Mathematics Part I Solutions Solutions for Class 9 Maths Chapter 4 Ratio And Proportion are provided here with simple step-by-step explanations. These solutions for Ratio And Proportion are extremely popular among Class 9 students for Maths Ratio And Proportion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 9 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.
Page No 61:
Question 1:
Answer:
(i)
72 : 60 = = 6 : 5 (HCF of 72 and 60 = 12)
Thus, the reduced form of 72 : 60 is 6 : 5.
(ii)
38 : 57 = = 2 : 3 (HCF of 38 and 57 = 19)
Thus, the reduced form of 38 : 57 is 2 : 3.
(iii)
52 : 78 = = 2 : 3 (HCF of 52 and 78 = 26)
Thus, the reduced form of 52 : 78 is 2 : 3.
Page No 61:
Question 2:
(i) 700 rs, 308 rs (ii) 14 rs, 12 rs. 40 paise. (iii) 5 litre, 2500 ml (iv) 3 years 4 months, 5years 8 months (v) 3.8 kg, 1900 gm
(vi) 7 minutes 20 seconds, 5 minutes 6 seconds.
Answer:
(i)
Rs 700 : Rs 308 = = 25 : 11 (HCF of 700 and 308 = 28)
(ii)
14 rupees = 1400 p
12 rupees 40 paise = 1200 + 40 = 1240 p
∴ 14 rupees : 12 rupees 40 paise = 1400 p : 1240 p = = 35 : 31 (HCF of 1400 and 1240 = 40)
(iii)
5 litre = 5000 mL (1 L = 1000 mL)
∴ 5 litre : 2500 mL = 5000 mL : 2500 mL = = 2 : 1 (HCF of 5000 and 2500 = 2500)
(iv)
3 years 4 months = 3 years + 4 months = 3 × 12 + 4 = 36 + 4 = 40 months (1 year = 12 months)
5 years 8 months = 5 years + 8 months = 5 × 12 + 8 = 60 + 8 = 68 months
∴ 3 years 4 months : 5 years 8 months = 40 months : 68 months = = 10 : 17 (HCF of 40 and 68 = 4)
(v)
3.8 kg = 3.8 × 1000 = 3800 g (1 kg = 1000 g)
∴ 3.8 kg : 1900 g = 3800 g : 1900 g = = 2 : 1 (HCF of 3800 and 1900 = 1900)
(iv)
7 minutes 20 seconds = 7 minutes + 20 seconds = 7 × 60 + 20 = 420 + 20 = 440 seconds (1 minute = 60 seconds)
5 minutes 6 seconds = 5 minutes + 6 seconds = 5 × 60 + 6 = 300 + 6 = 306 seconds
∴ 7 minutes 20 seconds : 5 minutes 6 seconds = 440 seconds : 306 seconds = = 220 : 153 (HCF of 220 and 153 = 2)
Page No 61:
Question 3:
Answer:
(i)
75 : 100 = = 3 : 4 (HCF of 75 and 100 = 25)
(ii)
44 : 100 = = 11 : 25 (HCF of 44 and 100 = 4)
(iii)
6.25% = = 1 : 16 (HCF of 625 and 10000 = 625)
(iv)
52 : 100 = = 13 : 25 (HCF of 52 and 100 = 4)
(v)
0.64% = = 4 : 625 (HCF of 64 and 10000 = 16)
Page No 61:
Question 4:
Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required?
Answer:
Let the number of persons required to build the same house in 6 days be x.
The number of persons and the number of days required to build the house are in inverse variation. So, the product of number of persons and the number of days required to build the house is constant.
∴ x × 6 = 3 × 8
⇒ x = = 4
Thus, 4 persons are required to build the same house in 6 days.
Page No 61:
Question 5:
Answer:
(i)
15 : 25 = = 15 × 4% = 60%
(ii)
47 : 50 = = 47 × 2% = 94%
(iii)
= 7 × 10%= 70%
(iv)
= 91%
(v)
= 43.75%
Page No 61:
Question 6:
The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha's birth her mothers age was 27 year. Find the present ages of Abha and her mother.
Answer:
The ratio of the present ages of Abha and her mother is 2 : 5.
Let the present age of Abha and her mother be 2x years and 5x years, respectively.
∴ Abha's mother age at the time of Abha's birth = 5x − 2x = 3x years
It is given that at the time of Abha's birth her mothers age was 27 year.
∴ 3x = 27
⇒ x = 9
∴ Present age of Abha = 2x = 2 × 9 = 18 years
Present age of Abha's mother = 5x = 5 × 9 = 45 years
Thus, the present age of Abha and her mother is 18 years and 45 years, respectively.
Page No 61:
Question 7:
Answer:
Let after x years, the ratio of their ages will become 5 : 4.
Age of Vatsala after x years = (14 + x) years
Age of Sara after x years = (10 + x) years
Thus, the ratio of their ages will become 5 : 4 after 6 years.
Page No 61:
Question 8:
The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana's present age ?
Answer:
Let the present ages of Rehana and her mother be 2x years and 7x years, respectively.
After 2 years,
Age of Rehana = (2x + 2) years
Age of Rehana's mother = (7x + 2) years
It is given that after 2 years, the ratio of their ages will be 1 : 3.
∴ Rehana's present age = 2x = 2 × 4 = 8 years
Thus, the present age of Rehana is 8 years.
Page No 63:
Question 1:
Using the property , fill in the blanks substituting proper numbers in the following.
(i) (ii)
Answer:
(i)
28 = 7 × 4, 35 = 5 × 4, 3.5 = 7 × 0.5
Now,
(ii)
4.5 = 9 × 0.5, 42 = 14 × 3, 3.5 = 14 × 0.25
Now,
Page No 63:
Question 2:
(i) The ratio of radius to circumference of the circle.
Answer:
(i)
Let the radius of the circle be r units.
∴ Circumference of the circle = 2r units
Radius of the circle : Circumference of the circle = r : 2r = = 1 : 2
Thus, the ratio of radius to cirumference of the circle is 1 : 2.
(ii)
Radius of the circle = r units
∴ Circumference of the circle = 2r units
Area of the circle = r2 square units
Circumference of the circle : Area of the circle = 2r : r2 = = 2 : r
Thus, the ratio of circumference of circle with radius r to its area is 2 : r.
(iii)
Side of the square = 7 cm
∴ Length of diagonal of the square = × Side of the square = cm
Length of diagonal of the square : Side of the square = cm : 7 cm = = : 1
Thus, the ratio of diagonal of the square to its side is : 1.
(iv)
Length of the rectangle, l = 5 cm
Breadth of the rectangle, b = 3.5 cm
∴ Perimeter of the rectangle = 2(l + b) = 2 × (5 + 3.5) = 2 × 8.5 = 17 cm
Area of the rectangle = l × b = 5 × 3.5 = 17.5 cm2
Perimeter of the rectangle : Area of the rectangle = 17 : 17.5 = = 34 : 35
Thus, the ratio of perimeter to area of the rectangle is 34 : 35.
Page No 63:
Question 3:
Compare the following pairs of ratios.
(i) (ii) (iii) (iv) (v)
Answer:
(i)
Now,
(ii)
Now, 75 < 105
(iii)
Now, 605 > 306
(iv)
Now,
(v)
Now, 65.32 > 17.34
Page No 63:
Question 4:
(i) ABCD is a parallelogram. The ratio of A and B of this parallelogram is 5 : 4. Find the measure of B.
Answer:
(i)
Quadrilateral ABCD is a parallelogram.
Let the measure of A and B be 5x and 4x, respectively.
Now,
A + B = 180º (Adjacent angles of a parallelogram are supplementary)
∴ 5x + 4x = 180º
⇒ 9x = 180º
⇒ x = 20º
∴ Measure of B = 4x = 4 × 20º = 80º
Thus, the measure of B is 80º.
(ii)
Let the present ages of Albert and Salim be 5x years and 9x years, respectively.
5 years hence,
Age of Albert = (5x + 5) years
Age of Salim = (9x + 5) years
It is given that five year hence, the ratio of their ages will be 3 : 5.
∴ Present age of Albert = 5x = 5 × 5 = 25 years
Present age of Salim = 9x = 9 × 5 = 45 years
Thus, the present age of Albert is 25 years and the present age of Salim is 45 years.
(iii)
Let the length and breadth of the rectangle be 3x cm and x cm, respectively.
Perimeter of the rectangle = 36 cm
∴ 2(Length + Breadth) = 36 cm
⇒ 2(3x + x) = 36
⇒ 2 × 4x = 36
⇒ 8x = 36
⇒ x = 4.5
∴ Length of the rectangle = 3x = 3 × 4.5 = 13.5 cm
Breadth of the rectangle = x = 4.5 cm
Thus, the length and breadth of the rectangle is 13.5 cm and 4.5 cm, respectively.
(iv)
Let the two numbers be 31x and 23x.
Sum of the two numbers = 216
∴ 31x + 23x = 216
⇒ 54x = 216
⇒ x = = 4
∴ One number = 31x = 31 × 4 = 124
Other number = 23x = 23 × 4 = 92
Thus, the two numbers are 92 and 124.
(v)
Let the two numbers be 10x and 9x.
Product of the two numbers = 360
∴ 10x × 9x = 360
⇒ 90x2 = 360
⇒ x2 = 4
⇒ x = 2
∴ One number = 10x = 10 × 2 = 20
Other number = 9x = 9 × 2 = 18
Thus, the two numbers are 18 and 20.
Page No 64:
Question 5:
If a : b = 3 : 1 and b : c = 5 : 1 then find the value of
(i) (ii)
Answer:
a : b = 3 : 1
.....(1)
b : c = 5 : 1
.....(2)
From (1) and (2), we have
a = 3 × 5c = 15c
(i)
(ii)
Page No 64:
Question 6:
If then find the ratio .
Answer:
∴ a : b = 2 : 125
Page No 64:
Question 7:
( x + 3) : ( x + 11) = ( x 2) : ( x + 1) then find the value of x.
Answer:
Thus, the value of x is 5.
Page No 70:
Question 1:
If then find the values of the following ratios.
(i) (ii) (iii) (iv)
Answer:
(i)
(ii)
(iii)
(iv)
Page No 70:
Question 2:
If then find the values of following ratios.
(i) (ii) (iii) (iv)
Answer:
(i)
Applying componendo and dividendo, we get
(ii)
(iii)
(iv)
Page No 70:
Question 3:
If then find the value of the ratio
Answer:
Applying componendo and dividendo, we get
Now,
Page No 70:
Question 4:
Solve the following equations.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Answer:
(i)
Multiplying both sides by , we get
Using dividendo, we get
This equation is true for x = 0. Therefore, x = 0 is a solution of the given equation.
If x ≠ 0, then x2 ≠ 0.
Dividing both sides by x2, we get
Thus, the solutions of the given equation are x = 0 and x = 8.
(ii)
Multiplying both sides by , we get
Using dividendo, we get
Thus, the solution of the give equation is x = 9.
(iii)
Applying componendo and dividendo, we get
Thus, the solution of the given equation is x = 2.
(iv)
Applying componendo and dividendo, we get
Squaring on both sides, we get
Thus, the solution of the given equation is x = 6.
(v)
Applying dividendo, we get
Taking square root on both sides, we get
Thus, the solution of the given equation is .
(vi)
Applying componendo and dividendo, we get
Taking cube root on both sides, we get
Thus, the solution of the given equation is x = 3.
Page No 73:
Question 1:
Fill in the blanks of the following
(i)
(ii)
Answer:
(i)
Also,
(ii)
Also,
Page No 73:
Question 2:
5 then find the values of the following expressions.
(i)
(ii)
Answer:
(i)
​
(ii)
Page No 73:
Question 3:
(ii) If and then show that the value of each ratio is equal to 1.
(iii) If and then show that .
(iv) If then show that .
(v) If then show that every ratio = .
Answer:
(i)
Let
Similarly,
From (1), (2) and (3), we have
(ii)
(Theorem of equal ratios)
(iii)
(Theorem of equal ratios)
(iv)
(By invertendo)
Now,
Also,
From (1), (2) and (3), we have
(v)
Page No 74:
Question 4:
Solve
(i)
(ii)
Answer:
(i)
If x = 0, then ⇒ , which is not true.
So, x = 0 is not a solution of the given equation.
Now,
Thus, the solution of the given equation is x = 2.
(ii)
If y = 0, then ⇒ , which is not true.
So, y = 0 is not a solution of the given equation.
Now,
Thus, the solution of the given equation is y = 1.
Page No 77:
Question 1:
Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?
Answer:
Le the number x be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion.
∴ (12 − x), (16 − x), (21 − x) are in continued proportion.
Thus, the number −4 must be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion.
Page No 77:
Question 2:
If (28) is the mean proportional of (23) and (19) then find the vaue of x.
Answer:
It is given that (28 − x) is the mean proportional of (23 − x) and (19 − x).
Thus, the value of x is .
Page No 77:
Question 3:
Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.
Answer:
Let the remaining two numbers be x and y.
So, the numbers x, 12, y are in continued proportion.
​∴ xy = (12)2 = 144 .....(1)
Also,
x + y = 26 .....(2)
Solving (1) and (2), we get
​​
⇒ x − 8 = 0 or x − 18 = 0
⇒ x = 8 or x = 18
When x = 8, y = 26 − 8 = 18 [Using (2)]
When x = 18, y = 26 − 18 = 8
Thus, the numbers are 8, 12, 18 or 18, 12, 8.
Page No 77:
Question 4:
If (a + b + c) (a b + c) = a2 + b2 + c2 show that a, b, c are in continued proportion.
Answer:
Therefore, a, b, c are in continued proportion.
Page No 77:
Question 5:
(ii) (a2 + b2) (b2 + c2 ) = (ab + bc)2
Answer:
(i)
(ii)
(iii)
Page No 77:
Question 6:
Find mean proportional of
Answer:
If b is the mean proportional of a and c, then or .
Mean proportional of and
Page No 77:
Question 1:
(ii) What is the ratio of 1 mm to 1 cm ?
(iii * ) The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin’s age to Mohasin’s age ?
(iv) 24 Bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get ?
(v) What is the mean proportional of 4 and 25 ?
(A) 6 (B) 8 (C) 10 (D) 12
Answer:
(i)
6 : 5 = y : 20
Thus, the value of y is 24.
Hence, the correct answer is option (B).
(ii)
Ratio of 1 mm to 1 cm
= 1 mm : 1 cm
= 1 mm : 10 mm (1 cm = 10 mm)
= 1 : 10
Hence, the correct answer is option (C).
(iii)
Age of Nitin = 24 years
Age of Mohasin = 36 years
∴ Ratio of Nitin's age to Mohansin's age
= Age of Nitin : Age of Mohasin
= 24 years : 36 years
= 2 : 3
Thus, the ratio of Nitin's age to Mohasin's age is 2 : 3.
Hence, the correct answer is option (B).
(iv)
The number of bananas received by Shubham and Anil is the ratio 3 : 5.
Let the number of bananas received by Shubham and Anil be 3x and 5x, respectively.
∴ 3x + 5x = 24
⇒ 8x = 24
⇒ x = 3
∴ Number of bananas received by Shubham = 3x = 3 × 3 = 9
Thus, Shubham gets 9 bananas.
Hence, the correct answer is option (D).
(v)
Mean proportional of 4 and 25
Thus, the mean proportional of 4 and 25 is 10.
Hence, the correct answer is option (C).
Page No 78:
Question 2:
Answer:
(i)
21 : 48 = = 7 : 16 (HCF of 21 and 48 = 3)
Thus, the reduced form of 21 : 48 is 7 : 16.
(ii)
36 : 90 = = 2 : 5 (HCF of 36 and 90 = 18)
Thus, the reduced form of 36 : 90 is 2 : 5.
(iii)
65 : 117 = = 5 : 9 (HCF of 65 and 113 = 13)
Thus, the reduced form of 65 : 117 is 5 : 9.
(iv)
138 : 161 = = 6 : 7 (HCF of 138 and 161 = 23)
Thus, the reduced form of 138 : 161 is 6 : 7.
(v)
114 : 133 = = 6 : 7 (HCF of 114 and 133 = 19)
Thus, the reduced form of 114 : 133 is 6 : 7.
Page No 78:
Question 3:
Answer:
(i)
Let the radius of the circle be r units.
Diameter of the circle = 2 × Radius of the circle = 2r units
∴ Ratio of radius to diameter of a circle = Radius of the circle : Diameter of the circle = r : 2r = 1 : 2
(ii)
Length of the rectangle, l = 4 cm
Breadth of the rectangle, b = 3 cm
Now,
Diagonal of the rectangle = = 5 cm
∴ Ratio of diagonal to the length of the rectangle = Diagonal of the rectangle : Length of the rectangle = 5 cm : 4 cm = 5 : 4
(iii)
Side of the squre = 4 cm
Perimeter of the square = 4 × Side of the squre = 4 × 4 cm = 16 cm
Area of the squre = (Side of the square)2 = (4 cm)2 = 16 cm2
∴ Ratio of perimeter to area of the square = Perimeter of the square : Area of the squre = 16 : 16 = 1 : 1
Page No 78:
Question 4:
Answer:
The numbers a, b, c are in continued proportion if .
(i)
Since , so the numbers 2, 4, 8 are in continued proportion.
(ii)
Since , so the numbers 1, 2, 3 are not in continued proportion.
(iii)
Since , so the numbers 9, 12, 16 are in continued proportion.
(iv)
Since , so the numbers 3, 5, 8 are not in continued proportion.
Page No 78:
Question 5:
a, b, c are in continued proportion. If a = 3 and c = 27 then find b.
Answer:
It is given that 3, b, 27 are in continued proportion.
Thus, the value of b is 9.
Page No 78:
Question 6:
Convert the following ratios into percentages..
(i) 37 : 500 (ii) (iii) (iv) ( v)
Answer:
(i)
37 : 500 =
(ii)
(iii)
(iv)
(v)
​
Page No 78:
Question 7:
(i) 1024 MB, 1.2 GB [(1024 MB = 1 GB)] (ii) 17 Rupees, 25 Rupees 60 paise (iii) 5 dozen, 120 units
Answer:
(i)
Ratio of 1024 MB to 1.2 GB
= 1024 MB : 1.2 × 1024 MB
= 5 : 6
(ii)
17 Rupees = 1700 paise (Re 1 = 100 paise)
25 Rupees 60 paise = 25 Rupees + 60 paise = 2500 paise + 60 paise = 2560 paise
∴ Ratio of 17 Rupees to 25 Rupees 60 paise
= 1700 paise : 2560 paise
= 85 : 128
(iii)
Ratio of 5 dozen to 120 units
= 5 × 12 units : 120 units
= 60 units : 120 units
= 1 : 2
(iv)
1 m2 = 10000 cm2
4 m2 = 4 × 10000 cm2 = 40000 cm2
∴ Ratio of 4 m2 to 800 cm2
= 40000 cm2 : 800 cm2
= 50 : 1
(v)
1.5 kg = 1.5 × 1000 g = 1500 g
∴ Ratio of 1.5 kg to 2500 g
= 1500 g : 2500 g
= 3 : 5
Page No 78:
Question 8:
If then find the values of the following expressions.
(i) (ii) (iii) (iv)
Answer:
(i)
Applying componendo, we get
(ii)
Applying componendo and dividendo, we get
(iii)
Applying componendo, we get
(iv)
Applying invertendo, we get
Applying componendo and dividendo, we get
Applying invertendo, we get
Page No 78:
Question 9:
If a, b, c, d are in proportion , then prove that
(i)
(ii)
(iii)
Answer:
It is given that a, b, c, d are in proportion.
(i)
From (1) and (2), we get
(ii)
From (1) and (2), we get
(iii)
From (1) and (2), we get
Page No 79:
Question 10:
If a, b, c are in continued proportion , then prove that
(i)
(ii)
Answer:
a, b, c are in continued proportion.
(i)
From (1) and (2), we get
(ii)
From (1) and (2), we get
Page No 79:
Question 11:
Solve :
Answer:
If x = 0, then , which is not true.
So, x = 0 is not a solution of the given equation.
Now,
Thus, the solution of the given equation is x = 9.
Page No 79:
Question 12:
If then prove that every ratio = .
Answer:
Page No 79:
Question 13:
(13 * ) If then prove that .
Answer:
Again applying theorem of equal ratios, we get
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