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4081. The average age of 12 RPF SI trainees is 75 years. A new trainee aged 52 years joins the batch. Find the new average age. (SI Set 4)
- A. 75
- B. 78.23
- C. 71.23
- D. 73.23
Answer: 73.23
Explanation: Old total = 12 x 75 = 900. New total = 900 + 52 = 952. New average = 952/13 = 73.23.
4082. If 15 railway signal poles cost the same as 12 units cost 13 units of a fourth quantity (15:12 :: 13:x), find x. (SI Set 4)
- A. 16.25
- B. 10.4
- C. 13.4
- D. 5.4
Answer: 10.4
Explanation: 15:12 :: 13:x => x = (12 x 13)/15 = 10.4.
4083. A sum of ₹640 is divided between two RPF constables in the ratio 5:3 as duty allowance. Find the share of the first constable. (SI Set 4)
- A. 383
- B. 423
- C. 400
- D. 240
Answer: 400
Explanation: First share = [5/(5+3)] x 640 = 400.
4084. Find the compound interest on ₹40000 for 2 years at 20% per annum compounded annually, for a railway pension-fund deposit. (SI Set 4)
- A. 17781
- B. 17600
- C. 16000
- D. 17410
Answer: 17600
Explanation: A = P(1+R/100)^T = 40000(1+20/100)^2 = 57600. CI = A - P = 57600 - 40000 = 17600.
4085. A railway cooperative society lends ₹47443 to a staff member at 9% per annum simple interest for 3 years. Find the simple interest. (SI Set 4)
- A. 12666.61
- B. 12996.61
- C. 17079.48
- D. 12809.61
Answer: 12809.61
Explanation: SI = (P x R x T)/100 = (47443 x 9 x 3)/100 = 12809.61.
4086. A railway equipment supplier offers successive discounts of 25% and 12% on a marked price of ₹3123. Find the final selling price. (SI Set 4)
- A. 2061.18
- B. 2126.18
- C. 1967.49
- D. 2020.18
Answer: 2061.18
Explanation: SP = 3123 x (1-25/100) x (1-12/100) = 3123 x 0.75 x 0.88 = 2061.18.
4087. A marked price of a railway staff uniform kit is ₹2965. A discount of 20% is offered during a department sale. What is the selling price? (SI Set 4)
- A. 2283
- B. 2445
- C. 2223.75
- D. 2372
Answer: 2372
Explanation: SP = MP x (1 - 20/100) = 2965 x 0.8 = 2372.
4088. A railway canteen vendor bought a stock of items for ₹5677 and sold it at a loss of 10%. Find the selling price. (SI Set 4)
- A. 5012.3
- B. 6244.7
- C. 5109.3
- D. 5207.3
Answer: 5109.3
Explanation: SP = CP x (1 - 10/100) = 5677 x 0.9 = 5109.3.
4089. The monthly ration budget of an RPF barrack was ₹4062. It was increased by 18%. What is the new budget? (SI Set 4)
- A. 4793.16
- B. 3330.84
- C. 4782.16
- D. 4830.16
Answer: 4793.16
Explanation: New value = 4062 x (1 + 18/100) = 4793.16.
4090. In an RPF SI recruitment drive, 3958 candidates appeared for the written exam. If 42% of them qualified for the physical efficiency test, how many candidates qualified? (SI Set 4)
- A. 1701.36
- B. 1747.36
- C. 1644.36
- D. 1662.36
Answer: 1662.36
Explanation: 42% of 3958 = (42/100) x 3958 = 1662.36.
4091. Find the difference between the compound interest and simple interest on ₹25000 for 2 years at 15% per annum (an RPF finance-cell training problem). (SI Set 3)
- A. 514.5
- B. 625.5
- C. 562.5
- D. 1125
Answer: 562.5
Explanation: SI = (25000×15×2)/100 = 7500. CI = 25000[(1+15/100)²-1] = 8062.5. Difference = 8062.5 - 7500 = 562.5. (Shortcut: Diff = P(R/100)² for 2 years.)
4092. A sum of ₹35948 earns simple interest at 5% p.a. for 3 years under Scheme A, versus 4% p.a. for 4 years under Scheme B (an RPF welfare-fund comparison). Find the difference between the two interests. (SI Set 3)
- A. 11143.88
- B. 397.48
- C. 359.48
- D. 235.48
Answer: 359.48
Explanation: SI(A) = (35948×5×3)/100 = 5392.2. SI(B) = (35948×4×4)/100 = 5751.68. Difference = |5392.2 - 5751.68| = 359.48.
4093. Three RPF maintenance workers can individually repair a section of boundary fencing in 19, 10, and 25 days respectively. Working together, how many days will they take to finish the job? (SI Set 3)
- A. 5.19
- B. 18
- C. 3.69
- D. 8.19
Answer: 5.19
Explanation: Combined rate = 1/19 + 1/10 + 1/25 per day. Time = 1/(1/19+1/10+1/25) = 5.19 days.
4094. Two RPF joggers start at the same point on a circular track of length 600 m and run in opposite directions at 10 m/s and 3 m/s. After how many seconds will they meet for the first time? (SI Set 3)
- A. 53.15
- B. 32.15
- C. 46.15
- D. 85.71
Answer: 46.15
Explanation: Relative speed (opposite directions) = 10+3 = 13 m/s. Time to meet = Circumference/Relative speed = 600/13 = 46.15 s.
4095. An RPF vehicle travels from post A to post B at 30 km/h and returns the same distance at 60 km/h. Find its average speed for the whole journey. (SI Set 3)
- A. 45
- B. 35
- C. 40
- D. 44
Answer: 40
Explanation: Average speed (equal distances) = 2×S1×S2/(S1+S2) = 2×30×60/(30+60) = 40 km/h.
4096. Find the average of 8 consecutive even numbers starting from 18, used in an RPF fitness-score tally. (SI Set 3)
- A. 9
- B. 65
- C. 76
- D. 25
Answer: 25
Explanation: Consecutive even numbers from 18 to 32. Average = (first+last)/2 = (18+32)/2 = 25.
4097. An RPF SI officer's father's age is currently 5 times the officer's age of 12 years. Find the ratio of father's age to the officer's age after 14 years. (SI Set 3)
- A. 2.85
- B. 3.35
- C. 5
- D. 2.55
Answer: 2.85
Explanation: Father's current age = 5 × 12 = 60. After 14 years: Father = 74, Officer = 26. Ratio = 74/26 = 2.85.
4098. Two RPF canteen contract partners invest ₹16896 and ₹36547 respectively in a joint catering contract for the same period. If the total profit is ₹18826, find the first partner's share. (SI Set 3)
- A. 5567.84
- B. 5951.84
- C. 12874.16
- D. 6391.84
Answer: 5951.84
Explanation: Profit ratio = Capital ratio = 16896:36547. First partner's share = [16896/(16896+36547)] × 18826 = 5951.84.
4099. An RPF canteen mixes two grades of rice costing ₹31/kg and ₹65/kg to get a mixture worth ₹57/kg. Find the ratio in which they must be mixed. (SI Set 3)
- A. 5:13
- B. 4:13
- C. 13:4
- D. 4:14
Answer: 4:13
Explanation: By alligation: ratio = (CP2-Mean):(Mean-CP1) = (65-57):(57-31) = 8:26 = 4:13 (simplified).
4100. An RPF stores bag contains 4 red, 8 blue, and 4 green signal flags. One flag is drawn at random. Find the probability that it is red. (SI Set 3)
- A. 0.25
- B. 0.09
- C. 0.47
- D. 0.35
Answer: 0.25
Explanation: Total flags = 4+8+4 = 16. Probability(red) = 4/16 = 0.25.
4101. From a group of 9 RPF SI candidates, in how many ways can a team (unordered) of 4 be selected? (SI Set 3)
- A. 155
- B. 126
- C. 3024
- D. 107
Answer: 126
Explanation: nCr = n!/(r!(n-r)!) = 9!/(4!×5!) = 126.
4102. The bar graph shows the number of RPF SI cases resolved per year: 2021=521, 2022=771, 2023=1510, 2024=1266, 2025=1045. Find the difference in cases resolved between 2021 and 2025. (SI Set 3)
- A. 516
- B. 524
- C. 538
- D. 549
Answer: 524
Explanation: Difference = |521 - 1045| = 524.
4103. A pie chart shows the RPF annual budget of ₹3600000 divided among 5 heads by central angle: Equipment=69°, Training=44°, Welfare=83°, Infrastructure=109°, Communication=55°. Find the budget allocated to Infrastructure. (SI Set 3)
- A. ₹550000
- B. ₹1146103
- C. ₹1090000
- D. ₹1056393
Answer: ₹1090000
Explanation: Share = (109°/360°) × 3600000 = 1090000.
4104. The table shows RPF SI recruits across 5 zones: North=126, South=274, East=341, West=88, Central=117. Find the ratio of recruits in West zone to South zone in simplest form. (SI Set 3)
- A. 85:137
- B. 16:137
- C. 44:137
- D. 426:137
Answer: 44:137
Explanation: Ratio = 88:274 = 44:137 (dividing by HCF 2).
4105. An RPF watchtower is 23 m tall. From a point on the ground, the angle of elevation to the top of the tower is 30°. Find the horizontal distance from the point to the base of the tower. (tan 30° = 0.577) (SI Set 3)
- A. 34.86
- B. 44.86
- C. 39.86
- D. 13.27
Answer: 39.86
Explanation: tan(30°) = height/distance => distance = height/tan(30°) = 23/0.577 = 39.86 m.
4106. An RPF storage container is a cuboid of dimensions 17 m × 9 m × 4 m. Find its total surface area. (SI Set 3)
- A. 448
- B. 579
- C. 602
- D. 514
Answer: 514
Explanation: TSA = 2(lw+wh+hl) = 2(153+36+68) = 514 sq m.
4107. A circular RPF watchtower base has a radius of 35 m. Find its circumference. (Use π = 22/7) (SI Set 3)
- A. 220
- B. 241
- C. 209
- D. 3850
Answer: 220
Explanation: Circumference = 2πr = 2 × (22/7) × 35 = 220 m.
4108. A rectangular RPF parade ground is 38 m long and 24 m wide. Find its area. (SI Set 3)
- A. 946
- B. 912
- C. 906
- D. 124
Answer: 912
Explanation: Area = length × width = 38 × 24 = 912 sq m.
4109. If x² - 14x + 33 = 0 are the roots representing two patrol-team sizes, find the larger root. (SI Set 3)
- A. 11
- B. 14
- C. 3
- D. 1
Answer: 11
Explanation: Factorizing: x² - 14x + 33 = (x - 3)(x - 11) = 0. Roots are 3 and 11. Larger root = 11.
4110. In an RPF duty-roster puzzle, if 2x + 47 = 69, find the value of x. (SI Set 3)
- A. 11
- B. 13
- C. 9
- D. 34.5
Answer: 11
Explanation: 2x = 69 - 47 = 22. x = 22/2 = 11.
4111. Simplify: 7 × (2/9) + 7 = ? (SI Set 3)
- A. 7.56
- B. 13.56
- C. 14.22
- D. 8.56
Answer: 8.56
Explanation: 7 × (2/9) = 1.56. Adding 7: 1.56 + 7 = 8.56.
4112. Simplify: 35 + 10 × 6 − 9 + 5 = ? (SI Set 3)
- A. 96
- B. 266
- C. 89
- D. 91
Answer: 91
Explanation: By BODMAS: 10 × 6 = 60 first. Then 35 + 60 - 9 + 5 = 91.
4113. In an RPF stores ledger, 62840 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 3)
- A. 13
- B. 7
- C. 11
- D. 12
Answer: 12
Explanation: 62840 mod 19 = 7. Number to add = 19 - 7 = 12 (0 if already divisible).
4114. Find the LCM of the patrol shift durations 43 minutes and 35 minutes used for syncing two RPF post schedules. (SI Set 3)
- A. 1
- B. 1505
- C. 1509
- D. 1502
Answer: 1505
Explanation: LCM(43, 35) = (43 x 35)/HCF(43,35) = (43 x 35)/1 = 1505.
4115. An inlet pipe at a railway water tank can fill it in 13 hours, while an outlet pipe (leak) can empty it in 19 hours. If both are opened together, in how many hours will the tank be filled? (SI Set 3)
- A. 16
- B. 41.17
- C. 39.67
- D. 42.17
Answer: 41.17
Explanation: Net rate = 1/13 - 1/19 per hour. Time to fill = 1/(1/13 - 1/19) = 41.17 hours.
4116. RPF constable A can clear a stretch of railway track of debris alone in 24 days, and constable B can do it alone in 20 days. Working together, in how many days will they finish the job? (SI Set 3)
- A. 12.91
- B. 9.41
- C. 22
- D. 10.91
Answer: 10.91
Explanation: Combined time = (A x B)/(A+B) = (24 x 20)/(24+20) = 10.91 days.
4117. A boat used for river patrol by RPF has a speed of 15 km/h in still water. The current speed is 4 km/h. Find the time taken to cover 42 km upstream. (SI Set 3)
- A. 5.82
- B. 3.32
- C. 3.82
- D. 2.8
Answer: 3.82
Explanation: Upstream speed = 15 - 4 = 11 km/h. Time = 42/11 = 3.82 hours.
4118. Two trains of lengths 138 m and 148 m run on parallel tracks in opposite directions at 36 km/h and 36 km/h respectively. Find the time they take to cross each other completely. (SI Set 3)
- A. 286
- B. 9.3
- C. 14.3
- D. 20.3
Answer: 14.3
Explanation: Relative speed (opposite dir) = (36+36) x 5/18 = 20 m/s. Total length = 138+148 = 286 m. Time = 286/20 = 14.3 s.
4119. A railway goods train 163 m long is moving at a speed of 63 km/h. How much time will it take to cross a stationary signal pole? (SI Set 3)
- A. 8.36
- B. 4.31
- C. 9.31
- D. 13.31
Answer: 9.31
Explanation: Speed = 63 x 5/18 = 17.5 m/s. Time = Length/Speed = 163/17.5 = 9.31 seconds.
4120. An RPF patrol vehicle covers a distance of 363 km at a constant speed of 65 km/h. How much time does it take? (SI Set 3)
- A. 5.58
- B. 4.08
- C. 7.08
- D. 4.84
Answer: 5.58
Explanation: Time = Distance/Speed = 363/65 = 5.58 hours.
4121. The average daily patrol distance of 7 RPF beat officers is 66 km. One officer who covered only 53 km is excluded. Find the new average distance of the remaining officers. (SI Set 3)
- A. 68.17
- B. 66
- C. 70.17
- D. 63.17
Answer: 68.17
Explanation: Total = 7 x 66 = 462. New total = 462 - 53 = 409. New average = 409/6 = 68.17.
4122. The average age of 5 RPF SI trainees is 74 years. A new trainee aged 84 years joins the batch. Find the new average age. (SI Set 3)
- A. 74
- B. 75.67
- C. 70.67
- D. 77.67
Answer: 75.67
Explanation: Old total = 5 x 74 = 370. New total = 370 + 84 = 454. New average = 454/6 = 75.67.
4123. If 13 railway signal poles cost the same as 8 units cost 13 units of a fourth quantity (13:8 :: 13:x), find x. (SI Set 3)
- A. 8
- B. 4
- C. 21.13
- D. 13
Answer: 8
Explanation: 13:8 :: 13:x => x = (8 x 13)/13 = 8.
4124. A sum of ₹186 is divided between two RPF constables in the ratio 2:4 as duty allowance. Find the share of the first constable. (SI Set 3)
- A. 124
- B. 99
- C. 17
- D. 62
Answer: 62
Explanation: First share = [2/(2+4)] x 186 = 62.
4125. Find the compound interest on ₹20000 for 2 years at 5% per annum compounded annually, for a railway pension-fund deposit. (SI Set 3)
- A. 2002
- B. 2050
- C. 2000
- D. 2215
Answer: 2050
Explanation: A = P(1+R/100)^T = 20000(1+5/100)^2 = 22050. CI = A - P = 22050 - 20000 = 2050.
4126. A railway cooperative society lends ₹87347 to a staff member at 5% per annum simple interest for 3 years. Find the simple interest. (SI Set 3)
- A. 13102.05
- B. 17469.4
- C. 13042.05
- D. 13217.05
Answer: 13102.05
Explanation: SI = (P x R x T)/100 = (87347 x 5 x 3)/100 = 13102.05.
4127. A railway equipment supplier offers successive discounts of 10% and 8% on a marked price of ₹2054. Find the final selling price. (SI Set 3)
- A. 1700.71
- B. 1684.28
- C. 1654.71
- D. 1768.71
Answer: 1700.71
Explanation: SP = 2054 x (1-10/100) x (1-8/100) = 2054 x 0.9 x 0.92 = 1700.71.
4128. A marked price of a railway staff uniform kit is ₹5250. A discount of 25% is offered during a department sale. What is the selling price? (SI Set 3)
- A. 3980.5
- B. 3675
- C. 3937.5
- D. 3868.5
Answer: 3937.5
Explanation: SP = MP x (1 - 25/100) = 5250 x 0.75 = 3937.5.
4129. A railway canteen vendor bought a stock of items for ₹8356 and sold it at a loss of 10%. Find the selling price. (SI Set 3)
- A. 7472.4
- B. 7601.4
- C. 7520.4
- D. 9191.6
Answer: 7520.4
Explanation: SP = CP x (1 - 10/100) = 8356 x 0.9 = 7520.4.
4130. The monthly ration budget of an RPF barrack was ₹2079. It was decreased by 25%. What is the new budget? (SI Set 3)
- A. 1515.25
- B. 1580.25
- C. 2598.75
- D. 1559.25
Answer: 1559.25
Explanation: New value = 2079 x (1 - 25/100) = 1559.25.
4131. In an RPF SI recruitment drive, 2210 candidates appeared for the written exam. If 35% of them qualified for the physical efficiency test, how many candidates qualified? (SI Set 3)
- A. 795.5
- B. 773.5
- C. 736.5
- D. 833.5
Answer: 773.5
Explanation: 35% of 2210 = (35/100) x 2210 = 773.5.
4132. Find the difference between the compound interest and simple interest on ₹8000 for 2 years at 20% per annum (an RPF finance-cell training problem). (SI Set 2)
- A. 320
- B. 342
- C. 640
- D. 286
Answer: 320
Explanation: SI = (8000×20×2)/100 = 3200. CI = 8000[(1+20/100)²-1] = 3520. Difference = 3520 - 3200 = 320. (Shortcut: Diff = P(R/100)² for 2 years.)
4133. A sum of ₹45887 earns simple interest at 6% p.a. for 4 years under Scheme A, versus 4% p.a. for 2 years under Scheme B (an RPF welfare-fund comparison). Find the difference between the two interests. (SI Set 2)
- A. 7303.92
- B. 7341.92
- C. 14683.84
- D. 7419.92
Answer: 7341.92
Explanation: SI(A) = (45887×6×4)/100 = 11012.88. SI(B) = (45887×4×2)/100 = 3670.96. Difference = |11012.88 - 3670.96| = 7341.92.
4134. Three RPF maintenance workers can individually repair a section of boundary fencing in 22, 13, and 24 days respectively. Working together, how many days will they take to finish the job? (SI Set 2)
- A. 19.67
- B. 5.1
- C. 6.1
- D. 10.1
Answer: 6.1
Explanation: Combined rate = 1/22 + 1/13 + 1/24 per day. Time = 1/(1/22+1/13+1/24) = 6.1 days.
4135. Two RPF joggers start at the same point on a circular track of length 400 m and run in opposite directions at 8 m/s and 4 m/s. After how many seconds will they meet for the first time? (SI Set 2)
- A. 38.33
- B. 20.33
- C. 33.33
- D. 100
Answer: 33.33
Explanation: Relative speed (opposite directions) = 8+4 = 12 m/s. Time to meet = Circumference/Relative speed = 400/12 = 33.33 s.
4136. An RPF vehicle travels from post A to post B at 50 km/h and returns the same distance at 60 km/h. Find its average speed for the whole journey. (SI Set 2)
- A. 49.55
- B. 55
- C. 54.55
- D. 61.55
Answer: 54.55
Explanation: Average speed (equal distances) = 2×S1×S2/(S1+S2) = 2×50×60/(50+60) = 54.55 km/h.
4137. Find the sum of squares of the first 14 positive integers (used to compute a variance check in an RPF statistics drill). (SI Set 2)
- A. 1085
- B. 1045
- C. 981
- D. 1015
Answer: 1015
Explanation: Sum of squares = n(n+1)(2n+1)/6 = 14(14+1)(28+1)/6 = 1015.
4138. An RPF SI officer's father's age is currently 4 times the officer's age of 10 years. Find the ratio of father's age to the officer's age after 7 years. (SI Set 2)
- A. 2.76
- B. 4
- C. 2.46
- D. 3.26
Answer: 2.76
Explanation: Father's current age = 4 × 10 = 40. After 7 years: Father = 47, Officer = 17. Ratio = 47/17 = 2.76.
4139. Two RPF canteen contract partners invest ₹36414 and ₹29461 respectively in a joint catering contract for the same period. If the total profit is ₹7319, find the first partner's share. (SI Set 2)
- A. 4207.75
- B. 3273.25
- C. 3917.75
- D. 4045.75
Answer: 4045.75
Explanation: Profit ratio = Capital ratio = 36414:29461. First partner's share = [36414/(36414+29461)] × 7319 = 4045.75.
4140. An RPF canteen mixes two grades of rice costing ₹51/kg and ₹79/kg to get a mixture worth ₹73/kg. Find the ratio in which they must be mixed. (SI Set 2)
- A. 4:11
- B. 11:3
- C. 3:12
- D. 3:11
Answer: 3:11
Explanation: By alligation: ratio = (CP2-Mean):(Mean-CP1) = (79-73):(73-51) = 6:22 = 3:11 (simplified).
4141. 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 spade? (SI Set 2)
- A. 0.08
- B. 0.47
- C. 0.25
- D. 0.41
Answer: 0.25
Explanation: Favorable outcomes = 13. Total = 52. Probability = 13/52 = 0.25.
4142. From a group of 5 RPF SI candidates, in how many ways can a team (unordered) of 2 be selected? (SI Set 2)
- A. 35
- B. 20
- C. 10
- D. 1
Answer: 10
Explanation: nCr = n!/(r!(n-r)!) = 5!/(2!×3!) = 10.
4143. The bar graph shows the number of RPF SI cases resolved per year: 2021=927, 2022=2056, 2023=917, 2024=1437, 2025=1960. Find the difference in cases resolved between 2022 and 2023. (SI Set 2)
- A. 1123
- B. 1166
- C. 1143
- D. 1139
Answer: 1139
Explanation: Difference = |2056 - 917| = 1139.
4144. A pie chart shows the RPF annual budget of ₹1800000 divided among 5 heads by central angle: Equipment=100°, Training=103°, Welfare=120°, Infrastructure=29°, Communication=8°. Find the budget allocated to Welfare. (SI Set 2)
- A. ₹563927
- B. ₹666533
- C. ₹145000
- D. ₹600000
Answer: ₹600000
Explanation: Share = (120°/360°) × 1800000 = 600000.
4145. The table shows RPF SI recruits across 5 zones: North=315, South=299, East=124, West=399, Central=227. Find the ratio of recruits in West zone to Central zone in simplest form. (SI Set 2)
- A. 354:227
- B. 129:227
- C. 399:227
- D. 468:227
Answer: 399:227
Explanation: Ratio = 399:227 = 399:227 (dividing by HCF 1).
4146. An RPF watchtower is 27 m tall. From a point on the ground, the angle of elevation to the top of the tower is 60°. Find the horizontal distance from the point to the base of the tower. (tan 60° = 1.732) (SI Set 2)
- A. 46.76
- B. 24.59
- C. 15.59
- D. 11.59
Answer: 15.59
Explanation: tan(60°) = height/distance => distance = height/tan(60°) = 27/1.732 = 15.59 m.
4147. A cubical RPF equipment crate has a side of 11 m. Find its volume. (SI Set 2)
- A. 1369
- B. 1331
- C. 1309
- D. 1427
Answer: 1331
Explanation: Volume = side³ = 11³ = 1331 cu m.
4148. A circular RPF watchtower base has a radius of 49 m. Find its area. (Use π = 22/7) (SI Set 2)
- A. 7509
- B. 7546
- C. 308
- D. 7595
Answer: 7546
Explanation: Area = πr² = (22/7) × 49 × 49 = 7546 sq m.
4149. A rectangular RPF parade ground is 13 m long and 9 m wide. Find its area. (SI Set 2)
- A. 117
- B. 104
- C. 143
- D. 44
Answer: 117
Explanation: Area = length × width = 13 × 9 = 117 sq m.
4150. If x² - 11x + 28 = 0 are the roots representing two patrol-team sizes, find the larger root. (SI Set 2)
- A. 3
- B. 4
- C. 7
- D. 10
Answer: 7
Explanation: Factorizing: x² - 11x + 28 = (x - 4)(x - 7) = 0. Roots are 4 and 7. Larger root = 7.
4151. In an RPF duty-roster puzzle, if 3x + 14 = 32, find the value of x. (SI Set 2)
- A. 4
- B. 6
- C. 11
- D. 10.67
Answer: 6
Explanation: 3x = 32 - 14 = 18. x = 18/3 = 6.
4152. Simplify: 10 × (9/15) + 8 = ? (SI Set 2)
- A. 17
- B. 13
- C. 18.6
- D. 14
Answer: 14
Explanation: 10 × (9/15) = 6. Adding 8: 6 + 8 = 14.
4153. Simplify: 45 + 11 × 5 − 9 + 22 = ? (SI Set 2)
- A. 115
- B. 113
- C. 293
- D. 106
Answer: 113
Explanation: By BODMAS: 11 × 5 = 55 first. Then 45 + 55 - 9 + 22 = 113.
4154. In an RPF stores ledger, 78434 cartridge boxes need to be arranged into equal rows for a divisor of 17. What is the least number of boxes that must be added to make the total exactly divisible by 17? (SI Set 2)
- A. 1
- B. 6
- C. 13
- D. 4
Answer: 4
Explanation: 78434 mod 17 = 13. Number to add = 17 - 13 = 4 (0 if already divisible).
4155. Find the LCM of the patrol shift durations 48 minutes and 19 minutes used for syncing two RPF post schedules. (SI Set 2)
- A. 1
- B. 912
- C. 914
- D. 910
Answer: 912
Explanation: LCM(48, 19) = (48 x 19)/HCF(48,19) = (48 x 19)/1 = 912.
4156. An inlet pipe at a railway water tank can fill it in 20 hours, while an outlet pipe (leak) can empty it in 13 hours. If both are opened together, in how many hours will the tank be filled? (SI Set 2)
- A. 36.64
- B. 40.14
- C. 16.5
- D. 37.14
Answer: 37.14
Explanation: Net rate = 1/20 - 1/13 per hour. Time to fill = 1/(1/20 - 1/13) = 37.14 hours.
4157. RPF constable A can clear a stretch of railway track of debris alone in 15 days, and constable B can do it alone in 23 days. Working together, in how many days will they finish the job? (SI Set 2)
- A. 11.08
- B. 8.08
- C. 19
- D. 9.08
Answer: 9.08
Explanation: Combined time = (A x B)/(A+B) = (15 x 23)/(15+23) = 9.08 days.
4158. A boat used for river patrol by RPF has a speed of 17 km/h in still water. The current speed is 2 km/h. Find the time taken to cover 100 km downstream. (SI Set 2)
- A. 5.26
- B. 5.88
- C. 4.76
- D. 7.26
Answer: 5.26
Explanation: Downstream speed = 17 + 2 = 19 km/h. Time = 100/19 = 5.26 hours.
4159. Two trains of lengths 199 m and 177 m run on parallel tracks in opposite directions at 60 km/h and 45 km/h respectively. Find the time they take to cross each other completely. (SI Set 2)
- A. 12.89
- B. 90.24
- C. 10.89
- D. 17.89
Answer: 12.89
Explanation: Relative speed (opposite dir) = (60+45) x 5/18 = 29.17 m/s. Total length = 199+177 = 376 m. Time = 376/29.17 = 12.89 s.
4160. A railway goods train 231 m long is moving at a speed of 60 km/h. How much time will it take to cross a stationary signal pole? (SI Set 2)
- A. 11.86
- B. 12.37
- C. 13.86
- D. 18.86
Answer: 13.86
Explanation: Speed = 60 x 5/18 = 16.67 m/s. Time = Length/Speed = 231/16.67 = 13.86 seconds.
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