Maths Class 7 RS Aggarwal Solutions

RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9A
• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9B
• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9C
• RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method CCE Test Paper

Question 1.
Solution:
Cost of 15 oranges = Rs. 110
Cost of 1 orange = Rs. \(\frac { 110 }{ 15 }\)
and cost of 39 oranges = Rs. \(\frac { 110 }{ 15 }\) x 39
= Rs. 22 x 13 = Rs. 286

Question 2.
Solution:
In Rs. 260, the sugar is bought = 8 kg
and in Re. 1, the sugar is bought = \(\frac { 8 }{ 260 }\) kg
Then in Rs. 877.50, the sugar will be bought = \(\frac { 8 }{ 260 }\) x 877.50 kg
= \(\frac { 8 }{ 260 }\) x \(\frac { 87750 }{ 100 }\)
= 27 kg

Question 3.
Solution:
In Rs. 6290, silk is purchased = 37 m
and in Re. 1, silk is purchased = \(\frac { 37 }{ 6290 }\) m
and in Rs. 4420, silk will be purchased 37
= \(\frac { 37 }{ 6290 }\) x 4420 m = 26 m

Question 4.
Solution:
Rs. 1110 is wages for = 6 days.
Re. 1 will be wages for = \(\frac { 6 }{ 1110 }\) days
and Rs. 4625 will be wages for

Question 5.
Solution:
In 42 litres of petrol, a car covers = 357 km
and in 1 litre, car will cover = \(\frac { 357 }{ 42 }\) km
and in 12 litres, car will cover = \(\frac { 357 }{ 42 }\) x 12 = 102 km

Question 6.
Solution:
Cost of travelling 900 km is = Rs. 2520
and cost of 1 km will be = Rs. \(\frac { 2520 }{ 900 }\)
andcostof360kmwillbe = Rs. \(\frac { 2520 }{ 900 }\) x 360 = Rs. 1008

Question 7.
Solution:
To cover a distance of 51 km, time is taken = 45 minutes

Question 8.
Solution:
If weight is 85.5 kg, then length of iron rod = 22.5 m
If weight is 1 kg, then length of rod will be

Question 9.
Solution:
In 162 grams, sheets are = 6

Question 10.
Solution:
1152 bars of soap can be packed in 8 cartons
1 bar of soap coil be packed in

Question 11.
Solution:
In 44 mm of thickness, cardboards are = 16
In 1 mm of thickness, cardboards will be

Question 12.
Solution:
If length of shadow is 8.2 m, then
height of flag staff is = 7 m
If length of shadow is 1 m, then height will

Question 13.
Solution:
16.25 m long wall is build by = 15 men

Question 14.
Solution:
1350 litres of milk cm be consumed by = 60 patients
1 litres of milk can be consumed by = \(\frac { 60 }{ 1350 }\) patients
and 1710 litres of milk can be consumed

Question 15.
Solution:
2.8 cm extension is produced by = 150 g.
1 cm extension will be produced by = \(\frac { 150 }{ 2.8 }\) g
and 19.6 cm extension will be produced by

Hope given RS Aggarwal Solutions Class 7 Chapter 9 Unitary Method Ex 9A are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:
The sum of ratio terms is = 2 + 3 + 5 = 10
Then, we have :
A’s share = ₹ \(\frac { 2 }{ 10 }\) x 1100 = ₹ 220
B’s share = ₹ \(\frac { 3 }{ 10 }\) x 1100 = ₹ 330
C’s share = ₹ \(\frac { 5 }{ 10 }\) x 1100 = ₹ 550

Question 3.
Solution:
Product of the extremes = 25 x 6 = 150
Product of the means = 36 x 5 = 180
The product of the extremes is not equal to that of the means.
Hence, 25, 36, 5 and 6 are not in proportion.

Question 4.
Solution:
x : 18 :: 18 : 108
⇒ x x 108 = 18 x 18
(Product of extremes = Product of means)
⇒ 108x = 324
⇒ x = 3
Hence, the value of x is 3.

Question 5.
Solution:
Suppose that the numbers are 5x and 7x
Then, 5x + 7x = 84
12x = 84
x = 7
Hence, the numbers are (5 x 7) = 35 and (7 x 7) = 49.

Question 6.
Solution:
Suppose that the present ages of A and B are 4x yrs and 3x yrs, respectively
Eight years ago, age of A = (4x – 8) yrs
Eight years ago, age of B = (3x – 8) yrs
Then,
(4x – 8) : (3x – 8) = 10 : 7
⇒ \(\frac { 4x – 8 }{ 3x – 8 }\) = \(\frac { 10 }{ 7 }\)
⇒ 28x – 56 = 30x – 80
⇒ 2x = 24
⇒ x = 12

Question 7.
Solution:
Distance covered in 60 min = 54 km
Distance covered in 1 min = \(\frac { 54 }{ 60 }\) km
Distance covered in 40 min = \(\frac { 54 }{ 60 }\) x 40 = 36 km

Question 8.
Solution:
Suppose that the third proportional to 8 and 12 is x
Then, 8 : 12 :: 12 : x
⇒ 8x = 144 (Product of extremes = Product of means)
x = 18
Hence, the third proportional is 18.

Question 9.
Solution:
40 men can finish the work in 60 days
1 man can finish the work in 60 x 40 days [Less men, more days]
75 men will finish the work in = \(\frac { 60 x 40 }{ 75 }\) = 32 days
Hence, 75 men will finish the same work in 32 days.

Mark (✓) against the correct answer in each of the following :
Question 10.
Solution:
(d) 6 : 4 : 3
A = \(\frac { 3 }{ 2 }\) B
C = \(\frac { 3 }{ 4 }\) B
A : B : C = \(\frac { 3 }{ 2 }\) B : B : \(\frac { 3 }{ 4 }\) B
= 6 : 4 : 3

Question 11.
Solution:
(a) 2 : 3 : 4

Question 12.
Solution:
(c) 11 : 3

Question 13.
Solution:
(a) 3
Let us assume that the number to be subtracted is x
Then, (15 – x) : (19 – x) = 15 : 3
⇒ \(\frac { 15 – x }{ 19 – x }\) = \(\frac { 3 }{ 4 }\)
⇒ 60 – 4x = 57 – 3x
⇒ x = 3

Question 14.
Solution:
(b) ₹ 360
Sum of the ratio terms = 4 + 3 = 7
B’s share = ₹ 840 x \(\frac { 3 }{ 7 }\) = ₹ 360

Question 15.
Solution:
(c) 40 years
Suppose that the present ages of A and B are 5x yrs and 2x yrs, respectively
After 5 years, the ages of A and B will be (5x + 5) yrs and (2x + 5) yrs, respectively
Then, (5x+ 5) : (2x + 5) = 15 : 7
⇒ \(\frac { 5x+ 5 }{ 2x + 5 }\) = \(\frac { 15 }{ 7 }\)
Cross multiplying; we get:
⇒ 35x + 35 = 30x + 75
⇒ 5x = 40
⇒ x = 8
Then, the present age of A is 5 x 8 = 40 yrs.

Question 16.
Solution:
(a) 896
Suppose that the number of boys in the school is x
Then, x : 320 = 9 : 5
⇒ 5x = 2880
⇒ x = 576
Hence, total strength of the school = 576 + 320 = 896

Question 17.
Solution:

Question 18.
Solution:
(i) True
Suppose that the men proportional is x
Then, 0.4 : x :: x : 0.9
⇒ 0.9 x 0.4 = x x x (Product of extremes = Product of means)
⇒ x² = 0.36
⇒ x = 0.6
(ii) False
Suppose that the third proportional is x.
Then, 9: 12 :: 12 : x
⇒ 9x = 144 (Product of extremes = Product of means)
⇒ x = 16
(iii) True
8 : x :: 48 : 18
⇒ 144 = 48x (Product of extremes = Product of means)
⇒ x = 3
(iv) True

Hope given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper

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)
\(\frac { A }{ 3 }\) = \(\frac { B }{ 4 }\) = \(\frac { C }{ 5 }\) = 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 \(\frac { 3 + x }{ 5 + x }\) = \(\frac { 5 }{ 6 }\)
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,
\(\frac { 3x + 10 }{ 5x + 10 }\) = \(\frac { 5 }{ 7 }\)
(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
\(\frac { 15 – x }{ 19 – x }\) = \(\frac { 3 }{ 4 }\)
⇒ 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 = \(\frac { 420 x 3 }{ 7 }\) = 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 = \(\frac { 160 x 8 }{ 5 }\) = 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 = \(\frac { 12 x 12 }{ 9 }\) = \(\frac { 144 }{ 9 }\) = 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
\(\frac { 3x + 6 }{ 8x + 6 }\) = \(\frac { 4 }{ 9 }\)
⇒ 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

Hope given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper

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 = \(\frac { 2 x 27 }{ 9 }\) = 2 x 3 = 6

Question 4.
Solution:
8 : x : : 16 : 35
x x 16 = 8 x 35
x = \(\frac { 8 x 35 }{ 6 }\) = \(\frac { 35 }{ 2 }\) = 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 = \(\frac { 35 }{ 2 }\) = 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
\(\frac { 5 + x }{ 9 + x }\) = \(\frac { 7 + x }{ 12 + x }\)
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
\(\frac { 10 – x }{ 12 – x }\) = \(\frac { 19 – x }{ 24 – x }\)
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= \(\frac { 6 x 20 }{ 8 }\) = 15 m
Height of pole = 15 m

Hope given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C
• RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper

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
\(\frac { 7x + 7 }{ 11x + 7 }\) = \(\frac { 2 }{ 3 }\)
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
\(\frac { 5x – 3 }{ 9x – 3 }\) = \(\frac { 1 }{ 2 }\)
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
\(\frac { 8x + 6 }{ 3x + 6 }\) = \(\frac { 9 }{ 4 }\)
(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
\(\frac { 4x + 5 }{ 3x + 11 }\) = \(\frac { 13 }{ 17 }\)
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
\(\frac { x }{ y }\) = \(\frac { 6 }{ 11 }\)
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)

Hope given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.