RS Aggarwal Solutions for Class 7 Maths Chapter 8–Ratio and Proportion

Class 7: Maths Chapter 8 solutions. Complete Class 7 Maths Chapter 8 Notes.

RS Aggarwal Solutions for Class 7 Maths Chapter 8–Ratio and Proportion

RS Aggarwal 7th Maths Chapter 8, Class 7 Maths Chapter 8 solutions

Ex 8A Solutions

Question 1.
Solution:
(i) 24 : 40
HCF of 24, 40 = 8
24 : 40 = 24 ÷ 8 : 40 ÷ 8 = 3 : 5 (Dividing by 8)
(ii) 13.5 : 15 or 135 : 150
HCF of 135 and 150 = 15
135 ÷ 15 : 150 ÷ 15 (Dividing by 15)
= 9 : 10

HCF of 25, 65, 80 = 5
Dividing by 5,
5 : 13 : 16

Question 2.
Solution:
(i) 75 paise : 3 rupees = 75 paise : 300 paise
(converting into same kind)
HCF of 75, 300 = 75
75 : 300 = 75 ÷ 75 : 300 ÷ 75 (Dividing by 75) = 1 : 4
(ii) 1 m 5 cm : 63 cm = 105 cm : 63 cm
(converting into same kind)
HCF of 105 and 63 = 21
105 ÷ 21 : 63 ÷ 21 (Dividing by 21)
= 5 : 3
(iii) 1 hour 5 minutes : 45 minutes = 65 minutes : 45 minutes
(converting into minutes)
13 : 9 (dividing by 5)
= 13 : 9
(iv) 8 months : 1 year = 8 months : 12 months
(converting into the same kind)
HCF of 8 and 12 = 4
Dividing by 4
8 ÷ 4 : 12 ÷ 4
= 2 : 3
(v) 2 kg 250 g : 3 kg = 2250 g : 3000 g (converting into the same kind)
HCF of 2250 and 3000 = 750
Dividing by 750,
2250 ÷ 750 : 3000 ÷ 750 = 3 : 4
(vi) 1 km : 750 m = 1000 m : 750 m
(converting into metre)
= 4 : 3 (dividing by 250)
= 4 : 3

Question 3.
Solution:

Question 4.
Solution:
A : B = 5 : 8, B : C = 16 : 25

Question 5.
Solution:
A : B = 3 : 5, B : C = 10 : 13

A : B : C = 6 : 10 : 13

Question 6.
Solution:
A : B = 5 : 6 and B : C = 4 : 7

Question 7.
Solution:
Total amount = Rs. 360
Sum of ratios = 7 + 8 = 15

Question 8.
Solution:
Total amount = Rs. 880

Question 9.
Solution:
Total amount = Rs. 5600
Ratio in A : B : C = 1 : 3 : 4

Question 10.
Solution:
Let x be added to each term Then
9 + x : 16 + x = 2 : 3

Question 11.
Solution:
Let x be subtracted from each term Then
(17 – x) : (33 – x) = 7 : 15

Question 12.
Solution:
Ratio in two numbers = 7 : 11
Let first number = 7x
Then second number = 11x
Then adding 7 to each number, the ratio is 2 : 3
7x+711x+7 = 23
By cross multiplying:
3 (7x + 7) = 2 (11x + 7)
⇒ 21x + 21 = 22x + 14
⇒ 21 – 14 = 22x – 21x
⇒ x = 7
First number = 7x = 7 x 7 = 49
and second number = 11x = 11 x 7 = 77
Hence numbers are 49, 77

Question 13.
Solution:
The ratio in two numbers = 5 : 9
Let the first number = 5x
Then second number = 9x
By subtracting 3 from each number the ratio is 1 : 2
5x–39x–3 = 12
By cross multiplication,
2 (5x – 3) = 1 (9x – 3)
⇒ 10x – 6 = 9x – 3
⇒ 10x – 9x = -3 + 6
⇒ x = 3
First number = 5x = 5 x 3 = 15
and second .number = 9x = 9 x 3 = 27
Hence numbers are 15, 27

Question 14.
Solution:
Ratio in two numbers = 3 : 4
LCM = 180
Let first number = 3x
Then second number = 4x
Now LCM = 3 x 4 x x = 12x
12x = 180
⇒ x = 15
Numbers will be 3 x 15 = 45 and 4 x 15 = 60

Question 15.
Solution:
Ratio in present ages of A and B = 8 : 3
Let A’s age = 8x
Then B’s age = 3x
6 years hence,
A’s age will be = 8x + 6
and B’s will be = 3x + 6
8x+63x+6 = 94
(By cross multiplication)
4 (8x + 6) = 9 (3x + 6)
⇒ 32x + 24 = 27x + 54
⇒ 32x – 27x = 54 – 24
⇒ 5x = 30
⇒ x = 6
A’s present age = 8x = 8 x 6 = 48 years
and B’s age = 3x = 3 x 6 = 18 years

Question 16.
Solution:
Ratio in copper and zinc = 9 : 5
Let alloy = x gm

Question 17.
Solution:
Ratio in boys and girls = 8 : 3
and total number of girls = 375
Let number of boys = 8x
Then number of girls = 3x
3x = 375
⇒ x = 125
Number of boys = 8x = 8 x 125 = 1000

Question 18.
Solution:
Ratio in income and savings = 11 : 2
Let income = 11x
Then savings = 2x
But savings = Rs. 2500
2x = 2500
⇒ x = 1250
Then income = 1250 x 11 = Rs. 13750
and expenditure = Total income – savings = 13750 – 2500 = Rs. 11250

Question 19.
Solution:
Total amount = Rs. 750
Ratio in rupee, 50 P and 25 P coins =5 : 8 : 4
Let number of rupees = 5x
Number of 50 P coins = 8x
and number of 25 coins = 4x
According to the condition,

Number of 1 Re coins = 5x = 5 x 75 = 375
Number of 50 P coins = 8x = 8 x 75 = 600
and number of 25 P coins = 4x = 4 x 75 = 300

Question 20.
Solution:
(4x + 5) : (3x + 11) = 13 : 17
4x+53x+11 = 1317
By cross multiplication,
68x + 85 = 39x + 143
⇒ 68x – 39x = 143 – 85
⇒ 29x = 58
x = 2
Hence x = 2

Question 21.
Solution:
x : y = 3 : 4

Question 22.
Solution:
x : y = 6 : 11
xy = 611
Now (8x – 3y) : (3x + 2y)

Question 23.
Solution:
Sum of two numbers = 720
Ratio of two numbers = 5 : 7
Let first number = 5x
Then second number = 7x
5x + 7x = 720
⇒ 12x = 720
⇒ x = 60
First number = 5x = 5 x 60 = 300
and second number = 7x = 7 x 60 = 420

Question 24.
Solution:
(i) (5 : 6) or (7 : 9)

Question 25.
Solution:
(i) (5 : 6), (8 : 9), (11 : 18)

Ex 8B Solutions

Question 1.
Solution:
We know that a, b, c, d are in proportion if ad = bc
Now 30, 40, 45, 60 are in proportion
if 30 x 60 = 40 x 45
if 1800= 1800
which is true
30, 40, 45, 60 are in proportion.

Question 2.
Solution:
We know that if a, b, c, d are in proportion if ad = bc
Now 36, 49, 6, 7 are in proportion
if 36 x 7 = 49 x 6
if 252 = 294
But 252 ≠ 294
36, 49, 6, 7 are not in proportion

Question 3.
Solution:
2 : 9 : : x : 27
9 x x = 2 x 27
x = 2×279 = 2 x 3 = 6

Question 4.
Solution:
8 : x : : 16 : 35
x x 16 = 8 x 35
x = 8×356 = 352 = 17.5

Question 5.
Solution:
x : 35 : : 48 : 60
x x 60 = 35 x 48
x = 7 x 4 = 28

Question 6.
Solution:
Let x be the fourth proportional, then

x = 352 = 17.5
Fourth proportional = 17.5

Question 7.
Solution:
36, 54, x are in continued proportion
36 : 54 : : 54 : x
⇒ 36 x x = 54 x 54

Question 8.
Solution:
27, 36, x are in continued proportion
27 : 36 :: 36 : x
27 x x = 36 x 36

Question 9.
Solution:
Let x be the third proportional, then
(i) 8 : 12 : : 12 : x

Question 10.
Solution:
Third proportional = 28 then
7 : x :: x : 28
⇒ 7 x 28 = x x x
⇒ x² = 28 x 7 = 196
⇒ x = √196 = 14
x = 14

Question 11.
Solution:
Let x be the mean proportional, then
(i) 6 : x :: x : 24
⇒ x² = 6 x 24 = 144
x = √144 = 12
Mean proportional = 12
(ii) 3 : x : : x : 27
⇒ x² = 3 x 27 = 81
x = √81 = 9
Mean proportional = 9
(iii) 0.4 : x :: x : 0.9
⇒ x² = 0.4 x 0.9
x = √o.36 = 0.6
Mean proportional = 0.6

Question 12.
Solution:
Let x be added to each of the given numbers then
5 + x, 9 + x, 7 + x, 12 + x are in proportion
5+x9+x = 7+x12+x
By cross multiplication :
(5 + x) (12 + x) = (7 + x) (9 + x)
⇒ 60 + 5x + 12x + x² = 63 + 7x + 9x + x²
⇒ 60 + 17x + x² = 63 + 16x + x²
⇒ 17x + x² – 16x – x² = 63 – 60
⇒ x = 3
Required number = 3

Question 13.
Solution:
Let x be subtracted from each of the given number, then
10 – x, 12 – x, 19 – x and 24 – x are in proportion
10–x12–x = 19–x24–x
By cross multiplication :
(10 – x) (24 – x) = (19 – x) (12 – x)
⇒ 240 – 10x – 24x + x² = 228 – 19x – 12x + x²
⇒ 240 – 34x + x² = 228 – 31x + x²
⇒ -34x + x² + 31x – x² = 228 – 240
⇒ -3x = -2
⇒ 3x = 12
⇒ x = 4
Required number = 4

Question 14.
Solution:
Scale of map = 1 : 5000000
Distance between two town on the map = 4 cm

Question 15.
Solution:
Height of a tree = 6 cm
and its shadow at same time = 8 m
Shadow of a pole = 20 m
Let height of pole = x m
6 : 8 = x : 20
⇒ x= 6×208 = 15 m
Height of pole = 15 m

Ex 8C Solutions

Objective questions :
Mark (✓) against the correct answers in each of the following :
Question 1.
Solution:
(d) a : b = 3 : 4, b : c = 8 : 9

Question 2
Solution:
(a) A : B = 2 : 3, B : C = 4 : 5

Question 3.
Solution:
(d)

Question 4.
Solution:
(b) 15% of A = 20% of B

Question 5.
Solution:
(a)

Question 6.
Solution:
(b) A : B = 5 : 7, B : C = 6 : 11
LCM of 7, 6 = 42

Question 7.
Solution:
(c) 2A = 3B = 4C = x

Question 8.
Solution:
(a)
A3 = B4 = C5 = 1(suppose)
A = 3, B = 4, C = 5
A : B : C = 3 : 4 : 5

Question 9.
Solution:
(b)

Question 10.
Solution:
(c)

Question 11.
Solution:
(c) (3a + 5b) : (3a – 5b) = 5 : 1

Question 12.
Solution:
(c) 7 : x :: 35 : 45

x = 9

Question 13.
Solution:
(b) Let x to be added to each term of 3 : 5
Then 3+x5+x = 56
By cross multiplication
18 + 6x = 25 + 5x
6x – 5x = 25 – 18
x = 7
7 is to be added

Question 14.
Solution:
(d) Ratio in two numbers = 3 : 5
Let first number = 3x
Then second number = 5x
According to the condition,
3x+105x+10 = 57
(By cross multiplication)
25x + 50 = 21x + 70
25x – 21x = 70 – 50
4x = 20
x = 5
First number = 3 x 5 = 15
and second number = 5 x 5 = 25
Sum of numbers = 15 + 25 = 40

Question 15.
Solution:
(a)
Let x be subtracted from each of the term
15–x19–x = 34
⇒ 4 (15 – x) = 3 (19 – x)
⇒ 60 – 4x = 57 – 3x
⇒ -4x + 3x = 57 – 60
⇒ -x = -3
x = 3
Required number = 3

Question 16.
Solution:
(a)
Amount = Rs. 420
and ratio = 3 : 4
Sum of ratios = 3 + 4 = 7
A’s share = 420×37 = Rs. 60 x 3 = Rs. 180

Question 17.
Solution:
(d)
Let number of boys = x, then
x : 160 : : 8 : 5
⇒ x x 5 = 160 x 8
x = 160×85 = 32 x 8 = 256
Number of total students of the school = 256 + 160 = 416

Question 18.
Solution:
(a)

Question 19.
Solution:
(c)
Let x be the third proportional to 9 and 12 then
9 : 12 :: x : 12
⇒ 9 x x = 12 x 12
⇒ x = 12×129 = 1449 = 16
Third proportional = 16

Question 20.
Solution:
Answer = (b)
Mean proportional of 9 and 16 = √(9 x 16) = √144 = 12

Question 21.
Solution:
(a)
Let age of A = 3x
and age of B = 8x
6 years hence, their ages will be 3x + 6 and 8x + 6
3x+68x+6 = 49
⇒ 9 (3x + 6) = 4 (8x + 6)
⇒ 27x + 54 = 32x + 24
⇒ 32x – 27x = 54 – 24
⇒ 5x = 30
⇒ x = 6
A’s age = 3x = 3 x 6 = 18 years

RS Aggarwal Solutions for Class 7 Maths Chapter 8: Download PDF

RS Aggarwal Solutions for Class 7 Maths Chapter 8–Ratio and Proportion

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Chapterwise RS Aggarwal Solutions for Class 7 Maths :

• Chapter 1–Integers
• Chapter 2–Fractions
• Chapter 3–Decimals
• Chapter 4–Rational Numbers
• Chapter 5–Exponents
• Chapter 6–Algebraic Expressions
• Chapter 7–Linear Equations in One Variable
• Chapter 8–Ratio and Proportion
• Chapter 9–Unitary Method
• Chapter 10–Percentage
• Chapter 11–Profit and Loss
• Chapter 12–Simple Interest
• Chapter 13–Lines and Angles
• Chapter 14–Properties of Parallel Lines
• Chapter 15–Properties of Triangles
• Chapter 16–Congruence
• Chapter 17–Constructions
• Chapter 18–Reflection and Rotational Symmetry
• Chapter 19–Three-Dimensional Shapes
• Chapter 20–Mensuration
• Chapter 21–Collection and Organisation of Data (Mean, Median and Mode)
• Chapter 22–Bar Graphs
• Chapter 23–Probability

About RS Aggarwal Class 7 Book

Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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