Q1. How many factors (divisors) does 252 have?
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Answer: C. 18
252 = 2^2 × 3^2 × 7; number of factors = 3 × 3 × 2 = 18.
SSC CGL Quantitative Aptitude Intermediate Practice Set 12
25 intermediate questions. Tap an option to check it, or open “Show answer” for the explanation.
Topics: Number System, Simple Interest, Problems on Ages, Average, Simplification, Compound Interest, Percentage, Time and Work, Ratio and Proportion, Mensuration, Speed, Time and Distance, Boats and Streams, LCM and HCF, Statistics
Score: 0 / 0
Q2. At 10% per annum simple interest, a sum earns ₹9250 as interest in 5 years. Find the sum (in ₹).
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Answer: D. 18500
P = 9250 × 100 / (10 × 5) = ₹18500.
Q3. The present ages of A and B are in the ratio 2:5. After 4 years, their ages will be in the ratio 6:13. What is A's present age?
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Answer: D. 14 years
13(2x + 4) = 6(5x + 4) ⇒ x = 7. A's age = 2 × 7 = 14 years.
Q4. A sum of ₹16000 amounts to ₹20800 in 3 years at simple interest. What is the rate of interest per annum?
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Answer: D. 10%
SI = 20800 − 16000 = ₹4800. R = 4800 × 100 / (16000 × 3) = 10%.
Q5. A batsman has an average of 59 runs after 12 innings. How many runs must he score in the next innings to raise his average by 4?
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Answer: B. 111
New total = 13 × 63 = 819; old total = 12 × 59 = 708. Runs = 111.
Q6. Find the value of 53² − 25².
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Answer: B. 2184
(53 + 25)(53 − 25) = 78 × 28 = 2184.
Q7. Find the compound interest on ₹14000 for 2 years at 20% per annum, compounded annually (in ₹).
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Answer: B. 6160
Amount = 14000 × (120/100)² = ₹20160. CI = 20160 − 14000 = ₹6160.
Q8. The present ages of A and B are in the ratio 2:5. After 9 years, their ages will be in the ratio 5:8. What is A's present age?
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Answer: C. 6 years
8(2x + 9) = 5(5x + 9) ⇒ x = 3. A's age = 2 × 3 = 6 years.
Q9. A's salary is 100% more than B's salary. By what percent is B's salary less than A's?
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Answer: D. 50%
Let B = 100, so A = 200. B is less by 100/200 × 100 = 50%.
Q10. The average age of 10 students in a class is 40 years. When the teacher's age is included, the average increases by 1 year. What is the teacher's age?
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Answer: C. 51 years
Teacher's age = (10 + 1) × 41 − 10 × 40 = 451 − 400 = 51 years.
Q11. 24 men can complete a work in 11 days. In how many days can 6 men complete the same work?
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Answer: C. 44 days
24 × 11 = 6 × D ⇒ D = 264/6 = 44 days.
Q12. If A:B = 7:6 and B:C = 7:2, find A:B:C.
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Answer: B. 49:42:12
Multiply the first ratio by 7 and the second by 6: A:B:C = 49:42:12 = 49:42:12.
Q13. A, B and C can do a work alone in 12, 15 and 20 days respectively. How many days will they take working together?
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Answer: C. 5 days
1/12 + 1/15 + 1/20 = 1/5. So they need 5 days.
Q14. Find the volume of a cone with base radius 21 cm and height 29 cm. (Take π = 22/7)
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Answer: D. 13398 cm³
V = 1/3 × 22/7 × 21² × 29 = 13398 cm³.
Q15. Two cities are 390 km apart. Two cars start at the same time from the two cities towards each other at 25 km/h and 53 km/h. After how many hours will they meet?
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Answer: D. 5 hours
Time = 390/(25 + 53) = 5 hours.
Q16. A train 100 m long runs at 90 km/h. How long will it take to cross a platform 200 m long?
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Answer: C. 12 seconds
Speed = 90 × 5/18 = 25 m/s. Distance = 100 + 200 = 300 m. Time = 300/25 = 12 s.
Q17. A boat goes downstream at 21 km/h and upstream at 19 km/h. What is the speed of the stream?
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Answer: B. 1 km/h
(21 − 19)/2 = 1 km/h.
Q18. A boat goes downstream at 16 km/h and upstream at 8 km/h. What is the speed of the stream?
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Answer: A. 4 km/h
(16 − 8)/2 = 4 km/h.
Q19. What is the least number which, when divided by 24, 10 and 6, leaves a remainder of 3 in each case?
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Answer: C. 123
LCM(24, 10, 6) = 120. Required number = 120 + 3 = 123.
Q20. How many factors (divisors) does 120 have?
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Answer: A. 16
120 = 2^3 × 3 × 5; number of factors = 4 × 2 × 2 = 16.
Q21. Find the median of: 4, 3, 35, 9, 40, 33, 31, 24, 8.
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Answer: B. 24
In order: 3, 4, 8, 9, 24, 31, 33, 35, 40. Median = 24.
Q22. What is the unit digit of 8^121?
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Answer: C. 8
Unit digits of 8^1, ^2, ^3, ^4 are 8, 4, 2, 6. 121 leaves remainder 1 on division by 4, so take the 1st value in the cycle: 8.
Q23. What is the smallest 4-digit number exactly divisible by 37?
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Answer: B. 1036
1000 ÷ 37 leaves remainder 1. 1000 + 36 = 1036.
Q24. A batsman has an average of 60 runs after 19 innings. How many runs must he score in the next innings to raise his average by 2?
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Answer: A. 100
New total = 20 × 62 = 1240; old total = 19 × 60 = 1140. Runs = 100.
Q25. Find the value of 72² − 42².
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Answer: B. 3420
(72 + 42)(72 − 42) = 114 × 30 = 3420.