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Download Time Speed and Distance Questions for CMAT<\/a><\/p>\n Enroll to CMAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>A train 350 m long takes 36 seconds to cross a man running at a speed of 5 km\/h in the direction opposite to that of train. What is the speed of the train?<\/p>\n a)\u00a030 km\/h<\/p>\n b)\u00a040 km\/h<\/p>\n c)\u00a024 km\/h<\/p>\n d)\u00a034 km\/h<\/p>\n e)\u00a0Other than those given as options<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the train be x m\/s Question 2:\u00a0<\/b>A boat running downstream covers a distance of 24 km in 4 hours, while for covering the same distance upstreams it takes 6 hours. What is the speed of the boat in still water?<\/p>\n a)\u00a05.5 kmph<\/p>\n b)\u00a06 kmph<\/p>\n c)\u00a03.5 kmph<\/p>\n d)\u00a0Data inadequate<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the boat be S and the speed of the stream be R.<\/p>\n So, 24 \/ 4 = 6 = S + R<\/p>\n 24 \/ 6 = 4 = S – R<\/p>\n So, 2S = 10 => S = 5 and R = 1<\/p>\n So, speed of the boat in still water = 1 kmph<\/p>\n Question 3:\u00a0<\/b>A motor starts with the speed of 70 kmph with its speed increasing every two hours by 10 kmph. In how many hours will it cover 345 km?<\/p>\n a)\u00a0$2\\frac{1}{4}$hours<\/p>\n b)\u00a0$4\\frac{1}{2}$hours<\/p>\n c)\u00a0$4$ hours $5$ minutes<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Distance = Speed $\\times$ Time<\/p>\n Initial speed = 70 kmph<\/p>\n In first 2 hours, motor covers a distance = $70 \\times 2 = 140~km$<\/p>\n Speed after 2 hours = 70 + 10 = 80 kmph<\/p>\n In the next 2 hours, it covers a distance = $80 \\times 2 = 160~km$<\/p>\n Distance covered = 300 km<\/p>\n Now, the speed is 90 kmph<\/p>\n Remaining 45 km is covered in $\\frac{45}{90}$ = 0.5 hr<\/p>\n Total time taken to cover 345 km = 2 + 2 + 0.5 = 4.5 hr<\/p>\n Question 4:\u00a0<\/b>A truck covers a certain distance in 12 h at the speed of 70 km\/h. What is the average speed of a car which travels a distance of 120 km more than the truck in the same time ?<\/p>\n a)\u00a076 km\/h<\/p>\n b)\u00a085 km\/h<\/p>\n c)\u00a082 km\/h<\/p>\n d)\u00a078 km\/h<\/p>\n e)\u00a0None of these<\/p>\n 4)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n The distance travelled by the truck in 12 hours equals 12*70 = 840 km So, the average speed of the car equals $\\frac{960}{12} = 80$ kmph<\/p>\n Question 5:\u00a0<\/b>A 420 m long train crosses a pole in 70 seconds. What is the speed of the train ?<\/p>\n a)\u00a05 m\/s<\/p>\n b)\u00a07 m\/s<\/p>\n c)\u00a04.5 m\/s<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Time taken by the train to cross the pole is 70 seconds. Hence, the speed of the train is $\\frac{420}{70}=6$ seconds.<\/p>\n Question 6:\u00a0<\/b>A boat running downstreams covers a distance of 16 km in 2 hours while for covering the same distance upstream it takes 4 hours. What is the speed of the boat in still water?<\/p>\n a)\u00a04 kmph<\/p>\n b)\u00a06 kmph<\/p>\n c)\u00a08 kmph<\/p>\n d)\u00a0Data inadequate<\/p>\n e)\u00a0None of these<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the boat in still water be B and the speed of the water be W. From the information given, the speed of the boat downstream is 16\/2 = 8 kmph Hence, B+W = 8 and B-W =4 Question 7:\u00a0<\/b>A train running at the speed of 66 kmph crosses a signal pole in 18 seconds.\u00a0What is the length of the train?<\/p>\n a)\u00a0330 meters<\/p>\n b)\u00a0300 metres<\/p>\n c)\u00a0360 metres<\/p>\n d)\u00a0320 metres<\/p>\n e)\u00a0None of these<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n 66 kmph = 66 *5\/18 m\/s<\/p>\n It takes 18 secs to cross the post. So, length = 66 * 18* 5\/18 = 330 metres.<\/p>\n Question 8:\u00a0<\/b>Two cars A and B are running in the same direction. Car \u2018A\u2019 had already covered a distance of 60 kms, when car \u2018B\u2019 started running. The cars meet each other in 3 hours after car \u2018B\u2019 started running. What was the speed of car \u2018A\u2019 ?<\/p>\n a)\u00a040 kmph<\/p>\n b)\u00a060 kmph<\/p>\n c)\u00a045 kmph<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of car A be ‘x’ kmph and the speed of car B be ‘y’ kmph. Distance covered by B in three hours is 3y So, 3y = 3x + 60 or y=x+20 Question 9:\u00a0<\/b>A train running at a speed of 60 kmph crosses a platform double its length in 32.4 seconds. What is the length of the platform?<\/p>\n a)\u00a0108 metres<\/p>\n b)\u00a0240 meters<\/p>\n c)\u00a0360 meters<\/p>\n d)\u00a090 meters<\/p>\n e)\u00a0Cannot be determined<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the length of the train be ‘x’ metres Question 10:\u00a0<\/b>A person has to travel from point B in certain time. Travelling at a speed of 5 kmph he reaches 48 minutes late and while travelling at a speed of 8 kmph he reaches 15 minutes early. What is the distance from point A to point B?<\/p>\n a)\u00a015 kms<\/p>\n b)\u00a091 kms<\/p>\n c)\u00a012 kms<\/p>\n d)\u00a018 kms<\/p>\n e)\u00a014 kms<\/p>\n 10)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the distance between A and B be ‘d’ km and time taken be ‘t’ hrs. Question 11:\u00a0<\/b>Ram and Shyam are travelling from point A to B, which are 60km apart. Travelling at a certain speed Ram takes one hour more than Shyam to reach point B. If Ram doubles his speed he will take 30 minutes less than Shyam to reach point B. At what speed was Ram driving from point A to B?<\/p>\n a)\u00a015 kmph<\/p>\n b)\u00a035 kmph<\/p>\n c)\u00a030 kmph<\/p>\n d)\u00a025 kmph<\/p>\n e)\u00a020 kmph<\/p>\n 11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of Ram be ‘v’ kmph Question 12:\u00a0<\/b>The respective ratio between speed of the boat upstream and speed of the boat downstream is 3 : 4. What is the speed of the boat in still water if it covers 70 km downstream in 3 hours 30 minutes? (in km\/h)<\/p>\n a)\u00a018<\/p>\n b)\u00a018.5<\/p>\n c)\u00a017<\/p>\n d)\u00a017.5<\/p>\n e)\u00a016<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of boat in still water and speed of river be B km\/hr and R km\/hr respectively<\/p>\n Speed of boat in downstream = (B+R)km\/hr<\/p>\n Speed of boat in upstream = (B-R) km\/hr<\/p>\n It is given that $\\frac{B-R}{B+R} = \\frac{3}{4}$<\/p>\n we get , B = 7 R<\/p>\n Now it is given that boat covers 70 km in 3.5 hours downstream<\/p>\n so ,<\/p>\n 70 = (B+R)3.5<\/p>\n B+R = 20<\/p>\n R = 2.5 km\/hr<\/p>\n B = 17.5 km\/hr<\/p>\n Question 13:\u00a0<\/b>The time taken by a boat to travel a distance downstream is half the time taken by it to travel the same distance upstream. What is the speed of the boat downstream if it travels 7.5 km upstream in 1 hour 30 minutes ? (in km\/h)<\/p>\n a)\u00a07.5<\/p>\n b)\u00a05<\/p>\n c)\u00a09<\/p>\n d)\u00a010<\/p>\n e)\u00a0None of these<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of boat in still water and speed of river be B km\/hr and R km\/hr respectively .<\/p>\n Speed of boat in downstream = B+ R km\/hr<\/p>\n Speed of boat in upstream = B-R km\/hr = $\\frac{7.5}{1.5}$= 5km\/hr<\/p>\n Let the same distance travelled in upstream and downstream be D km<\/p>\n So ,<\/p>\n $\\frac{D}{B-R}$ =2\u00d7 $\\frac{D}{B+R}$<\/p>\n B+R = 10 km\/hr<\/p>\n Question 14:\u00a0<\/b>A car starts at 11 am from point A towards point B at 36 kmph while another car starts at 1 pm from point B towards A at 44 kmph. They cover a distance of 592 km till meeting. At what time will they meet each other ?<\/p>\n a)\u00a08 pm<\/p>\n b)\u00a06 : 30 pm<\/p>\n c)\u00a07 : 30 pm<\/p>\n d)\u00a05 : 30 pm<\/p>\n e)\u00a0None of these<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n distance travelled by car which started from point A from 11 am to 1 pm is = 36 \u00d72 = 72 km<\/p>\n Now it is given that total distance is 592 km<\/p>\n Distance left to be covered = 592 – 72 = 520 km<\/p>\n Speed of car started from B = 44 km\/hr<\/p>\n Soeed of car started from A = 36km\/hr<\/p>\n Relative speed = 80 km\/hr<\/p>\n Time taken = 520\/80 = 6.5 hours<\/p>\n So the two cars will meet at 7.30 pm<\/p>\n Question 15:\u00a0<\/b>Two trains crosses each other in 14 sec when they are moving in opposite direction, and when they are moving in same direction they crosses each other in 3 min 2 sec. Find the speed of the faster train by what percent more than the speed of the slower train ?<\/p>\n a)\u00a016.67%<\/p>\n b)\u00a017.33%<\/p>\n c)\u00a016.33%<\/p>\n d)\u00a017.67%<\/p>\n e)\u00a018.33%<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n let the speed of faster train be F m\/s<\/p>\n And slower train be S m\/s<\/p>\n When they move in same direction relative speed = (F -S) m\/s<\/p>\n Time taken = 14 sec<\/p>\n When they move in opposite direction = (F+S) m\/s<\/p>\n Time taken in this case = 182 sec<\/p>\n In both cases they are covering same distance<\/p>\n So , (F+S )\u00d7 14 = (F-S ) \u00d7 182<\/p>\n 12 F = 14 S<\/p>\n F = $\\frac{7}{6}$ S<\/p>\n so faster train is 1\/6 more than the slower one which means it is 16.67% faster than the slower train.<\/p>\n Question 16:\u00a0<\/b>The time taken by a boat to travel\u037e `x’ km upstream is twice the time taken by the same boat to travel `x’ km downstream. If speed of the boat in still water is 12 km\/h. what is the speed of current ? (in km\/h)<\/p>\n a)\u00a03<\/p>\n b)\u00a04<\/p>\n c)\u00a03.5<\/p>\n d)\u00a04.5<\/p>\n e)\u00a0None of these<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Distance travelled by boat = $x$ km<\/p>\n Let speed of current = $y$ km\/h<\/p>\n Speed of boat = 12 km\/h<\/p>\n => Downstream speed = $(12 + y)$ km\/h and upstream speed = $(12 – y)$ km\/h<\/p>\n Acc. to ques,\u00a0time taken by a boat to travel\u037e `x’ km upstream is twice the time taken by the same boat to travel `x’ km downstream<\/p>\n => $\\frac{x}{12 – y} = 2 \\times \\frac{x}{12 + y}$<\/p>\n => $\\frac{1}{12 – y} = \\frac{2}{12 + y}$<\/p>\n => $12 + y = 24 – 2y$<\/p>\n => $y + 2y = 3y = 24 – 12$<\/p>\n => $y = \\frac{12}{3} = 4$ km\/h<\/p>\n Question 17:\u00a0<\/b>The ratio between speed of a boat downstream and speed of the boat upstream is 7 : 5 respectively. If the boat travels a distance of 63 km downstream in 3 hours, what is the speed of the boat in still water ? (in km\/h)<\/p>\n a)\u00a020 km\/h<\/p>\n b)\u00a016 km\/h<\/p>\n c)\u00a014 km\/h<\/p>\n d)\u00a012 km\/h<\/p>\n e)\u00a018 km\/h<\/p>\n 17)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Le the speed of boat be B km\/hr and speed of river be R km\/hr<\/p>\n Speed of boat in downstream = (B + R) km\/hr<\/p>\n Speed of boat in upstream = (B- R)km\/hr<\/p>\n It is given that $\\frac{B+R}{B-R}$ = $\\frac{7}{5}$<\/p>\n B = 6R………..(1)<\/p>\n Distance travelled downstream = 63 km<\/p>\n Time taken = 3 hr<\/p>\n 63 = (B+R)3<\/p>\n B+R = 21 ……….(2)<\/p>\n From equation 1 and 2<\/p>\n R = 18 km\/hr<\/p>\n Question 18:\u00a0<\/b>The speed of a man is 3\/4 th the speed of a bicycle. The bicycle covers 192 m. in 8 seconds. How much time will the man take to cover 54 m. ?<\/p>\n a)\u00a03 seconds<\/p>\n b)\u00a04 seconds<\/p>\n c)\u00a07 seconds<\/p>\n d)\u00a05 seconds<\/p>\n e)\u00a0None of these<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of Bicycle = $\\frac{192}{8}$ = 24 m\/s<\/p>\n Speed of Man = $\\frac{3}{4}\\times24$<\/p>\n Speed of Man = 18 m\/s<\/p>\n Distance to be covered by Man = 54 m<\/p>\n Time taken by Man = $\\frac{54}{18}$ = 3 sec<\/p>\n Question 19:\u00a0<\/b>A bus covers 572 kms in .13 hours. What is the speed of the bus ?<\/p>\n a)\u00a040 km\/hr<\/p>\n b)\u00a044 km\/hr<\/p>\n c)\u00a043 km\/hr<\/p>\n d)\u00a047 km\/hr<\/p>\n e)\u00a0None of these<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Using,<\/p>\n Distance = Speed x Time<\/p>\n 572 = Speed x 13<\/p>\n Speed = 44 km\/hr<\/p>\n Question 20:\u00a0<\/b>Two pipes can fill a tank in 10 hours and 16 hours respectively. A third pipe can empty the tank in 32 hours. If all the three pipes are opened simultaneously then in how much time the tank will be full ? (in hours)<\/p>\n a)\u00a0$7\\frac{11}{21}$<\/p>\n b)\u00a0$7\\frac{13}{21}$<\/p>\n c)\u00a0$8\\frac{4}{21}$<\/p>\n d)\u00a0$6\\frac{5}{14}$<\/p>\n e)\u00a0$8\\frac{9}{14}$<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total quantity = 160 units Speed pipe 2 = 10 $\\frac{units}{hours}$<\/p>\n Speed pipe 3 = 5 $\\frac{units}{hours}$<\/p>\n Time reqd = Total Units \/ [Speed (pipe 1 + pipe 2) – Speed Pipe 3]<\/p>\n Time = 160\/21 = 7 $\\frac{13}{21}$ hours<\/p>\n
\nSpeed of the man = 5km\/hr = 1.388m\/s
\nRelative speed of man when he is running in opposite direction = (x+1.388)
\nHowever train crossess him in 36 seconds = 350\/36 =\u00a0 9.722 m\/s
\nx+1.388 = 9.722
\nx = 8.334 m\/s
\nx = 30 km\/hr<\/p>\n
\nHence, the car travels 840 + 120 = 960 kms in 12 hours.<\/p>\n
\nThe length of the train is 420 metres.<\/p>\n
\nSo, speed of the boat downstream is B+W and the speed of the boat upstream is B-W.<\/p>\n
\nSpeed of the boat upstream is 16\/4 = 4 kmph<\/p>\n
\nSo, B = 6 kmph and W = 2 kmph<\/p>\n
\nBy the time car B\u00a0started running, car A covered 60 Km.<\/p>\n
\nDistance covered by A in three hours is 3x<\/p>\n
\nHence, the speed of B is 20 kmph more than the speed of A. As we don’t know the speed of car B, we can’t find the exact speed of car A. We can just infer that the speed of car A is 20 kmph less than the speed of car\u00a0B.<\/p>\n
\nLength of the platform = 2x
\nDistance travelled by the train = x + 2x = 3x
\nSpeed of the train = 60 kmph = $60 \\times \\frac{5}{18}$ m\/sec
\nDistance = Speed $\\times$ Time
\n$3x = 60 \\times \\frac{5}{18} \\times 32.4$
\n$x = 180$
\nThe length of the platform = 2x = 360 m<\/p>\n
\nIn both the cases, distance travelled is the same.
\nDistance = Speed $\\times$ Time
\nCase (i) Time taken = 48 min late = (t + 0.8) hrs
\n$d = 5 \\times (t + 0.8)$
\nCase (ii) Time taken = 15 min early = (t – 0.25) hrs
\n$d = 8 \\times (t – 0.25)$
\nOn equating them, we get
\n5t + 4 = 8t – 2
\nt = 2
\nDistance, d = 5 (2 + 0.8) = 14 km<\/p>\n
\nTime taken by Shyam be ‘t’ hrs
\nDistance = Speed $\\times$ Time
\n60 = v (t + 1)
\nAfter doubling the speed,
\n60 = 2v(t – 0.5)
\nSimplifying them, we get
\nv (t + 1) = 2v (t – 0.5)
\nt = 2
\nv = 60\/3 = 20 kmph<\/p>\n
\nSpeed pipe 1 = 16 $\\frac{units}{hours}$<\/p>\n