Ratio and Proportion Class 10 RS Aggarwal Exe-7A Goyal Brothers ICSE Maths Solutions Ch-7. Step by step solutions of exercise-7A questions as latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## Ratio and Proportion Class 10 RS Aggarwal Exe-7A Goyal Brothers ICSE Maths Solutions Ch-7

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-7 | Ratio and Proportion |

Writer/ Book | RS Aggarwal |

Topics | Solution of Exe-7A Questions |

Academic Session | 2024-2025 |

### Solution of Exe-7A Questions

Ratio and Proportion Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-7

In this exercise you will learn the important property and comparison of ratio with practice questions and their step by step solutions.

#### Imp Features of Ratio

- The ratio should exist between the quantities of the same kind.
- While comparing two things, the units should be similar.
- There should be significant order of terms.
- The comparison of two ratios can be performed,
- multiplying / dividing same term the ratio does not change

**Que-1: Find the ratio between :**

**(i) 60 paise and Rs 1.35 (ii) 1.8m and 75cm (iii) 1.5kg and 600g (iv) 35 min and 1*(3/4) hrs.**

**Sol: **(i) 60p ; 1.35 Rs = 135p

60 p / 135 p

12 / 27

4 / 9

4 : 9.

(ii) 1.8m = 180cm ; 75cm

= 180 : 75

= 12 : 5.

(iii) 1.5kg = 1500g ; 600g

= 1500 : 600

= 5 : 2.

(iv) 1*(3/4) hrs = 7/4 hrs

7/4 hrs = 1.75 hrs.

35 min = 0.58 hrs.

= 0.58 : 1.75

= 1 : 3.

**Que-2: If A:B = 5:6 and B:C = 9:11, find (i) A:C (ii) A:B:C**

**Sol: **(i) Given

A/B = 5/6 …(1)

B/C = 9/11 …(2)

On multiplying (1) & (2)

A/C = 45/66 = 15/22 …(3)

(ii) Let A = x

By (1) B = 6x/5 & by (3) C = 22x/15

Therefore

A:B:C :: 1:6/5:22/15

=> A:B:C :: 15:18:22

**Que-3: If P:Q = 7:4 and Q:R = 5:14, find (i) R:P (ii) P:Q:R **

**Sol: **5:14. lcm of 4and 5 is 20

7×5:4×5

5×4:14×4

P:Q:R

35:20:56

(i) R : P = 56 : 35 = 8:5

(ii) P : Q : R = 35:20:56.

**Que-4: If 3A = 5B = 6C, find A:B:C**

**Sol: **Given information: 3A = 5B = 6C

Consider we need to find the ratio A:B:C.

Coefficients of A,B and C are 3, 5 and 6 respectively.

LCM (3,6,5) = 30.

Divide the given equation by 30.

3A/30 = 5B/30 = 6C/30

Cancel out common factors.

A/10 = B/6 = C/5.

Therefore, the ratio A:B:C is 10:6:5.

**Que-5: If A/2 = B/3 = C/6, find A:B:C.**

**Sol: **let’s set each of these equal to a common variable k. This gives us:

A/2 = k

A = 2k

B/3 = k

B = 3k

C/6 = k

C = 6k

A:B:C = 2k:3k:6k

The k cancels out, leaving us with the ratio:

A:B:C = 2:3:6.

**Que-6: If a:b = 8:5, find (7a+5b) : (8a-9b)**

**Sol: **a:b = 8:5

a = 8

b = 5

(7a + 5b):(8a – 9b)

(7(8) + 5(5)):(8(8) – 9(5)

(56 + 25):(64 – 45)

81:19.

**Que-7: If x:y = 3:2, find (5x-3y) : (7x+2y)**

**Sol: **x:y = 3:2

x = 3, y = 2

(5x-3y) : (7x+2y)

(5×3-3×2) : (7×3+2×2)

(15-6) : (21+4)

9 : 25.

**Que-8: If x:y = 10:3, find (3x²+2y²) : (3x²-2y²)**

**Sol: **x:y = 10:3

x = 10 and y = 3.

(3×10²+2×3²) : (3×10²-2×3²)

(300+18) : (300-18)

318 : 282

= 318/282

= 53/47

= 53 : 47.

**Que-9: If a:b = 2:5, find (3a²-2ab+5b²) : (a²+7ab-2b²)**

**Sol: **a:b = 2:5

a = 2 and b = 5

(3×2²-2x2x5+5×5²) : (2²+7x2x5-2×5²)

(12-20+125) : (4+70-50)

117 : 24

39 : 8.

**Que-10: If (5x+2y) : (7x+4y) = 13:20, find x:y**

**Sol: **(5x+2y) : (7x+4y) = 13 : 20

(5x+2y)/(7x+4y) = 13/20

20(5x+2y) = 13(7x+4y)

100x+40y = 91x+52y

9x = 12y

x/y = 12/9

x : y = 4 : 3.

**Que-11: If (3x+5y) : (3x-5y) = 7:3, find x:y.**

**Sol: **(3x+5y) : (3x-5y) = 7:3

(3x+5y)/(3x-5y) = 7/3

⇒ 9x + 15y = 21x – 35y …[By cross multiplication]

⇒ 21x – 9x = 15y + 35y

⇒ 12x = 50y

x/y = 50/12 = 25/6

x : y = 25 : 6.

**Que-12: If (6x²-xy) : (2xy-y²) = 6:1, find x:y**

**Sol: **(6x² – xy) : (2xy – y²) = 6 : 1

⇒ (6x² – xy) / (2xy – y²) = 6 / 1

⇒ 1 (6x² – xy) = 6 (2xy – y²)

⇒ 6x² – xy = 12xy – 6y²

⇒ 6x² – 13xy + 6y² = 0

⇒ 6x² – (9 + 4)xy + 6y² = 0

⇒ 6x² – 9xy – 4xy + 6y² = 0

⇒ 3x (2x – 3y) – 2y (2x – 3y) = 0

⇒ (2x – 3y) (3x – 2y) = 0

So, either 2x – 3y = 0 or, 3x – 2y = 0

⇒ 2x = 3y & 3x = 2y

⇒ x/y = 3/2 & x/y = 2/3

⇒ x : y = 3 : 2 & x : y = 2 : 3

**Que-13: If (4x²-3y²) : (2x²+5y²) = 12:19, find x:y.**

**Sol: **( 4x² – 3y²) : ( 2x² + 5y² ) = 12 : 19

( 4x² – 3y² ) 19 = ( 2x² + 5y² )12

76x² – 57y² = 24x² + 60y²

76x² – 24x² = 60y² +57y²

52x² = 117y²

x² / y² = 117 / 52

( x/ y )² = ( 13 × 9 ) / ( 13 × 4 )

= 9 / 4

( x / y )² = ( 3 / 2 )²

Therefore ,

x / y = 3 / 2

x : y = 3 : 2

**Que-14: If x²+4y² = 4xy, find x:y.**

**Sol: **x²+4y² = 4xy

x²−4xy+4y² = 0

⟹ (x−2y)² = 0

⟹ x = 2y

⟹ x/y = 2/1

x : y = 2 : 1.

**Que-15: If 10x²-23xy+9y² = 0, find x:y.**

**Sol: **10x²-23xy+9y² = 0

=> 10x²-5xy-18xy+9y² = 0

=> 5x(2x-1y)-9y(2x-1y) = 0

=> (2x-y)(5x-9y) = 0

2x-y = 0 (or) 5x-9y = 0

2x = 1y 5x = 9y

x/y=1/2 x/y=9/5

=> x : y = 1:2 or 9:5

**Que-16: A ratio is equal to 3:4. If its consequent is 144, what is its antecedents?**

**Sol: **Let the antecedent be ‘x’

ATQ. 3/4 = x/144

x = (144×3)/4

x = 108.

**Que-17: Two numbers are in the ratio 8:13. If 14 is added to each of the numbers, the ratio becomes 2:3. Find the numbers.**

**Sol: **Ratio of the numbers = 8 : 13

Let the two numbers be 8x and 13x .

When 14 is added to these numbers the ratio becomes 2 : 3 .

(8x+14)/(3x+14) = 2/3

3(8x+14) = 2(3x+14)

24x+42 = 6x+28

2x = 14

x = 7

Now, the number are

8x = 8 x 7 = 56

13x = 13 x 7 = 91.

**Que-18: Two numbers are in the ratio 5:7. If 8 is subtracted from each, the ratio becomes 3:5. Find the numbers.**

**Sol: **Ratio between the original numbers= 5:7

Let the original numbers be 5x and 7x **
**8 is subtracted from each and ratio becomes 3:5

i.e 5x -8 : 7x – 8 = 3:5

=》 5x – 8 / 7x – 8 = 3/5

=》 5( 5x- 8) = 3(7x -8) ( cross multiplication)

=》 25x – 40 = 21x – 24

=》 25x- 21x = -24 + 40

=》 4x = 16

=》 x = 16/4

=》 x = 4

Now the number are :

5x = 5 x 4 = 20.

7x = 7 x 4 = 28.

**Que-19: What least number must be added to each term of the ratio 5:7 to make it 8:9 ?**

**Sol: **Let the number be x.

ATQ, the equation will be

(5+x)/(7+x) = 8/9

45+9x = 56+8x

x = 11.

**Que-20: Out of the monthly income of Rs 45000, Rahul spends Rs 31500 and the rest he saves. Find the ratio of his**

**(i) income to expenditure (ii) income to saving (iii) saving to expenditure.**

**Sol: **monthly income = ₹ 45000

Rahul’s expenditure = ₹ 31500

Rahul’s savings = ₹ 13,500

(i) income to expenditure = 45000/31500

= 10/7

(ii) income to savings = 45000/13500

= 10/3

(iii) savings to expenditure = 13500/31500

= 3/7

**Que-21: The cost of making an umbrella is divided between material, labour and overheads in the ratio 6:4:1. If the material costs Rs 132, find the cost of production of an umbrella.**

**Sol: **The costs 6x, 4x, x will be in the same ratio as 6:4:1.

The sum of the cost is the cost of making an umbrella which is

6x + 4x + x = 11x

We are given in the question that the material cost is 132 rupees. So we have,

6x = 132

⇒ x = 132/6 = 22

We are asked to find the total cost. The total cost in rupees is

11x = 11×22 = 242

Therefore, the total cost is Rs.242.

**Que-22: Divide Rs 6720 in the ratio 5:3 .**

**Sol: **The ratio is 5 : 3

Sum of the ratio = 5+3 = 8

First part = 6720 x 5/8 = 4200

Second part = 6720 x 3/8 = 2520.

**Que-23: Divide Rs 11620 among A,B and C in the ratio 35:28:20.**

**Sol: **Sum of ratio terms = (35 + 28 + 20) = 83

A’s share = Rs.(11620×(35/83)) = Rs.4900

B’s share = Rs. (11620×(28/83)) = Rs.3920

C’s share = Rs.(11620×(20/83)) = Rs.2800

**Que-24: Divide Rs 782 among P,Q and R in the ratio (1/2):(2/3):(3/4).**

**Sol: **Given number = 782

{1/2+2/3+3/4}⋅x = 782

{(12+16+18)/24} × x = 782

{46/24}⋅x = 782

{23/12}⋅x = 782

x = 408

now the first part = 1/2⋅x = (1/2⋅408) = 204

second part = 2/3⋅x = (2/3⋅408) = 272

third part = 3/4⋅x = (3/4⋅408) = 306

**Que-25: If Rs 5100 be divided among A,B,C in such a way that A gets (2/3) of what B gets and B gets (1/4) of what C gets, find their respective shares.**

**Sol: **Share of C is x

(2/2)×(1/4x)+(x/4)+x = 5100

{(2x×3x)+12x}/12 = 5100

17x/12 = 5100

x = (5100 x 12)/17

x = 3600

C = 3600

B = 360/4 = 900

A = (2×900)/3 = 600.

**Que-26: Divide Rs 8300 among A,B and C such that 4 times A’s share, 5 times B’s share and 7 times C’s share may all be equal.**

**Sol: **Let A’s share be x

B’s share be y

C’s share be z

ATQ, 4x = 5y = 7z

y = 4x/5 and z = 4x/7

x+y+z = 8300

(x)+(4x/5)+(4x/7) = 8300

(35x+28x+20x)/35 = 8300

83x/35 = 8300

x = (8300 x 35)/83

x = 3500

y = (4 x 3500)/5 = 2800

z = (4 x 3500)/7 = 2000.

**Que-27: A sum of money is divided between A and B in the ratio 6:11. If B’s share is Rs 7315, find (i) A’ share (ii) the total amount of money.**

**Sol: **The ratio of money between A and B = 6 : 11

Let share of A = 6x

Share of B = 11x

According to question- B’s share is 7315

(i) So, 11x = 7315

x = 7315//11

x = 665

Hence, share of A = 6x

= 6×665

Share of A = 3,990 Rs

(ii) Share of B = 7,315 Rs

Total amount = Share of A + Share of B

= 7315 + 3990

Total amount = 11,305 Rs

**Que-28: The ages of Tanvy and Divya are in the ratio 5:7. Five years hence, their ages will be in the ratio 3:4. Find their present ages.**

**Sol: **Age of Tanvi: Age of Divya = 5:7

Therefore, age of Tanvi = 5x

And, age of Divya = 7x

5 years hence, their ages of Tanvi and Divya will be (5x+5) and (7x+5) respectively.

Now, according to the question,

⇒3/4 = 5x+5/7x+5

⇒3(7x+5) = 4(5x+5)

⇒21x+15 = 20x+20

⇒21x-20x = 20-15

⇒x = 5

Thus, Age of Tanvi = 5x = 25.

And, Age of Divya = 7x = 35.

**Que-29: One year ago, the ratio Amit’s and Arun’s ages was 6:7 respectively. Four years hence, their ages will be in the ratio 7:8. How does is Amit ?**

**Sol: **Age of Tanvi: Age of Divya = 5:7

Therefore, age of Tanvi = 5x

And, age of Divya = 7x

5 years hence, their ages of Tanvi and Divya will be (5x+5) and (7x+5) respectively.

Now, according to the question,

⇒3/4 = 5x+5/7x+5

⇒3(7x+5) = 4(5x+5)

⇒21x+15 = 20x+20

⇒21x-20x = 20-15

⇒x = 5

Thus, Age of Tanvi = 5x = 25.

And, Age of Divya = 7x = 35.

**Que-30: Reena reduces her weight in the ratio 5:4. What is her weight now, if originally it was 70 kg ?**

**Sol: **Let us assume that, the new weight of Reena is x kg.

→ 70 : x = 5 : 4

→ 70 * 4 = 5 * x

→ 280 = 5 * x

→ 280 = 5x

→ 280/5 = x

→ 56 = x

→ x = 56

**Que-31: 68 kg of a mixture contains milk and water in the ratio 27:7. How much more water is to be added to this mixture to get a new mixture containing milk and water in the ratio 3:1 ?**

**Sol: **68 litre of mixture contains milk and water in the ratio 27 : 7

i.e 54 litre milk and 14 litre water

X litre of water is added to get new mixture in ratio 3:1

milk/water = 54/14+x = 3/1

54 = 42+3x

X = 4 litre

**Que-32: A mixtures contains milk and water in the ratio 5:1. On adding 5 litres of water , the ratio of milk to water becomes 5:2. Find the quantity of milk in the original mixture.**

**Sol: **A mixture contains milk and water in the ratio 5 : 1

On adding 5 liters of water, the ratio of milk and water becomes 5 : 2

Let the quantity of milk and water in the original mixture be 5x and x respectively

According to the qestion,

(5x)/(x + 5) = 5/2

⇒ 10x = 5x + 25

⇒ 5x = 25

⇒ x = 5

∴ Quantity of milk in the original mixture is 5x = 5 × 5 = 25 litres

**Que-33: In an examination, the ratio of passes to failures was 4:1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5:1. How many students appeared for the examination ?**

**Sol: **Suppose x candidates passed and y failed; therefore,

x/y = 4/1

x = 4y (1)

In the second case; no. of students appeared = x+y−30

and no. of those who passed = x−20

No of failed

N = x+y−30−(x−20)

N = y−10

from the question

(x−20)/(y−10) = 5/1

x−20 = 5y−50

x−5y = −30

from ( 1)

4y−5y = −30

y = 30

x = 120

So, x+y = 150

**Que-34: Find the angles of a triangles which are in the ratio 5:4:3.**

**Sol: **Given that in a △ABC, ∠A : ∠B : ∠C = 5:4:3

Let the angles of the given triangle be 5x°, 4x°, 3x°.

We know that the sum of the angles of a triangle is 180°

5x°+4x°+3x° = 180°

⇒ 12x° = 180°

x° = 110°/12 = 15°

So, the angles of the triangle are 70°, 60°, 45°.

**Que-35: The sides of the triangles are in the ratio (1/2):(1/3):(1/4) and its perimeter is 91 cm. Find the lengths of the sides of the triangles.**

**Sol: **LCM of 2,3,4 = 12

therefore ratio = 6:4:3

let the sides be 6x,4x,3x

6x+4x+3x = 91

13x = 91

x = 7

now, sides

6x = 6*7 = 42cm

4x = 4*7 = 28cm

3x = 3*7 = 21cm

**Que-36: In a school, the boys and girls are in the ratio 9:5. If there are 425 girls, what is the total number of students in the school ?**

**Sol: **Let the total no. Of boys be 9x

No. Of girls be 5x

According to question

5x = 425

X = 425/5

X= 85

Now,

Boys = 9x

= 9*85

= 765

Now,

Girls = 425

Boys = 765

Total = 425 + 765

= 1190.

**Que-37: Compare the following ratios :**

**(i) (7:9) and (11:16) (ii) (19:25) and (17:20) (iii) (1/2:1/5) and (5:2)**

**Sol: **(i) 7/9 = 11/16

cross multiplication

7×16 = 11×9

112 = 99

therefore 7:9 is greater than 11:16

(ii) 17/20 and 19/25

LCM of 20 and 25 = 100

17/20 × 5/5 = 85/100

19/25 × 4/4 = 76/100

By comparing 85/100 and 76/100, we got 85/100 > 76/100

Answer : 17/20 is the greater fraction.

(iii) (1/2 : 1/5) = 5 : 2

1/2 = 1/5

5 : 2 = 5 : 2.

**Que-38: Arrange the following ratios in descending order of magnitudes:**

**(i) (5:6), (8:9), (13:18) and (19:24)
(ii) (6:7), (13:14), (19:21) and (23:28)
(iii) (7:12), (9:16), (13:20) and (5:8).**

**Sol: **(i) 5/6, 8/9, 13/18, 19/24

LCM of denominator is 72 :

{(5×12)/(6×12)}, {(8×8)/(9×8)}, {(13×4)/(18×4)}, {(19×3)/(24×3)}

60/72, 64/72, 52/72, 57/72

64/72 > 60/72 > 57/72 > 52/72

(8:9) > (5:6) > (19:24) > (13:18)

(ii) 6/7, 13/14, 19/21, 23/28

LCM of denominator is 84

{(6×12),(7×12)}, {(13×6)/(14×6)}, {(19×4)/(21×4)}, {(23×3)/(28×3)}

72/84, 78/84, 76/84, 69/84

78/84 > 76/84 > 72/84 > 69/84

(13:14) > (19:21) > (6:7) > (23:28)

(iii) 7/12, 9/16, 13/20, 5/8

LCM of denominator is 240.

{(7×20)/(12×20)}, {(9×15)/(16×15)}, {(13×12)/(20×12)}, {(5×30)/(8×30)}

140/240, 135/240, 156/240, 150/240

156/240 > 150/240 > 140/240 > 135/240.

(13:20) > (5:8) > (7:12) > (9:16).

**Que-39: Arrange the following ratios in ascending order of magnitudes :**

**(i) (4:9), (6:11), (7:13) and (27:50)
(ii) (2:3), (8:15), (11:12) and (7:16)
(iii) (3:5), (4:9), (5:11) and (10:17)**

**Sol: **(i) 4/9, 6/11, 7/13, 27/50

LCM of denominator is 64,350.

{(4×7150)/(9×7150)}, {(6×5850)/(11×5850)}, {(7×4950)/(13×4950)}, {(27×1287)/(50×1287)}

28600/64350, 35100/64350, 34650/64350, 34668/64350

28600/64350 < 34650/64350 < 34668/64350 < 35100/64350

4:9 < 7:13 < 27:50 < 6:11.

(ii) 2/3, 8/15, 11/12, 7/16

0.66, 0.53, 0.92, 0.44

0.44 < 0.53 < 0.66 < 0.92

7:16 < 8:15 < 2/3 < 11/12

(iii) 3/5, 4/9, 5/11, 10/17

0.6, 0.44, 0.45, 0.56

0.44 < 0.45 < 0.56 < 0.6

4:9 < 5:11 < 10:17 < 3:5.

**Que-40: If (3a+2b) : (5a+3b) = 18:29, find (a:b)**

**Sol: **(3a+2b) : (5a+3b)

⇒ 29(3a + 2b) = 18(5a + 3b)

⇒ 87a + 58b = 90a + 54b

⇒ 58b – 54b = 90a – 87a

⇒ 4b = 3a

a : b = 4 : 3.

— : End of Ratio and Proportion Class 10 RS Aggarwal Exe-7A Goyal Brothers ICSE Maths Solutions :-

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