SSC GD 2025 PYQs

Question 1

What is the third proportional to 16 and 48?

Accepted Answer

If a, b, c are in continued proportion, then a:b = b:c. So, c = b²/a. Here, c = 48²/16 = 2304/16 = 144.

Question 2

The price (per litre) of petrol increases by 60%. By what percent should its consumption be reduced such that the expenditure on it increases by 12% only?

Accepted Answer

Let original price be 100 and consumption be 100. Original expenditure = 100*100 = 10000. New expenditure = 10000 * 1.12 = 11200. New price = 100 * 1.60 = 160. New consumption = 11200/160 = 70. Reduction = (100-70)/100 * 100 = 30%.

Question 3

If the cost price of a camera is 75% of its selling price, then the profit per cent is:

Accepted Answer

Let SP be 100. Then CP is 75. Profit = SP - CP = 100 - 75 = 25. Profit percentage = (Profit/CP) * 100 = (25/75) * 100 = 33⅓%.

Question 4

The mean proportional between 2.8 and 0.7 is:

Accepted Answer

The mean proportional between two numbers 'a' and 'b' is given by √(a*b). Calculate √(2.8 * 0.7) = √(1.96) = 1.4.

Question 5

A dealer marks 20% above the cost price and allows a 20% discount. What is the profit or loss per cent?

Accepted Answer

Assume cost price is 100. Marked price is 120 (20% above CP). Discount of 20% on MP is 20% of 120 = 24. Selling price is 120 - 24 = 96. Loss is 100 - 96 = 4, so 4% loss.

Question 6

The greatest number, which divides 406 and 1388 to leave 1 and 3 respectively as remainders, is:

Accepted Answer

Subtract the remainders from the numbers: 406-1=405 and 1388-3=1385. Find the HCF of 405 and 1385. The HCF is 5, which is the greatest number.

Question 7

A does half as much work as B in three-fourth of the time. If together they take 66 days to complete the work, how many days shall B alone take to do it?

Accepted Answer

Let B's work rate be W and time be T. A's work is W/2 in 3T/4. So A's rate is (W/2)/(3T/4) = 2W/3T. Thus, A's rate is 2/3 of B's rate. If B takes x days, A takes 3x/2 days. Together, 1/x + 2/3x = 1/66. Solving gives x = 110 days.

Question 8

If at same rate of interest, in 2 years, the simple interest is ₹42 and compound interest is ₹51, then what is the principal (in ₹)?

Accepted Answer

The difference between CI and SI for 2 years is ₹51 - ₹42 = ₹9. This difference is the interest on the first year's simple interest. First year SI = ₹42/2 = ₹21. Rate = (9/21)*100 = 42.85%. Principal = (21*100)/42.85 = ₹49.

Question 9

Nitin has ₹1617 with him. He divided it amongst his sons Pravin and Rishi and asked them to invest it at 10% rate of interest compounded annually. It was seen that Pravin and Rishi got same amount after 17 and 18 years respectively. How much (in ₹) did Nitin give to Pravin?

Accepted Answer

Let Pravin's share be P and Rishi's be R. Total amount is 1617. P(1+0.10)^17 = R(1+0.10)^18. This simplifies to P = R(1.10). Since P+R = 1617, substitute R = P/1.10. P + P/1.10 = 1617. Solve for P.

Question 10

If the radius of a sphere is tripled, what is the ratio of the volume of the original sphere to that of the new sphere?

Accepted Answer

The volume of a sphere is (4/3)πr³. If the radius is tripled (3r), the new volume is (4/3)π(3r)³ = (4/3)π27r³. The ratio of original to new volume is r³ : 27r³, which simplifies to 1:27.

Question 11

The price of fuel decreases by 55%, 10% and 20% in three successive months, but increases by 65% in the fourth month. What is the percentage increase/decrease in the price of fuel in the fourth month as compared to its original price?

Accepted Answer

Assume original price is 100. After successive changes: 100 * (0.45) * (0.90) * (0.80) * (1.65) = 53.46. The final price is 53.46, so the decrease is 100 - 53.46 = 46.54%.

Question 12

A shopkeeper bought an article for ₹500. At what price (in ₹) should he sell the article to make 28% profit?

Accepted Answer

To find the selling price, calculate 28% of the cost price (500 * 0.28 = 140). Add this profit to the cost price (500 + 140 = 640).

Question 13

Ashish has ₹1617 with him. He divided it amongst his sons Arun and Mahesh and asked them to invest it at 10% rate of interest compounded annually. It was seen that Arun and Mahesh got same amount after 14 and 15 years respectively. How much (in ₹) did Ashish give to Arun?

Accepted Answer

Let Arun get A and Mahesh get M. A(1+0.10)^14 = M(1+0.10)^15. So, A = M(1.1). Since A+M = 1617, substitute M = A/1.1. A + A/1.1 = 1617 => 2.1A/1.1 = 1617 => A = 1617 * 1.1 / 2.1 = 847.

Question 14

Find the surface area (in cm2) of a sphere whose diameter is 21 cm.

Accepted Answer

The diameter is 21 cm, so the radius is 10.5 cm. Surface area of a sphere = 4πr². Substitute r=10.5 and π=22/7 to get 4 * (22/7) * (10.5)² = 1386 cm².

Question 15

Determine the largest four-digit number that is exactly divisible by 15, 25, 40, and 75.

Accepted Answer

Find the LCM of 15, 25, 40, 75, which is 600. Divide the largest four-digit number (9999) by 600. Subtract the remainder (399) from 9999 to get 9600.

Question 16

A shopkeeper has one spherical laddoo of radius 7 cm. With the same material, how many laddoos of radius 3.5 cm can be made?

Accepted Answer

The volume of a sphere is (4/3)πr³. The ratio of volumes of the large laddoo to a small laddoo is (7³)/(3.5³) = (7/3.5)³ = 2³ = 8. So, 8 small laddoos can be made.

Question 17

400 apples were bought at ₹1200 per hundred and were sold at a profit of ₹800. Find the selling price (in ₹) per dozen of apples.

Accepted Answer

Cost price of 400 apples = (400/100) * 1200 = ₹4800. Total selling price = 4800 + 800 = ₹5600. Selling price per dozen = (5600/400) * 12 = ₹168.

Question 18

The selling price of a fan is ₹7,392. If profit percentage is 32%, then what is the cost price (in ₹) of the fan?

Accepted Answer

Given selling price (SP) is ₹7392 and profit is 32%. Cost price (CP) = SP / (1 + Profit%). So, CP = 7392 / (1 + 0.32) = 7392 / 1.32 = ₹5600.

Question 19

Divide ₹4,800 among Ram, Shyam and Kartik in the ratio of 4 : 8 : 3. Find the amount received by Shyam.

Accepted Answer

The total ratio parts are 4 + 8 + 3 = 15. Shyam's share is (8/15) * ₹4,800 = ₹2,560.

Question 20

The average weight of Rupesh, Mandar and Ketan is 45 kg. If the average weight of Rupesh and Mandar be 43 kg and that of Mandar and Ketan be 45 kg, then the weight of Mandar (in kg ) is:

Accepted Answer

Total weight of R, M, K = 45 * 3 = 135 kg. Total weight of R, M = 43 * 2 = 86 kg. Total weight of M, K = 45 * 2 = 90 kg. Mandar's weight = (R+M) + (M+K) - (R+M+K) = 86 + 90 - 135 = 41 kg.