SSC GD 2025 PYQs

Question 1

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 2

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 3

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 4

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 5

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 6

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 7

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 8

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 9

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 10

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.

Question 11

The average weight of Mandar, Ketan and Tushar is 45 kg. If the average weight of Mandar and Ketan be 43 kg and that of Ketan and Tushar be 48 kg, then the weight of Ketan (in kg ) is:

Accepted Answer

Total weight of M+K+T = 3 * 45 = 135 kg. M+K = 2 * 43 = 86 kg. K+T = 2 * 48 = 96 kg. Add (M+K) and (K+T) to get M+2K+T = 182 kg. Subtract (M+K+T) from this to find K = 182 - 135 = 47 kg.

Question 12

To do a certain task, A and B work alternately on alternate days, with B starting on the first day. A alone can complete the task in 22 days. If the work is completed in11‌12days, then in how many days will B alone complete the‌34part of the original task?

Accepted Answer

A's 1-day work is 1/22. In 11.5 days, B works 6 days and A works 5.5 days. Calculate A's total work, then B's total work. Find B's 1-day work and then time for 3/4 task.

Question 13

Zakir travels from City A to City B. If Zakir drives his car at‌23of his normal speed, then he reaches City B 10 minutes late. Find the time (in minutes) that Zakir would have taken to travel from City A to City B if he drove at his normal speed.

Accepted Answer

If speed is 2/3 of normal, time taken will be 3/2 of normal. The extra time (3/2 - 1) = 1/2 of normal time is 10 minutes. So, normal time is 20 minutes.

Question 14

Ashok has ₹1612 with him. He divided it amongst his sons Raj and Varun and asked them to invest it at 8% rate of interest compounded annually. It was seen that Raj and Varun got same amount after 15 and 16 years respectively. How much (in ₹) did Ashok give to Raj?

Accepted Answer

Let Raj's share be R and Varun's be V. Amounts are equal: R(1+0.08)^15 = V(1+0.08)^16. This simplifies to R = V(1.08). Since R+V = 1612, substitute V = R/1.08. Solve for R.

Question 15

A can do a piece of work in 32 days and B in 48 days. They work together for 8 days and then A goes away. In how much time (in days) will B finish the 60% of the remaining work?

Accepted Answer

A's 1-day work = 1/32, B's 1-day work = 1/48. Together, they do (1/32 + 1/48) = 5/96 work per day. In 8 days, they complete 8 * (5/96) = 5/12 of the work. Remaining work = 1 - 5/12 = 7/12. B finishes 60% of remaining work: (7/12) * (60/100) = 7/20. Time taken by B = (7/20) / (1/48) = 16.8 days or 16 4/5 days.

Question 16

Two trains are moving in opposite directions at speeds of 60 km/h and 110 km/h. The length of one train is 360 m. The time taken by them to cross each other is 21 seconds. The length (in m) of the other train, correct to 2 decimal places, is:

Accepted Answer

Relative speed = 60 + 110 = 170 km/h = 170 * (5/18) m/s. Total distance = (360 + L2) m. Time = 21 s. So, (360 + L2) = 170 * (5/18) * 21. Solving for L2 gives 631.66 m.

Question 17

The simple interest on a principal amount (in ₹) is ₹318 for a period of 2 years at the rate of 6% per annum. The principal amount (in ₹) is:

Accepted Answer

Use the simple interest formula: SI = (P * R * T) / 100. Substitute SI=318, R=6, T=2. Solve for P: P = (318 * 100) / (6 * 2) = 31800 / 12 = 2650.

Question 18

What will be the difference between the sale price (in ₹) of a book with a marked price ₹ 1,500 under the following discount schemes?(i) Two successive discounts of 20%each(ii) Two successive discounts of 30% and 10%

Accepted Answer

Calculate the effective discount for scheme (i) as 1 - (0.8 * 0.8) = 0.36. For scheme (ii), 1 - (0.7 * 0.9) = 0.37. The difference in discounts is 0.01. Multiply by marked price ₹1500 to get ₹15.

Question 19

What will be the volume (V) of the cuboid, if its length is doubled, height is halved and breadth is tripled?

Accepted Answer

Original volume V = l * b * h. New dimensions are 2l, 3b, h/2. New volume = (2l) * (3b) * (h/2) = 6lbh/2 = 3lbh = 3V.

Question 20

Evaluate: (-9) - (-60) ÷ (-15) + (-4) × 6

Accepted Answer

Apply BODMAS/PEMDAS rule. First, division: (-60) ÷ (-15) = 4. Then, multiplication: (-4) × 6 = -24. Finally, subtraction and addition: -9 - 4 + (-24) = -13 - 24 = -37.