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

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

Publications | Goyal Brothers Prakashan |

Subject | Maths |

Class | 8th |

writer | RS Aggarwal |

Book Name | Foundation |

Ch-9 | Ratio and Proportion |

Exe-9C | Multiple Choice Questions |

Edition | 2024-2025 |

### MCQs on Ratio and Proportion With Solutions / Answer

The MCQs problems on Ratio and Proportion is very common now days not only in academic exam but also for various one day exam. Hence various type questions with solutions has been given for your complete practice.

**Exercise- 9C**

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

Multiple Choice Questions :

**Que-1: In a ratio which is equal to 5:8, the antecedent is 40, then the consequent is :**

(a) 25 (b) 48 (c) 64 (d) none of these

**Sol: **(c) 64

**Reason: **Let the consequent be x

5/8 = 40/x

x = (8×40)/5

x = 64

**Que-2: 0.5 of a number is equal to 0.07 of another. The ratio of the numbers is :**

(a) 1:14 (b) 5:7 (c) 50:7 (d) 7:50

**Sol: **(d) 7:50

**Reason: **5x/10=7y/100

5x =7y/10

50x=7y

x/y=7/50

**Que-3: If 2x =3y = 4z, then x:y:z is :**

(a) 2:3:4 (b) 4:3:2 (c) 3:4:2 (d) 6:4:3

**Sol: **(d) 6:4:3

**Reason: **Since 2x = 3y = 4z, we see that:

k = 2x, k = 3y, and k = 4z, for some k.

Solving these for x, y, and z yields:

x = k/2, y = k/3, and z = k/4.

Therefore:

x:y:z = (k/2):(k/3):(k/4)

= 6k:4k:3k, by multiplying all parts of the proportion by 12, the LCD

= 6:4:3, by dividing all parts by k.

**Que-4: If x:y = 7:9 and y:z = 5:4, then x:y:z is**

(a) 7:45:36 (b) 35:45:36 (c) 28:36:35 (d) none of these

**Sol: **(b) 35:45:36

**Reason: **Given,

x/y = 7/9 and y/z = 5/4.

To find the required x:y:z, we take the LCM of y, that is 9 and 5.

LCM of 9 and 5 = 45.

We change the ratios such that y always contain 45 in both the cases.

So, x:y = 7:9

= (7*5):(9*5)

= 35/45

Again, y:z = 5:4

= (5*9):(4*9)

= 45/36

So, x:y:z = 35:45:36

**Que-5: The fourth proportional to 3,5 and 21 is**

(a) 35 (b) 7/5 (c) 5/7 (d) 12.6

**Sol: **(a) 35

**Reason: **Let the fourth proportional be x

3:5 :: 21:x

3/5 = 21/x

x = (21×5)/3

x = 35

**Que-6: Mean proportional between 7 and 28 is**

(a) 17.5 (b) 12 (c) 14 (d) 16

**Sol: **(c) 14

**Reason: **Let the mean proportion be x

7:x :: x:28

7/x = x/28

x² = 7×28

x = √196

x = 14

**Que-7: Third proportional to 9 and 12 is**

(a) 6√3 (b) 10.5 (c) 16 (d) none of these

**Sol: **(c) 16

**Reason: **Third proportional = (b × b) / a

where a = 9 and b = 12

Third proportional = (12 × 12) / 9

= 144 / 9

= 16

**Que-8: A fraction bears the same ratio to 1/27 as 3/7 does to 5/9. The fraction is**

(a) 7/45 (b) 1/35 (c) 45/7 (d) 5/21

**Sol: **(b) 1/35

**Reason: **Let the fraction be x, then according to the question

⇒ x : 1/27 = 3/7 : 5/9

⇒ x × 27 = 3/7 × 9/5

⇒ 27x = 27/35

∴ x = 1/35

**Que-9: What must be added to each term of the ratio 49:68 so that it becomes 3:4?**

**(a) 8 (b) 9 (c) 5 (d) 3**

**Sol: **(a) 8

**Reason: **(49+x)/(68+x) = 3/4

⇒ 196+4x = 204+3x

4x-3x = 204-196

⇒ x = 8

**Que-10: What least number must be added to each one of 6,14,18,38 to make them in proportion?**

(a) 1 (b) 2 (c) 3 (d) 4

**Sol: **(b) 2

**Reason: **6+x : 14+x :: 18+x : 38+x

⇒ (6+x)(38+x) = (14+x)(18+x)

⇒ 228 + 44x + x² = 252 + 32x + x²

⇒ 44x – 32x = 252 – 228

⇒ 12x = 24

⇒ x = 24/12 = 2

**Que-11: What least number must be subtracted from each of the numbers 14,17,34,42 so that the remainders may be proportional?**

**(a) 0 (b) 1 (c) 2 (d) 7**

**Sol: **(c) 2

**Reason: **Let the number subtracted be x.

so (7−x):(17−x) :: (17−x):(47−x)

→(7−x)/(17−x) = (17−x)/(47−x)

→(7−x)(47−x) = (17−x)²

→329−47x−7x+x² = 289−34x+x²

→329−289 = 34x+54x

→20x = 40

→x = 2

**Que-12: If the ratio of A’s money to B’s money is 4:5 and that of B’s money to C’s money is 2:3 and A has Rs800, then C has :**

(a) Rs1000 (b) Rs1200 (c) Rs1500 (d) Rs2000

**Sol: **(c) Rs1500

**Reason: **⇒ A:B = 4:5 = 8:10 and B:C = 2:3 = 10:15∴

A:B:C = 8:10:15

⇒ If A has Rs.8, C has Rs.15.

⇒ If A has Rs.800

⇒ Then, C has Rs.(15×100) = Rs.1500

**Que-13: A sum of Rs7000 is divided among A,B and C in such a way that the shares of A and B are in the ratio 2:3 and those of B and C are in the ratio 4:5. The amount received by C is :**

(a) Rs2600 (b) Rs2800 (c) Rs3000 (d) Rs3900

**Sol: **(c) Rs3000

**Reason: **A:B = 2:3 = 8:12

B:C = 4:5 = 12:15

Now, sum of money of b is same in both,

Let, sum of money of :

a = 8x ; b = 12x ; c = 15x

Their sum = 7000

→ 8x + 12x + 15x = 7000

→ 35x = 7000

→ x = 200

Hence,

Money of a = 8x = 8(200) = ₹1600

Money of b = 12x = 12(200) =₹ 2400

Money of c = 25x = 15(200) = ₹3000

**Que-14: A sum of Rs53 is divided among A,B and C in such a way that A gets Rs7 more than what B gets and B gets Rs8 more than what C gets. The ratio of their shares is**

(a) 16:9:18 (b) 25:18:10 (c) 18:25:10 (d) 15:8:30

**Sol: **(b) 25:18:10

**Reason: **Let x be the share C gets.

So B gets x+8 and C gets x+15

x+x+8+x+15 = 53

So x = 10

So ratio of shares = 25:18:10

**Que-15: A bag contains Rs600 in the form of one-rupee, 50-paisa and 25-paisa coins in the ratio 3:4:12. The number of 25-paisa coins is**

(a) 600 (b) 900 (c) 1200 (d) 1376

**Sol: **(b) 900

**Reason: **Let the number of Rs.1 ,50 paisa and 25 paisa be 3,4,12

The value of one rupee coins = Rs.1×3 = 3Rs.

The value of 50 paisa coins = 0.50×4 = 2Rs

The value of 25 paisa coins = 0.25×12 = 3Rs

Total value = 3+2+3 = 8

If the total value is Rs 8 there are 12 coins of 25 paisa

If the total value is 600 then the number of 25 paisa coins

= (600/8) × 12 = 900

**Que-16: A certain was divided between P and Q in the ratio 4:3. If Q’s share was Rs4800, the total amount was :**

**(a) Rs11200 (b) Rs6400 (c) Rs19200 (d) Rs39200**

**Sol: **(a) Rs11200

**Reason: **Let, the amount of a and b = Rs. 4x and Rs. 3x

(obtained from the ratio 4:3)

According to the data mentioned in the question,

3x = 4800

x = 4800/3

x = 1600

a’s share will be :

= Rs. 4x

= Rs. (4 × 1600)

= Rs. 6400

Total amount :

= a’s share + b’s share

= Rs. (6400 + 4800)

= Rs. 11200

**Que-17: The ratio of number of boys and girls in a school of 720 students is 7:5. How many more girls should be admitted to make the ratio 1:1 ?**

(a) 90 (b) 120 (c) 220 (d) 240

**Sol: **(b) 120

**Reason: **Ratio of boys and girls in the art class = 7 ∶ 5

Total students = 720

∴ Total number of boys = 720 × 7/12

⇒ 60 × 7

⇒ 420

∴ Total number of girls = 720 – 420

⇒ 300

Now, number of girls that should join to make ratio 1 ∶ 1

⇒ 420 – 300

⇒ 120

**Que-18: The prices of scooter and a moped are in the ratio 9:5. If a scooter costs Rs6800 more than a moped, the price of scooter is**

(a) Rs13600 (b) Rs15300 (c) Rs17000 (d) none of these

**Sol: **(b) Rs15300

**Reason: **Let the price of the scooter and mop be 9x and 5x.

As given, the difference between the two = 4200

⇒ 9x – 5x = 6800

⇒ 4x = 6800

⇒ x = 1700

∴ The price of a mop will be = 9x

⇒ 9 × 1700 = 15300

**Que-19: The ratio of zinc and copper in a brass piece is 13:7. How much zinc will be there in 100kg of such a piece?**

(a) 20kg (b) 35kg (c) 55kg (d) 65kg

**Sol: **(d) 65kg

**Reason: **Let total units is 20 kg

So, in 20 kg of brass zinc will be 7 kg and copper is 13 kg

In 100 kg of brass zinc will be

⇒ 100 × (7/20)

⇒ 35 kg

∴ The zinc in 100 kg of brass is 35 kg.

**Que-20: Two numbers are in the ratio 3:5. If 9 be subtracted from each, then they are in the ratio 12:23. The second number is :**

(a) 52 (b) 53 (c) 54 (d) 55

**Sol: **(d) 55

**Reason: **Let the two numbers be 3x & 5x.

According to question,

(3x – 9)/(5x – 9) = 12/23

x = 11

3x = 33

5x = 55

**Que-21: Two whole numbers whose sum is 64 cannot be in the ratio :**

(a) 5:3 (b) 7:1 (c) 3:4 (d) 9:7

**Sol: **(c) 3:4

**Reason: **For dividing 64 into two whole numbers, the sum of the terms of the ratio must be a factor of 64.

So, they cannot be in the ratio 3:4.

Hence, option C is correct.

**Que-22: Out of the ratios (7:20); (13:25); (17:30) and (11:15), the smallest is :**

**(a) 7:20 (b) 13:25 (c) 17:30 (d) 11:15**

**Sol: **(a) 7:20

**Reason: **7/20 = 0.35

13/25 = 0.52

17/30 = 0.56

11/15 = 0.73

So, the smallest ratio is 7:20.

**Que-23: In a ratio which is equal to 5:8, the antecedent is 40, then the consequent is :**

(a) 25 (b) 48 (c) 64 (d) none of these

**Sol: **(c) 64

**Reason: **Let the consequent be x

5/8 = 40/x

x = (8×40)/5

x = 64.

**–: **Ratio and Proportion MCQS Class 8 RS Aggarwal Exe-9C Goyal Brothers ICSE Maths Solutions** :–**

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