Ratio and Proportion Class 8 RS Aggarwal Exe-9A Goyal Brothers ICSE Maths Solutions Ch-9. We provide step by step Solutions of council prescribe textbook / publication to develop skill and confidence. Visit official Website **CISCE** for detail information about ICSE Board Class-8 Mathematics.

## Ratio and Proportion Class 8 RS Aggarwal Exe-9A Goyal Brothers ICSE Maths Solutions Ch-9

Board | ICSE |

Publications | Goyal Brothers Prakashan |

Subject | Maths |

Class | 8th |

writer | RS Aggarwal |

Book Name | Foundation |

Ch-9 | Ratio and Proportion |

Exe-9A | Ratio |

Academic Session | 2024-2025 |

### RATIO

the comparison of two quantities by the method of division is very efficient. We can say that the comparison or simplified form of two quantities of the same kind is referred to as a ratio. This relation gives us how many times one quantity is equal to the other quantity. In simple words, the ratio is the number that can be used to express one quantity as a fraction of the other ones.

The two numbers in a ratio can only be compared when they have the same unit. We make use of ratios to compare two things. The sign used to denote a ratio is ( : ) .

#### Expressing RATIO in three different forms

- a to b
- a : b
- a/b

#### 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 in natural number
- The comparison of two ratios can be performed, if the ratios are equivalent like the fractions
- It must be in Simplest form

**Exercise- 9A**

( Ratio and Proportion Class 8 RS Aggarwal Exe-9A Goyal Brothers ICSE Maths Solutions Ch-9 )

**Que-1: Express each of the following ratios in simplest form :**

(i) 18:30 (ii) 7.5:9 (iii) 6*(2/3):7*(1/2) (iv) (1/6):(1/9):(1/12) (v) 5:7:(9/2) (vi) 3*(1/5):5*(1/3):6*(2/3)

**Sol: **(i) 18/30 = 3/5 = 3:5

(ii) 7.5/9 = 75/90 = 5/6 = 5:6

(iii) (20/3)/(15/2) = 40/45 = 8:9

(iv) (1/6):(1/9):(1/12) = multiply by 36

= 6:4:3

(v) 5:7:(9/2) = multiply by 2

10:14:9

(vi) 3*(1/5):5*(1/3):6*(2/3) = (16/5):(16/3):(20/3)

Multiply by 15

= 48:80:100 = 12:20:25

**Que-2: Express each of the following ratios in simplest form :**

(i) 75 paisa : 4 rupees (ii) 1m 6cm : 72 cm (iii) 1h 15min : 45min (iv) 2kg 750g : 3kg (v) 1 year 9 months : 2 years 4 months

**Sol: **(i) 75 paisa : 4 rupees

1 rupee = 100 paisa

= 75 : 4×100

= 75:400

= 3:16

(ii) 1m 6cm : 72 cm

(110+8)cm : 72 cm [1m = 100cm]

= 108 : 72

= 3:2

(iii) 1h 15min : 45min

(60+15)min : 45 min [1hour = 60 min]

= 75 : 45

= 5:3

(iv) 2kg 750g : 3kg

(2000+750)g : 3000g [1kg = 1000g]

= 2750 : 3000

= 11:12

(v) 1 year 9 months : 2 years 4 months

(12+9)months : (24+4)months [1 year = 12 months]

= 21 : 28

= 3:4

**Que-3: Which ratio is greater ?**

(i) (4:9) or (3:7) (ii) [2*(1/3):3*(1/3)] or (3.6 : 4.8) (iii) [1/2 : 1/3] or [1/6 : 1/4] (iv) [3*(1/3):4*(1/6)] or [0.9 : 1]

**Sol: **(i) 4/9 = 0.444

3/7 = 0.429

0.444 > 0.429

= 4/9 > 3/7

(ii) [2*(1/3):3*(1/3)]

[2*(1/3)]/[3*(1/3)]

(2/3)/(3/3) = 2/3 = 0.666

3.6/4.8 = 36/48

= 3/4 = 0.75

0.75 > 0.66

3.6 : 4.8 is greater

(iii) [1/2 : 1/3]

(1/2)/(1/3) = 3/2 = 1.5

[1/6 : 1/4]

(1/6)/(1/4) = 4/6 = 0.666

1.5 > 0.666

[1/2 : 1/3] is greater.

(iv) [3*(1/3):4*(1/6)]

[3*(1/3)]/[4*(1/6)]

(10/3)/(25/6) = 4/5 = 0.8

0.9 : 1

0.9/1 = 0.9

0.9 > 0.8

0.9 : 1 is greater

**Que-4: Arrange the following ratios in ascending order of magnitude :**

(i) (2:3). (5:9) and (11:15) (ii) (5:7), (9:14), (20:21) and (3:8)

**Sol: **(i)2/3; 5/9; 11/15

HCM 45

(2*15)/(3*15); (5*5)/(9*5); (11*3)/(15*3)

30/45 ; 25/45; 33/45

25/45 < 30/45 < 33/45

5/9 < 2/3 < 11/15

(ii) 5:7, 9:14, 20:21, 3:8.

lcm = 7×2×3×2×2 = 168.

120:168, 108:168, 160:168, 63:168.

A.O = 63:168 < 108:168 < 120:168 < 160:168.

there for (3:8)<(9:14)<(5:7)<(20:21)

**Que-5: Divide Rs142.20 between Gagan and Mukesh in the ratio 1/4 : 1/5.**

**Sol: **We are given that the amount has been divided between Gagan ad Mukesh in a ratio as 1/4:1/5

Now, Let the ratio = x

so, Gagan’s ratio = 1/4x

Mukesh’s ratio =1/5x

Total Amount = 142.20

1/4x+1/5x = 142.20

9x/20 = 142.20

x = (142.20*20)/9

x = 316

Now we get, the Gagan ratio as 1/4(316)=79

Mukesh ratio as 1/5(316)= 63.2

Therefore the ratio of Gagan and Mukesh are Rs79 and Rs63.2 respectively.

**Que-6: Divide Rs3726 among A,B,C in the ratio 1/3 : 1/4 : 1/6.**

**Sol: **A: B:C = 1/3: 1/4 : 1:6

Multiply all by 12

A: B:C = 4: 3 : 2

So a+b+c = 4+3+2 = 9

A =

3726/ 9 * 4 = 414 = 1656

B= 414* 3 = 1242

C= 414* 2 = 828

**Que-7: Divide Rs810 among A,B,C in the ratio 1/4 : 2/5 : 1*(3/8)**

**Sol: **1/4 : 2/5 : 1*(3/8)

(1/4)+(2/5)+(11/8)

[1/4×40]:[2/5×40]:[11/8×40]

= 10+16+55 = 81

share of a = (10/81)×810 = ₹100

share of b = (16/81)×810 = ₹160

share of c = (55/81)×810 = ₹550

**Que-8: Divide Rs1050 between Geeta and Renu in the ratio 2*(2/3) : 6*(2/3).**

**Sol: **Given as, Geeta and Renu amount is in the ratio of 8 /3 : 20/ 3 ,i.e., 2:5

So,

=> 2X + 5X = 1050

=> 7X = 1050

=> X = 150

Geeta has amount 2X= 2×150 = ₹ 300

Renu has amount 5X= 5×150 = ₹ 750

**Que-9: Divide Rs747 among A,B,C such that 4A=5B=7C.**

**Sol: **4A = 5B = 7C = k

So, we have:

A = k/4

B = k/5

C = k/7

We know that the sum of the amounts given to A, B, and C is Rs 747, so:

k/4 + k/5 + k/7 = 747

To solve for k, we can find the least common multiple (LCM) of 4, 5, and 7, which is 140. Multiplying both sides of the equation by 140, we get:

35k + 28k + 20k = 747 x 140

83k = 104580

k = 1260

Now we can find the amounts given to A, B, and C:

A = k/4 = 1260/4 = Rs 315

B = k/5 = 1260/5 = Rs 252

C = k/7 = 1260/7 = Rs 180

**Que-10: A bag contains one rupee, 50p and 25p coins in the ratio 5:6:8 amounting to Rs210. Find the number of coins of each type.**

**Sol: **Let the number of 1 rupee coins = 5x

Number of 50 paise coins = 6x

And, number of 25 paise coins = 8x

Therefore, amount of 1 rupee coins in Rs = 5x

Amount of 50 paise coins in Rs = 3x

Amount of 25 paise coins in Rs = 2x

Now, total amount = 210 rupees

Thus, 5x+3x+2x = 210

i.e. 10x = 210

i.e. x = 21

Therefore, number of 1 rupee coin = 5×21

=105

Number of 50 paise coins = 6×21

=126

And, number of 25 paise coins = 8×21

=168

**Que-11: Find A:B:C when :**

(i) A:B = 2:5 and B:C = 7:9 (ii) A:B = 3:4 and B:C = 6:11 (iii) A:B = 1/2 : 1/3 and B:C = 1/4 : 1/5

**Sol: **(i) A : B = 2:5 or 14:35 (multiplied by 7)

B:C = 7:9 or 35:45 (multiplied by 5)

Now B’s value is same in both of the ratios.

A:B:C = 14:35:45

(ii) A:B = 3:4

B:C = 6:11

lcm of 4 and 6 is 12

so multiplying the ratio 3:4 with 3

gives 9:12

multiplying the ratio 6:11 with 2

gives 12:22

now new ratios are 9:12 and 12:22

A:B:C are 9:12:22

(iii) Given a : b = 3 : 2

c : b = 4 : 5 or b : c = 5 : 4

= (5×2)/5 : (4×2)/5

= 2 : 8/5

∴ a:b:c = 3 : 2 : 8/5

= 15 : 10 : 8

**Que-12: If A:B = 4:9 and A:C = 2:3, find B:C and A:B:C.**

**Sol: **(i) A:B = 4:9

A:C = 2:3 = 4:6

B:C = 9:6 = 3:2

(ii) A:B = 4:9

B:C = 3:2 = 9:6

A:B:C = 4:9:6

**Que-13: If A:C = 5:8 and B:C = 5:6, find A:B and A:B:C.**

**Sol: **A:C= 5:8

B:C= 5:6

Make C equal in both, So,

Take LCM of 8 and 6 which is 24.

A:C= 15:24 ( by multiplying by 3)

B:C= 20:24 ( by multiplying by 4)

From here, A:B= 15:20

or simplifying gives A:B= 3:4

Now A:B= 15:20 and B:C= 20:24

So A:B:C = 15:20:24

**Que-14: Two numbers are in the ratio 6:11. On adding 2 to the first and 7 to the second, their ratio becomes 8:15. Find the numbers.**

**Sol: **Let the two numbers be 6x and 11x as their ratio is 6:11

Given 2 is added to first and 7 is added to second

Hence the numbers become, (6x + 2) and (11x + 7)

Given (6x + 2) : (11x + 7) = 8:15

Hence 15(6x + 2) = 8(11x + 7)

90x + 30 = 88x + 56

2x = 26

Therefore, x = 13

Hence the numbers are 6x = 6(13) = 78

11x = 11(13) = 143

**Que-15: A sum of money is divided between Mohan and Sohan in the ratio 5:7. If Sohan’s share is Rs665, find the total amount.**

**Sol: **Let the sum of money Rs x.

The sum of money is divided in the ratio of 5:7

∴ Rohan’s share = (7/12) of x.

Rohan’s share (given) =Rs 665

∴ (7/12)x = 665

x = (12×665)/7

x = 1140

**Que-16: Two numbers are in the ratio 7:4. If their difference is 72, find the numbers.**

**Sol: **Let the common ratio be x.

Let 7x be the first number.

Let 4x be the second number.

Given that the difference of two numbers = 72.

7x – 4x = 72

3x = 72

x = 72/3

x = 24.

So, first number = 7 * 24 = 168.

and second number = 4 * 24 = 96.

Therefore the numbers are 168 and 96.

**Que-17: A certain sum of money is divided among A,B,C in the ratio 5:6:7. If A’s share is Rs175, find the total amount and the share of each one of B and C.**

**Sol: **Ratio of A : B : C = 5 :

6 : 7 (Given)

Let x be the constant ratio:

Ratio of A : B : C = 5x :

6x : 7x

Given that A’s share is ₹175

5x = ₹175

x = 175 ÷ 5

x = ₹35

B = 6x = 6(35) = ₹210**
**C = 7x= 7(35) = ₹245

**Total amount = 175 + 210 + 245 = ₹630**

**Que-18: The ratio of the number of boys to the number of girls in a school of 1440 students is 7:5. If 40 new boys are admitted, find how many new girls may be admitted to make this ratio 4:3.**

**Sol: **Let the no. of boys = 7x

No. of girls = 5x

7x+5x =1440

12x = 1440

⇒ x = 120

∴ No. of boys = 120×7 = 840

No. of girls = 5×120 = 600

Let y be the no. of girls to be added

⇒ 4/3 = (840+40)/(600+y)

⇒ 2400+4y = 2520+120

⇒ 4y = 2640−2400 = 240

⇒ y =240/4 = 60

∴ no. of girls that needs to be added = 60

**Que-19: The sum of three numbers is 212. If the ratio of the first to the second is 13:16 and that of the second to the third is 2:3, then find the numbers.**

**Sol: **sum of the three numbers is 212.

the ratios are 13:16 and 2:3.

we have to make this in 3 ratios because they are in 4 ratios and the total is the sum of three numbers.

so,we write this as 13/16 and 2/3.

we take the LCM of 16 and 2=16.

THEN,

13/16×1/1 and 2/3×8/8

13/16 and 16/24.

SO, WE WRITE THIS AS THE

13:16:24.

SUM OF THE RATIOS =13+16+24

=53.

SUM OF NUMBERS IS =212.

SO,

FIRST NUMBER=212×13/53

= 52.

SECOND NUMBER=212×16/53

= 64.

THIRD NUMBER=212×24/53

= 96.

**Que-20: Find the number which when added to each term of the ratio 27:35, change the ratio to 4:5.**

**Sol: **Let the number to be added be denoted by ‘ x ‘

So according to the question we get,

27 + x / 35 + x = 4 / 5

= 5 ( 27 + x ) = 4 ( 35 + x )

= 135 + 5x = 140 + 4x

= 5x – 4x = 140 – 135

= x = 5

**Que-21: The present ages of Mr.Sen and his son are in the ratio 17:9. If the ratio of their ages, 9 years ago was 7:3, then find their present ages.**

**Sol: **Let present age of sen and his son be 17X and 9X .

Before 9 years age of sen = ( 17X – 9 ) years.

Before 9 years age of his son = ( 9X – 9) years.

According to question,

17X-9 : 9X – 9 = 7 : 3

3 ( 17X – 9 ) = 7 ( 9X – 9 )

51X – 27 = 63X – 63

51X – 63X = -63 + 27

-12X = -36

X = -36/-12 = 3

Present age of sen = 17X = 17 × 3 = 51 years.

Present age of his son = 9X = 9 × 3 = 27 years.

**Que-22: A sample of vinegar was prepared by dissolving 705ml of acetic acid in 4.7 litres of water. Find the ratio by volume of acetic acid to water in this sample.**

**Sol: **1000ml = 1litre.

We convert the 4.7 litres to ml.

4.7 × 1000=4700ml of water.

Ratio of Acetic acid to water :

705 / 4700 = 3/20

The ratio is thus:

3 : 20

**Que-23: The salaries of A,B and C are in the ratio 2:3:5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be the new ratio of their salaries?**

**Sol: **Let the constant be x

Then ,Salaries of A,B,C are 2x,3x,5x respectively.

Increment in salary of A = 15%

Therefore A’s new salary = Rs.(2x+(15/100)×2x) = Rs.230x/100

Increment in B’s salary = 10%

Therefore, B’s new salary = Rs.(3x+(10/100)×3x) = Rs.330x/100

Increment in C’s salary = 20%

Therefore C’s new salary = Rs.(5x+(20/100)×5x) = Rs.6x

Therefore our ratio is 23:33:60

**Que-24: Anuj got 58 marks out of 75 in Physics and 97 marks out of 120 in Maths. In which of the two subjects did he perform better?**

**Sol: **In Physics, Anuj scored 58 out of 75, so the ratio is:

Ratio in Physics = 58 / 75

In Maths, he scored 97 out of 120, so the ratio is:

Ratio in Maths = 97 / 120

Ratio in Physics = (58 / 75) * (4/4) = 232/300

Ratio in Maths = (97 / 120) * (5/5) = 485/600

Now, we can compare the two ratios:

– In Physics: 232/300

– In Maths: 485/600

To compare, we can calculate the percentages:

– Physics: (232/300) * 100 ≈ 77.33%

– Maths: (485/600) * 100 ≈ 80.83%

Anuj performed better in Mathematics because he achieved a higher percentage (80.83%) compared to Physics (77.33%).

**Que-25: A mixture consists of only two components A and B. In 60 litres of this mixture, the components A and B are present in the ratio 2:1. What quantity of component B has to be added to this mixture so that the new ratio is 1:2 ?**

**Sol: **Original volume = 60 litres.

a and b are ratio 2: 1 i.e. 2/3 parts of a and 1/3 parts of b

Component a = 2/3 * 60 = 40 litres.

Component b = 1/3 * 60 = 20 litres.

Now for the ratio to be 1 : 2,

i.e Component A /Component B = 1/2

let x be the quantity of b to be added.

=> 40 / (20 + x) = 1/2

(20+x) * 1 = 40 * 2

20 + x = 80

x = 60.

**–: Ratio and Proportion Class 8 RS Aggarwal Exe-9A Goyal Brothers ICSE Maths Solutions Ch-9 Questions :-**

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