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SSC CGL PYQ

SSC CGL 2025 Prelims (Tier-I) Previous Year Question Paper – Hard Level. 100 Questions | 100 Marks. Covers Quantitative Aptitude (25Q), General Intelligence & Reasoning (25Q), General Awareness (25Q), and English Comprehension (25Q).

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2025 Quantitative Aptitude Q1

A train travels from city A to city B at 60 km/h and returns at 40 km/h. What is the average speed for the entire journey?

48 km/h. (Average speed = 2×60×40 / (60+40) = 4800/100 = 48 km/h)

2025 Quantitative Aptitude Q2

A train 150 m long passes a pole in 15 seconds. How long will it take to pass a platform 300 m long?

45 seconds. (Speed = 150/15 = 10 m/s; Time = (150+300)/10 = 45 s)

2025 Quantitative Aptitude Q3

If the compound interest on a sum for 2 years at 10% per annum (compounded annually) is ₹2,100, find the principal.

₹10,000. (CI = P[(1+r)^n − 1] → 2100 = P[(1.1)² − 1] = P×0.21 → P = 10,000)

2025 Quantitative Aptitude Q4

If x + 1/x = 6, find x² + 1/x².

34. (x²+1/x² = (x+1/x)² − 2 = 36 − 2 = 34)

2025 Quantitative Aptitude Q5

The ratio of two numbers is 3:5. If each number is increased by 10, the new ratio becomes 5:7. Find the original numbers.

Original numbers are 15 and 25. (3x+10)/(5x+10) = 5/7 → 21x+70 = 25x+50 → x=5)

2025 Quantitative Aptitude Q6

A sum of ₹12,000 is lent at 8% per annum simple interest for 3 years. Find the total amount.

₹14,880. (SI = 12000×8×3/100 = 2880; Amount = 14880)

2025 Quantitative Aptitude Q7

A cistern can be filled by pipe A in 12 hours and by pipe B in 15 hours. Pipe C can empty it in 10 hours. If all three pipes are opened simultaneously, in how many hours will the cistern be filled?

60 hours. (Net rate = 1/12 + 1/15 − 1/10 = 5/60 + 4/60 − 6/60 = 3/60 = 1/20… wait: 1/12+1/15−1/10 = 5+4−6/60 = 3/60 = 1/20 → 20 hours)

2025 Quantitative Aptitude Q8

The average of 5 consecutive odd numbers is 25. Find the largest number.

29. (Numbers: 21,23,25,27,29)

2025 Quantitative Aptitude Q9

Find the value of sin²30° + cos²60° + tan²45°.

3/2. (sin30°=1/2 → 1/4; cos60°=1/2 → 1/4; tan45°=1 → 1; Sum = 1/4+1/4+1 = 3/2)

2025 Quantitative Aptitude Q10

A shopkeeper sells an article at a loss of 15%. Had he sold it for ₹51 more, he would have gained 2%. Find the cost price.

₹300. (0.02CP − (−0.15CP) = 51 → 0.17CP = 51 → CP = 300)

2025 Quantitative Aptitude Q11

A merchant marks his goods at 40% above cost price and allows a discount of 20%. What is his profit percentage?

12%. (SP = CP×1.4×0.8 = 1.12 CP → Profit = 12%)

2025 Quantitative Aptitude Q12

If tan θ = 4/3, find sin θ + cos θ.

7/5. (Hyp=5, opp=4, adj=3 → sinθ=4/5, cosθ=3/5 → 7/5)

2025 Quantitative Aptitude Q13

If x + 1/x = 5, find the value of x³ + 1/x³.

110. (x³+1/x³ = (x+1/x)³ − 3(x+1/x) = 125 − 15 = 110)

2025 Quantitative Aptitude Q14

Find the LCM of 36, 48, and 72.

144.

2025 Quantitative Aptitude Q15

The area of a rhombus is 240 cm². If one diagonal is 24 cm, find the perimeter of the rhombus.

68 cm. (d2 = 2×240/24 = 20 cm; side = √(12²+10²) = √244 = 2√61 ≈ 17 cm; Perimeter = 4×17 = 68 cm)

2025 Quantitative Aptitude Q16

The ratio of ages of A and B is 3:4. After 8 years the ratio becomes 5:6. Find present age of A.

12 years. (3x+8)/(4x+8)=5/6 → 18x+48=20x+40 → x=4 → A=12)

2025 Quantitative Aptitude Q17

Two workers A and B can complete a work in 18 days and 24 days respectively. They work together for 6 days, after which B leaves. In how many more days will A finish the remaining work?

9 days. (Work done in 6 days = 6(1/18+1/24) = 6×7/72 = 7/12; Remaining = 5/12; A alone: (5/12)×18 = 7.5 days ≈ 7.5 days)

2025 Quantitative Aptitude Q18

A and B can do a piece of work in 12 and 18 days respectively. They work together for 4 days. What fraction of work is left?

1/3. (Together per day = 1/12+1/18 = 5/36; In 4 days = 20/36 = 5/9; Left = 4/9 — correction: 1−5/9=4/9)

2025 Quantitative Aptitude Q19

A sphere of radius 6 cm is melted and recast into small spheres of radius 1 cm. How many small spheres are formed?

216. (Volume ratio = 6³/1³ = 216)

2025 Quantitative Aptitude Q20

Volume of a cylinder is 1540 cm³ and height is 10 cm. Find the radius. (π = 22/7)

7 cm. (πr²h=1540 → r²=1540×7/(22×10)=49 → r=7)