Using the formula and practice file

There are 200 original practice questions: twenty sets of ten, covering arithmetic, elementary mathematics, statistics and mental ability. They are practice items, not claimed PYQs. Most have short numerical answers so you must perform the working before opening the solution. Units and assumptions are stated in each item.

Do one set in 12–18 minutes, then review every error. For a first pass use paper; for a second pass shorten the working only after the method is reliable.

High-yield formula sheet

Topic	Formula / method
Percentages	Change = original × rate/100; successive changes multiply factors.
Ratio and proportion	Share = total × part/sum; inverse proportion keeps the product constant.
Profit, loss and discounts	Profit% = profit/CP × 100; successive discounts multiply net-price factors.
Simple interest	SI = PRT/100; A = P + SI.
Compound interest	Annual A = P(1+r/100)^n; for two years CI−SI = P(r/100)^2.
Time and work	Work rates add; combined time for two workers = ab/(a+b).
Pipes and cisterns	Net filling rate = inlet rates − outlet rates.
Speed, trains and relative motion	km/h to m/s: multiply 5/18; crossing time = total length/relative speed.
Boats and streams	Downstream = boat + stream; upstream = boat − stream.
Averages	Total = count × mean; replacement changes total by new minus old.
Mixtures	Weighted mean uses quantities; replacement retains the fraction (1−removed/total).
Number system, HCF and LCM	For two positive integers, HCF × LCM = product.
Algebra and equations	Preserve equality by performing the same operation on both sides.
Arithmetic progressions	a_n = a+(n−1)d; S_n = n[2a+(n−1)d]/2.
Mensuration	Rectangle area = lb; circle area = πr²; cylinder volume = πr²h.
Probability	P(event) = favourable/equally likely outcomes; independent probabilities multiply.
Permutations and combinations	nPr = n!/(n−r)!; nCr = n!/[r!(n−r)!].
Statistics	Population variance = Σ(x−mean)²/N; CV = SD/mean ×100.
Series and coding	Identify the rule and verify it against every displayed step.
Directions and logical inference	Use coordinates for directions; do not infer existence from a universal statement alone.

Set 1: Percentages

Method: Change = original × rate/100; successive changes multiply factors.

Q1. A salary of ₹12000 rises by 15%. Find the new salary.

Answer and working

13,800

Multiply the original by 1.15.

Q2. A price rises by 20% and then falls by 10%. Its initial value is ₹600. Find the final price.

Answer and working

648

Multiply by 1.20 × 0.90 = 1.08: a net rise of 8%.

Q3. A salary of ₹14000 rises by 15%. Find the new salary.

Answer and working

16,100

Multiply the original by 1.15.

Q4. A price rises by 20% and then falls by 10%. Its initial value is ₹700. Find the final price.

Answer and working

756

Multiply by 1.20 × 0.90 = 1.08: a net rise of 8%.

Q5. A salary of ₹16000 rises by 15%. Find the new salary.

Answer and working

18,400

Multiply the original by 1.15.

Q6. A price rises by 20% and then falls by 10%. Its initial value is ₹800. Find the final price.

Answer and working

864

Multiply by 1.20 × 0.90 = 1.08: a net rise of 8%.

Q7. A salary of ₹18000 rises by 15%. Find the new salary.

Answer and working

20,700

Multiply the original by 1.15.

Q8. A price rises by 20% and then falls by 10%. Its initial value is ₹900. Find the final price.

Answer and working

972

Multiply by 1.20 × 0.90 = 1.08: a net rise of 8%.

Q9. A salary of ₹20000 rises by 15%. Find the new salary.

Answer and working

23,000

Multiply the original by 1.15.

Q10. A price rises by 20% and then falls by 10%. Its initial value is ₹1000. Find the final price.

Answer and working

1,080

Multiply by 1.20 × 0.90 = 1.08: a net rise of 8%.

Set 2: Ratio and proportion

Method: Share = total × part/sum; inverse proportion keeps the product constant.

Q11. Divide ₹8800 between A and B in the ratio 3:5. Find B’s share.

Answer and working

5,500

B takes 5 of the 8 equal parts.

Q12. 5 identical machines finish work in 60 hours. How long will 10 machines take at the same efficiency?

Answer and working

30 hours

Machines double, so time halves; machines × time is constant.

Q13. Divide ₹9600 between A and B in the ratio 3:5. Find B’s share.

Answer and working

6,000

B takes 5 of the 8 equal parts.

Q14. 6 identical machines finish work in 72 hours. How long will 12 machines take at the same efficiency?

Answer and working

36 hours

Machines double, so time halves; machines × time is constant.

Q15. Divide ₹10400 between A and B in the ratio 3:5. Find B’s share.

Answer and working

6,500

B takes 5 of the 8 equal parts.

Q16. 7 identical machines finish work in 84 hours. How long will 14 machines take at the same efficiency?

Answer and working

42 hours

Machines double, so time halves; machines × time is constant.

Q17. Divide ₹11200 between A and B in the ratio 3:5. Find B’s share.

Answer and working

7,000

B takes 5 of the 8 equal parts.

Q18. 8 identical machines finish work in 96 hours. How long will 16 machines take at the same efficiency?

Answer and working

48 hours

Machines double, so time halves; machines × time is constant.

Q19. Divide ₹12000 between A and B in the ratio 3:5. Find B’s share.

Answer and working

7,500

B takes 5 of the 8 equal parts.

Q20. 9 identical machines finish work in 108 hours. How long will 18 machines take at the same efficiency?

Answer and working

54 hours

Machines double, so time halves; machines × time is constant.

Set 3: Profit, loss and discounts

Method: Profit% = profit/CP × 100; successive discounts multiply net-price factors.

Q21. An item costs ₹500 and sells at 25% profit. Find selling price.

Answer and working

625

SP = CP × 1.25.

Q22. Marked price is ₹2500. Discounts are 10% and then 20%. Find final price.

Answer and working

1,800

Net factor = 0.90 × 0.80 = 0.72; equivalent discount is 28%.

Q23. An item costs ₹600 and sells at 25% profit. Find selling price.

Answer and working

750

SP = CP × 1.25.

Q24. Marked price is ₹3000. Discounts are 10% and then 20%. Find final price.

Answer and working

2,160

Net factor = 0.90 × 0.80 = 0.72; equivalent discount is 28%.

Q25. An item costs ₹700 and sells at 25% profit. Find selling price.

Answer and working

875

SP = CP × 1.25.

Q26. Marked price is ₹3500. Discounts are 10% and then 20%. Find final price.

Answer and working

2,520

Net factor = 0.90 × 0.80 = 0.72; equivalent discount is 28%.

Q27. An item costs ₹800 and sells at 25% profit. Find selling price.

Answer and working

1,000

SP = CP × 1.25.

Q28. Marked price is ₹4000. Discounts are 10% and then 20%. Find final price.

Answer and working

2,880

Net factor = 0.90 × 0.80 = 0.72; equivalent discount is 28%.

Q29. An item costs ₹900 and sells at 25% profit. Find selling price.

Answer and working

1,125

SP = CP × 1.25.

Q30. Marked price is ₹4500. Discounts are 10% and then 20%. Find final price.

Answer and working

3,240

Net factor = 0.90 × 0.80 = 0.72; equivalent discount is 28%.

Set 4: Simple interest

Method: SI = PRT/100; A = P + SI.

Q31. Find simple interest on ₹11000 at 8% a year for three years.

Answer and working

2,640

SI = P × 8 × 3 / 100 = 0.24P.

Q32. ₹6000 earns ₹1800 simple interest in five years. Find the annual rate.

Answer and working

6%

R = 100I/(PT) = 100 × 0.30 / 5 = 6%.

Q33. Find simple interest on ₹12000 at 8% a year for three years.

Answer and working

2,880

SI = P × 8 × 3 / 100 = 0.24P.

Q34. ₹7000 earns ₹2100 simple interest in five years. Find the annual rate.

Answer and working

6%

R = 100I/(PT) = 100 × 0.30 / 5 = 6%.

Q35. Find simple interest on ₹13000 at 8% a year for three years.

Answer and working

3,120

SI = P × 8 × 3 / 100 = 0.24P.

Q36. ₹8000 earns ₹2400 simple interest in five years. Find the annual rate.

Answer and working

6%

R = 100I/(PT) = 100 × 0.30 / 5 = 6%.

Q37. Find simple interest on ₹14000 at 8% a year for three years.

Answer and working

3,360

SI = P × 8 × 3 / 100 = 0.24P.

Q38. ₹9000 earns ₹2700 simple interest in five years. Find the annual rate.

Answer and working

6%

R = 100I/(PT) = 100 × 0.30 / 5 = 6%.

Q39. Find simple interest on ₹15000 at 8% a year for three years.

Answer and working

3,600

SI = P × 8 × 3 / 100 = 0.24P.

Q40. ₹10000 earns ₹3000 simple interest in five years. Find the annual rate.

Answer and working

6%

R = 100I/(PT) = 100 × 0.30 / 5 = 6%.

Set 5: Compound interest

Method: Annual A = P(1+r/100)^n; for two years CI−SI = P(r/100)^2.

Q41. Find the amount on ₹12000 at 10% compounded annually for two years.

Answer and working

14,520

A = P × 1.1² = 1.21P.

Q42. Find CI minus SI for two years at 5% on ₹44000.

Answer and working

110

Difference = P × 0.05² = 0.0025P.

Q43. Find the amount on ₹14000 at 10% compounded annually for two years.

Answer and working

16,940

A = P × 1.1² = 1.21P.

Q44. Find CI minus SI for two years at 5% on ₹48000.

Answer and working

120

Difference = P × 0.05² = 0.0025P.

Q45. Find the amount on ₹16000 at 10% compounded annually for two years.

Answer and working

19,360

A = P × 1.1² = 1.21P.

Q46. Find CI minus SI for two years at 5% on ₹52000.

Answer and working

130

Difference = P × 0.05² = 0.0025P.

Q47. Find the amount on ₹18000 at 10% compounded annually for two years.

Answer and working

21,780

A = P × 1.1² = 1.21P.

Q48. Find CI minus SI for two years at 5% on ₹56000.

Answer and working

140

Difference = P × 0.05² = 0.0025P.

Q49. Find the amount on ₹20000 at 10% compounded annually for two years.

Answer and working

24,200

A = P × 1.1² = 1.21P.

Q50. Find CI minus SI for two years at 5% on ₹60000.

Answer and working

150

Difference = P × 0.05² = 0.0025P.

Set 6: Time and work

Method: Work rates add; combined time for two workers = ab/(a+b).

Q51. A completes work in 6 days and B in 3 days. Find their joint time.

Answer and working

2 days

Joint rate = 1/(6k) + 1/(3k) = 1/(2k).

Q52. A is twice as efficient as B. Together they finish a job in 4 days. Find A’s solo time.

Answer and working

6 days

A supplies 2/3 of the joint rate, so solo time is 3/2 of joint time.

Q53. A completes work in 12 days and B in 6 days. Find their joint time.

Answer and working

4 days

Joint rate = 1/(6k) + 1/(3k) = 1/(2k).

Q54. A is twice as efficient as B. Together they finish a job in 8 days. Find A’s solo time.

Answer and working

12 days

A supplies 2/3 of the joint rate, so solo time is 3/2 of joint time.

Q55. A completes work in 18 days and B in 9 days. Find their joint time.

Answer and working

6 days

Joint rate = 1/(6k) + 1/(3k) = 1/(2k).

Q56. A is twice as efficient as B. Together they finish a job in 12 days. Find A’s solo time.

Answer and working

18 days

A supplies 2/3 of the joint rate, so solo time is 3/2 of joint time.

Q57. A completes work in 24 days and B in 12 days. Find their joint time.

Answer and working

8 days

Joint rate = 1/(6k) + 1/(3k) = 1/(2k).

Q58. A is twice as efficient as B. Together they finish a job in 16 days. Find A’s solo time.

Answer and working

24 days

A supplies 2/3 of the joint rate, so solo time is 3/2 of joint time.

Q59. A completes work in 30 days and B in 15 days. Find their joint time.

Answer and working

10 days

Joint rate = 1/(6k) + 1/(3k) = 1/(2k).

Q60. A is twice as efficient as B. Together they finish a job in 20 days. Find A’s solo time.

Answer and working

30 days

A supplies 2/3 of the joint rate, so solo time is 3/2 of joint time.

Set 7: Pipes and cisterns

Method: Net filling rate = inlet rates − outlet rates.

Q61. A pipe fills a tank in 4 hours; a drain empties it in 12 hours. Both run. Find filling time.

Answer and working

6 hours

Net rate = 1/(4k) − 1/(12k) = 1/(6k).

Q62. Two inlets fill a tank in 6 and 12 hours. An outlet empties it in 12 hours. Find net filling time.

Answer and working

6 hours

The second inlet and outlet cancel; the first inlet remains.

Q63. A pipe fills a tank in 8 hours; a drain empties it in 24 hours. Both run. Find filling time.

Answer and working

12 hours

Net rate = 1/(4k) − 1/(12k) = 1/(6k).

Q64. Two inlets fill a tank in 12 and 24 hours. An outlet empties it in 24 hours. Find net filling time.

Answer and working

12 hours

The second inlet and outlet cancel; the first inlet remains.

Q65. A pipe fills a tank in 12 hours; a drain empties it in 36 hours. Both run. Find filling time.

Answer and working

18 hours

Net rate = 1/(4k) − 1/(12k) = 1/(6k).

Q66. Two inlets fill a tank in 18 and 36 hours. An outlet empties it in 36 hours. Find net filling time.

Answer and working

18 hours

The second inlet and outlet cancel; the first inlet remains.

Q67. A pipe fills a tank in 16 hours; a drain empties it in 48 hours. Both run. Find filling time.

Answer and working

24 hours

Net rate = 1/(4k) − 1/(12k) = 1/(6k).

Q68. Two inlets fill a tank in 24 and 48 hours. An outlet empties it in 48 hours. Find net filling time.

Answer and working

24 hours

The second inlet and outlet cancel; the first inlet remains.

Q69. A pipe fills a tank in 20 hours; a drain empties it in 60 hours. Both run. Find filling time.

Answer and working

30 hours

Net rate = 1/(4k) − 1/(12k) = 1/(6k).

Q70. Two inlets fill a tank in 30 and 60 hours. An outlet empties it in 60 hours. Find net filling time.

Answer and working

30 hours

The second inlet and outlet cancel; the first inlet remains.

Set 8: Speed, trains and relative motion

Method: km/h to m/s: multiply 5/18; crossing time = total length/relative speed.

Q71. A train 120 metres long passes a pole at 72 km/h. Find the time.

Answer and working

6 seconds

72 km/h = 20 m/s; divide train length by 20.

Q72. Two vehicles 30 km apart approach at 40 and 50 km/h. Find meeting time in minutes.

Answer and working

20 minutes

Relative speed = 90 km/h; time = 30k/90 hours = 20k minutes.

Q73. A train 140 metres long passes a pole at 72 km/h. Find the time.

Answer and working

7 seconds

72 km/h = 20 m/s; divide train length by 20.

Q74. Two vehicles 60 km apart approach at 40 and 50 km/h. Find meeting time in minutes.

Answer and working

40 minutes

Relative speed = 90 km/h; time = 30k/90 hours = 20k minutes.

Q75. A train 160 metres long passes a pole at 72 km/h. Find the time.

Answer and working

8 seconds

72 km/h = 20 m/s; divide train length by 20.

Q76. Two vehicles 90 km apart approach at 40 and 50 km/h. Find meeting time in minutes.

Answer and working

60 minutes

Relative speed = 90 km/h; time = 30k/90 hours = 20k minutes.

Q77. A train 180 metres long passes a pole at 72 km/h. Find the time.

Answer and working

9 seconds

72 km/h = 20 m/s; divide train length by 20.

Q78. Two vehicles 120 km apart approach at 40 and 50 km/h. Find meeting time in minutes.

Answer and working

80 minutes

Relative speed = 90 km/h; time = 30k/90 hours = 20k minutes.

Q79. A train 200 metres long passes a pole at 72 km/h. Find the time.

Answer and working

10 seconds

72 km/h = 20 m/s; divide train length by 20.

Q80. Two vehicles 150 km apart approach at 40 and 50 km/h. Find meeting time in minutes.

Answer and working

100 minutes

Relative speed = 90 km/h; time = 30k/90 hours = 20k minutes.

Set 9: Boats and streams

Method: Downstream = boat + stream; upstream = boat − stream.

Q81. A boat’s still-water speed is 11 km/h and current is 1 km/h. Find upstream speed.

Answer and working

10 km/h

Subtract current: (10+k) − k = 10.

Q82. Downstream speed is 18 and upstream speed is 14 km/h. Find stream speed.

Answer and working

2 km/h

Stream = half the difference = 4/2 = 2.

Q83. A boat’s still-water speed is 12 km/h and current is 2 km/h. Find upstream speed.

Answer and working

10 km/h

Subtract current: (10+k) − k = 10.

Q84. Downstream speed is 20 and upstream speed is 16 km/h. Find stream speed.

Answer and working

2 km/h

Stream = half the difference = 4/2 = 2.

Q85. A boat’s still-water speed is 13 km/h and current is 3 km/h. Find upstream speed.

Answer and working

10 km/h

Subtract current: (10+k) − k = 10.

Q86. Downstream speed is 22 and upstream speed is 18 km/h. Find stream speed.

Answer and working

2 km/h

Stream = half the difference = 4/2 = 2.

Q87. A boat’s still-water speed is 14 km/h and current is 4 km/h. Find upstream speed.

Answer and working

10 km/h

Subtract current: (10+k) − k = 10.

Q88. Downstream speed is 24 and upstream speed is 20 km/h. Find stream speed.

Answer and working

2 km/h

Stream = half the difference = 4/2 = 2.

Q89. A boat’s still-water speed is 15 km/h and current is 5 km/h. Find upstream speed.

Answer and working

10 km/h

Subtract current: (10+k) − k = 10.

Q90. Downstream speed is 26 and upstream speed is 22 km/h. Find stream speed.

Answer and working

2 km/h

Stream = half the difference = 4/2 = 2.

Set 10: Averages

Method: Total = count × mean; replacement changes total by new minus old.

Q91. The mean of 6 numbers is 20. One number 10 is replaced by 16. Find the new mean.

Answer and working

21

The total rises by the count (5+k), so mean rises by one.

Q92. Four values are 1, 3, 5, 7. Find their mean.

Answer and working

4

The arithmetic progression’s mean is the average of first and last.

Q93. The mean of 7 numbers is 20. One number 10 is replaced by 17. Find the new mean.

Answer and working

21

The total rises by the count (5+k), so mean rises by one.

Q94. Four values are 2, 4, 6, 8. Find their mean.

Answer and working

5

The arithmetic progression’s mean is the average of first and last.

Q95. The mean of 8 numbers is 20. One number 10 is replaced by 18. Find the new mean.

Answer and working

21

The total rises by the count (5+k), so mean rises by one.

Q96. Four values are 3, 5, 7, 9. Find their mean.

Answer and working

6

The arithmetic progression’s mean is the average of first and last.

Q97. The mean of 9 numbers is 20. One number 10 is replaced by 19. Find the new mean.

Answer and working

21

The total rises by the count (5+k), so mean rises by one.

Q98. Four values are 4, 6, 8, 10. Find their mean.

Answer and working

7

The arithmetic progression’s mean is the average of first and last.

Q99. The mean of 10 numbers is 20. One number 10 is replaced by 20. Find the new mean.

Answer and working

21

The total rises by the count (5+k), so mean rises by one.

Q100. Four values are 5, 7, 9, 11. Find their mean.

Answer and working

8

The arithmetic progression’s mean is the average of first and last.

Set 11: Mixtures

Method: Weighted mean uses quantities; replacement retains the fraction (1−removed/total).

Q101. Mix 1 litres at ₹40/litre with 3 litres at ₹60/litre. Find mean cost.

Answer and working

₹55/litre

Weighted cost = (40k+180k)/(4k) = 55.

Q102. A 10-litre vessel of milk has one-fifth removed and replaced by water twice. Find milk remaining.

Answer and working

32/5 litres

Milk retained = 10k × (4/5)² = 6.4k litres.

Q103. Mix 2 litres at ₹40/litre with 6 litres at ₹60/litre. Find mean cost.

Answer and working

₹55/litre

Weighted cost = (40k+180k)/(4k) = 55.

Q104. A 20-litre vessel of milk has one-fifth removed and replaced by water twice. Find milk remaining.

Answer and working

64/5 litres

Milk retained = 10k × (4/5)² = 6.4k litres.

Q105. Mix 3 litres at ₹40/litre with 9 litres at ₹60/litre. Find mean cost.

Answer and working

₹55/litre

Weighted cost = (40k+180k)/(4k) = 55.

Q106. A 30-litre vessel of milk has one-fifth removed and replaced by water twice. Find milk remaining.

Answer and working

96/5 litres

Milk retained = 10k × (4/5)² = 6.4k litres.

Q107. Mix 4 litres at ₹40/litre with 12 litres at ₹60/litre. Find mean cost.

Answer and working

₹55/litre

Weighted cost = (40k+180k)/(4k) = 55.

Q108. A 40-litre vessel of milk has one-fifth removed and replaced by water twice. Find milk remaining.

Answer and working

128/5 litres

Milk retained = 10k × (4/5)² = 6.4k litres.

Q109. Mix 5 litres at ₹40/litre with 15 litres at ₹60/litre. Find mean cost.

Answer and working

₹55/litre

Weighted cost = (40k+180k)/(4k) = 55.

Q110. A 50-litre vessel of milk has one-fifth removed and replaced by water twice. Find milk remaining.

Answer and working

32 litres

Milk retained = 10k × (4/5)² = 6.4k litres.

Set 12: Number system, HCF and LCM

Method: For two positive integers, HCF × LCM = product.

Q111. Find the HCF of 12 and 18.

Answer and working

6

Factor out k; HCF of 12 and 18 is 6.

Q112. Find the LCM of 4 and 6.

Answer and working

12

LCM of 4k and 6k is 12k; use product/HCF.

Q113. Find the HCF of 24 and 36.

Answer and working

12

Factor out k; HCF of 12 and 18 is 6.

Q114. Find the LCM of 8 and 12.

Answer and working

24

LCM of 4k and 6k is 12k; use product/HCF.

Q115. Find the HCF of 36 and 54.

Answer and working

18

Factor out k; HCF of 12 and 18 is 6.

Q116. Find the LCM of 12 and 18.

Answer and working

36

LCM of 4k and 6k is 12k; use product/HCF.

Q117. Find the HCF of 48 and 72.

Answer and working

24

Factor out k; HCF of 12 and 18 is 6.

Q118. Find the LCM of 16 and 24.

Answer and working

48

LCM of 4k and 6k is 12k; use product/HCF.

Q119. Find the HCF of 60 and 90.

Answer and working

30

Factor out k; HCF of 12 and 18 is 6.

Q120. Find the LCM of 20 and 30.

Answer and working

60

LCM of 4k and 6k is 12k; use product/HCF.

Set 13: Algebra and equations

Method: Preserve equality by performing the same operation on both sides.

Q121. Solve 3x + 2 = 17.

Answer and working

5

3x = 15k, hence x = 5k.

Q122. The sum of two numbers is 10; their difference is 2. Find the larger.

Answer and working

6

Larger = (sum + difference)/2 = 6k.

Q123. Solve 3x + 4 = 34.

Answer and working

10

3x = 15k, hence x = 5k.

Q124. The sum of two numbers is 20; their difference is 4. Find the larger.

Answer and working

12

Larger = (sum + difference)/2 = 6k.

Q125. Solve 3x + 6 = 51.

Answer and working

15

3x = 15k, hence x = 5k.

Q126. The sum of two numbers is 30; their difference is 6. Find the larger.

Answer and working

18

Larger = (sum + difference)/2 = 6k.

Q127. Solve 3x + 8 = 68.

Answer and working

20

3x = 15k, hence x = 5k.

Q128. The sum of two numbers is 40; their difference is 8. Find the larger.

Answer and working

24

Larger = (sum + difference)/2 = 6k.

Q129. Solve 3x + 10 = 85.

Answer and working

25

3x = 15k, hence x = 5k.

Q130. The sum of two numbers is 50; their difference is 10. Find the larger.

Answer and working

30

Larger = (sum + difference)/2 = 6k.

Set 14: Arithmetic progressions

Method: a_n = a+(n−1)d; S_n = n[2a+(n−1)d]/2.

Q131. An AP starts at 1, with common difference 3. Find its tenth term.

Answer and working

28

a₁₀ = k + 9×3.

Q132. Find the sum of the first 5 positive integers.

Answer and working

15

Use n(n+1)/2 with n = 5k.

Q133. An AP starts at 2, with common difference 3. Find its tenth term.

Answer and working

29

a₁₀ = k + 9×3.

Q134. Find the sum of the first 10 positive integers.

Answer and working

55

Use n(n+1)/2 with n = 5k.

Q135. An AP starts at 3, with common difference 3. Find its tenth term.

Answer and working

30

a₁₀ = k + 9×3.

Q136. Find the sum of the first 15 positive integers.

Answer and working

120

Use n(n+1)/2 with n = 5k.

Q137. An AP starts at 4, with common difference 3. Find its tenth term.

Answer and working

31

a₁₀ = k + 9×3.

Q138. Find the sum of the first 20 positive integers.

Answer and working

210

Use n(n+1)/2 with n = 5k.

Q139. An AP starts at 5, with common difference 3. Find its tenth term.

Answer and working

32

a₁₀ = k + 9×3.

Q140. Find the sum of the first 25 positive integers.

Answer and working

325

Use n(n+1)/2 with n = 5k.

Set 15: Mensuration

Method: Rectangle area = lb; circle area = πr²; cylinder volume = πr²h.

Q141. A rectangle measures 3 m by 4 m. Find its diagonal.

Answer and working

5 m

Use Pythagoras: √(9k²+16k²) = 5k.

Q142. A cylinder has radius 7 cm and height 3 cm. Find volume using π=22/7.

Answer and working

462 cm³

V = (22/7) × 49 × (k+2) = 154(k+2).

Q143. A rectangle measures 6 m by 8 m. Find its diagonal.

Answer and working

10 m

Use Pythagoras: √(9k²+16k²) = 5k.

Q144. A cylinder has radius 7 cm and height 4 cm. Find volume using π=22/7.

Answer and working

616 cm³

V = (22/7) × 49 × (k+2) = 154(k+2).

Q145. A rectangle measures 9 m by 12 m. Find its diagonal.

Answer and working

15 m

Use Pythagoras: √(9k²+16k²) = 5k.

Q146. A cylinder has radius 7 cm and height 5 cm. Find volume using π=22/7.

Answer and working

770 cm³

V = (22/7) × 49 × (k+2) = 154(k+2).

Q147. A rectangle measures 12 m by 16 m. Find its diagonal.

Answer and working

20 m

Use Pythagoras: √(9k²+16k²) = 5k.

Q148. A cylinder has radius 7 cm and height 6 cm. Find volume using π=22/7.

Answer and working

924 cm³

V = (22/7) × 49 × (k+2) = 154(k+2).

Q149. A rectangle measures 15 m by 20 m. Find its diagonal.

Answer and working

25 m

Use Pythagoras: √(9k²+16k²) = 5k.

Q150. A cylinder has radius 7 cm and height 7 cm. Find volume using π=22/7.

Answer and working

1,078 cm³

V = (22/7) × 49 × (k+2) = 154(k+2).

Set 16: Probability

Method: P(event) = favourable/equally likely outcomes; independent probabilities multiply.

Q151. A bag has 2 red and 4 blue balls. Find probability of red on one draw.

Answer and working

1/3

Red/(red+blue) = (k+1)/(3k+3).

Q152. A fair die is rolled once. Find probability of a result at most 1.

Answer and working

1/6

There are k favourable faces among six equally likely faces.

Q153. A bag has 3 red and 6 blue balls. Find probability of red on one draw.

Answer and working

1/3

Red/(red+blue) = (k+1)/(3k+3).

Q154. A fair die is rolled once. Find probability of a result at most 2.

Answer and working

1/3

There are k favourable faces among six equally likely faces.

Q155. A bag has 4 red and 8 blue balls. Find probability of red on one draw.

Answer and working

1/3

Red/(red+blue) = (k+1)/(3k+3).

Q156. A fair die is rolled once. Find probability of a result at most 3.

Answer and working

1/2

There are k favourable faces among six equally likely faces.

Q157. A bag has 5 red and 10 blue balls. Find probability of red on one draw.

Answer and working

1/3

Red/(red+blue) = (k+1)/(3k+3).

Q158. A fair die is rolled once. Find probability of a result at most 4.

Answer and working

2/3

There are k favourable faces among six equally likely faces.

Q159. A bag has 6 red and 12 blue balls. Find probability of red on one draw.

Answer and working

1/3

Red/(red+blue) = (k+1)/(3k+3).

Q160. A fair die is rolled once. Find probability of a result at most 5.

Answer and working

5/6

There are k favourable faces among six equally likely faces.

Set 17: Permutations and combinations

Method: nPr = n!/(n−r)!; nCr = n!/[r!(n−r)!].

Q161. How many ways can two people be chosen from 4 distinct people?

Answer and working

6

Selection ignores order: nC2 = n(n−1)/2.

Q162. How many ordered president–secretary pairs can be formed from 4 people, with different holders?

Answer and working

12

Order matters: n choices for president and n−1 for secretary.

Q163. How many ways can two people be chosen from 5 distinct people?

Answer and working

10

Selection ignores order: nC2 = n(n−1)/2.

Q164. How many ordered president–secretary pairs can be formed from 5 people, with different holders?

Answer and working

20

Order matters: n choices for president and n−1 for secretary.

Q165. How many ways can two people be chosen from 6 distinct people?

Answer and working

15

Selection ignores order: nC2 = n(n−1)/2.

Q166. How many ordered president–secretary pairs can be formed from 6 people, with different holders?

Answer and working

30

Order matters: n choices for president and n−1 for secretary.

Q167. How many ways can two people be chosen from 7 distinct people?

Answer and working

21

Selection ignores order: nC2 = n(n−1)/2.

Q168. How many ordered president–secretary pairs can be formed from 7 people, with different holders?

Answer and working

42

Order matters: n choices for president and n−1 for secretary.

Q169. How many ways can two people be chosen from 8 distinct people?

Answer and working

28

Selection ignores order: nC2 = n(n−1)/2.

Q170. How many ordered president–secretary pairs can be formed from 8 people, with different holders?

Answer and working

56

Order matters: n choices for president and n−1 for secretary.

Set 18: Statistics

Method: Population variance = Σ(x−mean)²/N; CV = SD/mean ×100.

Q171. Find the median of 1, 2, 3, 9, 10.

Answer and working

3

Five ordered values have their third value as median.

Q172. Find population variance of 1, 3, 5.

Answer and working

8/3

Mean = k+2; squared deviations are 4, 0 and 4. Divide by N=3.

Q173. Find the median of 2, 3, 4, 10, 11.

Answer and working

4

Five ordered values have their third value as median.

Q174. Find population variance of 2, 4, 6.

Answer and working

8/3

Mean = k+2; squared deviations are 4, 0 and 4. Divide by N=3.

Q175. Find the median of 3, 4, 5, 11, 12.

Answer and working

5

Five ordered values have their third value as median.

Q176. Find population variance of 3, 5, 7.

Answer and working

8/3

Mean = k+2; squared deviations are 4, 0 and 4. Divide by N=3.

Q177. Find the median of 4, 5, 6, 12, 13.

Answer and working

6

Five ordered values have their third value as median.

Q178. Find population variance of 4, 6, 8.

Answer and working

8/3

Mean = k+2; squared deviations are 4, 0 and 4. Divide by N=3.

Q179. Find the median of 5, 6, 7, 13, 14.

Answer and working

7

Five ordered values have their third value as median.

Q180. Find population variance of 5, 7, 9.

Answer and working

8/3

Mean = k+2; squared deviations are 4, 0 and 4. Divide by N=3.

Set 19: Series and coding

Method: Identify the rule and verify it against every displayed step.

Q181. Find the next term: 1, 4, 7, 10, __.

Answer and working

13

Constant difference is +3.

Q182. With A=1, B=2, …, Z=26, a letter code is increased by 1. What is the code of D after the increase?

Answer and working

5

D has code 4; add the stated shift.

Q183. Find the next term: 2, 5, 8, 11, __.

Answer and working

14

Constant difference is +3.

Q184. With A=1, B=2, …, Z=26, a letter code is increased by 2. What is the code of D after the increase?

Answer and working

6

D has code 4; add the stated shift.

Q185. Find the next term: 3, 6, 9, 12, __.

Answer and working

15

Constant difference is +3.

Q186. With A=1, B=2, …, Z=26, a letter code is increased by 3. What is the code of D after the increase?

Answer and working

7

D has code 4; add the stated shift.

Q187. Find the next term: 4, 7, 10, 13, __.

Answer and working

16

Constant difference is +3.

Q188. With A=1, B=2, …, Z=26, a letter code is increased by 4. What is the code of D after the increase?

Answer and working

8

D has code 4; add the stated shift.

Q189. Find the next term: 5, 8, 11, 14, __.

Answer and working

17

Constant difference is +3.

Q190. With A=1, B=2, …, Z=26, a letter code is increased by 5. What is the code of D after the increase?

Answer and working

9

D has code 4; add the stated shift.

Set 20: Directions and logical inference

Method: Use coordinates for directions; do not infer existence from a universal statement alone.

Q191. A person walks 3 km east and 4 km north. Find straight-line distance from the start.

Answer and working

5 km

Perpendicular displacements form a 3k–4k–5k triangle.

Q192. Group 1: all auditors are graduates; some graduates are musicians. Must some auditors be musicians?

Answer and working

No

The “some graduates” could all be non-auditors. The conclusion does not follow.

Q193. A person walks 6 km east and 8 km north. Find straight-line distance from the start.

Answer and working

10 km

Perpendicular displacements form a 3k–4k–5k triangle.

Q194. Group 2: all auditors are graduates; some graduates are musicians. Must some auditors be musicians?

Answer and working

No

The “some graduates” could all be non-auditors. The conclusion does not follow.

Q195. A person walks 9 km east and 12 km north. Find straight-line distance from the start.

Answer and working

15 km

Perpendicular displacements form a 3k–4k–5k triangle.

Q196. Group 3: all auditors are graduates; some graduates are musicians. Must some auditors be musicians?

Answer and working

No

The “some graduates” could all be non-auditors. The conclusion does not follow.

Q197. A person walks 12 km east and 16 km north. Find straight-line distance from the start.

Answer and working

20 km

Perpendicular displacements form a 3k–4k–5k triangle.

Q198. Group 4: all auditors are graduates; some graduates are musicians. Must some auditors be musicians?

Answer and working

No

The “some graduates” could all be non-auditors. The conclusion does not follow.

Q199. A person walks 15 km east and 20 km north. Find straight-line distance from the start.

Answer and working

25 km

Perpendicular displacements form a 3k–4k–5k triangle.

Q200. Group 5: all auditors are graduates; some graduates are musicians. Must some auditors be musicians?

Answer and working

No

The “some graduates” could all be non-auditors. The conclusion does not follow.

Error log and practice targets

Error	Repair task
Wrong percentage base	Write the original quantity explicitly before multiplying
Unit mismatch	Convert all distances/times before calculation
Wrong average weight	Use total quantity/value rather than averaging averages
Work-rate confusion	Assume one total job and add/subtract daily fractions
Population/sample variance mismatch	Use the denominator the question asks for
Syllogism overreach	Draw a valid counterexample before accepting the conclusion

Keep a record of time, correct answers and reason for error. A high score on these drills does not by itself establish readiness for every difficulty level in the recruitment test.

Frequently asked questions

Are these 200 UPSC previous-year questions?

No. They are original CrackTarget drills; the separate solved-paper page covers sourced PYQs.

Does the variance set use population variance?

Yes, those items explicitly say population and divide by N. A sample-variance question normally uses N−1.

Sources and further reading

• UPSC Special Advertisement 52/2026: scheme and nine syllabus heads

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