Train Questions for SBI PO Prelims and Mains exam. Download SBI PO Train questions on with solutions.<\/p>\n
Download SBI PO Train Questions PDF<\/a><\/p>\n 105 SBI PO Mocks – Just Rs. 249<\/a><\/p>\n Question 1:\u00a0<\/b>A man covers one-third of total distance by train at 60 kmph. Half of the remaining distance he covers by bus at 40 kmph. At what speed he should cover the remaining distance so that he can attain 50 kmph for the entire journey?<\/p>\n a)\u00a048 kmph<\/p>\n b)\u00a055 kmph<\/p>\n c)\u00a0580\/11 kmph<\/p>\n d)\u00a0600\/11 kmph<\/p>\n e)\u00a052 kmph<\/p>\n Question 2:\u00a0<\/b>A local train can travel 50% faster than Rohan\u2019s car. Ram is travelling by the local train. Both start from Dadar at the same time and reach Borivali that\u2019s 30 kms away at the same time. On the way, however, the local train lost 5 minutes while stopping at the stations. Figure out the speed of the car (in km\/hr):<\/p>\n a)\u00a0150<\/p>\n b)\u00a0100<\/p>\n c)\u00a0140<\/p>\n d)\u00a0120<\/p>\n e)\u00a0None of the above.<\/p>\n Question 3:\u00a0<\/b>A cop sees a thief fleeing at a distance of 120 m and starts chasing him. A train of length 300 m running in the opposite direction as that of the thief and the cop, crosses the thief in 15 seconds and the police in 10 seconds. How many seconds after spotting the thief will the cop catch the thief?<\/p>\n a)\u00a024 seconds<\/p>\n b)\u00a018 seconds<\/p>\n c)\u00a06 seconds<\/p>\n d)\u00a012 seconds<\/p>\n e)\u00a015 seconds<\/p>\n Question 4:\u00a0<\/b>Two trains of lengths 172m and 228m are coming towards each other at speeds of 108 Km\/hr and 72 Km\/hr respectively. How long will they take to cross each other completely?<\/p>\n a)\u00a07.2 s<\/p>\n b)\u00a02.2 s<\/p>\n c)\u00a03.4 s<\/p>\n d)\u00a03.6 s<\/p>\n e)\u00a08 s<\/p>\n Question 5:\u00a0<\/b>A train travelling at 90 km\/hr crosses another train which is running in opposite direction in 20 seconds. The length of the second train is half the length of the first train and its speed is 20 percent less than the first train. Find the time which the first train will take to cross a platform of length 300 m.<\/p>\n a)\u00a020 seconds<\/p>\n b)\u00a025 seconds<\/p>\n c)\u00a036 seconds<\/p>\n d)\u00a040 seconds<\/p>\n e)\u00a050 seconds<\/p>\n SBI PO Free Mock Test (Latest Pattern)<\/a><\/p>\n Download SBI PO Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>Trains A, B and platform P are of same length. Train A takes 20 seconds to cross the platform completely while train B takes 30 seconds to cross the same. How much time will the trains take to cross each other if they are running in the opposite direction?<\/p>\n a)\u00a010 seconds<\/p>\n b)\u00a012 seconds<\/p>\n c)\u00a015 seconds<\/p>\n d)\u00a08 seconds<\/p>\n e)\u00a016 seconds<\/p>\n Question 7:\u00a0<\/b>Train B takes twice as much time as train A to cross the same platform. If the speeds of both the trains is the same and the length of platform is half the length of train A, what is the ratio of the length of train A to train B?<\/p>\n a)\u00a02:3<\/p>\n b)\u00a01:5<\/p>\n c)\u00a03: 5<\/p>\n d)\u00a02:5<\/p>\n e)\u00a01:3<\/p>\n Question 8:\u00a0<\/b>2 trains are of equal length. They take 120 seconds to cross each other while running in the same direction. If the ratio of the speeds of the 2 trains is 3:2, the time taken by the trains to cross each other completely while running in the opposite directions is<\/p>\n a)\u00a060 seconds<\/p>\n b)\u00a048 seconds<\/p>\n c)\u00a030 seconds<\/p>\n d)\u00a012 seconds<\/p>\n e)\u00a024 seconds<\/p>\n Question 9:\u00a0<\/b>Two trains of length 200 m and 160 m can cross each other in 30 seconds when they are running in the same direction. It is also known that if they are running in the opposite directions then they can cross each other in 15 seconds. What is the speed of the faster train?<\/p>\n a)\u00a072 km\/hr<\/p>\n b)\u00a064.8 km\/hr<\/p>\n c)\u00a090 km\/hr<\/p>\n d)\u00a083.33 km\/hr<\/p>\n e)\u00a054 km\/hr<\/p>\n Question 10:\u00a0<\/b>A train 200 m long can cross an electric pole in 10 seconds. How much time will it take to overtake another train which is running in the same direction on a parallel track? It is known that the length of the second train is twice that of first and its speed is half the speed of the first train.<\/p>\n a)\u00a01 minute<\/p>\n b)\u00a01 minute 20 seconds<\/p>\n c)\u00a01 minute 30 seconds<\/p>\n d)\u00a045 seconds<\/p>\n e)\u00a01 minute 45 seconds<\/p>\n SBI PO Important Questions (download PDF)<\/a><\/p>\n Question 11:\u00a0<\/b>A train X runs at a speed of 36 kmph. It crosses another train Y running in the opposite direction with the same speed in 80 seconds. Also, the train crosses a platform of length 500 m in 100 seconds. The length of train Y is<\/p>\n a)\u00a0800 m<\/p>\n b)\u00a0900 m<\/p>\n c)\u00a01000 m<\/p>\n d)\u00a01100 m<\/p>\n e)\u00a01200 m<\/p>\n Question 12:\u00a0<\/b>Two trains are of equal length. The speed of one train is 50% more than the other. If the 2 trains were running in the opposite directions, they will cross each other completely in 100 seconds. What will be the time taken by the 2 trains to cross each other if they are running in the same direction?<\/p>\n a)\u00a0150 seconds<\/p>\n b)\u00a0200 seconds<\/p>\n c)\u00a0250 seconds<\/p>\n d)\u00a0500 seconds<\/p>\n e)\u00a01000 seconds<\/p>\n Question 13:\u00a0<\/b>2 trains A and B start moving towards each other. Both of them cross a platform in 10 seconds. If train A is twice as long as train B and the length of the platform is half the length of train A, what will be the time taken by trains A and B to completely cross each other?<\/p>\n a)\u00a04 seconds<\/p>\n b)\u00a06 seconds<\/p>\n c)\u00a08 seconds<\/p>\n d)\u00a010 seconds<\/p>\n e)\u00a012 seconds<\/p>\n Question 14:\u00a0<\/b>A 200 m long train can cross a pole in 10 seconds. Another train of same length can cross a 400 m long platform in 60 seconds. If both the trains are travelling in opposite directions then how much time will they take to cross each other completely?<\/p>\n a)\u00a020 seconds<\/p>\n b)\u00a015 seconds<\/p>\n c)\u00a012 seconds<\/p>\n d)\u00a025 seconds<\/p>\n e)\u00a013.33 seconds<\/p>\n Question 15:\u00a0<\/b>A 260 meter long train crosses a 120 meter long wall in 19 seconds .What is the speed of the train?<\/p>\n a)\u00a027 km\/hr<\/p>\n b)\u00a049 km\/hr<\/p>\n c)\u00a072 km\/hr<\/p>\n d)\u00a070 km\/hr<\/p>\n e)\u00a0None of these<\/p>\n Question 16:\u00a0<\/b>A train running at speed of 120 kmph crosses a signal in 15 seconds .What is the length of the train in meters?<\/p>\n a)\u00a0300<\/p>\n b)\u00a0200<\/p>\n c)\u00a0500<\/p>\n d)\u00a0Cannot determined<\/p>\n e)\u00a0None of these<\/p>\n Question 17:\u00a0<\/b>A 240-metre long train running at the speed of 60 kmph will take how much time to cross another 270-metre long train running in opposite direction at the speed of 48 kmph?<\/p>\n a)\u00a017 seconds<\/p>\n b)\u00a03 seconds<\/p>\n c)\u00a012 seconds<\/p>\n d)\u00a08 seconds<\/p>\n e)\u00a0None of these<\/p>\n Question 18:\u00a0<\/b>At 60% of its usual speed, a train of length L metres crosses a platform 240 metre long in 15 seconds. At its usual speed, the train crosses a pole in 6 seconds. What is the value of L (in metre)?<\/p>\n a)\u00a0270<\/p>\n b)\u00a0225<\/p>\n c)\u00a0220<\/p>\n d)\u00a0480<\/p>\n e)\u00a0240<\/p>\n Question 19:\u00a0<\/b>At its usual speed, a train of length L metres crosses platform 300 metre long in 25 seconds. At 50% of its usual speed, the train crosses a vertical pole in 20 seconds. What is the value of L?<\/p>\n a)\u00a0160<\/p>\n b)\u00a0260<\/p>\n c)\u00a0200<\/p>\n d)\u00a0310<\/p>\n e)\u00a0350<\/p>\n Question 20:\u00a0<\/b>At its usual speed, a 150 metre long train crosses a platform of length L metres in 24 seconds. AT 75% of its usual speed, the train crosses a vertical pole in 12 seconds. What is the value of L?<\/p>\n a)\u00a0250<\/p>\n b)\u00a0225<\/p>\n c)\u00a0240<\/p>\n d)\u00a0260<\/p>\n e)\u00a0280<\/p>\n Download SBI PO Syllabus PDF<\/a><\/p>\n Free Banking Online Tests<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the distance be \u2018d\u2019 that has to be covered. Remaining distance = d – d\/3 = 2d\/3 Remaining distance = 2d\/3 – d\/3 = d\/3 For entire distance ‘d’<\/p>\n $\\Rightarrow$ 50 = $\\frac{d}{\\frac{d}{180}+\\frac{d}{120}+\\frac{d}{3x}}$ 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let assume Car\u2019s speed = S km\/hr. So, we can say that<\/p>\n $ \\frac{30}{S} – \\frac{30}{1.5S} = \\frac{1}{12}$<\/p>\n $\\frac{10}{S} = \\frac{1}{12}$<\/p>\n S = 120 kmph<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n We do not know the speed of the cop or the thief. But, using the train, we can find the difference between the relative speeds of the cop and the thief.<\/p>\n Sum of the speeds of the train and the thief = 300\/15 = 20 m\/s Difference between the speeds of the cop and the thief = 30-20 = 10 m\/s. 4)\u00a0Answer\u00a0(E)<\/strong><\/p>\n The speeds of the trains are 30 m\/s and 20 m\/s respectively. The length to traversed to completely cross each other = 400 m<\/p>\n Time = $\\frac{400}{50}$ = 8 seconds<\/p>\n EndGroup:<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Speed of first train = 90 km\/hr = 25 m\/s Take a free mock test for SBI PO<\/a><\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the length of train A, train B and platform be l. When they trains run towards each other, the relative speed will be the sum of the individual speeds. Total length the trains must travel to cross each other completely = length of train A + length of train B = l+l = 2l. 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the length of train A be l and the length of train B be d. 8)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Let the length of the trains be \u2018l\u2019. When they run in the opposite direction, the relative speed will be 3x+2x = 5x. Therefore, option E is the right answer.<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The total distance that the trains need to cover to completely cross each other is 200 + 160 = 360 m 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The train takes 10 seconds to cross the electric pole. Hence, the speed of the train will be 200\/10 = 20 m\/s 100 Free GK Tests for Banking exams<\/a><\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Speed of the train = 36 kmph = 36*5\/18 = 10 m\/s. 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the length of one train be \u2018l\u2019. If the trains run in the same direction, relative velocity will be 1.5s – s = 0.5s 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the length of train A be 2d. Time taken by train A to cross 2d+d = Time taken by train B to cross d+d = 10 seconds. Relative speed will be the sum of the speed of the 2 trains since they are moving towards each other.<\/p>\n Relative speed = 3d\/10 + 2d\/10 = 5d\/10 = d\/2.<\/p>\n Time taken = 3d\/(d\/2) = 6 seconds. Therefore, option B is the right answer.<\/p>\n 14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n The length of the train is 200 m and it takes 10 seconds to cross a pole. Hence, the speed of the train will be 200\/10 = 20 m\/s 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Length of the train is 260 metres<\/p>\n Length of the wall is 120 metres<\/p>\n total is 260+120 = 380 metres<\/p>\n Time taken is 19 seconds.<\/p>\n Hence, the speed is 380\/19 = 20 m\/s = 72 Km\/hr<\/p>\n Answer is option C<\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Speed of train = 120 kmph<\/p>\n = $(120 \\times \\frac{5}{18})$ m\/s = $\\frac{100}{3}$ m\/s<\/p>\n Let length of the train = $l$ meters<\/p>\n Using, speed = distance\/time<\/p>\n => $\\frac{100}{3} = \\frac{l}{15}$<\/p>\n => $l=\\frac{100}{3} \\times 15$<\/p>\n => $l=100 \\times 5=500$ meters<\/p>\n => Ans – (C)<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Length of first train = 240 m and second train = 270 m<\/p>\n Total length of the two trains = 240 + 270 = 510 m<\/p>\n Speed of first train = 60 kmph and second train = 48 kmph<\/p>\n Since, the trains are moving in opposite direction, thus relative speed = 60 + 48 = 108 kmph<\/p>\n = $(108 \\times \\frac{5}{18})$ m\/s = $30$ m\/s<\/p>\n Let time taken = $t$ seconds<\/p>\n Using, time = distance\/speed<\/p>\n => $t=\\frac{510}{30}=17$ seconds<\/p>\n => Ans – (A)<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let speed of the train = $10x$ m\/s<\/p>\n Length of train = $l$ m<\/p>\n Time taken to cross the pole = 6 sec<\/p>\n Using, $speed = \\frac{distance}{time}$<\/p>\n => $10x = \\frac{l}{6}$<\/p>\n => $x = \\frac{l}{60}$<\/p>\n Now, 60% of the speed = $\\frac{60}{100} \\times 10x = 6x$ m\/s<\/p>\n Length of platform = 240 m<\/p>\n Acc. to ques, => $6x = \\frac{240 + l}{15}$<\/p>\n => $6 \\times \\frac{l}{60} = \\frac{240 + l}{15}$<\/p>\n => $\\frac{l}{10} = \\frac{240 + l}{15}$<\/p>\n => $15l = 2400 + 10l$<\/p>\n => $15l – 10l = 5l = 2400$<\/p>\n => $l = \\frac{2400}{5} = 480$ m<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let usual speed of the train = $10x$ m\/s<\/p>\n Now, 50% of the speed = $\\frac{50}{100} \\times 10x = 5x$ m\/s<\/p>\n Length of train = $l$ m<\/p>\n Time taken to cross the pole = 20 sec<\/p>\n Using, $speed = \\frac{distance}{time}$<\/p>\n => $5x = \\frac{l}{20}$<\/p>\n => $x = \\frac{l}{100}$<\/p>\n Length of platform = 300 m<\/p>\n Acc. to ques, => $10x = \\frac{300 + l}{25}$<\/p>\n => $10 \\times \\frac{l}{100} = \\frac{300 + l}{25}$<\/p>\n => $\\frac{l}{10} = \\frac{300 + l}{25}$<\/p>\n => $25l = 3000 + 10l$<\/p>\n => $25l – 10l = 15l = 3000$<\/p>\n => $l = \\frac{3000}{15} = 200$ m<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let speed of the train = $20x$ m\/s<\/p>\n Now, 75% of the speed = $\\frac{75}{100} \\times 20x = 15x$ m\/s<\/p>\n Length of train = 150 m<\/p>\n Time taken to cross the pole = 12 sec<\/p>\n Using, $speed = \\frac{distance}{time}$<\/p>\n => $15x = \\frac{150}{12}$<\/p>\n => $x = \\frac{10}{12} = \\frac{5}{6}$<\/p>\n Length of platform = $l$ m<\/p>\n Acc. to ques, => $20x = \\frac{150 + l}{24}$<\/p>\n => $20 \\times \\frac{5}{6} = \\frac{150 + l}{24}$<\/p>\n => $\\frac{50}{3} = \\frac{150 + l}{24}$<\/p>\n => $150 + l = 400$<\/p>\n => $l = 400 – 150 = 250$ m<\/p>\n
\nSo the distance covered by the train = d\/3
\nTime taken to cover d\/3 distance by train = $\\frac{\\frac{d}{3}}{60}$ = d\/180 hr<\/p>\n
\nDistance covered by the bus = $\\frac{1}{2}*\\frac{2d}{3}$ = d\/3
\nTime taken to cover d\/3 distance by the bus = $\\frac{\\frac{d}{3}}{40}$ = d\/120 hr<\/p>\n
\nLet assume that the remaining distance is covered by x kmph
\nTime taken to cover this distance = $\\frac{\\frac{d}{3}}{x}$ = d\/3x hr<\/p>\n
\n$\\Rightarrow$ 50 = $\\frac{360x}{2x+3x+120}$
\n$\\Rightarrow$ x = 600\/11 kmph<\/p>\n
\nSo local train\u2019s speed will be = 1.5*S km\/hr
\nIt is given that both the vehicle covered same distance = 30 kms and Car took 5 minutes more as compare to the local train.<\/p>\n
\nSum of the speeds of the train and the cop = 300\/10 = 30 m\/s<\/p>\n
\nThe cop will take 120\/10 = 12 seconds to catch the thief after spotting him.
\nTherefore, option D is the right answer.<\/p>\n
\nThe relative speed = 50 m\/s<\/p>\n
\nLet the length of first train be 2l. So the length of second train will be l.
\nSpeed of second train = 90*.8 = 72 km\/hr = 20 m\/s
\nHence, total distance travelled in 20 s = 45*20 = 900 m
\n3l = 900 m
\n=> l = 300 m
\nHence, the length of first train is 600 m.
\nTime taken to cross a platform of length 300 m = 900 \/ 25 = 36 seconds.
\nHence, option C is the correct answer.<\/p>\n
\nTotal length a train has to traverse to cross the platform = length of the train + length of the platform = l+ l = 2l.
\nSpeed of train A = 2l\/20
\nSpeed of train B = 2l\/30<\/p>\n
\n=> Relative speed = 2l\/20 + 2l\/30 = 10l\/60<\/p>\n
\n=> Time taken by the trains to completely cross each other = 2l\/(10l\/60) = 12 seconds.
\nTherefore, option B is the right answer.<\/p>\n
\nLength of platform = l\/2 = 0.5l
\nSpeed of both the trains is same.
\nA takes twice as much time as B to cross the platform.
\n(l+0.5l)\/t = (d+0.5l)\/2t
\n2l+l = d + 0.5l
\n2.5l = d
\n=> l\/d = 2\/5.
\nTherefore, option D is the right answer.<\/p>\n
\nWhen they are running in the same direction, they take 120 seconds to cross each other.
\n=> (l+l)\/(3x-2x) = 120
\n2l\/x = 120 seconds.<\/p>\n
\n=>(l+l)\/(3x+2x) = ?
\n2l\/5x = ?
\nWe know that 2l\/x = 120.
\n=> 2l\/5x = 120\/5 = 24 seconds.<\/p>\n
\nNow, let the speed of the faster and slower trains be f and s
\nWhen they are travelling in same direction, the effective speed becomes f – s
\nHence, we have
\nf – s = 360\/30 = 12
\nSimilarly, when they are travelling in same direction the effective speed becomes f + s
\nHence,
\nf + s = 360\/15 = 24
\nSolving the two equations we get
\n2f = 36
\nf = 18 m\/s = 64.8 km\/hr<\/p>\n
\nNow the length of second train is 400 m and its speed is 10 m\/s.
\nHence, the effective relative speed is 10 m\/s and the total distance to be covered is 600 m.
\nHence, the required time will be 600\/10 = 60 seconds.<\/p>\n
\nLet the length of train X be x and length of train Y be y.
\nWe know that train X crosses a platform of length 500 m in 100 seconds.
\n(x+500)\/10 = 100
\nx+500 = 1000
\nx = 500 m.
\nBoth trains X and Y run at the same speed but in the opposite direction.
\nTherefore, (x+y)\/(10+10) = 80
\n500+y = 1600 m
\n=> y = 1100 m.
\nTherefore, option D is the right answer.<\/p>\n
\nLet the speed of one train be \u2018s\u2019 and the speed of the other train be 1.5s.
\nIt has been given that 2l\/(s+1.5s) = 100 (Running in opposite directions)
\n2l\/s = 250 seconds ———–(1).<\/p>\n
\n2l\/0.5s = 2*(2l\/s) = 2*250 = 500 seconds.
\nTherefore, option D is the right answer.<\/p>\n
\nLength of train B = d.
\nLength of the platform = d.<\/p>\n
\nTherefore, their speeds are in the ratio 3:2.
\nSpeed of train A = 3d\/10 and speed of train B = 2d\/10.
\nWhen trains A and B are moving towards each other, total distance to cross will be the sum of the lengths of the trains = d+2d = 3d.<\/p>\n
\nThe length of second train is also 200 m. The length of the platform is 400 m. So the total distance covered is 600 m.
\nTime taken is 60 seconds. So speed of the second train is 600\/60 = 10 m\/s
\nNow both the trains are moving in opposite directions. So their speeds will add up. Hence, required time to cross will be 400\/30
\n= 13.33 seconds<\/p>\n