Ratio And Proportion MCQs

Option A: 9

Option B: 12

Option C: 15

Option D: 18

Correct Answer: 15 ✔

Option A: 80:70:60

Option B: 8:7:5

Option C: 4:7:5

Option D: 7:8:5

Correct Answer: 8:7:5 ✔

Option A: 40

Option B: 50

Option C: 60

Option D: 30

Correct Answer: 50 ✔

A. 8:15:30
B. 5:18:28
C. 4:5:6
D. 2:3:5

P1:P2:P3 = (2*4):(3*5):(5*6)
= 8 : 15 : 30

Correct Answer: A. 8:15:30 ✔

A. 160
B. 200
C. 250
D. 240

Let the number of 50p, 1 rupee and 2 rupee coins be 2x,5x and 8x. 0.5*(2x)+5x+2*8x = 352 22x=352 x=16 Total no. of coins
= 2x+5x+8x
= 15x = 240

Correct Answer: D. 240 ✔

A. 2
B. 3
C. 5
D. 6

X:Y:Z=2:3:4
X=4
Then, Y=6, Z=8 XY=kZ 4*6=k*8 8k=24 k=3
Y=12, Z=8
X*12=3*8
X=2

Correct Answer: A. 2 ✔

A. 9
B. 15
C. 25
D. 50

5:x::x:45 x²=45*5 = 225 x=15

Correct Answer: B. 15 ✔

A. 2:3:5
B. 8:12:15
C. 2:8:5
D. 2:7:5

If A:B = m1:n1 ; B:C = m2:n2
Then, A:B:C = (m1*m2):(n1*m2):(n1*n2)
Therefore, X:Y:Z = (2*4):(3*4):(3*5)
= 8:12:15

Correct Answer: B. 8:12:15 ✔

A. 15
B. 10
C. 8
D. 5

a:b = 3:5 5a = 3b (a+2):(b+2) = 2:3 3(a+2) = 2(b+2) 3a+6 = 2b+4 3*(3b/5) = 2b-2 9b = 10b-10 b =10

Correct Answer: B. 10 ✔

A. 1200 m²
B. 1500 m²
C. 12000 m²
D. 15000 m²

Actual length
= 30*500 cm Actual breadth
= 20*500 cm Actual area
= 30*20*500*500 = 15000 m²

Correct Answer: D. 15000 m² ✔

A. 54 cm
B. 27 cm
C. 12 cm
D. 15 cm

Let the length and breadth be 5x cm and 4x cm.
5x*4x=180
20x²=180
x²=9,x=3 Length = 15cm Breadth = 12cm Perimeter
= 2(l+b)
= 2*27
= 54 cm

Correct Answer: A. 54 cm ✔

A. 16 : 2 : 3
B. 8 : 3 : 6
C. 16 : 2 : 9
D. 6 : 9 : 1

Let the profit of X be P1, that of Y be P2 and of Z be P3.
P1:P2:P3 = 20000*12 : 7500*4 : 15000*9 = 240 : 30 : 135 = 80 : 10 : 45
= 16 : 2 : 9

Correct Answer: C. 16 : 2 : 9 ✔

A. 1/3
B. 7/9
C. 5/12
D. None of these

M/P = M/N * N/O * O/P = ¾ * 5/7 * 7/9 = 5/ 12

Correct Answer: C. 5/12 ✔

A. 45 cm
B. 40 cm
C. 36 cm
D. 27 cm

Let the three sides be 4x,3x and 2x. 4x+3x+2x = 81 9x = 81 x = 9 Length of the largest side
= 4x = 36 cm

Correct Answer: C. 36 cm ✔

A. 96
B. 68
C. 72
D. 108

Let the number of boys and girls in the class be 11x and 13x. 4 boys and 8 girls join the class. Strength of the class
= 11x+13x+4+8
= 24x+12 (11x+4):(13x+8) = 4:5 5*(11x+4)=4*(13x+8) 55x+20=52x+32 3x=12 x=4 24x+12=108

Correct Answer: D. 108 ✔

A. 48
B. 42
C. 60
D. 70

7x+8x=90 15x=90 x=6 Number of boys=7x=42

Correct Answer: B. 42 ✔

A. 120
B. 80
C. 192
D. 48

If A invests amount C1 for T1 time and his share of profit is P1, and B invests amount C2 for T2 time and his share of profit is P2, then, C1 * T1 / C2 * T2 = P1/P2
If P is the Nasir’s share of profit, then Changaz gets (240 – P)
Therefore, 6000 * 12 / 3000 * 6 = (240 – P) / P = 72/ 18 = 4
4P = (240 – P)
5P = 240
P = 48

Correct Answer: D. 48 ✔

A. 5:4
B. 5:2
C. 2:5
D. 4:5

(x/100)*200 = (y/100)*250 y/x = 20/25 = 4/5 y:x=4:5

Correct Answer: D. 4:5 ✔

A. 66
B. 48
C. 24
D. 8

Let the number of 1 rupee, 2 rupee and 5 rupee coins be 3x, 5x and 4x. (1*3x)+(2*5x)+(5*4x)=264 3x+10x+20x=264 33x=264 x=8 No. of 1 rupee coins
= 3x = 24

Correct Answer: C. 24 ✔

A. Rs.1200,Rs.800
B. Rs.800,Rs.1200
C. Rs.1500,Rs.500
D. Rs.1100, Rs.900

3x+2x=2000 5x=2000 x=400 3x=1200 2x=800

Correct Answer: A. Rs.1200,Rs.800 ✔

A. 30, 19
B. 21, 37
C. 15, 34
D. 15, 27

Let the two numbers be 5x and 9x.
(5x-5)/(9x-5) = 5:11
(5x-5)*11 = (9x-5)*5
55x – 55 = 45x – 25
10x = 30
x = 3
Therefore, the numbers are 15 and 27.

Correct Answer: D. 15, 27 ✔

A. 8.96
B. 8
C. 4.5
D. 6.2

3.5/5/6 = 5.6/x ; x = 5.6 * 5.6 / 3.5 = 8.96

Correct Answer: A. 8.96 ✔

A. 360
B. 380
C. 375
D. 400

Flour and sugar have to mixed in the ratio 4:5 4:5::300:x 4x=1500 x=375 375 grams of sugar should be mixed.

Correct Answer: C. 375 ✔

A. 32:35
B. 67:56
C. 5:14
D. 5:7

X:Z=(X:Y)*(Y:Z) =(5:8)*(4:7) = (5/8)*(4/7) = 5/14

Correct Answer: C. 5:14 ✔

A. 3:1
B. 4:1
C. 3:4
D. 4:3

Let the total quantity of mixture be 1. If quantity of 1st type is x then quantity of the 2nd type will be (1-x)
Therefore, 13x + 19(1-x) = 14.2
13x -19x + 19 = 14.2
4.8 = 6x
x = 0.8 ; (1-x) = 0.2
Therefore, the ratio is 0.8 : 0.2 = 4:1

Correct Answer: B. 4:1 ✔

A. Rs.100
B. Rs.700
C. Rs.1600
D. Rs.2000

Let the amount be divided into 7x,4x,3x and 2x.
7x-2x = 500 5x=500 x = 100

Total amount = 7x+4x+3x+2x = 16x = Rs.1600

Correct Answer: C. Rs.1600 ✔

A. 1:4
B. 3:4
C. 1:3
D. 2:3

x = z+(20/100)z = 1.2z y = 1.8z x:y = 1.2z:1.8z = 12:18 = 2:3

Correct Answer: D. 2:3 ✔

A. 18
B. 15
C. 12
D. 5

2x/25 = 6/(x/3) 2x/25 = 18/x 2x*x = 18*25 x² = 9*25 x=3*5 x=15

Correct Answer: B. 15 ✔

A. 1:2
B. 3:5
C. 1:3
D. 3:4

72 should be divisible by addition of the above ratio.
Using trial and error method
For (a): – 1+2 = 3, 72 is divisible by 3
For (b): – 3+5 = 8, 72 is divisible by 8
For (a): – 1+3 = 4, 72 is divisible by 4
For (a): – 3+4 = 7, 72 is not divisible by 7
Therefore, the answer is option (d)

Correct Answer: D. 3:4 ✔

A. 2:3
B. 3:2
C. 3:4
D. 4:3

x+y = 750 y-x = 150 Solving,
y = 450
x = 300 x:y = 300:450 = 2:3

Correct Answer: A. 2:3 ✔

A. 3:2
B. 1:2
C. 1:5
D. 2:1

p = q+(50/100)q = 1.5q p:q = 1.5q:q = 1.5:1 = 3:2

Correct Answer: A. 3:2 ✔

A. 8
B. 6
C. 12
D. 18

Let the numbers be 3x,5x and 6x. (3x)²+(5x)²+(6x)²=280 9x²+25x²+36x²=280 70x²=280 x²=4 x=2 The greatest number is 6x=12.

Correct Answer: C. 12 ✔

A. 9:19
B. 19:9
C. 15:28
D. 28:15

Let A1 be the area of flat and A2 be that of the commercial space
Total cost = area * rate
Therefore, cost of flat = A1*4500 ; cost of commercial space = A2*9500
Both the above costs are same
A1*4500 = A2*9500
A1:A2 = 9500:4500 = 19:9

Correct Answer: B. 19:9 ✔

A. 44
B. 55
C. 66
D. 77

Let the numbers be a, b and c. a+b+c = 172 a:b = 8:11 b:c = 5:7 11a = 8b a = 8b/11 7b = 5c c = 7b/5 (8b/11)+b+(7b/5) = 172 40b+55b+77b = 172*55 172b = 172*55 b=55

Correct Answer: B. 55 ✔

A. Rs.18000
B. Rs.13000
C. Rs.15000
D. Rs.10000

4S+5D+3C=130000 3C=2D 3D=2S S=3D/2 [4*(3D/2)]+5D+2D = 130000 6D+5D+2D = 130000 13D = 130000 D = Rs.10000 S=3D/2 = Rs.15000

Correct Answer: C. Rs.15000 ✔

A. 4
B. 6
C. 3
D. 2

Let the numbers be 2x and 3x. (2x)³+(3x)³=280 8x³+27x³=280 35x³=280 x³=8 x=2 The numbers are 4 and 6.

Correct Answer: A. 4 ✔

A. 38
B. 35
C. 10
D. 28

Let z = 4y-1
When x = 14, y = 2, z = (4*2) – 1 = 7
Now, x varies directly as z = 4y-1
When y = 5, z = (4*5) – 1 = 19
x 14 7
y – 19
Therefore, x = (14*19)/7 = 38

Correct Answer: A. 38 ✔

A. 6:10
B. 13:20
C. 3:4
D. 35:33

Let the number of seats be 5x and 8x. New ratio
= (1.3*5x):(1.25*8x)
= 6.5x:10x
= 13:20

Correct Answer: B. 13:20 ✔

A. 2:5
B. 4:5
C. 6:5
D. 8:5

A:C=(A:B)*(B:C) =(2/3)*(3/4)= 1/2 C:D=(C:A)*(A:D) =(2)*(4/5) = 8:5

Correct Answer: D. 8:5 ✔

A. 63
B. 31
C. 21
D. 27

3:7::9:x 3x=7*9=63 x=21

Correct Answer: C. 21 ✔

A. 2 : 5
B. 3 : 7
C. 5 : 3
D. 7 : 3

Correct Answer: C. 5 : 3 ✔

A. 2 : 3 : 4
B. 6 : 7 : 8
C. 6 : 8 : 9
D. None of these

5:7:8 = 50:70:80
50*140/100=70 ; 70*150/100=105 ; 80*175/100=140
70: 105: 140 or
14: 21: 28 or
2: 3: 4

Correct Answer: A. 2 : 3 : 4 ✔

A. 50
B. 100
C. 150
D. 200

Correct Answer: C. 150 ✔

A. 27
B. 33
C. 49
D. 55

Correct Answer: C. 49 ✔

A. 2 : 5
B. 3 : 5
C. 4 : 5
D. 6 : 7

Let the third number be x.
Then, first number = 120% of x = 120x/100 = 6x/5 second numbet = 150% of x = 150x/100 = 3x/2 Ratio of first two numbers = 6x/5 : 3x/2 =12x : 15x = 4 : 5.

Correct Answer: C. 4 : 5 ✔

A. Rs. 500
B. Rs. 1500
C. Rs. 2000
D. None of these

Correct Answer: C. Rs. 2000 ✔

A. 8 : 9
B. 17 : 18
C. 21 : 22
D. Cannot be determined

Originally, let the number of boys and girls in the college be 7x and 8x respectively. Their increased number is (120% of 7x) and (110% of 8x), i . e . ( 120/100 X 7x ) and ( 110/100 X 8x ) i . e . 42x/5 and 44x/5 Required ratio = 42x/5 : 44x/5 = 21 : 22

Correct Answer: C. 21 : 22 ✔

A. 20
B. 30
C. 48
D. 58

Let the three parts be A, B, C Then,
A : B = 2 : 3 and B : C = 5 : 8 (5 X 3/5 ) : (8 X 3/5 ) = 3 : 24/5
A:B:C= 2 : 3 : 24/5 10:15:24
B = (98 X 15/49 ) = 30.

Correct Answer: B. 30 ✔