Time Speed Distance is an important topic in the Quant section of the SNAP Exam. You can also download this Free Time Speed Distance Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Time Speed Distance questions in the SNAP exam. Utilize this PDF practice set, <\/strong>which is one of the best sources for practicing.<\/p>\n Download Time Speed Distance Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\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 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/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 Question 2:\u00a0<\/b>A boat takes 2 hours to travel from point A to B in still water .To find out it\u2019s speed up-stream ,which of the following information is needed. a)\u00a0All are required<\/p>\n b)\u00a0Even these we cannot found the answer<\/p>\n c)\u00a0Only i,iii, and either ii or iv<\/p>\n d)\u00a0Only i and iii<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Time taken by boat\u00a0to travel from point A to B in still water\u00a0= 2 hours<\/p>\n To find the upstream speed, we definitely need the speed of stream, thus statement (iii) is mandatory.<\/p>\n Also, the distance between points A and B or the speed of boat in still water is needed.<\/p>\n Thus, statements (i) and (iii) are required to find the upstream speed of the boat.<\/p>\n => Ans – (D)<\/p>\n Question 3:\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 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/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 Question 4:\u00a0<\/b>A bus covers a distance of 2,924 km,in 43 hours .what is the bus speed?<\/p>\n a)\u00a072 km\/hr<\/p>\n b)\u00a060 km\/hr<\/p>\n c)\u00a068 km\/hr<\/p>\n d)\u00a0cannot determined<\/p>\n e)\u00a0none of these<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of bus = $s$ km\/hr<\/p>\n Distance covered = 2924 km<\/p>\n Time taken = 43 hours<\/p>\n Using speed = distance\/time<\/p>\n => $s=\\frac{2924}{43}=68$ km\/hr<\/p>\n => Ans – (C)<\/p>\n Question 5:\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 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/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 Question 6:\u00a0<\/b>A man takes 2.2 times as long to row a distance upstream as to row the same distance downstream. If he can row 55 km downstream in 2 hours 30 minutes, what is the speed of the boat in still water? (in km\/h)<\/p>\n a)\u00a040<\/p>\n b)\u00a08<\/p>\n c)\u00a016<\/p>\n d)\u00a024<\/p>\n e)\u00a032<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of boat in still water = $x$ km\/hr<\/p>\n => Speed of current = $y$ km\/hr<\/p>\n Let distance travelled = $d$ km<\/p>\n Acc. to ques, => $2.2 (\\frac{d}{x + y}) = \\frac{d}{x – y}$<\/p>\n => $2.2x – 2.2y = x + y$<\/p>\n => $2.2x – x = y + 2.2y$<\/p>\n => $3x = 8y$ ————-(i)<\/p>\n Also, the man takes 2 hrs 30 mins in travelling 55 km downstream.<\/p>\n => $\\frac{55}{x + y} = 2 + \\frac{1}{2}$<\/p>\n => $\\frac{55}{x + y} = \\frac{5}{2}$<\/p>\n => $x + y = 22$<\/p>\n Multiplying both sides by 8, and using eqn(i), we get\u00a0:<\/p>\n => $8x + 3x = 22 \\times 8$<\/p>\n => $x = \\frac{22 \\times 8}{11} = 16$ km\/hr<\/p>\n Question 7:\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 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/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 Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 8:\u00a0<\/b>A boat takes a total time of twelve hours to travel 105 kms upstream and the same distance downstream. The speed of the boat in still water is six times of the speed of the current. What is the speed of the boat in still water? (in km\/hr)<\/p>\n a)\u00a012<\/p>\n b)\u00a030<\/p>\n c)\u00a018<\/p>\n d)\u00a024<\/p>\n e)\u00a036<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of current = $x$ km\/hr<\/p>\n => Speed of boat in still water = $6x$ km\/hr<\/p>\n Acc. to ques, => $\\frac{105}{7x} + \\frac{105}{5x} = 12$<\/p>\n => $\\frac{15}{x} + \\frac{21}{x} = 12$<\/p>\n => $\\frac{36}{x} = 12$<\/p>\n => $x = \\frac{36}{12} = 3$<\/p>\n $\\therefore$ Speed of boat in still water = $6 \\times 3 = 18$ km\/hr<\/p>\n Question 9:\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 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/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 Question 10:\u00a0<\/b>A boat takes a total time of eight hours to travel 63 kms upstream and the same distance downstream. The speed of the current is ${1 \\over 8}$th of the speed of the boat in still water. What is the speed of the boat in still water? (in km\/hr)<\/p>\n a)\u00a032<\/p>\n b)\u00a024<\/p>\n c)\u00a016<\/p>\n d)\u00a08<\/p>\n e)\u00a038<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of current = $x$ km\/hr<\/p>\n => Speed of boat in still water = $8x$ km\/hr<\/p>\n Acc. to ques, => $\\frac{63}{9x} + \\frac{63}{7x} = 8$<\/p>\n => $\\frac{7}{x} + \\frac{9}{x} = 8$<\/p>\n => $\\frac{16}{x} = 8$<\/p>\n => $x = \\frac{16}{8} = 2$<\/p>\n $\\therefore$ Speed of boat in still water = $8 \\times 2 = 16$ km\/hr<\/p>\n Question 11:\u00a0<\/b>A completes ${5 \\over 6}$ th of a given task in 10 days and is then replaced by B. The entire task is completed in 13 days. What is the respective ratio of the number of days in which A and B independently can complete the entire task?<\/p>\n a)\u00a02 : 7<\/p>\n b)\u00a03 : 8<\/p>\n c)\u00a01 : 4<\/p>\n d)\u00a02 : 3<\/p>\n e)\u00a06 : 11<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let total work to be done = 6 units<\/p>\n => A completes $\\frac{5}{6} \\times 6 = 5$ units in 10 days<\/p>\n => A’s efficiency = $\\frac{5}{10} = \\frac{1}{2}$ units \/day<\/p>\n Now, B finishes $\\frac{1}{6}$th of the task in 3 days<\/p>\n => B completes $\\frac{1}{6} \\times 6 = 1$ units in 3 days<\/p>\n => B’s efficiency = $\\frac{1}{3}$ units \/day<\/p>\n Now, time taken by A alone to complete the entire task = $\\frac{6}{\\frac{1}{2}} = 12$ days<\/p>\n Time taken by B alone to complete the entire task = $\\frac{6}{\\frac{1}{3}} = 18$ days<\/p>\n $\\therefore$ Required ratio = $\\frac{12}{18} = 2 : 3$<\/p>\n Question 12:\u00a0<\/b>A boat takes six hours to travel a certain distance downstream and five hours to travel a certain distance upstream. The distance travelled upstream is half of the travelled downstream. If the speed of the current is 4 km\/hr, what is the speed of the boat in still water? (in km\/hr)<\/p>\n a)\u00a016<\/p>\n b)\u00a020<\/p>\n c)\u00a024<\/p>\n d)\u00a010<\/p>\n e)\u00a018<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of boat in still water = $x$ km\/hr<\/p>\n Let distance travelled downstream = $2d$ km<\/p>\n => Distance travelled upstream = $d$ km<\/p>\n Using, $time = \\frac{distance}{speed}$<\/p>\n => $6 = \\frac{2d}{x + 4}$ ———(i)<\/p>\n and $5 = \\frac{d}{x – 4}$ ———(ii)<\/p>\n Dividing eqn(i) from (ii), we get :<\/p>\n => $\\frac{6}{5} = \\frac{\\frac{2d}{x + 4}}{\\frac{d}{x – 4}}$<\/p>\n => $\\frac{6}{5} = \\frac{2 (x – 4)}{x + 4}$<\/p>\n => $6x + 24 = 10x – 40$<\/p>\n => $10x – 6x = 24 + 40 = 64$<\/p>\n => $x = \\frac{64}{4} = 16$ km\/hr<\/p>\n Question 13:\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 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/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 Question 14:\u00a0<\/b>A bus covers a distance of 2,924 kms. in 43 hours. What is the speed of the bus?<\/p>\n a)\u00a072 kmph<\/p>\n b)\u00a060 kmph<\/p>\n c)\u00a068 kmph<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of bus = $\\frac{Distance}{Time}$<\/p>\n = $\\frac{2924}{43}$<\/p>\n = 68 kmph<\/p>\n Question 15:\u00a0<\/b>A train crosses a 300 metre long platform in 38 seconds while it crosses a signal pole in 18 seconds. What is the speed of the train in kmph ?<\/p>\n a)\u00a0Cannot be determined<\/p>\n b)\u00a072<\/p>\n c)\u00a048<\/p>\n d)\u00a054<\/p>\n e)\u00a0None of these<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the length of the train be\u00a0$l$ metre and speed of the train be $x$ m\/s<\/p>\n => $\\frac{l}{s} = 18$<\/p>\n => $l = 18s$<\/p>\n Also, $\\frac{l + 300}{s} = 38$<\/p>\n => $18s + 300 = 38s$<\/p>\n => $20s = 300$<\/p>\n => $s = \\frac{300}{20} = 15$ m\/s<\/p>\n $\\therefore$ Speed of train in kmph = $15 \\times \\frac{18}{5}$<\/p>\n = 54 kmph<\/p>\n Question 16:\u00a0<\/b>A car runs at the speed of 40 when not serviced and runs at 65 kmph when serviced. After servicing the car covers a certain distance in 5 hours. How much approximate time will the car take to cover the same distance when not serviced ?<\/p>\n a)\u00a010<\/p>\n b)\u00a07<\/p>\n c)\u00a012<\/p>\n d)\u00a08<\/p>\n e)\u00a06<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n After servicing, the distance covered by car in 5 hours = 65 * 5 = 325 km<\/p>\n Without servicing, speed of car = 40 kmph<\/p>\n => Required time = $\\frac{Distance}{Speed}$<\/p>\n = $\\frac{325}{40}$ = 8.125 hr<\/p>\n $\\approx$ 8 hours<\/p>\n Question 17:\u00a0<\/b>A car runs at the speed of 50 kms per hour when not serviced and runs at 60 km.\/hr. when serviced. After servicing the car covers a certain distance in 6 hours. How much time will the car take to cover the same distance when not serviced?<\/p>\n a)\u00a08.2 hours<\/p>\n b)\u00a06.5 hours<\/p>\n c)\u00a08 hours<\/p>\n d)\u00a07.2 hours<\/p>\n e)\u00a0None of these<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n After servicing, the distance covered by car in 6 hours = 60 * 6 = 360 km<\/p>\n Without servicing, speed of car = 50 kmph<\/p>\n => Required time = $\\frac{Distance}{Speed}$<\/p>\n = $\\frac{360}{50}$ = 7.2 hr<\/p>\n Question 18:\u00a0<\/b>Yesterday Priti type an essay of 5000 words at the speed of 60 words per minute: Today she type the same essay faster and her speed was 15% more than yesterday. What is the approximate <\/strong>difference in the time she took to type yesterday and the time she took to type today?<\/p>\n a)\u00a020 minutes<\/p>\n b)\u00a030 minutes<\/p>\n c)\u00a010 minutes<\/p>\n d)\u00a040 minutes<\/p>\n e)\u00a01 hour<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Priti’s second day typing speed = $\\frac{115}{100} \\times 60$ = 69 wpm<\/p>\n Required difference in time<\/p>\n = $(\\frac{5000}{60} – \\frac{5000}{69})$<\/p>\n = $5000(\\frac{69 – 60}{60 \\times 69})$<\/p>\n $\\approx$ 10 minutes<\/p>\n Question 19:\u00a0<\/b>The average speed of a train is 3 times the average speed of a car. The car covers a distance of 520 kms in 8 hours. How much distance will the train cover in 13 hours?<\/p>\n a)\u00a02553 kms<\/p>\n b)\u00a02585 kms<\/p>\n c)\u00a02355 kms<\/p>\n d)\u00a02535 kms<\/p>\n e)\u00a0None of these<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of car = $\\frac{Distance}{Time}$<\/p>\n = $\\frac{520}{8} = 65$ kmph<\/p>\n => Speed of train = $65 \\times 3 = 195$ kmph<\/p>\n $\\therefore$ Distance covered by train in 13 hours<\/p>\n = $13 \\times 195 = 2535$ km<\/p>\n Question 20:\u00a0<\/b>A boatman can row a boat downstream at 13 kmph and upstream at 9 kmph. What will be the speed of boat in still water ? (in kmph).<\/p>\n a)\u00a012<\/p>\n b)\u00a010.5<\/p>\n c)\u00a011<\/p>\n d)\u00a010<\/p>\n e)\u00a011.5<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of boat in still water<\/p>\n = $\\frac{1}{2}$ (downstream + upstream)<\/p>\n = $\\frac{1}{2} (13 + 9) = \\frac{22}{2}$<\/p>\n = $11$ kmph<\/p>\n
\ni. Distance between point A and B
\nIi.Time taken to travel down stream from B to A
\niii. Speed of the stream of the water
\niv. Effective speed of Boat while traveling Downstream from B to A<\/p>\n