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3761. The Nizam's Guaranteed State Railway, operating in the former princely state of Hyderabad, was unique among colonial-era railway companies because it was:
- A. Financed and owned by an Indian princely state rather than directly by the British government or a British company
- B. The first fully electrified railway in India
- C. Operated only goods trains, no passenger services
- D. Built entirely on narrow gauge
Answer: Financed and owned by an Indian princely state rather than directly by the British government or a British company
Explanation: The Nizam's Guaranteed State Railway was financed by the Nizam of Hyderabad's state government, making it one of the few major lines owned by an Indian princely state rather than British colonial or private British capital; it later merged into Central Railway/Nizam State Railway zone arrangements before South Central Railway's formation.
3762. The Assam Bengal Railway, constructed in the late 19th century, was primarily built to connect the Brahmaputra valley/Assam tea gardens to which port?
- A. Calcutta port
- B. Vishakhapatnam port
- C. Cochin port
- D. Chittagong port
Answer: Chittagong port
Explanation: Assam Bengal Railway was built to provide an outlet for Assam's tea and other produce via Chittagong port (now in Bangladesh), avoiding the circuitous Brahmaputra river route to Calcutta; it later became part of Northeast Frontier Railway's predecessor systems in independent India.
3763. The Sindh, Punjab and Delhi Railway, an early colonial-era company, is historically significant primarily because it:
- A. Built the Attock Bridge over the Indus and linked Lahore to Delhi via Multan and Karachi, forming part of the later North Western Railway
- B. Constructed the first railway line in Madras Presidency
- C. Built the Darjeeling Himalayan narrow-gauge line
- D. Operated India's first electric suburban train in Bombay
Answer: Built the Attock Bridge over the Indus and linked Lahore to Delhi via Multan and Karachi, forming part of the later North Western Railway
Explanation: The Sindh, Punjab and Delhi Railway built the strategically important Attock Bridge over the Indus and connected Lahore-Multan-Karachi-Delhi; after mergers it formed the core of North Western Railway, now largely in Pakistan post-partition, with the Indian remnant later merging into Northern Railway.
3764. A sum of ₹20581 earns simple interest at 6% p.a. for 5 years under Scheme A, versus 4% p.a. for 3 years under Scheme B (an RPF welfare-fund comparison). Find the difference between the two interests. (SI Set 11)
- A. 8644.02
- B. 3704.58
- C. 3659.58
- D. 3808.58
Answer: 3704.58
Explanation: SI(A) = (20581×6×5)/100 = 6174.3. SI(B) = (20581×4×3)/100 = 2469.72. Difference = |6174.3 - 2469.72| = 3704.58.
3765. Three RPF maintenance workers can individually repair a section of boundary fencing in 22, 20, and 16 days respectively. Working together, how many days will they take to finish the job? (SI Set 11)
- A. 6.33
- B. 5.33
- C. 8.33
- D. 19.33
Answer: 6.33
Explanation: Combined rate = 1/22 + 1/20 + 1/16 per day. Time = 1/(1/22+1/20+1/16) = 6.33 days.
3766. Two RPF joggers start at the same point on a circular track of length 300 m and run in opposite directions at 5 m/s and 3 m/s. After how many seconds will they meet for the first time? (SI Set 11)
- A. 150
- B. 57.5
- C. 37.5
- D. 24.5
Answer: 37.5
Explanation: Relative speed (opposite directions) = 5+3 = 8 m/s. Time to meet = Circumference/Relative speed = 300/8 = 37.5 s.
3767. An RPF vehicle travels from post A to post B at 60 km/h and returns the same distance at 70 km/h. Find its average speed for the whole journey. (SI Set 11)
- A. 60.62
- B. 64.62
- C. 69.62
- D. 65
Answer: 64.62
Explanation: Average speed (equal distances) = 2×S1×S2/(S1+S2) = 2×60×70/(60+70) = 64.62 km/h.
3768. Find the average of 7 consecutive even numbers starting from 14, used in an RPF fitness-score tally. (SI Set 11)
- A. 20
- B. 84
- C. 1
- D. 53
Answer: 20
Explanation: Consecutive even numbers from 14 to 26. Average = (first+last)/2 = (14+26)/2 = 20.
3769. An RPF SI officer's father's age is currently 5 times the officer's age of 17 years. Find the ratio of father's age to the officer's age after 6 years. (SI Set 11)
- A. 4.46
- B. 5
- C. 3.66
- D. 3.96
Answer: 3.96
Explanation: Father's current age = 5 × 17 = 85. After 6 years: Father = 91, Officer = 23. Ratio = 91/23 = 3.96.
3770. Two RPF canteen contract partners invest ₹34888 and ₹39531 respectively in a joint catering contract for the same period. If the total profit is ₹17197, find the first partner's share. (SI Set 11)
- A. 8599.04
- B. 8062.04
- C. 9134.96
- D. 7675.04
Answer: 8062.04
Explanation: Profit ratio = Capital ratio = 34888:39531. First partner's share = [34888/(34888+39531)] × 17197 = 8062.04.
3771. An RPF canteen mixes two grades of rice costing ₹49/kg and ₹72/kg to get a mixture worth ₹51/kg. Find the ratio in which they must be mixed. (SI Set 11)
- A. 2:21
- B. 21:3
- C. 21:2
- D. 22:2
Answer: 21:2
Explanation: By alligation: ratio = (CP2-Mean):(Mean-CP1) = (72-51):(51-49) = 21:2 = 21:2 (simplified).
3772. Two fair dice are rolled together by an RPF training instructor during a probability drill. What is the probability that the sum of the numbers is 7? (SI Set 11)
- A. 0.17
- B. 0.45
- C. 0.33
- D. 0.02
Answer: 0.17
Explanation: Number of favorable outcomes (sum=7) = 6. Total outcomes = 36. Probability = 6/36 = 0.17.
3773. From a group of 7 RPF SI candidates, in how many ways can a team (unordered) of 4 be selected? (SI Set 11)
- A. 840
- B. 46
- C. 17
- D. 35
Answer: 35
Explanation: nCr = n!/(r!(n-r)!) = 7!/(4!×3!) = 35.
3774. The bar graph shows the number of RPF SI cases resolved per year: 2021=1908, 2022=887, 2023=2017, 2024=964, 2025=1541. Find the difference in cases resolved between 2023 and 2025. (SI Set 11)
- A. 476
- B. 464
- C. 483
- D. 507
Answer: 476
Explanation: Difference = |2017 - 1541| = 476.
3775. A pie chart shows the RPF annual budget of ₹3600000 divided among 5 heads by central angle: Equipment=103°, Training=107°, Welfare=111°, Infrastructure=30°, Communication=9°. Find the budget allocated to Training. (SI Set 11)
- A. ₹1087684
- B. ₹1110000
- C. ₹1040217
- D. ₹1070000
Answer: ₹1070000
Explanation: Share = (107°/360°) × 3600000 = 1070000.
3776. The table shows RPF SI recruits across 5 zones: North=257, South=122, East=381, West=358, Central=248. What percentage of the total recruits belong to the North zone? (SI Set 11)
- A. 18.81
- B. 31.81
- C. 11.81
- D. 47.81
Answer: 18.81
Explanation: Total = 257+122+381+358+248 = 1366. Percentage for North = (257/1366) × 100 = 18.81%.
3777. An RPF watchtower is 25 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 11)
- A. 25
- B. 20
- C. 34
- D. 19
Answer: 25
Explanation: tan(45°) = height/distance => distance = height/tan(45°) = 25/1 = 25 m.
3778. An RPF storage container is a cuboid of dimensions 11 m × 9 m × 8 m. Find its volume. (SI Set 11)
- A. 734
- B. 792
- C. 894
- D. 849
Answer: 792
Explanation: Volume = l × w × h = 11 × 9 × 8 = 792 cu m.
3779. A circular RPF watchtower base has a radius of 42 m. Find its circumference. (Use π = 22/7) (SI Set 11)
- A. 264
- B. 310
- C. 5544
- D. 223
Answer: 264
Explanation: Circumference = 2πr = 2 × (22/7) × 42 = 264 m.
3780. A rectangular RPF parade ground is 38 m long and 6 m wide. Find its area. (SI Set 11)
- A. 217
- B. 263
- C. 88
- D. 228
Answer: 228
Explanation: Area = length × width = 38 × 6 = 228 sq m.
3781. If x² - 17x + 72 = 0 are the roots representing two patrol-team sizes, find the larger root. (SI Set 11)
- A. 12
- B. 9
- C. 8
- D. 5
Answer: 9
Explanation: Factorizing: x² - 17x + 72 = (x - 9)(x - 8) = 0. Roots are 9 and 8. Larger root = 9.
3782. In an RPF duty-roster puzzle, if 7x + 14 = 119, find the value of x. (SI Set 11)
- A. 10
- B. 17
- C. 16
- D. 15
Answer: 15
Explanation: 7x = 119 - 14 = 105. x = 105/7 = 15.
3783. Simplify: 9 × (5/11) + 6 = ? (SI Set 11)
- A. 15.09
- B. 10.09
- C. 15.45
- D. 9.09
Answer: 10.09
Explanation: 9 × (5/11) = 4.09. Adding 6: 4.09 + 6 = 10.09.
3784. Simplify: 45 + 8 × 8 − 8 + 20 = ? (SI Set 11)
- A. 115
- B. 129
- C. 436
- D. 121
Answer: 121
Explanation: By BODMAS: 8 × 8 = 64 first. Then 45 + 64 - 8 + 20 = 121.
3785. In an RPF stores ledger, 93824 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 11)
- A. 16
- B. 17
- C. 14
- D. 7
Answer: 16
Explanation: 93824 mod 23 = 7. Number to add = 23 - 7 = 16 (0 if already divisible).
3786. Find the LCM of the patrol shift durations 19 minutes and 42 minutes used for syncing two RPF post schedules. (SI Set 11)
- A. 801
- B. 794
- C. 798
- D. 1
Answer: 798
Explanation: LCM(19, 42) = (19 x 42)/HCF(19,42) = (19 x 42)/1 = 798.
3787. An inlet pipe at a railway water tank can fill it in 22 hours, while an outlet pipe (leak) can empty it in 14 hours. If both are opened together, in how many hours will the tank be filled? (SI Set 11)
- A. 18
- B. 41.5
- C. 37.5
- D. 38.5
Answer: 38.5
Explanation: Net rate = 1/22 - 1/14 per hour. Time to fill = 1/(1/22 - 1/14) = 38.5 hours.
3788. RPF constable A can clear a stretch of railway track of debris alone in 17 days, and constable B can do it alone in 10 days. Working together, in how many days will they finish the job? (SI Set 11)
- A. 5.8
- B. 7.3
- C. 6.3
- D. 13.5
Answer: 6.3
Explanation: Combined time = (A x B)/(A+B) = (17 x 10)/(17+10) = 6.3 days.
3789. A boat used for river patrol by RPF has a speed of 10 km/h in still water. The current speed is 2 km/h. Find the time taken to cover 114 km downstream. (SI Set 11)
- A. 9.5
- B. 8.5
- C. 10.5
- D. 11.4
Answer: 9.5
Explanation: Downstream speed = 10 + 2 = 12 km/h. Time = 114/12 = 9.5 hours.
3790. Two trains of lengths 159 m and 125 m run on parallel tracks in opposite directions at 36 km/h and 18 km/h respectively. Find the time they take to cross each other completely. (SI Set 11)
- A. 56.8
- B. 12.93
- C. 18.93
- D. 23.93
Answer: 18.93
Explanation: Relative speed (opposite dir) = (36+18) x 5/18 = 15 m/s. Total length = 159+125 = 284 m. Time = 284/15 = 18.93 s.
3791. A railway goods train 257 m long is moving at a speed of 45 km/h. How much time will it take to cross a stationary signal pole? (SI Set 11)
- A. 20.56
- B. 16.56
- C. 24.56
- D. 17.72
Answer: 20.56
Explanation: Speed = 45 x 5/18 = 12.5 m/s. Time = Length/Speed = 257/12.5 = 20.56 seconds.
3792. An RPF patrol vehicle covers a distance of 626 km at a constant speed of 45 km/h. How much time does it take? (SI Set 11)
- A. 15.41
- B. 11.38
- C. 13.91
- D. 12.91
Answer: 13.91
Explanation: Time = Distance/Speed = 626/45 = 13.91 hours.
3793. The average daily patrol distance of 7 RPF beat officers is 74 km. One officer who covered only 56 km is excluded. Find the new average distance of the remaining officers. (SI Set 11)
- A. 73
- B. 74
- C. 77
- D. 78
Answer: 77
Explanation: Total = 7 x 74 = 518. New total = 518 - 56 = 462. New average = 462/6 = 77.
3794. The average age of 11 RPF SI trainees is 94 years. A new trainee aged 48 years joins the batch. Find the new average age. (SI Set 11)
- A. 94
- B. 90.17
- C. 91.17
- D. 89.17
Answer: 90.17
Explanation: Old total = 11 x 94 = 1034. New total = 1034 + 48 = 1082. New average = 1082/12 = 90.17.
3795. If 5 railway signal poles cost the same as 15 units cost 3 units of a fourth quantity (5:15 :: 3:x), find x. (SI Set 11)
- A. 0.5
- B. 1
- C. 11
- D. 9
Answer: 9
Explanation: 5:15 :: 3:x => x = (15 x 3)/5 = 9.
3796. A sum of ₹612 is divided between two RPF constables in the ratio 4:5 as duty allowance. Find the share of the first constable. (SI Set 11)
- A. 251
- B. 302
- C. 272
- D. 340
Answer: 272
Explanation: First share = [4/(4+5)] x 612 = 272.
3797. Find the compound interest on ₹20000 for 2 years at 20% per annum compounded annually, for a railway pension-fund deposit. (SI Set 11)
- A. 8838
- B. 8000
- C. 8769
- D. 8800
Answer: 8800
Explanation: A = P(1+R/100)^T = 20000(1+20/100)^2 = 28800. CI = A - P = 28800 - 20000 = 8800.
3798. A railway cooperative society lends ₹50415 to a staff member at 4% per annum simple interest for 2 years. Find the simple interest. (SI Set 11)
- A. 6049.8
- B. 4159.2
- C. 4033.2
- D. 3739.2
Answer: 4033.2
Explanation: SI = (P x R x T)/100 = (50415 x 4 x 2)/100 = 4033.2.
3799. A railway equipment supplier offers successive discounts of 25% and 5% on a marked price of ₹9392. Find the final selling price. (SI Set 11)
- A. 6708.8
- B. 6651.8
- C. 6691.8
- D. 6574.4
Answer: 6691.8
Explanation: SP = 9392 x (1-25/100) x (1-5/100) = 9392 x 0.75 x 0.95 = 6691.8.
3800. A marked price of a railway staff uniform kit is ₹7680. A discount of 12% is offered during a department sale. What is the selling price? (SI Set 11)
- A. 6758.4
- B. 6374.4
- C. 6782.4
- D. 6670.4
Answer: 6758.4
Explanation: SP = MP x (1 - 12/100) = 7680 x 0.88 = 6758.4.
3801. A railway canteen vendor bought a stock of items for ₹2062 and sold it at a profit of 12%. Find the selling price. (SI Set 11)
- A. 2309.44
- B. 1814.56
- C. 2282.44
- D. 2366.44
Answer: 2309.44
Explanation: SP = CP x (1 + 12/100) = 2062 x 1.12 = 2309.44.
3802. The monthly ration budget of an RPF barrack was ₹891. It was decreased by 10%. What is the new budget? (SI Set 11)
- A. 817.9
- B. 801.9
- C. 980.1
- D. 757.9
Answer: 801.9
Explanation: New value = 891 x (1 - 10/100) = 801.9.
3803. In an RPF SI recruitment drive, 3024 candidates appeared for the written exam. If 65% of them qualified for the physical efficiency test, how many candidates qualified? (SI Set 11)
- A. 2043.6
- B. 1965.6
- C. 1938.6
- D. 1973.6
Answer: 1965.6
Explanation: 65% of 3024 = (65/100) x 3024 = 1965.6.
3804. 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 10)
- A. 160
- B. 80
- C. 56
- D. 117
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.)
3805. A sum of ₹34296 earns simple interest at 5% p.a. for 2 years under Scheme A, versus 4% p.a. for 5 years under Scheme B (an RPF welfare-fund comparison). Find the difference between the two interests. (SI Set 10)
- A. 3429.6
- B. 3336.6
- C. 10288.8
- D. 3542.6
Answer: 3429.6
Explanation: SI(A) = (34296×5×2)/100 = 3429.6. SI(B) = (34296×4×5)/100 = 6859.2. Difference = |3429.6 - 6859.2| = 3429.6.
3806. Three RPF maintenance workers can individually repair a section of boundary fencing in 20, 11, and 25 days respectively. Working together, how many days will they take to finish the job? (SI Set 10)
- A. 7.53
- B. 5.53
- C. 4.03
- D. 18.67
Answer: 5.53
Explanation: Combined rate = 1/20 + 1/11 + 1/25 per day. Time = 1/(1/20+1/11+1/25) = 5.53 days.
3807. 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 4 m/s. After how many seconds will they meet for the first time? (SI Set 10)
- A. 48.86
- B. 42.86
- C. 37.86
- D. 100
Answer: 42.86
Explanation: Relative speed (opposite directions) = 10+4 = 14 m/s. Time to meet = Circumference/Relative speed = 600/14 = 42.86 s.
3808. 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 10)
- A. 57.6
- B. 55.6
- C. 60.6
- D. 62.5
Answer: 57.6
Explanation: Average speed (equal distances) = 2×S1×S2/(S1+S2) = 2×45×80/(45+80) = 57.6 km/h.
3809. An RPF control room logs incident serial numbers from 1 to 24 in a single shift. Find the sum of all serial numbers. (SI Set 10)
- A. 336
- B. 300
- C. 289
- D. 366
Answer: 300
Explanation: Sum = n(n+1)/2 = 24(24+1)/2 = 300.
3810. An RPF SI officer's father's age is currently 6 times the officer's age of 19 years. Find the ratio of father's age to the officer's age after 10 years. (SI Set 10)
- A. 4.78
- B. 3.98
- C. 6
- D. 4.28
Answer: 4.28
Explanation: Father's current age = 6 × 19 = 114. After 10 years: Father = 124, Officer = 29. Ratio = 124/29 = 4.28.
3811. Two RPF canteen contract partners invest ₹13984 and ₹6932 respectively in a joint catering contract for the same period. If the total profit is ₹3117, find the first partner's share. (SI Set 10)
- A. 2801.96
- B. 1033.04
- C. 1760.96
- D. 2083.96
Answer: 2083.96
Explanation: Profit ratio = Capital ratio = 13984:6932. First partner's share = [13984/(13984+6932)] × 3117 = 2083.96.
3812. 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 10)
- A. 2:3
- B. 2:4
- C. 3:2
- D. 3:3
Answer: 2:3
Explanation: By alligation: ratio = (CP2-Mean):(Mean-CP1) = (60-56):(56-50) = 4:6 = 2:3 (simplified).
3813. An RPF stores bag contains 6 red, 6 blue, and 6 green signal flags. One flag is drawn at random. Find the probability that it is red. (SI Set 10)
- A. 0.21
- B. 0.46
- C. 0.62
- D. 0.33
Answer: 0.33
Explanation: Total flags = 6+6+6 = 18. Probability(red) = 6/18 = 0.33.
3814. From a group of 12 RPF SI candidates, in how many ways can a team (unordered) of 2 be selected? (SI Set 10)
- A. 132
- B. 54
- C. 92
- D. 66
Answer: 66
Explanation: nCr = n!/(r!(n-r)!) = 12!/(2!×10!) = 66.
3815. The bar graph shows the number of RPF SI cases resolved per year: 2021=2426, 2022=1490, 2023=2359, 2024=1252, 2025=1854. Find the percentage change in cases resolved from 2024 to 2025. (SI Set 10)
- A. 62.08
- B. 33.08
- C. 79.08
- D. 48.08
Answer: 48.08
Explanation: % change = [(1854 - 1252)/1252] × 100 = 48.08%.
3816. A pie chart shows the RPF annual budget of ₹9000000 divided among 5 heads by central angle: Equipment=44°, Training=114°, Welfare=93°, Infrastructure=82°, Communication=27°. Find the budget allocated to Infrastructure. (SI Set 10)
- A. ₹2050000
- B. ₹2070348
- C. ₹675000
- D. ₹2010426
Answer: ₹2050000
Explanation: Share = (82°/360°) × 9000000 = 2050000.
3817. The table shows RPF SI recruits across 5 zones: North=265, South=86, East=347, West=374, Central=392. What percentage of the total recruits belong to the West zone? (SI Set 10)
- A. 25.55
- B. 45.55
- C. 28.55
- D. 19.55
Answer: 25.55
Explanation: Total = 265+86+347+374+392 = 1464. Percentage for West = (374/1464) × 100 = 25.55%.
3818. An RPF watchtower is 47 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 10)
- A. 49
- B. 35
- C. 45
- D. 47
Answer: 47
Explanation: tan(45°) = height/distance => distance = height/tan(45°) = 47/1 = 47 m.
3819. An RPF storage container is a cuboid of dimensions 5 m × 14 m × 11 m. Find its total surface area. (SI Set 10)
- A. 558
- B. 683
- C. 521
- D. 619
Answer: 558
Explanation: TSA = 2(lw+wh+hl) = 2(70+154+55) = 558 sq m.
3820. A circular RPF watchtower base has a radius of 28 m. Find its circumference. (Use π = 22/7) (SI Set 10)
- A. 231
- B. 2464
- C. 150
- D. 176
Answer: 176
Explanation: Circumference = 2πr = 2 × (22/7) × 28 = 176 m.
3821. A rectangular RPF parade ground is 29 m long and 26 m wide. Find its area. (SI Set 10)
- A. 792
- B. 754
- C. 110
- D. 735
Answer: 754
Explanation: Area = length × width = 29 × 26 = 754 sq m.
3822. If x² - 13x + 36 = 0 are the roots representing two patrol-team sizes, find the larger root. (SI Set 10)
- A. 10
- B. 1
- C. 9
- D. 4
Answer: 9
Explanation: Factorizing: x² - 13x + 36 = (x - 4)(x - 9) = 0. Roots are 4 and 9. Larger root = 9.
3823. In an RPF duty-roster puzzle, if 9x + 41 = 320, find the value of x. (SI Set 10)
- A. 26
- B. 31
- C. 32
- D. 35.56
Answer: 31
Explanation: 9x = 320 - 41 = 279. x = 279/9 = 31.
3824. Simplify: 9 × (3/8) + 8 = ? (SI Set 10)
- A. 10.88
- B. 12.38
- C. 17.38
- D. 11.38
Answer: 11.38
Explanation: 9 × (3/8) = 3.38. Adding 8: 3.38 + 8 = 11.38.
3825. Simplify: 29 + 6 × 4 − 6 + 15 = ? (SI Set 10)
- A. 67
- B. 149
- C. 62
- D. 56
Answer: 62
Explanation: By BODMAS: 6 × 4 = 24 first. Then 29 + 24 - 6 + 15 = 62.
3826. In an RPF stores ledger, 77767 cartridge boxes need to be arranged into equal rows for a divisor of 13. What is the least number of boxes that must be added to make the total exactly divisible by 13? (SI Set 10)
- A. 17
- B. 10
- C. 1
- D. 12
Answer: 12
Explanation: 77767 mod 13 = 1. Number to add = 13 - 1 = 12 (0 if already divisible).
3827. Find the LCM of the patrol shift durations 37 minutes and 20 minutes used for syncing two RPF post schedules. (SI Set 10)
- A. 747
- B. 1
- C. 735
- D. 740
Answer: 740
Explanation: LCM(37, 20) = (37 x 20)/HCF(37,20) = (37 x 20)/1 = 740.
3828. An inlet pipe at a railway water tank can fill it in 11 hours, while an outlet pipe (leak) can empty it in 14 hours. If both are opened together, in how many hours will the tank be filled? (SI Set 10)
- A. 50.33
- B. 52.33
- C. 51.33
- D. 12.5
Answer: 51.33
Explanation: Net rate = 1/11 - 1/14 per hour. Time to fill = 1/(1/11 - 1/14) = 51.33 hours.
3829. RPF constable A can clear a stretch of railway track of debris alone in 15 days, and constable B can do it alone in 17 days. Working together, in how many days will they finish the job? (SI Set 10)
- A. 7.97
- B. 10.97
- C. 5.97
- D. 16
Answer: 7.97
Explanation: Combined time = (A x B)/(A+B) = (15 x 17)/(15+17) = 7.97 days.
3830. 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 92 km downstream. (SI Set 10)
- A. 4
- B. 6
- C. 6.13
- D. 3
Answer: 4
Explanation: Downstream speed = 15 + 8 = 23 km/h. Time = 92/23 = 4 hours.
3831. Two trains of lengths 113 m and 109 m run on parallel tracks in opposite directions at 60 km/h and 27 km/h respectively. Find the time they take to cross each other completely. (SI Set 10)
- A. 24.22
- B. 4.18
- C. 9.18
- D. 12.18
Answer: 9.18
Explanation: Relative speed (opposite dir) = (60+27) x 5/18 = 24.17 m/s. Total length = 113+109 = 222 m. Time = 222/24.17 = 9.18 s.
3832. 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 10)
- A. 12.37
- B. 15.86
- C. 13.86
- D. 6.86
Answer: 13.86
Explanation: Speed = 60 x 5/18 = 16.67 m/s. Time = Length/Speed = 231/16.67 = 13.86 seconds.
3833. An RPF patrol vehicle covers a distance of 669 km at a constant speed of 45 km/h. How much time does it take? (SI Set 10)
- A. 13.37
- B. 15.37
- C. 12.16
- D. 14.87
Answer: 14.87
Explanation: Time = Distance/Speed = 669/45 = 14.87 hours.
3834. The average daily patrol distance of 11 RPF beat officers is 90 km. One officer who covered only 20 km is excluded. Find the new average distance of the remaining officers. (SI Set 10)
- A. 90
- B. 101
- C. 92
- D. 97
Answer: 97
Explanation: Total = 11 x 90 = 990. New total = 990 - 20 = 970. New average = 970/10 = 97.
3835. The average age of 5 RPF SI trainees is 60 years. A new trainee aged 88 years joins the batch. Find the new average age. (SI Set 10)
- A. 60
- B. 64.67
- C. 59.67
- D. 68.67
Answer: 64.67
Explanation: Old total = 5 x 60 = 300. New total = 300 + 88 = 388. New average = 388/6 = 64.67.
3836. If 9 railway signal poles cost the same as 13 units cost 6 units of a fourth quantity (9:13 :: 6:x), find x. (SI Set 10)
- A. 8.67
- B. 3.67
- C. 4.15
- D. 18.67
Answer: 8.67
Explanation: 9:13 :: 6:x => x = (13 x 6)/9 = 8.67.
3837. A sum of ₹680 is divided between two RPF constables in the ratio 3:5 as duty allowance. Find the share of the first constable. (SI Set 10)
- A. 255
- B. 271
- C. 209
- D. 425
Answer: 255
Explanation: First share = [3/(3+5)] x 680 = 255.
3838. Find the compound interest on ₹15000 for 2 years at 5% per annum compounded annually, for a railway pension-fund deposit. (SI Set 10)
- A. 1699.5
- B. 1500
- C. 1343.5
- D. 1537.5
Answer: 1537.5
Explanation: A = P(1+R/100)^T = 15000(1+5/100)^2 = 16537.5. CI = A - P = 16537.5 - 15000 = 1537.5.
3839. A railway cooperative society lends ₹54037 to a staff member at 4% per annum simple interest for 2 years. Find the simple interest. (SI Set 10)
- A. 6484.44
- B. 4593.96
- C. 4108.96
- D. 4322.96
Answer: 4322.96
Explanation: SI = (P x R x T)/100 = (54037 x 4 x 2)/100 = 4322.96.
3840. A railway equipment supplier offers successive discounts of 15% and 5% on a marked price of ₹5989. Find the final selling price. (SI Set 10)
- A. 4836.12
- B. 4871.12
- C. 4791.2
- D. 4811.12
Answer: 4836.12
Explanation: SP = 5989 x (1-15/100) x (1-5/100) = 5989 x 0.85 x 0.95 = 4836.12.
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