Download SSC CGL Train questions with answers PDF based on previous papers very useful for SSC CGL exams. 20 Very important Train on objective questions (MCQ’s) for SSC exams.<\/p>\n
Download Train Problems For SSC CGL<\/a><\/p>\n 25 SSC CGL Mocks – Just Rs. 149<\/a><\/p>\n FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n Instructions<\/b><\/p>\n Question 1:\u00a0<\/b>Two trains of length 180 meters with velocity 30 m\/s and 60 m\/s travel in opposite direction and if the initial distance between them is 0.54 kilometres then what is the time taken for their tails to cross each other ?<\/p>\n a)\u00a05 sec<\/p>\n b)\u00a07 sec<\/p>\n c)\u00a08 sec<\/p>\n d)\u00a010 sec<\/p>\n Question 2:\u00a0<\/b>A man travels with a speed of 15 m\/s to reach the destination 5 seconds late and travels with 20 m\/s to reach the destination 5 seconds early.What is the original distance to be travelled ?<\/p>\n a)\u00a05.85 km<\/p>\n b)\u00a05.15 km<\/p>\n c)\u00a05.25 km<\/p>\n d)\u00a05.35 km<\/p>\n Question 3:\u00a0<\/b>A man travels half of the distance with speed \u2018v\/2\u2019 and one fourth of the distance with \u2018v\u2019 and remaining distance with \u20182v\u2019.What is the average speed of the journey ?<\/p>\n a)\u00a08v\/13<\/p>\n b)\u00a08v\/15<\/p>\n c)\u00a08v\/17<\/p>\n d)\u00a08v\/19<\/p>\n Question 4:\u00a0<\/b>A boat takes 4 hours to travel same distance \u2018x\u2019 upstream and downstream.If the speed of the boat is 4 km\/hr and stream is 2 km\/hr then find the value of x ?<\/p>\n a)\u00a06 km<\/p>\n b)\u00a08 km<\/p>\n c)\u00a04 km<\/p>\n d)\u00a010 km<\/p>\n Question 5:\u00a0<\/b>Two trains start from stations A and B which are 280 km apart towards each other with their speeds in the ratio of 2:5.If they meet after 1 hour 20 minutes then what is the difference between the speeds ?<\/p>\n a)\u00a045 km\/hr<\/p>\n b)\u00a060 km\/hr<\/p>\n c)\u00a075 km\/hr<\/p>\n d)\u00a090 km\/hr<\/p>\n SSC CGL Previous Papers Download PDF<\/a><\/p>\n SSC CHSL PREVIOUS PAPERS DOWNLOAD<\/a><\/p>\n Instructions<\/b><\/p>\n Question 6:\u00a0<\/b>Two trains of length 180 meters with velocity 30 m\/s and 60 m\/s travel in opposite direction and if the initial distance between them is 0.54 kilometres then what is the time taken for their tails to cross each other ?<\/p>\n a)\u00a05 sec<\/p>\n b)\u00a07 sec<\/p>\n c)\u00a08 sec<\/p>\n d)\u00a010 sec<\/p>\n Question 7:\u00a0<\/b>A train travelled from A to B at a speed of 60 km\/hr and returned from B to A at a speed of 40 km\/hr. Find the average speed of the train.<\/p>\n a)\u00a050 km\/hr<\/p>\n b)\u00a048 km\/hr<\/p>\n c)\u00a042 km\/hr<\/p>\n d)\u00a056 km\/hr<\/p>\n Question 8:\u00a0<\/b>A train travels a certain distance in a certain time. If it covers triple the distance in double the time, then find the ratio between original speed and current speed.<\/p>\n a)\u00a01:3<\/p>\n b)\u00a02:3<\/p>\n c)\u00a03:5<\/p>\n d)\u00a02:5<\/p>\n Question 9:\u00a0<\/b>A train crosses a platform of length 204 metres in 36 seconds. It crosses a man standing on the platform in 12 seconds. Find the speed of the train.<\/p>\n a)\u00a0102 m\/sec<\/p>\n b)\u00a0127 m\/sec<\/p>\n c)\u00a060 m\/sec<\/p>\n d)\u00a0168 m\/sec<\/p>\n Instructions<\/b><\/p>\n Question 10:\u00a0<\/b>Two trains start from stations A and B which are 200 km apart towards each other with their speeds in the ratio of 1:2.If they meet after 2 hours 40 minutes then what is the difference between the speeds ?<\/p>\n a)\u00a015 km\/hr<\/p>\n b)\u00a020 km\/hr<\/p>\n c)\u00a025 km\/hr<\/p>\n d)\u00a030 km\/hr<\/p>\n 18000+ Questions – Free SSC Study Material<\/a><\/p>\n Question 11:\u00a0<\/b>A train travels from Delhi to Mumbai at a speed of 60 km\/hr and returns at a speed of 75 km\/hr. Find its average speed.<\/p>\n a)\u00a066.67 km\/hr<\/p>\n b)\u00a045 km\/hr<\/p>\n c)\u00a060 km\/hr<\/p>\n d)\u00a067.5 km\/hr<\/p>\n Question 12:\u00a0<\/b>A train of length 120 m crossed a platform in 36 seconds. It crosses a man standing on the platform in 24 seconds running at the same speed. Find the length of the platform.<\/p>\n a)\u00a0120 m<\/p>\n b)\u00a060 m<\/p>\n c)\u00a0104 m<\/p>\n d)\u00a072 m<\/p>\n Question 13:\u00a0<\/b>A train crosses a pole in 12 seconds and a platform of length 160 m in 32 seconds running at the same speed. Find the length of the train.<\/p>\n a)\u00a0108 m<\/p>\n b)\u00a096 m<\/p>\n c)\u00a0120 m<\/p>\n d)\u00a084 m<\/p>\n Question 14:\u00a0<\/b>A train travels a distance of 480 km at uniform speed. Due to breakdown, its speed is reduced by 20 km\/hr and hence it travels the destination 4 hours late. Find the initial speed of the train.<\/p>\n a)\u00a060 km\/hr<\/p>\n b)\u00a040 km\/hr<\/p>\n c)\u00a025 km\/hr<\/p>\n d)\u00a035 km\/hr<\/p>\n Question 15:\u00a0<\/b>A train travels at a speed of 125 km\/hr without stoppages and at 100 km\/hr with stoppages. How many minutes per hour does the train stop?<\/p>\n a)\u00a025 min<\/p>\n b)\u00a020 min<\/p>\n c)\u00a015 min<\/p>\n d)\u00a012 min<\/p>\n SSC CGL Free Mock Test<\/a><\/p>\n SSC CHSL Free Mock Test<\/a><\/p>\n Question 16:\u00a0<\/b>A train travelled at a speed of 15 km\/hr and reached destination 20 minutes late. Had the speed increased to 20 km\/hr, it would reach the destination 20 minutes early. Find the distance travelled by the train.<\/p>\n a)\u00a030 km<\/p>\n b)\u00a028 km<\/p>\n c)\u00a040 km<\/p>\n d)\u00a045 km<\/p>\n Question 17:\u00a0<\/b>A person walked at a speed of 4 kmph while going to office and returned at a speed of 3 kmph. Find the average speed.<\/p>\n a)\u00a03.6 kmph<\/p>\n b)\u00a03.4 kmph<\/p>\n c)\u00a03.8 kmph<\/p>\n d)\u00a03.5 kmph<\/p>\n Question 18:\u00a0<\/b>A train travelled from Chennai to Hyderabad at a speed of 54 kmph and returned to Chennai at a speed of 36 kmph. Find its average speed.<\/p>\n a)\u00a042.5 kmph<\/p>\n b)\u00a043 kmph<\/p>\n c)\u00a043.2 kmph<\/p>\n d)\u00a045 kmph<\/p>\n Question 19:\u00a0<\/b>A train travels from Delhi to Agra at a speed of 72 km\/hr and returns at a speed of 90 km\/hr. Find the average speed.<\/p>\n a)\u00a078 km\/hr<\/p>\n b)\u00a081 km\/hr<\/p>\n c)\u00a080 km\/hr<\/p>\n d)\u00a092 km\/hr<\/p>\n Question 20:\u00a0<\/b>A train travels a distance of 480 km at uniform speed. Due to breakdown, its speed is reduced by 10 km\/hr and hence it travels the destination 8 hours late. Find the initial speed of the train.<\/p>\n a)\u00a030 km\/hr<\/p>\n b)\u00a020 km\/hr<\/p>\n c)\u00a025 km\/hr<\/p>\n d)\u00a035 km\/hr<\/p>\n 1500+ Free SSC Questions & Answers<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n As the both are travelling in opposite direction the relative velocity will be sum of the velocities i.e 30+60=90 m\/s 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n let the original time be t seconds 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the total distance be x 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Upstream speed =4-2=2 km\/hr 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n let the speeds be 2x and 5x. 100+ Free GK Tests for SSC Exams<\/a><\/p>\n Download Free GK PDF<\/a><\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n As the both are travelling in opposite direction the relative velocity will be sum of the velocities i.e 30+60=90 m\/s 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the distance between A and B be 120 km each (LCM of 60 and 40). 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the distance travelled by the train be D km. 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the length of the train be \u2018x\u2019 metres. Here, Speeds are equal. \u21d2 $x+204 = 3x$ 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n let the speeds be x and 2x. 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the total distance be 300 km (LCM of 60 and 75). 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Speed of the train to cross the man = 120\/24 = 5 m\/sec 15000 Questions – Free SSC Study Material<\/a><\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the length of the train be \u2018T\u2019 metres. \u21d2 $8T = 3T + 480$ 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given that the distance $= 480$ km 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Due to stoppages, train travelled 25 km less in an hour. 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the distance travelled be \u2018D\u2019 km. \u21d2 $\\dfrac{3D-2D}{60} = \\dfrac{2}{3}$<\/p>\n \u21d2 $\\dfrac{D}{60} = \\dfrac{2}{3}$<\/p>\n \u21d2 $D = 40 km$<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the onward distance and return distance be 12 km each (LCM of 4 and 3) 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the onward distance and return distance be 108 km each (LCM of 54 and 36). 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the distance between Delhi and Agra be 360 km (LCM of 72 and 90) 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given that the distance $= 480$ km
\nTotal distance to be travelled is length of the trains i.e 180+180=360 m and also the distance between them i.e 0.54 km=540 m
\nTotal=540+360=900 m
\nTime =900\/90
\n=10 seconds<\/p>\n
\nSo we have 15(t+5)=20(t-5)
\n15t+75=20t-100
\n5t=175 sec
\nt=35 sec
\nTotal distance=15*35
\n=5250 m
\n=5.25 km<\/p>\n
\nTotal time=(x\/(v\/2))+(x\/(4v))+(x\/(4*2v)
\n=19x\/8v
\nAverage speed=Total distance\/Total time
\n=x\/(19x\/8v)
\n=8v\/19<\/p>\n
\nDownstream speed=4+2=6 km\/hr
\nTherefore we have $\\frac{x}{6}+\\frac{x}{2}$=6
\n4x\/6 =4
\nx=6 km<\/p>\n
\nAs both the trains are moving towards each other their relative velocity will be 2x+5x=7x
\nTherefore 280\/7x =4\/3
\nx=120\/4
\nx=30 km\/hr
\ndIfference=5x-2x
\n=3x
\n=90 km\/hr<\/p>\n
\nTotal distance to be travelled is length of the trains i.e 180+180=360 m and also the distance between them i.e 0.54 km=540 m
\nTotal=540+360=900 m
\nTime =900\/90
\n=10 seconds<\/p>\n
\nTime taken by the train to travel 120 km at 60 km\/hr = 2 hours
\nTime taken by the train to travel 120 km at 40 km\/hr = 3 hours
\nTotal distance = 120+120 = 240 km
\nTotal time = 2+3 = 5 hours
\nAverage speed $= \\dfrac{\\text{Total Distance}}{\\text{Total Time}} = \\dfrac{240}{5} = 48 km\/hr$<\/p>\n
\nLet the time taken by the train to travel D km be T hours.
\nOriginal speed = $\\dfrac{D}{T}$ km\/hr
\nNew Distance = 3D km
\nNew time = 2T hours
\nNew speed = $\\dfrac{3D}{2T}$ km\/hr
\nRatio between original speed and new speed = $\\dfrac{D}{T} : \\dfrac{3D}{2T} = 1 : \\dfrac{3}{2} = 2:3$<\/p>\n
\nThen, Speed required to travel (x+204) metres in 36 seconds = $\\dfrac{x+204}{36}$ m\\sec
\nSpeed required to travel x metres in 12 seconds = $\\dfrac{x}{12}$ m\/sec<\/p>\n
\n\u21d2 $\\dfrac{x+204}{36} = \\dfrac{x}{12}$<\/p>\n
\n\u21d2 $2x = 204$
\n\u21d2 $x = 102$ metres
\nHence, the Length of the train = 102 metres.<\/p>\n
\nAs both the trains are moving towards each other their relative velocity will be x+2x=3x
\nTherefore 200\/3x =8\/3
\nx=200\/8
\nx=25 km\/hr
\ndIfference=2x-x
\n=x
\n=25 km\/hr<\/p>\n
\nTime required to travel 300 km at 60 km\/hr = 300\/60 = 5 hours
\nTime required to travel 300 km at 75 km\/hr = 300\/75 = 4 hours
\nTotal time to travel 600 km = 9 hours
\nTotal distance = 600 km
\nAverage speed = 600\/9 = 200\/3 = 66.67 km\/hr<\/p>\n
\nSpeed of the train to cross the platform = (120+P)\/36
\n$\\dfrac{120+P}{36} = 5$
\n\u21d2 $120+P = 180$
\n\u21d2 $P = 60$<\/p>\n
\nSpeed of the train to cross the pole in 12 seconds = T\/12 m\/sec
\nSpeed of the train to cross the platform in 32 seconds = (T+160)\/32 m\/sec
\nHere, Speeds are equal.
\n\u21d2 $\\dfrac{T}{12} = \\dfrac{T+160}{32}$<\/p>\n
\n\u21d2 $5T = 480$
\n\u21d2 $T = 96 m$
\nHence, The Length of the train = 96 m<\/p>\n
\nLet the speed of the train $= x$ km\/hr
\nThen, time in which it can travel $480$ km $= \\dfrac{480}{x}$ hr
\nNow, Speed is reduced by $20$ km\/hr
\nReduced speed $= (x-20)$ km\/hr
\nTime in which it can travel $480$ km at reduced speed $= \\dfrac{480}{(x-20)}$
\nDifference in time = 4 hours
\n\u21d2 $\\dfrac{480}{(x-20)} – \\dfrac{480}{x} = 4$
\n\u21d2 $\\dfrac{480(x-x+20)}{x^2-20x} = 4$
\n\u21d2 $9600 = 4(x^2-20x)$
\n\u21d2 $x^2-20x-2400 = 0$
\n\u21d2 $x^2-60x+40x-2400 = 0$
\n\u21d2 $x(x-60)+40(x-60) = 0$
\n\u21d2 $(x+40)(x-60) = 0$
\nx+40 = 0 | x-60 = 0
\nx = -40 | x = 60
\nSpeeds cannot be negative.
\nHence, Initial speed = 60 km\/hr<\/p>\n
\nTime required to travel 25 km in an hour $ = \\dfrac{25}{125}\\times 60 = 12$ minutes.<\/p>\n
\nTime taken to travel D km at 15 kmph = D\/15 hours
\nTime taken to travel D km at 20 kmph = D\/20 hours
\nGiven the difference between time = 40 minutes = $\\dfrac{40}{60} = \\dfrac{2}{3}$
\n\u21d2 $\\dfrac{D}{15} – \\dfrac{D}{20} = \\dfrac{2}{3}$<\/p>\n
\nTime taken to travel 12 km at a speed of 4 kmph = 12\/4 = 3 hours
\nTime taken to travel 12 km at a speed of 3 kmph = 12\/3 = 4 hours
\nTotal time taken by the train to travel 24 km = 3+4 = 7 hours
\nAverage speed = (Total Distance)\/(Total Time) = 24\/7 = 3.4 kmph<\/p>\n
\nTime taken to travel 108 km at a speed of 54 kmph = 108\/54 = 2 hours
\nTime taken to travel 108 km at a speed of 36 kmph = 108\/36 = 3 hours
\nTotal time taken by the train to travel 216 km = 2+3 = 5 hours
\nAverage speed = (Total Distance)\/(Total Time) = 216\/5 = 43.2 kmph<\/p>\n
\nTime taken to travel from Delhi to Agra at 72 km\/hr = 360\/72 = 5 hours
\nTime taken to travel from Agra to Delhi at 90 km\/hr = 360\/90 = 4 hours
\nTotal time = 5+4 = 9 hours
\nTotal distance = 360+360 = 720 km
\nAverage speed = $= \\dfrac{\\text{Total distance}}{\\text {Total time}} = \\dfrac{720}{9} = 80 \\text{km\/hr}$<\/p>\n
\nLet the speed of the train $= x$ km\/hr
\nThen, time in which it can travel $480$ km $= \\dfrac{480}{x}$ hr
\nNow, Speed is reduced by $10$ km\/hr
\nReduced speed $= (x-10)$ km\/hr
\nTime in which it can travel $480$ km at reduced speed $= \\dfrac{480}{(x-10)}$
\nDifference in time = 8 hours
\n\u21d2 $\\dfrac{480}{(x-10)} – \\dfrac{480}{x} = 8$
\n\u21d2 $\\dfrac{480(x-x+10)}{x^2-10x} = 8$
\n\u21d2 $4800 = 8(x^2-10x)$
\n\u21d2 $x^2-10x-600 = 0$
\n\u21d2 $x^2-30x+20x-600 = 0$
\n\u21d2 $x(x-30)+20(x-30) = 0$
\n\u21d2 $(x+20)(x-30) = 0$
\nx+20 = 0 | x-30 = 0
\nx = -20 | x = 30
\nSpeeds cannot be negative.
\nHence, Initial speed = 30 km\/hr<\/p>\n