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.
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