**Ratio and Proportional MCQ** Type Questions ICSE Class-10 Maths for Sem-1. These MCQ / Objective Type Questions is based on latest reduced syllabus according 2021-22 session on bifurcated pattern. Main motto of MCQ Type Question is cracking the next upcoming exam of council. Visit official website CISCE for detail information about ICSE Board Class-10 Maths.

## ICSE Class-10 Maths **Ratio and Proportional MCQ** Type Questions for Sem-1

Board | ICSE |

Class | 10th ( x ) |

Subject | Maths |

Chapter | Ratio and Proportional |

Syllabus | on bifurcated syllabus (after reduction) |

bifurcated pattern |
Semester-1 |

Session | 2021-22 |

Topic | MCQ / Objective Type Question |

### Maths **Ratio and Proportional MCQ** Type Questions of ICSE Class-10 for Sem-1

**Question-1** Given that (a^{3} + 3ab^{2}) / (b^{3 }+ 3a^{2}b) = 63 / 62 . Using componendo and dividendo , The value of a: b will be

(i) a : b = 2 : 3 ; (ii) a : b = 63 : 62 (iii) a : b = 62 : 63 (iv) a : b = 3 : 2

**Ans:- (iv) a : b = 3 : 2**

**Hint**

Given that (a^{3} + 3ab^{2})/(b^{3 }+ 3a^{2}b) = 63/62

By componendo and dividendo,

(a^{3} + 3ab^{2} + b^{3 }+ 3a^{2}b)/(a^{3} + 3ab^{2 }– b^{3} – 3a^{2}b) = (63 + 62)/(63 – 62) = 125/1

⇒ {(a + b)^{3}}/{(a – b)^{3}} = (5/1)^{3}

⇒ (a + b)/(a – b) = 5

⇒ a + b = 5a – 5b

⇒ 5a – a – 5b – b = 0

⇒ 4a – 6b = 0

⇒ 4a = 6b

⇒ a/b = 6/4

⇒ a/b = 3/2

a : b = 3 : 2

**Question- 2** If 3A = 4B = 6C, find A: B: C

(i) 5: 3: 2 ; (ii) 4: 1: 2 (iii) 4: 5: 2 (iv) 4: 3: 2

**Ans:- (iv) 4: 3: 2**

**Hint**

3A = 4B

A/B = 4/3

A: B = 4: 3

Similarly 4B = 6C

B/C = 6/4 = 3/2

B: C = 3: 2

So we get

A: B: C = 4: 3: 2

**Question-3** If (x – 9) : (3x + 6) is the duplicate ratio of 4 : 9, the value of x

(i) 33 ; (ii) 25 (iii) 48 (iv) 96

**Ans:- (ii) 25**

**hint**

(x – 9)/ (3x + 6) = (4/9)^{2}

(x – 9)/ (3x + 6) = 16/81

81x – 729 = 48x + 96

81x – 48x = 96 + 729

33x = 825

x = 825/33

= 25

**Question-4 **A certain sum was divided among A, B and C in the ratio 7 : 5 : 4. If B got Rs 500 more than C, find the total sum divided.

(i) 8000 ; (ii) 6000 (iii) 7000 (iv) 9000

**Ans:- (i) 8000**

**hint**

Ratio between A, B and C = 7 : 5 : 4

Let A’s share = 7x

B’s share = 5x

and C’s share = 4x

Total sum = 7x + 5x + 4x = 16x

5x – 4x = 500

x = 500

So the total sum = 16x = 16 × 500 = Rs 8000

**Question- 5 **the mean proportion of:(a – b) and (a^{3} – a^{2}b), a ˃ b

(i) a (b – a)

(ii) a (a – b)

(iii) -a (a – b)

(iv) a (-a – b)

**Ans (ii) a (a – b)**

**Hint**

let x as the mean proportion of (a – b) and (a^{3} – a^{2}b), a ˃ b

(a – b): x :: (a^{3} – a^{2}b)

x^{2} = (a – b) (a^{3} – a^{2}b)

x^{2} = (a – b) a^{2} (a – b)

x^{2} = a^{2} (a – b)^{2}

x = a (a – b)

Hence, mean proportion is a (a – b).

**Question -6** If a, 12, 16 and b are in continued proportion value of a and b will be

(i) a = 9 and b = 3

(ii) a = 64 and b = 3

(iii) a =64/3 and b = 9

(iv) a = 9 and b = 64/3

**Answer- (iv) a = 9 and b = 64/3 **

**hint**

∵ a, 12, 16, b are in continued proportion, then

a/12 = 12/16 = 16/b

a/12 = 12/16

16a = 144 (By cross multiplication)

a = 144/16 = 9

and

12/16 = 16/b

12b = 16 × 16 = 256 (By cross multiplication)

b = 256/12 = 64/3 = 21 1/3

Hence, a = 9 and b = 64/3

**Question-7** The following numbers, K + 3, K + 2, 3K – 7 and 2K – 3 are in proportion. value of k

(i) K = 5 or K = – 1

(ii) K = -5 or K = 1

(iii) K = 5 or K = 1

(iv) K = -5 or K = – 1

**Answer-(i) K = 5 or K = – 1**

**Hint**

(K + 3) / (K + 2) = (3K – 7) / (2K – 3 )

⇒ (K + 3) (2K – 3) = (K + 2) (3K – 7).

⇒ 2K^{2} – 3K + 6K – 9 = 3K^{2 }– 7K + 6K – 14

⇒ K^{2} – 4K – 5 = 0

⇒ (K – 5) (K + 1) = 0

⇒ K = 5 or K = – 1

**Question- 8** If x + 5 is the mean proportion between x + 2 and x + 9, find the value of x.

(i) 7 (ii) 8 (iii) 9 (iv) 10

**Answer-(i) 7**

**Hint**

∵ x + 5 is the mean proportion between x + 2 and x + 9, then

(x + 5)² = (x + 2) (x + 9)

⇒ x² + 10x + 25 = x² + 11x + 18,

and ⇒ x² + 10x – x² – 11x = 18 – 25

so ⇒ – x = – 7

∵ x = 7 Ans.

**Question -9 **The two numbers such that the mean proportional between them is 28 and the third proportional to them is 224. The two numbers are

(i) 28 x 224 /2 and (28 /224)/2

(ii) 28/2 and 224/2

(iii) 784 and 224

(iv) 14 and 56

**Answer-(iv) 14 and 56**

**Hint**

Let the two numbers are a and b.

∵ 28 is the mean proportional

∵ a : 28 : : 28 : b

ab = 28^{2} = 784

Here a = 784/b …… (1)

224 is the third proportional

a: b :: b: 224

b^{2} = 224a ….. (2)

substituting the value of a in equation (2)

b^{2} = 224 × 784/b

b^{3} = 224 × 784

b^{3} = 175616 = 56^{3}

b = 56

substituting the value of b in equation (1)

a = 784/56 = 14

Hence 14 and 56 are the two numbers

**Question -10** Rs. 8400 is divided among A, B, C and D in such a way that the shares of A and B, B and C, and C and D are in the ratios of 2:3, 4:5 and 6:7 respectively. The share of A is

(i) Rs. 1280 ; (ii) Rs. 1920 (iii) Rs. 2400 (iv) Rs. 2800

**Ans- (i) Rs. 1280**

**Question -11** The ratio of the present age of father to that of son is 7:2. After 10 years their ages will be in the ratio of 9:4. The present ages of the father is

**Ans-(i). 35 years**

**Question -12**The three numbers are in the ratio 1/2 : 2/3 : 3/4. The difference between greatest and smallest numbers is 36. the numbers are

**Ans-(i). 72, 84, 108**

**Question -13**If A:B = 2:3, B:C = 4:5 and C:D = 6:7, then A:B:C:D is

**Ans :- 16:24:30:35**

**Question -14**. A ratio equivalent to 3 : 7 is:

(i) 3 : 9; (ii) 6 : 10; (iii) 9 : 21; (iv) 18 : 49

**Ans:- (iii) 9 : 21;**

**Question -15**. The ratio 35 : 84 in simplest form is:

(i) 5 : 7; (ii) 7 : 12; (iii) 5 : 12; (iv) none of these

**Ans:- (iii) 5 : 12;**

**Question -16 **In a class there are 20 boys and 15 girls. The ratio of boys to girls is:

(i) 4 : 3; (ii) 3 : 4; (iii) 4 : 5; (iv) none of these

**Ans:-(i) 4 : 3;**

**Question -17 **Two numbers are in the ratio 7 : 9. If the sum of the numbers is 112, then the larger number is:

(i) 49; (ii) 72; (iii) 63; (iv) 42

**Ans:-(iii) 63;**

**Question -18** The ratio of 1.5 m to 10 cm is:

(i) 1 : 15; (ii) 15 : 10; (iii) 10 : 15; (iv) 15 : 1

**Ans:-(iv) 15 : 1**

**Question -19** The ratio of 1 hour to 300 seconds is:

(i) 1 : 12; (ii) 12 : 1; (iii) 1 : 5; (iv) 5 : 1

**Ans:-(ii) 12 : 1;**

**Question -20**. In 4 : 7 : : 16 : 28, 7 and 16 are called

(i) extreme terms; (ii) middle terms; (iii) b middle and c extreme term; (iv) none of these

**Ans:-(ii) middle terms;**

**Question -21**The first, second and fourth terms of a proportion are 16, 24 and 54 respectively. Then the third term is:

(i) 36; (ii) 28; (iii) 48; (iv) 32

**Ans:-(i) 36;**

**Question -22** If 12, 21, 72, 126 are in proportion, then:

(i) 12 × 21 = 72 × 126; (ii) 12 × 72 = 21 × 126; (iii) 12 × 126 = 21 × 72; (iv) none of these

**Ans:-(iii) 12 × 126 = 21 × 72;**

**Question -23**. If x, y and z are in proportion, then:

(i) x : y : : z : x; (ii) x : y : : y : z; (iii) x : y : : z : y; (iv) x : z : : y : z

**Ans:- (ii) x : y : : y : z**

**Question -24** 7 : 12 is equivalent to:

(i) 28 : 40; (ii) 42 : 71; (iii) 72 : 42; (iv) 42 : 72

**Ans:- (iv) 42 : 72**

**Question -25** The length and breadth of a rectangle are in the ratio 3 : 1. If the breadth is 7 cm, then the length of the rectangle is:

(i) 14 cm; (ii) 16 cm; (iii) 18 cm; (iv) 21 cm

**Ans:- (iv) 21 cm**

**Question -26 **The value of m, if 3, 18, m, 42 are in proportion is:

(i) 6; (ii) 54; (iii) 7; (iv) none of these

**Ans:- (ii) 54;**

**Question -27** Length and width of a field are in the ratio 5 : 3. If the width of the field is 42 m then its length is:

(i) 100 m; (ii) 80 m; (iii) 50 m; (iv) 70 m

**Ans:- (iv) 70 m**

#### More Questions on **Ratio and Proportional MCQ **of ML Aggarwal

#### Question -1

The ratio of 4 litres to 900 mL is

(a) 4 : 9

(b) 40 : 9

(c) 9 : 40

(d) 20 : 9

#### Answer-

4l : 900 mL,

= 4000 mL : 900 mL,

= 4000 : 900,

= 40 : 9

#### Question -2

When the number 210 is increased in the ratio 5 : 7, the the new number is

(a) 150

(b) 180

(c) 294

(d) 420

#### Answer –

210 is increased in the ratio 5 : 7, then

New increased number will be = 210 × 7/5

= 294

#### Question-3

Two numbers are in the ratio 7 : 9. If the sum of the numbers is 288, then the smaller number is

(a) 126

(b) 162

(c) 112

(d) 144

#### Answer –

Ratio in two number = 7 : 9

Sum of numbers = 288

Sum of ratios = 7 + 9 = 16

Smaller number = (288 × 7)/16

= 126

Question -4

The ratio of number of edges of a cube to the number of its faces is

(a) 2 : 1

(b) 1 : 2

(c) 3 : 8

(d) 8 : 3

#### Answer –

No. of edges of the cube = 12

No. of faces = 6

Ratio in edges a cube to the number of faces = 12 : 6

= 2 : 1

#### Question-5

If x, 12, 8 and 32 are in proportion, then the value of x is

(a) 6

(b) 4

(c) 3

(d) 2

#### Answer-

x, 12, 8, 32 are in proportion, then

x × 32 = 12 × 8 (∵ ad = bc)

⇒ x = (12 × 8)/32 = 3

x = 3

#### Question-6

The fourth proportional to 3, 4, 5 is

(a) 6

(b) 20/3

(c) 15/4

(d) 12/5

#### Answer-

The fourth proportion to 3, 4, 5 will be

= (4 × 5)/3

= 20/3

#### Question-7

The third proportional to and 5 is

(a) 4

(b) 8 ^{1}⁄_{2}

(c) 3

(d) none of these

#### Answer-

The third proportional to 6.1/4 and 5 is

⇒ 6. 1/4 : 5 : : 5 : x

⇒ 25/4 : 5 : : 5 : x

⇒ x = {(5 × 5)/25} × 4,

⇒ 4

Question-8

The mean proportional between ^{1}⁄_{2} and 128 is

(a) 64

(b) 32

(c) 16

(d) 8

#### Answer-

The mean proportional between ^{1}⁄_{2} and 128 is

= √(128/2)= √64 = 8 (d)

–: End of **Ratio and Proportional MCQ** Type Questions ICSE Class-10 Maths :-

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