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4001. A sum of ₹418 is divided between two RPF constables in the ratio 7:4 as duty allowance. Find the share of the first constable. (SI Set 6)
- A. 152
- B. 287
- C. 266
- D. 219
Answer: 266
Explanation: First share = [7/(7+4)] x 418 = 266.
4002. Find the compound interest on ₹40000 for 2 years at 12% per annum compounded annually, for a railway pension-fund deposit. (SI Set 6)
- A. 10000
- B. 10298
- C. 10176
- D. 9600
Answer: 10176
Explanation: A = P(1+R/100)^T = 40000(1+12/100)^2 = 50176. CI = A - P = 50176 - 40000 = 10176.
4003. A railway cooperative society lends ₹37845 to a staff member at 9% per annum simple interest for 2 years. Find the simple interest. (SI Set 6)
- A. 6812.1
- B. 7090.1
- C. 6678.1
- D. 10218.15
Answer: 6812.1
Explanation: SI = (P x R x T)/100 = (37845 x 9 x 2)/100 = 6812.1.
4004. A railway equipment supplier offers successive discounts of 10% and 8% on a marked price of ₹9195. Find the final selling price. (SI Set 6)
- A. 7539.9
- B. 7613.46
- C. 7578.46
- D. 7644.46
Answer: 7613.46
Explanation: SP = 9195 x (1-10/100) x (1-8/100) = 9195 x 0.9 x 0.92 = 7613.46.
4005. A marked price of a railway staff uniform kit is ₹3519. A discount of 15% is offered during a department sale. What is the selling price? (SI Set 6)
- A. 2991.15
- B. 2938.15
- C. 3069.15
- D. 2815.2
Answer: 2991.15
Explanation: SP = MP x (1 - 15/100) = 3519 x 0.85 = 2991.15.
4006. A railway canteen vendor bought a stock of items for ₹1569 and sold it at a loss of 22%. Find the selling price. (SI Set 6)
- A. 1309.82
- B. 1151.82
- C. 1223.82
- D. 1914.18
Answer: 1223.82
Explanation: SP = CP x (1 - 22/100) = 1569 x 0.78 = 1223.82.
4007. The monthly ration budget of an RPF barrack was ₹4373. It was decreased by 25%. What is the new budget? (SI Set 6)
- A. 3254.75
- B. 3279.75
- C. 3314.75
- D. 5466.25
Answer: 3279.75
Explanation: New value = 4373 x (1 - 25/100) = 3279.75.
4008. In an RPF SI recruitment drive, 4067 candidates appeared for the written exam. If 52% of them qualified for the physical efficiency test, how many candidates qualified? (SI Set 6)
- A. 2114.84
- B. 2198.84
- C. 2151.84
- D. 2084.84
Answer: 2114.84
Explanation: 52% of 4067 = (52/100) x 4067 = 2114.84.
4009. Find the difference between the compound interest and simple interest on ₹20000 for 2 years at 12% per annum (an RPF finance-cell training problem). (SI Set 5)
- A. 355
- B. 246
- C. 288
- D. 576
Answer: 288
Explanation: SI = (20000×12×2)/100 = 4800. CI = 20000[(1+12/100)²-1] = 5088. Difference = 5088 - 4800 = 288. (Shortcut: Diff = P(R/100)² for 2 years.)
4010. A sum of ₹44182 earns simple interest at 8% p.a. for 3 years under Scheme A, versus 9% p.a. for 5 years under Scheme B (an RPF welfare-fund comparison). Find the difference between the two interests. (SI Set 5)
- A. 30485.58
- B. 9208.22
- C. 9390.22
- D. 9278.22
Answer: 9278.22
Explanation: SI(A) = (44182×8×3)/100 = 10603.68. SI(B) = (44182×9×5)/100 = 19881.9. Difference = |10603.68 - 19881.9| = 9278.22.
4011. Three RPF maintenance workers can individually repair a section of boundary fencing in 14, 20, and 11 days respectively. Working together, how many days will they take to finish the job? (SI Set 5)
- A. 4.21
- B. 8.71
- C. 15
- D. 4.71
Answer: 4.71
Explanation: Combined rate = 1/14 + 1/20 + 1/11 per day. Time = 1/(1/14+1/20+1/11) = 4.71 days.
4012. Two RPF joggers start at the same point on a circular track of length 600 m and run in opposite directions at 8 m/s and 5 m/s. After how many seconds will they meet for the first time? (SI Set 5)
- A. 66.15
- B. 200
- C. 32.15
- D. 46.15
Answer: 46.15
Explanation: Relative speed (opposite directions) = 8+5 = 13 m/s. Time to meet = Circumference/Relative speed = 600/13 = 46.15 s.
4013. An RPF vehicle travels from post A to post B at 45 km/h and returns the same distance at 75 km/h. Find its average speed for the whole journey. (SI Set 5)
- A. 60.25
- B. 56.25
- C. 60
- D. 54.25
Answer: 56.25
Explanation: Average speed (equal distances) = 2×S1×S2/(S1+S2) = 2×45×75/(45+75) = 56.25 km/h.
4014. An RPF control room logs incident serial numbers from 1 to 30 in a single shift. Find the sum of all serial numbers. (SI Set 5)
- A. 536
- B. 465
- C. 433
- D. 471
Answer: 465
Explanation: Sum = n(n+1)/2 = 30(30+1)/2 = 465.
4015. An RPF SI officer's father's age is currently 3 times the officer's age of 16 years. Find the ratio of father's age to the officer's age after 6 years. (SI Set 5)
- A. 2.45
- B. 3
- C. 2.95
- D. 2.15
Answer: 2.45
Explanation: Father's current age = 3 × 16 = 48. After 6 years: Father = 54, Officer = 22. Ratio = 54/22 = 2.45.
4016. Two RPF canteen contract partners invest ₹21073 and ₹30486 respectively in a joint catering contract for the same period. If the total profit is ₹27975, find the first partner's share. (SI Set 5)
- A. 16541.16
- B. 11734.84
- C. 10985.84
- D. 11433.84
Answer: 11433.84
Explanation: Profit ratio = Capital ratio = 21073:30486. First partner's share = [21073/(21073+30486)] × 27975 = 11433.84.
4017. An RPF canteen mixes two grades of rice costing ₹50/kg and ₹60/kg to get a mixture worth ₹56/kg. Find the ratio in which they must be mixed. (SI Set 5)
- A. 3:2
- B. 2:3
- C. 3:3
- D. 2:4
Answer: 2:3
Explanation: By alligation: ratio = (CP2-Mean):(Mean-CP1) = (60-56):(56-50) = 4:6 = 2:3 (simplified).
4018. An RPF stores bag contains 7 red, 3 blue, and 5 green signal flags. One flag is drawn at random. Find the probability that it is red. (SI Set 5)
- A. 0.47
- B. 0.58
- C. 0.34
- D. 0.67
Answer: 0.47
Explanation: Total flags = 7+3+5 = 15. Probability(red) = 7/15 = 0.47.
4019. From a group of 6 RPF SI candidates, in how many ways can a team (unordered) of 3 be selected? (SI Set 5)
- A. 20
- B. 38
- C. 120
- D. 1
Answer: 20
Explanation: nCr = n!/(r!(n-r)!) = 6!/(3!×3!) = 20.
4020. The bar graph shows the number of RPF SI cases resolved per year: 2021=1475, 2022=1476, 2023=2271, 2024=1738, 2025=1011. Find the percentage change in cases resolved from 2021 to 2022. (SI Set 5)
- A. 0.07
- B. 0
- C. 14.07
- D. 34.07
Answer: 0.07
Explanation: % change = [(1476 - 1475)/1475] × 100 = 0.07%.
4021. A pie chart shows the RPF annual budget of ₹1800000 divided among 5 heads by central angle: Equipment=73°, Training=56°, Welfare=92°, Infrastructure=77°, Communication=62°. Find the budget allocated to Welfare. (SI Set 5)
- A. ₹385000
- B. ₹403200
- C. ₹509479
- D. ₹460000
Answer: ₹460000
Explanation: Share = (92°/360°) × 1800000 = 460000.
4022. The table shows RPF SI recruits across 5 zones: North=278, South=152, East=246, West=258, Central=178. Find the ratio of recruits in East zone to West zone in simplest form. (SI Set 5)
- A. 32:43
- B. 45:43
- C. 54:43
- D. 41:43
Answer: 41:43
Explanation: Ratio = 246:258 = 41:43 (dividing by HCF 6).
4023. An RPF watchtower is 48 m tall. From a point on the ground, the angle of elevation to the top of the tower is 45°. Find the horizontal distance from the point to the base of the tower. (tan 45° = 1) (SI Set 5)
- A. 40
- B. 53
- C. 48
- D. 42
Answer: 48
Explanation: tan(45°) = height/distance => distance = height/tan(45°) = 48/1 = 48 m.
4024. An RPF storage container is a cuboid of dimensions 15 m × 13 m × 4 m. Find its total surface area. (SI Set 5)
- A. 614
- B. 600
- C. 693
- D. 749
Answer: 614
Explanation: TSA = 2(lw+wh+hl) = 2(195+52+60) = 614 sq m.
4025. A circular RPF watchtower base has a radius of 28 m. Find its circumference. (Use π = 22/7) (SI Set 5)
- A. 2464
- B. 136
- C. 176
- D. 209
Answer: 176
Explanation: Circumference = 2πr = 2 × (22/7) × 28 = 176 m.
4026. A rectangular RPF parade ground is 30 m long and 36 m wide. Find its perimeter. (SI Set 5)
- A. 1080
- B. 159
- C. 124
- D. 132
Answer: 132
Explanation: Perimeter = 2(length + width) = 2(30+36) = 132 m.
4027. If x² - 20x + 96 = 0 are the roots representing two patrol-team sizes, find the larger root. (SI Set 5)
- A. 5
- B. 8
- C. 12
- D. 13
Answer: 12
Explanation: Factorizing: x² - 20x + 96 = (x - 12)(x - 8) = 0. Roots are 12 and 8. Larger root = 12.
4028. In an RPF duty-roster puzzle, if 5x + 37 = 67, find the value of x. (SI Set 5)
- A. 5
- B. 13.4
- C. 6
- D. 7
Answer: 6
Explanation: 5x = 67 - 37 = 30. x = 30/5 = 6.
4029. Simplify: 2 × (4/11) + 7 = ? (SI Set 5)
- A. 7.73
- B. 6.23
- C. 9.36
- D. 11.73
Answer: 7.73
Explanation: 2 × (4/11) = 0.73. Adding 7: 0.73 + 7 = 7.73.
4030. Simplify: 11 + 5 × 13 − 4 + 14 = ? (SI Set 5)
- A. 77
- B. 86
- C. 218
- D. 91
Answer: 86
Explanation: By BODMAS: 5 × 13 = 65 first. Then 11 + 65 - 4 + 14 = 86.
4031. In an RPF stores ledger, 98495 cartridge boxes need to be arranged into equal rows for a divisor of 23. What is the least number of boxes that must be added to make the total exactly divisible by 23? (SI Set 5)
- A. 9
- B. 14
- C. 17
- D. 13
Answer: 14
Explanation: 98495 mod 23 = 9. Number to add = 23 - 9 = 14 (0 if already divisible).
4032. Find the LCM of the patrol shift durations 59 minutes and 53 minutes used for syncing two RPF post schedules. (SI Set 5)
- A. 1
- B. 3127
- C. 3132
- D. 3125
Answer: 3127
Explanation: LCM(59, 53) = (59 x 53)/HCF(59,53) = (59 x 53)/1 = 3127.
4033. Pipe A can fill a railway water tank in 10 hours and pipe B can fill it in 11 hours. If both pipes are opened together, in how many hours will the tank be filled? (SI Set 5)
- A. 9.24
- B. 4.24
- C. 5.24
- D. 10.5
Answer: 5.24
Explanation: Combined time = (A x B)/(A+B) = (10 x 11)/(10+11) = 5.24 hours.
4034. RPF constable A can clear a stretch of railway track of debris alone in 29 days, and constable B can do it alone in 17 days. Working together, in how many days will they finish the job? (SI Set 5)
- A. 10.72
- B. 9.22
- C. 11.72
- D. 23
Answer: 10.72
Explanation: Combined time = (A x B)/(A+B) = (29 x 17)/(29+17) = 10.72 days.
4035. A boat used for river patrol by RPF has a speed of 15 km/h in still water. The current speed is 5 km/h. Find the time taken to cover 90 km upstream. (SI Set 5)
- A. 10
- B. 8
- C. 9
- D. 6
Answer: 9
Explanation: Upstream speed = 15 - 5 = 10 km/h. Time = 90/10 = 9 hours.
4036. Two trains of lengths 195 m and 184 m run on parallel tracks in opposite directions at 54 km/h and 36 km/h respectively. Find the time they take to cross each other completely. (SI Set 5)
- A. 18.16
- B. 12.16
- C. 15.16
- D. 75.8
Answer: 15.16
Explanation: Relative speed (opposite dir) = (54+36) x 5/18 = 25 m/s. Total length = 195+184 = 379 m. Time = 379/25 = 15.16 s.
4037. A railway goods train 232 m long is moving at a speed of 108 km/h. How much time will it take to cross a stationary signal pole? (SI Set 5)
- A. 10.73
- B. 1.73
- C. 7.25
- D. 7.73
Answer: 7.73
Explanation: Speed = 108 x 5/18 = 30 m/s. Time = Length/Speed = 232/30 = 7.73 seconds.
4038. An RPF patrol vehicle covers a distance of 173 km at a constant speed of 45 km/h. How much time does it take? (SI Set 5)
- A. 3.84
- B. 2.84
- C. 4.84
- D. 3.15
Answer: 3.84
Explanation: Time = Distance/Speed = 173/45 = 3.84 hours.
4039. The average daily patrol distance of 7 RPF beat officers is 74 km. One officer who covered only 72 km is excluded. Find the new average distance of the remaining officers. (SI Set 5)
- A. 74.33
- B. 69.33
- C. 74
- D. 75.33
Answer: 74.33
Explanation: Total = 7 x 74 = 518. New total = 518 - 72 = 446. New average = 446/6 = 74.33.
4040. The average age of 7 RPF SI trainees is 48 years. A new trainee aged 105 years joins the batch. Find the new average age. (SI Set 5)
- A. 55.13
- B. 48
- C. 60.13
- D. 50.13
Answer: 55.13
Explanation: Old total = 7 x 48 = 336. New total = 336 + 105 = 441. New average = 441/8 = 55.13.
4041. If 9 railway signal poles cost the same as 10 units cost 13 units of a fourth quantity (9:10 :: 13:x), find x. (SI Set 5)
- A. 21.44
- B. 11.44
- C. 11.7
- D. 14.44
Answer: 14.44
Explanation: 9:10 :: 13:x => x = (10 x 13)/9 = 14.44.
4042. A sum of ₹517 is divided between two RPF constables in the ratio 8:3 as duty allowance. Find the share of the first constable. (SI Set 5)
- A. 320
- B. 376
- C. 141
- D. 397
Answer: 376
Explanation: First share = [8/(8+3)] x 517 = 376.
4043. Find the compound interest on ₹10000 for 2 years at 8% per annum compounded annually, for a railway pension-fund deposit. (SI Set 5)
- A. 1522
- B. 1600
- C. 1664
- D. 1777
Answer: 1664
Explanation: A = P(1+R/100)^T = 10000(1+8/100)^2 = 11664. CI = A - P = 11664 - 10000 = 1664.
4044. A railway cooperative society lends ₹38757 to a staff member at 12% per annum simple interest for 5 years. Find the simple interest. (SI Set 5)
- A. 23333.2
- B. 27905.04
- C. 22965.2
- D. 23254.2
Answer: 23254.2
Explanation: SI = (P x R x T)/100 = (38757 x 12 x 5)/100 = 23254.2.
4045. A railway equipment supplier offers successive discounts of 25% and 12% on a marked price of ₹2523. Find the final selling price. (SI Set 5)
- A. 1649.18
- B. 1589.49
- C. 1713.18
- D. 1665.18
Answer: 1665.18
Explanation: SP = 2523 x (1-25/100) x (1-12/100) = 2523 x 0.75 x 0.88 = 1665.18.
4046. A marked price of a railway staff uniform kit is ₹8133. A discount of 12% is offered during a department sale. What is the selling price? (SI Set 5)
- A. 6750.39
- B. 7157.04
- C. 7103.04
- D. 7210.04
Answer: 7157.04
Explanation: SP = MP x (1 - 12/100) = 8133 x 0.88 = 7157.04.
4047. A railway canteen vendor bought a stock of items for ₹3073 and sold it at a loss of 12%. Find the selling price. (SI Set 5)
- A. 3441.76
- B. 2670.24
- C. 2704.24
- D. 2744.24
Answer: 2704.24
Explanation: SP = CP x (1 - 12/100) = 3073 x 0.88 = 2704.24.
4048. The monthly ration budget of an RPF barrack was ₹2322. It was increased by 35%. What is the new budget? (SI Set 5)
- A. 3164.7
- B. 1509.3
- C. 3134.7
- D. 3081.7
Answer: 3134.7
Explanation: New value = 2322 x (1 + 35/100) = 3134.7.
4049. In an RPF SI recruitment drive, 573 candidates appeared for the written exam. If 75% of them qualified for the physical efficiency test, how many candidates qualified? (SI Set 5)
- A. 401.75
- B. 438.75
- C. 505.75
- D. 429.75
Answer: 429.75
Explanation: 75% of 573 = (75/100) x 573 = 429.75.
4050. Find the difference between the compound interest and simple interest on ₹8000 for 2 years at 10% per annum (an RPF finance-cell training problem). (SI Set 4)
- A. 114
- B. 63
- C. 80
- D. 160
Answer: 80
Explanation: SI = (8000×10×2)/100 = 1600. CI = 8000[(1+10/100)²-1] = 1680. Difference = 1680 - 1600 = 80. (Shortcut: Diff = P(R/100)² for 2 years.)
4051. A sum of ₹36033 earns simple interest at 6% p.a. for 3 years under Scheme A, versus 9% p.a. for 2 years under Scheme B (an RPF welfare-fund comparison). Find the difference between the two interests. (SI Set 4)
- A. 0
- B. 119
- C. 3
- D. 12971.88
Answer: 0
Explanation: SI(A) = (36033×6×3)/100 = 6485.94. SI(B) = (36033×9×2)/100 = 6485.94. Difference = |6485.94 - 6485.94| = 0.
4052. Three RPF maintenance workers can individually repair a section of boundary fencing in 17, 11, and 13 days respectively. Working together, how many days will they take to finish the job? (SI Set 4)
- A. 8.41
- B. 3.91
- C. 13.67
- D. 4.41
Answer: 4.41
Explanation: Combined rate = 1/17 + 1/11 + 1/13 per day. Time = 1/(1/17+1/11+1/13) = 4.41 days.
4053. Two RPF joggers start at the same point on a circular track of length 800 m and run in opposite directions at 6 m/s and 3 m/s. After how many seconds will they meet for the first time? (SI Set 4)
- A. 78.89
- B. 266.67
- C. 93.89
- D. 88.89
Answer: 88.89
Explanation: Relative speed (opposite directions) = 6+3 = 9 m/s. Time to meet = Circumference/Relative speed = 800/9 = 88.89 s.
4054. An RPF vehicle travels from post A to post B at 45 km/h and returns the same distance at 80 km/h. Find its average speed for the whole journey. (SI Set 4)
- A. 53.6
- B. 62.5
- C. 57.6
- D. 63.6
Answer: 57.6
Explanation: Average speed (equal distances) = 2×S1×S2/(S1+S2) = 2×45×80/(45+80) = 57.6 km/h.
4055. Find the sum of squares of the first 14 positive integers (used to compute a variance check in an RPF statistics drill). (SI Set 4)
- A. 997
- B. 1057
- C. 1034
- D. 1015
Answer: 1015
Explanation: Sum of squares = n(n+1)(2n+1)/6 = 14(14+1)(28+1)/6 = 1015.
4056. An RPF SI officer's father's age is currently 6 times the officer's age of 8 years. Find the ratio of father's age to the officer's age after 8 years. (SI Set 4)
- A. 4
- B. 3.2
- C. 3.5
- D. 6
Answer: 3.5
Explanation: Father's current age = 6 × 8 = 48. After 8 years: Father = 56, Officer = 16. Ratio = 56/16 = 3.5.
4057. Two RPF canteen contract partners invest ₹32768 and ₹28419 respectively in a joint catering contract for the same period. If the total profit is ₹6743, find the first partner's share. (SI Set 4)
- A. 3410.14
- B. 3131.86
- C. 4280.14
- D. 3611.14
Answer: 3611.14
Explanation: Profit ratio = Capital ratio = 32768:28419. First partner's share = [32768/(32768+28419)] × 6743 = 3611.14.
4058. An RPF canteen mixes two grades of rice costing ₹50/kg and ₹73/kg to get a mixture worth ₹70/kg. Find the ratio in which they must be mixed. (SI Set 4)
- A. 20:3
- B. 3:21
- C. 3:20
- D. 4:20
Answer: 3:20
Explanation: By alligation: ratio = (CP2-Mean):(Mean-CP1) = (73-70):(70-50) = 3:20 = 3:20 (simplified).
4059. A card is drawn at random from a well-shuffled standard deck of 52 playing cards during an RPF aptitude session. What is the probability of drawing a red card? (SI Set 4)
- A. 0.75
- B. 0.5
- C. 0.34
- D. 0.69
Answer: 0.5
Explanation: Favorable outcomes = 26. Total = 52. Probability = 26/52 = 0.5.
4060. From a group of 11 RPF SI candidates, in how many ways can a committee with distinct ranked posts (ordered) of 3 be selected? (SI Set 4)
- A. 990
- B. 165
- C. 967
- D. 1017
Answer: 990
Explanation: nPr = n!/(n-r)! = 11!/8! = 990.
4061. The bar graph shows the number of RPF SI cases resolved per year: 2021=1924, 2022=1099, 2023=1553, 2024=1904, 2025=2005. Find the percentage change in cases resolved from 2021 to 2022. (SI Set 4)
- A. -36.88
- B. -42.88
- C. -16.88
- D. 0
Answer: -42.88
Explanation: % change = [(1099 - 1924)/1924] × 100 = -42.88%.
4062. A pie chart shows the RPF annual budget of ₹1800000 divided among 5 heads by central angle: Equipment=45°, Training=98°, Welfare=93°, Infrastructure=103°, Communication=21°. Find the budget allocated to Training. (SI Set 4)
- A. ₹465000
- B. ₹527425
- C. ₹490000
- D. ₹445626
Answer: ₹490000
Explanation: Share = (98°/360°) × 1800000 = 490000.
4063. The table shows RPF SI recruits across 5 zones: North=351, South=153, East=131, West=254, Central=246. Find the ratio of recruits in North zone to South zone in simplest form. (SI Set 4)
- A. 39:17
- B. 44:17
- C. 7:17
- D. 36:17
Answer: 39:17
Explanation: Ratio = 351:153 = 39:17 (dividing by HCF 9).
4064. An RPF watchtower is 19 m tall. From a point on the ground, the angle of elevation to the top of the tower is 45°. Find the horizontal distance from the point to the base of the tower. (tan 45° = 1) (SI Set 4)
- A. 22
- B. 17
- C. 11
- D. 19
Answer: 19
Explanation: tan(45°) = height/distance => distance = height/tan(45°) = 19/1 = 19 m.
4065. An RPF storage container is a cuboid of dimensions 20 m × 6 m × 7 m. Find its total surface area. (SI Set 4)
- A. 680
- B. 604
- C. 534
- D. 706
Answer: 604
Explanation: TSA = 2(lw+wh+hl) = 2(120+42+140) = 604 sq m.
4066. A circular RPF watchtower base has a radius of 49 m. Find its area. (Use π = 22/7) (SI Set 4)
- A. 7578
- B. 308
- C. 7503
- D. 7546
Answer: 7546
Explanation: Area = πr² = (22/7) × 49 × 49 = 7546 sq m.
4067. A rectangular RPF parade ground is 52 m long and 14 m wide. Find its area. (SI Set 4)
- A. 747
- B. 728
- C. 132
- D. 703
Answer: 728
Explanation: Area = length × width = 52 × 14 = 728 sq m.
4068. If x² - 12x + 32 = 0 are the roots representing two patrol-team sizes, find the larger root. (SI Set 4)
- A. 10
- B. 1
- C. 4
- D. 8
Answer: 8
Explanation: Factorizing: x² - 12x + 32 = (x - 8)(x - 4) = 0. Roots are 8 and 4. Larger root = 8.
4069. In an RPF duty-roster puzzle, if 7x + 54 = 334, find the value of x. (SI Set 4)
- A. 41
- B. 36
- C. 40
- D. 47.71
Answer: 40
Explanation: 7x = 334 - 54 = 280. x = 280/7 = 40.
4070. Simplify: 7 × (6/10) + 8 = ? (SI Set 4)
- A. 15.2
- B. 12.2
- C. 10.2
- D. 15.6
Answer: 12.2
Explanation: 7 × (6/10) = 4.2. Adding 8: 4.2 + 8 = 12.2.
4071. Simplify: 40 + 7 × 6 − 3 + 11 = ? (SI Set 4)
- A. 290
- B. 84
- C. 90
- D. 100
Answer: 90
Explanation: By BODMAS: 7 × 6 = 42 first. Then 40 + 42 - 3 + 11 = 90.
4072. In an RPF stores ledger, 53351 cartridge boxes need to be arranged into equal rows for a divisor of 19. What is the least number of boxes that must be added to make the total exactly divisible by 19? (SI Set 4)
- A. 1
- B. 4
- C. 0
- D. 18
Answer: 1
Explanation: 53351 mod 19 = 18. Number to add = 19 - 18 = 1 (0 if already divisible).
4073. Find the HCF of the patrol shift durations 54 minutes and 25 minutes used for syncing two RPF post schedules. (SI Set 4)
- A. 1350
- B. -6
- C. 1
- D. 3
Answer: 1
Explanation: HCF(54, 25) = 1 (by Euclidean algorithm / prime factorization).
4074. An inlet pipe at a railway water tank can fill it in 9 hours, while an outlet pipe (leak) can empty it in 21 hours. If both are opened together, in how many hours will the tank be filled? (SI Set 4)
- A. 15
- B. 18.75
- C. 14.75
- D. 15.75
Answer: 15.75
Explanation: Net rate = 1/9 - 1/21 per hour. Time to fill = 1/(1/9 - 1/21) = 15.75 hours.
4075. RPF constable A can clear a stretch of railway track of debris alone in 11 days, and constable B can do it alone in 27 days. Working together, in how many days will they finish the job? (SI Set 4)
- A. 8.82
- B. 6.82
- C. 7.82
- D. 19
Answer: 7.82
Explanation: Combined time = (A x B)/(A+B) = (11 x 27)/(11+27) = 7.82 days.
4076. A boat used for river patrol by RPF has a speed of 15 km/h in still water. The current speed is 8 km/h. Find the time taken to cover 79 km upstream. (SI Set 4)
- A. 5.27
- B. 10.29
- C. 11.29
- D. 13.29
Answer: 11.29
Explanation: Upstream speed = 15 - 8 = 7 km/h. Time = 79/7 = 11.29 hours.
4077. Two trains of lengths 110 m and 132 m run on parallel tracks in opposite directions at 54 km/h and 18 km/h respectively. Find the time they take to cross each other completely. (SI Set 4)
- A. 24.2
- B. 12.1
- C. 7.1
- D. 17.1
Answer: 12.1
Explanation: Relative speed (opposite dir) = (54+18) x 5/18 = 20 m/s. Total length = 110+132 = 242 m. Time = 242/20 = 12.1 s.
4078. A railway goods train 297 m long is moving at a speed of 36 km/h. How much time will it take to cross a stationary signal pole? (SI Set 4)
- A. 27.7
- B. 34.7
- C. 24.75
- D. 29.7
Answer: 29.7
Explanation: Speed = 36 x 5/18 = 10 m/s. Time = Length/Speed = 297/10 = 29.7 seconds.
4079. An RPF patrol vehicle covers a distance of 714 km at a constant speed of 75 km/h. How much time does it take? (SI Set 4)
- A. 11.02
- B. 8.52
- C. 9.52
- D. 8.4
Answer: 9.52
Explanation: Time = Distance/Speed = 714/75 = 9.52 hours.
4080. The average daily patrol distance of 7 RPF beat officers is 68 km. One officer who covered only 43 km is excluded. Find the new average distance of the remaining officers. (SI Set 4)
- A. 71.17
- B. 68
- C. 72.17
- D. 74.17
Answer: 72.17
Explanation: Total = 7 x 68 = 476. New total = 476 - 43 = 433. New average = 433/6 = 72.17.
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