SSC GD 2025 PYQs

Question 1

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 2

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 3

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 4

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 5

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 6

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 7

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 8

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 9

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 10

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.

Question 11

A shopkeeper bought an article for ₹500. At what price (in ₹) should he sell the article to make 26% profit?

Accepted Answer

To find the selling price, calculate 26% of the cost price (₹500), which is ₹130. Add this profit to the cost price: ₹500 + ₹130 = ₹630.

Question 12

The average weight of Gopal, Akshay and Atul is 46 kg . If the average weight of Gopal and Akshay be 40 kg and that of Akshay and Atul be 45 kg, then the weight of Akshay (in kg) is:

Accepted Answer

Total weight of Gopal, Akshay, Atul is 46*3 = 138 kg. Gopal + Akshay = 40*2 = 80 kg. Akshay + Atul = 45*2 = 90 kg. Summing the last two gives Gopal + 2*Akshay + Atul = 170 kg. Subtracting the first total gives Akshay = 170 - 138 = 32 kg.

Question 13

The LCM of 672 and 7056 is _____.

Accepted Answer

To find the LCM, observe that 7056 is exactly 10.5 times 672. Since 7056 is a multiple of 672, the LCM is simply the larger number, 7056. Oh, wait, 7056 is not a multiple of 672. Let's recheck. 7056 / 672 = 10.5. So, 7056 is not a multiple of 672. We need to find the prime factorization of both numbers. 672 = 2^5 * 3 * 7. 7056 = 2^4 * 3^2 * 7^2. LCM = 2^5 * 3^2 * 7^2 = 32 * 9 * 49 = 14112.

Question 14

Mahesh has ₹1617 with him. He divided it amongst his sons Vijay and Ajay and asked them to invest it at 10% rate of interest compounded annually. It was seen that Vijay and Ajay got same amount after 13 and 14 years respectively. How much (in ₹) did Mahesh give to Vijay?

Accepted Answer

Let Vijay's share be V and Ajay's be A. V(1+0.10)^13 = A(1+0.10)^14. So, V = A(1.10). Total V+A = 1617. Substitute A = V/1.10. V + V/1.10 = 1617. 2.1V/1.1 = 1617. V = 1617 * 1.1 / 2.1 = 847.

Question 15

The simple interest on a principal amount (in ₹) is ₹824 for a period of 4 years at the rate of 8% per annum. The principal amount (in ₹) is:

Accepted Answer

Use the simple interest formula: Principal = (Simple Interest × 100) / (Rate × Time). Substitute the given values: (824 × 100) / (8 × 4) = 82400 / 32 = 2575.

Question 16

A sum of ₹x amounts to ₹54571 in2‌12years at 12% p.a., interest compounded after every 10 months. What is the value of x?

Accepted Answer

The interest is compounded every 10 months. The rate for 10 months is (12/12)*10 = 10%. The total time is 2.5 years = 30 months, so there are 3 compounding periods (30/10). Using A = P(1 + r/100)^n, 54571 = x(1 + 10/100)^3. Solving for x gives 41000.

Question 17

Find the third proportional of 36 and 42.

Accepted Answer

For three numbers a, b, c to be in continued proportion, a:b = b:c. So, c = b²/a. Here, a=36, b=42. Thus, c = 42*42 / 36 = 1764 / 36 = 49.

Question 18

In an examination, there were three papers of Mathematics, two papers of English and one paper of Science. All papers were of 100 marks. S got 60% in Mathematics, 70% in English and 50% in Science. What was his percentage of marks in all papers?

Accepted Answer

Total marks obtained = (3*60) + (2*70) + (1*50) = 180 + 140 + 50 = 370. Total maximum marks = (3+2+1)*100 = 600. Percentage = (370/600)*100 = 61.67%.

Question 19

Evaluate: 16 + 10 ÷ 5 - 2 × 3

Accepted Answer

Apply BODMAS/PEMDAS rule: First division (10 ÷ 5 = 2), then multiplication (2 × 3 = 6). Finally, addition and subtraction from left to right: 16 + 2 - 6 = 18 - 6 = 12.

Question 20

The LCM of 56, 30, 108 and 120 is:

Accepted Answer

Find the prime factorization of each number: 56=2³x7, 30=2x3x5, 108=2²x3³, 120=2³x3x5. The LCM is the product of the highest powers of all prime factors: 2³x3³x5x7 = 8x27x5x7 = 7560.