According to the given condition,
Time taken to cover 54 km + Time taken to cover 63 km = 3 h
∴ x = 36 (Speed cannot be negative)
Hence, the first speed of the train is 36 km/h.
Page No 224:
Answer:
Distance covered by the train = 180 km
We know that distance covered
Also, given that if the speed is increased by 9km/h, time of travel gets reduced by 1 hour.
Put the value of (i) in (ii), we get
Ignore the negative value
So, time taken = 5 hours
From (i)
Hence, the speed is 36 km/h.
Page No 224:
Answer:
Let, speed of stream be x km/h
Speed of boat = 15 km/h
Distance from each side = 30 km
We know that time taken
Total speed of the boat while going upstream km/h
Time taken to go upstream hrs
Total speed of boat while going downstream km/h
Time taken to go downstream hrs
Total time of the journey hrs = 4.5 hrs
Ignore the negative value.
So, the speed of the stream = x = 5 km/h
Page No 225:
Answer:
Distance covered by the train = 300 km
We know that distance covered
Also, given that if the speed is increased by 5 km/h, time of travel gets reduced by 2 hours.
Put the value of (i) in (ii), we get
Ignore the negative value
Therefore, the original speed = 25 km/h
Page No 225:
Answer:
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Answer:
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Answer:
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Answer:
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Answer:
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Answer:
Page No 225:
Answer:
Let the time taken by one pipe to fill the tank be x minutes.
∴ Time taken by the other pipe to fill the tank = (x + 5) min
Suppose the volume of the tank be V.
Volume of the tank filled by one pipe in x minutes = V
∴ Volume of the tank filled by one pipe in 1 minute =
⇒ Volume of the tank filled by one pipe in minutes =
Similarly,
Volume of the tank filled by the other pipe in minutes =
Now,
Volume of the tank filled by one pipe in minutes + Volume of the tank filled by the other pipe in minutes = V
∴ x = 20 (Time cannot be negative)
Time taken by one pipe to fill the tank = 20 min
Time taken by other pipe to fill the tank = (20 + 5) = 25 min
Page No 225:
Answer:
Let the tap of smaller diameter fill the tank in x hours.
∴ Time taken by the tap of larger diameter to fill the tank = (x − 9) h
Suppose the volume of the tank be V.
Volume of the tank filled by the tap of smaller diameter in x hours = V
∴ Volume of the tank filled by the tap of smaller diameter in 1 hour =
⇒ Volume of the tank filled by the tap of smaller diameter in 6 hours =
Similarly,
Volume of the tank filled by the tap of larger diameter in 6 hours =
Now,
Volume of the tank filled by the tap of smaller diameter in 6 hours + Volume of the tank filled by the tap of larger diameter in 6 hours = V
For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possibe.
∴ x = 18
Time taken by the tap of smaller diameter to fill the tank = 18 h
Time taken by the tap of larger diameter to fill the tank = (18 − 9) = 9 h
Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.
Page No 225:
Answer:
Page No 226:
Answer:
Page No 226:
Answer:
Page No 226:
Answer:
Page No 226:
Answer:
Page No 226:
Answer:
Let the length of the side of the first and the second square be and y, respectively.
Putting the value of in (i), we get:
Page No 226:
Answer:
Page No 226:
Answer:
Page No 226:
Answer:
Let the altitude of the triangle be x cm.
Therefore, the base of the triangle will be (x + 10) cm.
Page No 226:
Answer:
Let the altitude of the triangle be x m.
Therefore, the base will be 3x m.
Area of a triangle =
The value of x cannot be negative.
Therefore, the altitude and base of the triangle are 8 m and (3 8 = 24 m), respectively.
Page No 226:
Answer:
Let the base be m.
Therefore, the altitude will be m.
Area of a triangle =
The value of cannot be negative.
Therefore, the base is 15 m and the altitude is {(15 + 7) = 22 m}.
Page No 226:
Answer:
Let one side of the right-angled triangle be m and the other side be m.
On applying Pythagoras theorem, we have:
The value of x cannot be negative.
Therefore, the base is 12 m and the other side is {(12 + 4) = 16 m}.
Page No 226:
Answer:
Let the base and altitude of the right-angled triangle be and y cm, respectively.
Therefore, the hypotenuse will be cm.
Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively.
Page No 226:
Answer:
Let the shortest side be m.
Therefore, according to the question:
Hypotenuse = m
Third side = m
On applying Pythagoras theorem, we get:
The length of the side cannot be 0; therefore, the shortest side is 3 m.
Therefore,
Hypotenuse = 5 m
Third side = (3 + 1) = 4 m
Page No 234:
Answer:
(d)
Page No 234:
Answer:
(b)
Page No 234:
Answer:
(c)
Page No 234:
Answer:
(b) −11
Page No 234:
Answer:
(b) −7
Page No 234:
Answer:
(c) 6
Page No 234:
Answer:
(c) 8
Page No 235:
Answer:
(c) 2 : 3
Page No 235:
Answer:
(d) 3
Page No 235:
Answer:
(d) 5
Page No 235:
Answer:
(d)
Page No 235:
Answer:
(b)
Page No 235:
Answer:
(a)
Page No 235:
Answer:
It is given that α and β are the roots of the equation .
and
Hence, the correct answer is option C.
Page No 235:
Answer:
(c) c = a
Page No 235:
Answer:
(d)
Page No 235:
Answer:
(c) 2 or −2
Page No 235:
Answer:
(a) 1 or 4
Page No 235:
Answer:
(d)
Page No 236:
Answer:
(a) > 0
Page No 236:
Answer:
We know that when discriminant, , the roots of the given quadratic equation are real and unequal.
Hence, the correct answer is option B.
Page No 236:
Answer:
(d) imaginary
Page No 236:
Answer:
(b) real, unequal and irrational
Page No 236:
Answer:
(d) either or
Page No 236:
Answer:
(c)
Page No 236:
Answer:
(c) −2 < k < 2
Page No 236:
Answer:
(b)
Page No 236:
Answer:
(a)
Page No 236:
Answer:
(c) 16 m
Page No 236:
Answer:
Let the breadth of the rectangular field be x m.
∴ Length of the rectangular field = (x + 8) m
Area of the rectangular field = 240 m2 (Given)
(Area = Length × Breadth)
∴ x = 12 (Breadth cannot be negative)
Thus, the breadth of the field is 12 m.
Hence, the correct answer is option C.
Page No 237:
Answer:
The given quadratic equation is .
Thus, the roots of the given equation are 2 and .
Hence, the correct answer is option B.
Page No 237:
Answer:
Let the required natural numbers be x and (8 − x).
It is given that the product of the two numbers is 15.
Hence, the required numbers are 3 and 5.
Page No 237:
Answer:
The given equation is .
Putting x = −3 in the given equation, we get
LHS = = RHS
∴ x = −3 is a solution of the given equation.
Page No 237:
Answer:
The given equation is .
Putting x = −2 in the given equation, we get
LHS = = RHS
∴ x = −2 is a solution of the given equation.
Page No 237:
Answer:
It is given that is a solution of the quadratic equation .
Hence, the value of k is .
Page No 237:
Answer:
The given quadratic equation is .
Hence, the roots of the given equation are 2 and .
Page No 237:
Answer:
The given quadratic equation is .
Hence, and are the solutions of the given equation.
Page No 237:
Answer:
It is given that the roots of the quadratic equation are equal.
Hence, the value of k is 8.
Page No 237:
Answer:
It is given that the quadratic equation has two equal roots.
For p = 0, we get 15 = 0, which is not true.
∴ p ≠ 0
Hence, the value of p is 3.
Page No 237:
Answer:
It is given that y = 1 is a root of the equation .
Also, y = 1 is a root of the equation .
Hence, the value of ab is 3.
Page No 237:
Answer:
Let the other zero of the given polynomial be .
Now,
Sum of the zeroes of the given polynomial = = 4
Hence, the other zero of the given polynomial is .
Page No 237:
Answer:
Let and be the roots of the equation .
(Given)
Hence, the value of k is 3.
Page No 237:
Answer:
It is given that the roots of the quadratic equation are equal.
For p = 0, we get 6 = 0, which is not true.
∴ p ≠ 0
Hence, the value of p is 6.
Page No 237:
Answer:
It is given that the quadratic equation has equal roots.
Hence, 0 and are the required values of k.
Page No 237:
Answer:
It is given that the quadratic equation has equal roots.
Hence, 0 and 4 are the required values of k.
Page No 237:
Answer:
Hence, 1 and are the roots of the given equation.
Page No 237:
Answer:
Hence, −a and are the roots of the given equation.
Page No 237:
Answer:
Hence, and are the roots of the given equation.
Page No 237:
Answer:
Hence, and are the roots of the given equation.
Page No 238:
Answer:
Hence, and are the roots of the given equation.
Page No 238:
Answer:
Hence, and are the roots of the given equation.
Page No 238:
Answer:
Hence, and are the roots of the given equation.
Page No 238:
Answer:
Hence, and are the roots of the given equation.
Page No 238:
Answer:
Hence, and are the roots of the given equation.
Page No 238:
Answer:
Hence, (2a + b) and (2a − b) are the roots of the given equation.
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