Here you can download a free Time Speed and Distance questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the Time Speed and Distance of answers for the given questions. These questions will help you to make practice and solve the Time Speed and Distance questions in the MAH MBA CET exam. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Time Speed and Distance MCQ<\/strong> PDF for MBA-CET 2022 for free.<\/p>\n Download Time Speed and Distance Questions for MAH-CET<\/a><\/p>\n Enroll to MAH-CET 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 Take Free MAH-CET mock tests here<\/a><\/strong><\/span><\/p>\n Enrol to 5 MAH CET Latest Mocks For Just Rs. 299<\/a><\/strong><\/span><\/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 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
\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