Download Time Speed and Distance Questions for MAH-CET<\/a><\/p>\n Enroll to MAH-CET Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>one pipe can fill a tank three times as fast as another pipe. If the two pipes together can fill the tank in 36 minutes, then the slower pipe will take how much time to fill the tank alone?<\/p>\n a)\u00a081 minutes<\/p>\n b)\u00a0108 minutes<\/p>\n c)\u00a0192 minutes<\/p>\n d)\u00a0144 minutes<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let slower pipe fill the tank at the rate x units\/ minute.<\/p>\n Faster pipe will fill tank at the rate of 3x units\/ minute<\/p>\n Number of units filled by both the pipes in a minute = 4x<\/p>\n Total time taken to fill tank = 36 minutes<\/p>\n capacity of tank = 36$\\times\\ $4x = 144x<\/p>\n Time taken by slower pipe to fill the tank =\u00a0$\\frac{144x}{x}$ = 144 minutes<\/p>\n Answer is option D.<\/p>\n Question 2:\u00a0<\/b>With an average speed of 40 km\/hr, a train reaches its destination in time. If it goes with an average speed of 30 km\/hr, it is late by 36 minutes. The total distance travelled by train is :<\/p>\n a)\u00a072 km<\/p>\n b)\u00a078 km<\/p>\n c)\u00a060 km<\/p>\n d)\u00a040 km<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let us consider with average speed of 40 km\/hr, train reached its destination in time t hrs<\/p>\n with average speed of 30 km\/hr, it takes t +$\\frac{36}{60}$ hrs<\/p>\n Distance will be same in both the cases<\/p>\n 40t = 30t + 18<\/p>\n 10t = 18<\/p>\n t =\u00a0$\\frac{18}{10}$<\/p>\n Distance = 40t = 40$\\times\\ \\frac{18}{10}$ = 72 km<\/p>\n Answer is option A.<\/p>\n Question 3:\u00a0<\/b>Mira and Amal walk along a circular track, starting from the same point at the same time. If they walk in the same direction, then in 45 minutes, Amal completes exactly 3 more rounds than Mira. If they walk in opposite directions, then they meet for the first time exactly after 3 minutes. The number of rounds Mira walks in one hour is<\/p>\n 3)\u00a0Answer:\u00a08<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Considering the distance travelled by Mira in one minute = M,<\/p>\n The distance traveled by Amal in one minute = A.<\/p>\n Given if they walk in the opposite direction it takes 3 minutes for both of them to meet. Hence 3*(A+M) = C. (1)<\/p>\n C is the circumference of the circle.<\/p>\n Similarly, it is mentioned that if both of them walk in the same direction Amal completes 3 more rounds than Mira :<\/p>\n Hence 45*(A-M) = 3C. (2)<\/p>\n Multiplying (1)*15 we have :<\/p>\n 45A + 45M = 15C.<\/p>\n 45A – 45M = 3C.<\/p>\n Adding the two we have A = $\\frac{18C}{90}$<\/p>\n Subtracting the two M = $\\frac{12C}{90}$<\/p>\n Since Mira travels $\\frac{12C}{90}$ in one minute, in one hour she travels :$\\frac{12C}{90}\\cdot60\\ =\\ 8C$<\/p>\n Hence a total of 8 rounds.<\/p>\n Alternatively,<\/p>\n Let the length of track be L Question 4:\u00a0<\/b>One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is<\/p>\n a)\u00a011.5<\/p>\n b)\u00a010<\/p>\n c)\u00a012.5<\/p>\n d)\u00a012<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let Rahul work at a units\/hr and Gautam at b units\/hour Question 5:\u00a0<\/b>Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is<\/p>\n a)\u00a0144<\/p>\n b)\u00a0140<\/p>\n c)\u00a0264<\/p>\n d)\u00a0120<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let A fill the tank at x liters\/hour and B drain it at y liters\/hour Take Free MAH-CET mock tests here<\/a><\/p>\n Question 6:\u00a0<\/b>Two trains A and B were moving in opposite directions, their speeds being in the ratio 5 : 3. The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other. It took another 69 seconds for the rear ends of the trains to cross each other. The ratio of length of train A to that of train B is<\/p>\n a)\u00a03:2<\/p>\n b)\u00a05:3<\/p>\n c)\u00a02:3<\/p>\n d)\u00a02:1<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Considering the length of train A = La, length of train B = Lb.<\/p>\n The speed of train A be 5*x, speed of train B be 3*x.<\/p>\n From the information provided :<\/p>\n The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other.<\/p>\n In this case, train A traveled a distance equivalent to the length of train B which is Lb at a speed of 5*x+3*x = 8*x because both the trains are traveling in the opposite direction.<\/p>\n Hence (8*x)*(46) = Lb.<\/p>\n In the information provided :<\/p>\n It took another 69 seconds for the rear ends of the trains to cross each other.<\/p>\n In the next 69 seconds<\/p>\n The train B traveled a distance equivalent to the length of train A in this 69 seconds.<\/p>\n Hence (8*x)*(69) = La.<\/p>\n La\/Lb = 69\/46 = 3\/2 = 3 : 2<\/p>\n Question 7:\u00a0<\/b>Two trains cross each other in 14 seconds when running in opposite directions along parallel tracks. The faster train is 160 m long and crosses a lamp post in 12 seconds. If\u00a0the speed of the other train is 6 km\/hr less than the faster one, its length, in m, is<\/p>\n a)\u00a0184<\/p>\n b)\u00a0192<\/p>\n c)\u00a0190<\/p>\n d)\u00a0180<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Speed of the faster train =\u00a0$\\frac{160}{12}=\\frac{40}{3}\\ $ m\/s<\/p>\n Speed of the slower train =\u00a0$\\frac{40}{3}-\\left(6\\times\\ \\frac{5}{18}\\right)=\\frac{35}{3}$ m\/s<\/p>\n Sum of speeds (when the trains\u00a0travel towards each other) =\u00a0$\\frac{40}{3}+\\frac{35}{3}=25$ m\/s<\/p>\n Let the slower train be $x$ metres long; then:\u00a0$\\frac{160+x}{25}=14$<\/p>\n On solving,\u00a0$x=190\\ m$<\/p>\n Question 8:\u00a0<\/b>Given: Anuj takes 1 day to complete a job. Bharat takes twice the time as Anuj to complete the same job. Chetan takes twice the time as Bharat to complete that job. Dhiraj takes twice the time as Chetan to complete that job. a)\u00a0(A), (B) and (C) only<\/p>\n b)\u00a0(A) and (D) only<\/p>\n c)\u00a0(A) and (C) only<\/p>\n d)\u00a0(A) only<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Anuj takes 1 day to complete Question 9:\u00a0<\/b>Four friends, Ashish, Brian, Chaitra, and Dorothy, decide to jog for 30 minutes inside a stadium with a circular running track that is 200 metres long. The friends run at different speeds. Ashish completes a lap exactly every 60 seconds. Likewise, Brian, Chaitra and Dorothy complete a lap exactly every 1 minute 30 seconds, 40 seconds and 1 minute 20 seconds respectively. The friends begin together at the start line exactly at 4 p.m. What is the total of the numbers of laps the friends would have completed when they next cross the start line together ?<\/p>\n a)\u00a043<\/p>\n b)\u00a036<\/p>\n c)\u00a0They will never be at the start line together again before 4:30 p.m.<\/p>\n d)\u00a047<\/p>\n e)\u00a028<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n All the four friends will meet at the starting point after LCM(60,90,40,80) = 720 seconds.<\/p>\n Number of laps by A in 720 seconds = 12<\/p>\n Number of laps by B in 720 seconds = 8<\/p>\n Number of laps by C in 720 seconds = 18<\/p>\n Number of laps by D in 720 seconds = 9<\/p>\n Together they complete = 47 laps<\/p>\n Question 10:\u00a0<\/b>Two circular tracks T1 and T2 of radii 100 m and 20 m, respectively touch at a point A.\u00a0Starting from A at the same time, Ram and Rahim are walking on track T1 and track T2\u00a0at speeds 15 km\/hr and 5 km\/hr respectively. The number of full rounds that Ram will\u00a0make before he meets Rahim again for the first time is<\/p>\n a)\u00a05<\/p>\n b)\u00a03<\/p>\n c)\u00a02<\/p>\n d)\u00a04<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n To complete one round Ram takes 100m\/15kmph and Rahim takes 20m\/5kmph<\/p>\n They meet for the first time after L.C.M of (100m\/15kmph , 20m\/5kmph) = 100m\/5kmph=20m\/kmph.<\/p>\n Distance traveled by Ram =20m\/kmph * 15kmph =300m.<\/p>\n So, he must have ran 300\/100=3 rounds.<\/p>\n Note:<\/strong><\/p>\n CAT gave both 2 and 3 as correct answers because of the word ‘before<\/strong>‘.<\/p>\n Question 11:\u00a0<\/b>Two trains cross each other in 14 seconds when running in opposite directions along parallel tracks. The faster train is 160 m long and crosses a lamp post in 12 seconds. If\u00a0the speed of the other train is 6 km\/hr less than the faster one, its length, in m, is<\/p>\n a)\u00a0184<\/p>\n b)\u00a0192<\/p>\n c)\u00a0190<\/p>\n d)\u00a0180<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Speed of the faster train =\u00a0$\\frac{160}{12}=\\frac{40}{3}\\ $ m\/s<\/p>\n Speed of the slower train =\u00a0$\\frac{40}{3}-\\left(6\\times\\ \\frac{5}{18}\\right)=\\frac{35}{3}$ m\/s<\/p>\n Sum of speeds (when the trains\u00a0travel towards each other) =\u00a0$\\frac{40}{3}+\\frac{35}{3}=25$ m\/s<\/p>\n Let the slower train be $x$ metres long; then:\u00a0$\\frac{160+x}{25}=14$<\/p>\n On solving,\u00a0$x=190\\ m$<\/p>\n Question 12:\u00a0<\/b>A and B are two points on a straight line. Ram runs from A to B while Rahim runs from B to A. After crossing each other. Ram and Rahim reach their destination in one minute and four minutes, respectively. if they start at the same time, then the ratio of Ram’s speed to Rahim’s speed is<\/p>\n a)\u00a0$\\frac{1}{2}$<\/p>\n b)\u00a0$\\sqrt{2}$<\/p>\n c)\u00a0$2$<\/p>\n d)\u00a0$2\\sqrt{2}$<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the speed of Ram be v(r) and the speed of Rahim be v(h) respectively. Let them meet after time “t” from the beginning.<\/p>\n Hence Ram will cover v(r)(t) during that time and Rahim will cover v(h)t respectively.<\/p>\n Now after meeting Ram reaches his destination in 1 min i.e. Ram covered v(h)t in 1 minute or v(r)(1)= v(h)(t)<\/p>\n Similarly Rahim reaches his destination in 4 min i.e. Rahim covered v(r)t in 4 minutes or v(h)(4)= v(r)(t)<\/p>\n Dividing both the equations we get\u00a0$\\frac{v\\left(r\\right)}{4v\\left(h\\right)}=\\frac{v\\left(h\\right)}{v\\left(r\\right)}\\ or\\ \\frac{v\\left(r\\right)}{v\\left(h\\right)}=2$ Hence the ratio is 2.<\/p>\n Question 13:\u00a0<\/b>In a car race, car A beats car B by 45 km. car B beats car C by 50 km. and car A beats car C by 90 km. The distance (in km) over which the race has been conducted is<\/p>\n a)\u00a0475<\/p>\n b)\u00a0450<\/p>\n c)\u00a0500<\/p>\n d)\u00a0550<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Now car A beats car B by 45km. Let the speed of car A be v(a) and speed of car B be v(b).<\/p>\n $\\frac{v\\left(a\\right)}{v\\left(b\\right)}=\\frac{m}{m-45}$ …..(1)where ‘”m” is the entire distance of the race track.<\/p>\n Moreover\u00a0$\\frac{v\\left(b\\right)}{v\\left(c\\right)}=\\frac{m}{m-50}$…….(2)<\/p>\n and finally\u00a0$\\frac{v\\left(a\\right)}{v\\left(c\\right)}=\\frac{m}{m-90}$……(3)<\/p>\n Multiplying (1) and (2) we get (3).\u00a0$\\frac{m}{m-90}=\\frac{m}{m-45}\\left(\\frac{m}{m-50}\\right)$<\/p>\n Solving we get m=450 which is the length of the entire race track<\/p>\n Question 14:\u00a0<\/b>The distance from B to C is thrice that from A to B. Two trains travel from A to C via B.\u00a0The speed of train 2 is double that of train 1 while traveling from A to B and their\u00a0speeds are interchanged while traveling from B to C. The ratio of the time taken by\u00a0train 1 to that taken by train 2 in travelling from A to C is<\/p>\n a)\u00a05:7<\/p>\n b)\u00a04:1<\/p>\n c)\u00a01:4<\/p>\n d)\u00a07:5<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the distance from A to B be “x”, then the distance from B to C will be 3x. Now the speed of Train 2 is double of Train 1. Let the speed of Train 1 be “v”, then the speed of Train 2 will be “2v” while travelling from A\u00a0to B.<\/p>\n Time taken by Train 1 = (x\/v)<\/p>\n Time taken by Train 2 = (x\/2v)<\/p>\n Now from B to C distance is “3x” and the speed of Train 2 is (v) and the speed of Train 1 is (2v).<\/p>\n Time taken by Train 1 = 3x\/2v<\/p>\n Time taken by Train 2 = 3x\/v<\/p>\n Total time taken by Train 1 = x\/v(1+(3\/2)) = (5\/2)(x\/v)<\/p>\n Total time taken by Train 2 = x\/v(3+(1\/2))= (7\/2)(x\/v)<\/p>\n Ratio of time taken =\u00a0$\\frac{5}{\\frac{2}{\\frac{7}{2}}}=\\frac{5}{7}$<\/p>\n Question 15:\u00a0<\/b>Anil, Sunil, and Ravi run along a circular path of length 3 km, starting from the same\u00a0point at the same time, and going in the clockwise direction. If they run at speeds of 15\u00a0km\/hr, 10 km\/hr, and 8 km\/hr, respectively, how much distance in km will Ravi have run\u00a0when Anil and Sunil meet again for the first time at the starting point?<\/p>\n a)\u00a04.8<\/p>\n b)\u00a04.6<\/p>\n c)\u00a05.2<\/p>\n d)\u00a04.2<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Anil and Sunil will meet at a first point after LCM (\u00a0$\\frac{3}{15},\\frac{3}{10}$) = 3\/5 hr<\/p>\n In the mean time, distance travelled by ravi = 8 * 3\/5 = 4.8 km<\/p>\n
\nand velocity of Mira be a and Amal be b
\nNow when they meet after 45 minutes Amal completes 3 more rounds than Mira
\nso we can say they met for the 3rd time moving in the same direction
\nso we can say they met for the first time after 15 minutes
\nSo we know Time to meet = Relative distance \/Relative velocity
\nso we get $\\frac{15}{60}=\\frac{L}{a-b}$ \u00a0 \u00a0 (1)
\nNow When they move in opposite direction
\nThey meet after 3 minutes
\nso we get $\\frac{3}{60}=\\frac{L}{a+b}$ \u00a0\u00a0 (2)
\nDividing (1) and (2)
\nwe get $\\frac{\\left(a+b\\right)}{\\left(a-b\\right)}=5$
\nor 4a =6b
\nor a = 3b\/2
\nNow substituting in (1)
\nwe get :
\n$\\frac{L}{b}\\times\\ 2=\\ \\frac{15}{60}$
\nso $\\frac{L}{b}\\ =\\frac{1}{8}$
\nSo we can say 1 round is covered in $\\frac{1}{8}$ hours
\nso in 1-hour total rounds covered = 8.<\/p>\n
\nNow as per the condition :
\n8a+6b =7.5a+7.5b
\nso we get 0.5a=1.5b
\nor a=3b
\nTherefore total work = 8a +6b = 8a +2a =10a
\nNow Rahul alone takes 10a\/10 = 10 hours.<\/p>\n
\nNow as per Condition 1 :
\nWe get Volume filled till 10pm = 8x-7y \u00a0 \u00a0 \u00a0 (1) .
\nHere A operates for 8 hours and B operates for 7 hours .
\nAs per condition 2
\nWe get Volume filled till 6pm = 4x-2y \u00a0 \u00a0 \u00a0\u00a0 (2)
\nHere A operates for 4 hours and B operates for 2 hours .
\nNow equating (1) and (2)
\nwe get 8x-7y =4x-2y
\nso we get 4x =5y
\ny =4x\/5
\nSo volume of tank =\u00a0$8x-7\\times\\ \\frac{4x}{5}=\\frac{12x}{5}$
\nSo time taken by A alone to fill the tank = $\\frac{\\frac{12x}{5}}{x}=\\frac{12}{5}hrs\\ $
\n= 144 minutes<\/p>\n
\n(A) Chetan and Dhiraj will take 8\/3 days to complete the work
\n(B) The second fastest pair to complete the work is Anuj and Dhiraj
\n(C) The second slowest pair to complete the work is Bharat and Dhiraj
\n(D) Bharat and Dhiraj will take 4\/3 days to complete the work<\/p>\n
\nBharat takes 2 days
\nChetan takes 4 days and Dhiraj takes 8 days to complete
\noption A chetan and Dhiraj together will take\u00a0$\\frac{4\\times\\ 8}{4+8}=\\frac{32}{12}=\\frac{8}{3\\ }days$so A is correct.
\nNow calculating for all we get AB in 2\/3 days AC in 4\/5 days AD in 8\/9 days BC in 8\/6 days BD in 16\/10 days
\nSo we can say BD is 2nd slowest pair
\nHence A and C are correct<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/image_lmGqNsE.png\")
<\/figure>\n