SSC GD Ratio and Proportion: Important Practice Questions with Solutions
Elevate your SSC GD Maths preparation with our comprehensive collection of Ratio and Proportion practice questions. This guide includes essential concepts, common question patterns, and detailed solutions designed to help you build speed, accuracy, and confidence for your upcoming examination.
Ratio and Proportion is a high-scoring and fundamental topic in the SSC GD Mathematics syllabus. Questions from this section frequently appear in exams and are typically based on core formulas, logical deduction, and arithmetic calculations. Mastering these concepts through consistent practice with expert-verified solutions is key to achieving a top score in the SSC GD selection process.
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SSC GD Ratio and Proportion Questions with Solutions
Q.1 In a school of 3528 students, the ratio of boys to girls is 5:4. How many additional girls must be enrolled to achieve a ratio of 1:1?
A. 312
B. 392
C. 472
D. 552
Answer: B
Solution:
Total students = 3528
Ratio = 5 : 4 (Total parts = 9)
Value of one part = 3528 ÷ 9 = 392
Number of boys = 5 × 392 = 1960
Number of girls = 4 × 392 = 1568
Required girls to add = 1960 − 1568 = 392
Q.2 If (x + z)/y = z/(x − y) = x/z, what is the value of x : y : z?
A. 4 : 3 : 6
B. 3 : 4 : 2
C. 4 : 3 : 2
D. 3 : 4 : 6
Answer: D
Solution: By solving the given proportionality equations, we derive the ratio 3 : 4 : 6.
Q.3 If k is added to each of 11, 27, 20, and 44, the resulting numbers are in proportion. Find the mean proportional between (k+2) and (3k−5).
A. 6
B. 15
C. 9
D. 12
Answer: D
Solution:
(11+k)/(27+k) = (20+k)/(44+k) → (11+k)(44+k) = (27+k)(20+k)
Solving this gives k = 7
Mean proportional = √(9 × 16) = √144 = 12
Q.4 The initial student ratio in sections A, B, and C is 5:4:7. After new admissions, the ratio becomes 10:9:11. Determine the total student count.
A. 210
B. 120
C. 150
D. 90
Answer: B
Solution: By solving the linear equations derived from the ratio changes, the total number of students is 120.
Q.5 If 38% of the first number equals two-sevenths of the second, what is the ratio between them?
A. 101 : 129
B. 103 : 132
C. 100 : 133
D. 96 : 131
Answer: C
Q.6 Given that 25% of A = 20% of B = 10% of C, find A : B : C.
A. 4 : 5 : 10
B. 10 : 5 : 4
C. 5 : 4 : 10
D. 2 : 3 : 5
Answer: A
Q.7 When x is subtracted from 55, 50, 23, and 22, the results are in proportion. Find the fourth proportional of 3, 7, and x.
A. 35
B. 32
C. 38
D. 36
Answer: A
Q.8
A bag contains coins of ₹10, ₹5, and ₹2 in a 1:2:3 ratio, totaling ₹390. How many coins are there of each type?
A. 10, 20, 30
B. 15, 30, 45
C. 20, 40, 60
D. 12, 24, 36
Answer: B
Q.9 The mean proportional between 36 and N is three times the mean proportional between 8 and 32. Calculate the value of N.
A. 47
B. 58
C. 51
D. 64
Answer: D
Q.10 Given that x varies inversely as the square of y, and x=6 when y=5, find x when y=4.
A. 9.175
B. 9.835
C. 9.375
D. 9.925
Answer: C
Q.11 When x is subtracted from each of 43, 38, 11, and 6, the resulting numbers are in proportion. Solve for x.
Solution:
(43 − x) / (38 − x) = (11 − x) / (6 − x)
Equating cross-products: (43 − x)(6 − x) = (38 − x)(11 − x)
258 − 49x + x² = 418 − 49x + x²
Solving this results in x = 4
Q.12 Given 2x = 3y = 4z, determine the ratio x : y : z.
Solution:
Let 2x = 3y = 4z = k
Then x = k/2, y = k/3, z = k/4
x : y : z = (k/2) : (k/3) : (k/4) = 6 : 4 : 3
Q.13 Divide ₹3777 among A, B, and C in the ratio 7 : 9 : 3.
Solution:
Sum of ratios = 7 + 9 + 3 = 19
A = (7/19) × 3777 = ₹1391
B = (9/19) × 3777 = ₹1789
C = (3/19) × 3777 = ₹597
Q.14 If x : y = 4 : 5 and y : z = 10 : 7, find x : y : z.
Solution:
x : y = 4 : 5, which is equivalent to 8 : 10
y : z = 10 : 7
Combining these, x : y : z = 8 : 10 : 7
Q.15 The current age ratio of A and B is 3 : 5. After 6 years, the ratio becomes 5 : 7. Find A's present age.
Solution:
Let the present ages be 3x and 5x.
(3x + 6) / (5x + 6) = 5 / 7
21x + 42 = 25x + 30 → 4x = 12 → x = 3
Age of A = 3 × 3 = 9 years.
Q.16 In a class with a boy-to-girl ratio of 7 : 9, there are 63 boys. Find the total number of girls.
Solution:
7 parts = 63 → 1 part = 9
Number of girls = 9 parts × 9 = 81.
Q.17 If a : b = 5 : 6 and b : c = 7 : 8, find the combined ratio a : b : c.
Solution:
a : b = 35 : 42
b : c = 42 : 48
Combining gives a : b : c = 35 : 42 : 48.
Q.18 Three numbers total 98, with a ratio of 2 : 3 : 7. Find the largest number.
Solution:
Sum of ratios = 12
Value of one part = 98 / 12 ≈ 8.1667
Largest number = 7 × 8.1667 ≈ 57.17
Q.19 If x : y = 3 : 4 and y : z = 8 : 9, find the combined ratio x : y : z.
Solution:
x : y = 6 : 8 (scaled by 2)
y : z = 8 : 9
Combined x : y : z = 6 : 8 : 9
Q.20 Distribute ₹14000 among A, B, and C in the ratio 7 : 9 : 3.
Solution:
Total parts = 19
A = (7/19) × 14000 ≈ ₹5158
B = (9/19) × 14000 ≈ ₹6632
C = (3/19) × 14000 ≈ ₹2210
Q.21 If 3.6 : 1.2 :: 1.2 : y, what is the value of y?
A. 0.60
B. 0.40
C. 0.90
D. 0.80
Answer: B
Solution:
3.6/1.2 = 1.2/y
3.6y = 1.44 → y = 0.4
Q.22 What number must be added to 8, 35, 7, and 31 for the resulting set to be proportional?
A. 2
B. 3
C. 5
D. 1
Answer: D
Solution:
(8+x)/(35+x) = (7+x)/(31+x)
(8+x)(31+x) = (35+x)(7+x) → 248+39x+x² = 245+42x+x² → 3x = 3 → x=1
Q.23 If 77% of a number is equal to three-fourths of another, what is the ratio of the first number to the second?
A. 75 : 77
B. 74 : 75
C. 73 : 82
D. 72 : 73
Answer: A
Solution:
(77/100)x = (3/4)y → x/y = (3/4) × (100/77) = 75/77
Q.25 The ratio of income for Seema and Darshan is 7 : 8, with savings of ₹15,000 and ₹9,000. Given their expenditure ratio is 11 : 16, find the total expenditure.
A. 68,000
B. 64,000
C. 64,125
D. 63,000
Answer: C
Solution: Solving (7x - 15000)/(8x - 9000) = 11/16 results in x = 2375. Total Expenditure = 64125.
Q.25 Marbles in boxes A, B, C, and D are in the ratio 6:3:7:5. If box B contains 400 fewer marbles than box D, find the total marble count.
A. 4000
B. 4100
C. 4200
D. 3900
Answer: C
Solution:
(5x - 3x) = 400 → 2x = 400 → x = 200. Total = (6+3+7+5) × 200 = 4200.
Q.26 Given X ∝ A² and A ∝ 1/Y. If X=39 when Y=20, find X when Y=2.
A. 3902
B. 3901
C. 3897
D. 3900
Answer: D
Solution: X ∝ 1/Y². Therefore, X₁Y₁² = X₂Y₂² → 39 × 20² = X × 2² → X = 3900.
Q.27 C is the third proportional to 29 and B. B is the sum of the first three even natural numbers. Find C.
A. 3.53
B. 3.85
C. 7.64
D. 4.97
Answer: D
Solution:
B = 2+4+6 = 12. For third proportional, 29/12 = 12/C → C = 144/29 ≈ 4.97.
Q.28 y varies directly as (x+3). If y=8 when x=1, find y when x=2.
A. 10
B. 5
C. 8
D. 2
Answer: A
Solution:
y = k(x+3). 8 = k(1+3) → k=2. For x=2, y = 2(2+3) = 10.
Q.29 Three numbers are in a 65 : 224 : 260 ratio. If the difference between the largest and smallest is 30, find the largest number.
A. 40
B. 39
C. 41
D. 42
Answer: A
Solution:
(260x - 65x) = 30 → 195x = 30 → x = 2/13. Largest = 260 × (2/13) = 40.
Q.30 Four numbers are in a 16 : 10 : 9 : 17 ratio. If their sum is 6292, find the sum of the first and third numbers.
A. 3025
B. 3070
C. 3031
D. 3074
Answer: A
Solution:
(16+10+9+17)x = 52x = 6292 → x=121. Sum of first and third = (16+9) × 121 = 3025.