Download Speed Time and Distance Questions for XAT PDF \u2013 XAT Speed Time and Distance questions pdf by Cracku. Practice XAT solved Speed Time and Distance Questions paper tests, and these are the practice question to have a firm grasp on the Speed Time and Distance topic in the XAT exam. Top 20 very Important Speed Time and Distance Questions for XAT based on asked questions in previous exam papers. \u00a0The XAT question papers contain actual questions asked with answers and solutions.<\/p>\n
Download Speed Time and Distance Questions for XAT<\/a><\/p>\n Enroll to XAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>A person crosses a 900-metre long street in 6 minutes.The speed of the person in km\/h is:<\/p>\n a)\u00a09<\/p>\n b)\u00a010<\/p>\n c)\u00a015<\/p>\n d)\u00a012<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Distance travelled = 900 m =\u00a0$\\frac{900}{1000}=0.9\\ km$<\/p>\n Time taken = 6 minutes =\u00a0$\\frac{6}{60}=\\frac{1}{10\\ }=0.1\\ hr$<\/p>\n Speed of person (km\/hr) =\u00a0$\\frac{Dis\\tan ce}{time\\ }=\\frac{0.9}{0.1}$<\/p>\n = 9 km\/hr<\/p>\n Hence, Option A is correct.<\/p>\n Question 2:\u00a0<\/b>A train can travel 40% faster then a car.Both the train and the car start from point A at the same time and reach point B, which is 70km away point from A,at the same time.On the way, however,the train lost about 15 minutes while stopping at stations. The speed of the car in km\/h is:<\/p>\n a)\u00a0120<\/p>\n b)\u00a080<\/p>\n c)\u00a090<\/p>\n d)\u00a0100<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n let the speed of car be x.<\/p>\n then the speed of train =\u00a0$x\\left(1+\\frac{40}{100}\\right)=1.4x$<\/p>\n Time taken by car to cover 70km =\u00a0$\\frac{70}{x}$<\/p>\n Time taken\u00a0 by train to cover 70km =\u00a0$\\frac{70}{1.4x}=\\frac{50}{x}$<\/p>\n According to question,<\/p>\n $\\therefore\\ \\frac{70}{x}-\\frac{50}{x}=\\frac{15}{60}$<\/p>\n $\\therefore\\ \\frac{20}{x}=\\frac{1}{4}$<\/p>\n so, Speed of car =\u00a0x = 80 km\/hr<\/p>\n Hence, Option B is correct<\/p>\n Question 3:\u00a0<\/b>A man and a woman, working together can do a work in 66 days. The ratio of their working efficiencies is 3 : 2. In how many days 6 men and 2 women working together can do the same work?<\/p>\n a)\u00a018<\/p>\n b)\u00a015<\/p>\n c)\u00a014<\/p>\n d)\u00a012<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work be 330 units.<\/p>\n Efficiency of a man and a woman together = $\\frac{330}{66}$ = 5 units\/day<\/p>\n The ratio of the working efficiencies of man and woman is 3 : 2.<\/p>\n Efficiency of a man = 3 units\/day<\/p>\n Efficiency of a woman = 2 units\/day<\/p>\n Efficiency of 6 men and 2 women together = (6 x 3) + (2 x 2) = 22 units\/day<\/p>\n Time required for 6 men and 2 women together to complete the work = $\\frac{330}{22}$ = 15 days<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 4:\u00a0<\/b>A man and a woman, working together can do a work in 66 days. The ratio of their working efficiencies is 3 : 2. In how many days 6 men and 2 women working together can do the same work?<\/p>\n a)\u00a018<\/p>\n b)\u00a015<\/p>\n c)\u00a014<\/p>\n d)\u00a012<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work be 330 units.<\/p>\n Efficiency of a man and a woman together = $\\frac{330}{66}$ = 5 units\/day<\/p>\n The ratio of the working efficiencies of man and woman is 3 : 2.<\/p>\n Efficiency of a man = 3 units\/day<\/p>\n Efficiency of a woman = 2 units\/day<\/p>\n Efficiency of 6 men and 2 women together = (6 x 3) + (2 x 2) = 22 units\/day<\/p>\n Time required for 6 men and 2 women together to complete the work = $\\frac{330}{22}$ = 15 days<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 5:\u00a0<\/b>A train is to cover 370 km at a uniform speed. After running 100 km,the train could run at a speed 5 km\/h less than its normal speed due to some technical fault. The train got delayed by 36 minutes. What is the normal speed of the train, in km\/h?<\/p>\n a)\u00a048<\/p>\n b)\u00a045<\/p>\n c)\u00a040<\/p>\n d)\u00a050<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the normal speed of the train = s km\/h<\/p>\n According to the problem,<\/p>\n $\\frac{100}{s}+\\frac{270}{s-5}=\\frac{370}{s}+\\frac{36}{60}$<\/p>\n $\\frac{270}{s-5}-\\frac{270}{s}=\\frac{3}{5}$<\/p>\n $90\\left[\\frac{s-s+5}{\\left(s-5\\right)s}=\\frac{1}{5}\\right]$<\/p>\n $s^2-5s-2250=0$<\/p>\n s = 50 or s = -45<\/p>\n ‘s’ cannot be negative.<\/p>\n So s = 50 km\/h<\/p>\n Normal speed of the train = s = 50 km\/h<\/p>\n Hence, the correct answer is Option D<\/p>\n Take XAT 2023 Mock Tests<\/a><\/p>\n Question 6:\u00a0<\/b>A train running at $40\\frac{1}{2}$ km\/h takes 24 seconds to cross a pole. How much time (in seconds) will it take to pass a 450 m long bridge?<\/p>\n a)\u00a056<\/p>\n b)\u00a052<\/p>\n c)\u00a060<\/p>\n d)\u00a064<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the length of the train = L<\/p>\n Speed of the train =\u00a0$40\\frac{1}{2}$ km\/h =\u00a0$\\frac{81}{2}$ km\/h =\u00a0$\\frac{81}{2}\\times\\frac{5}{18}$ m\/sec = $\\frac{45}{4}$ m\/sec<\/p>\n Train crosses a pole in 24 seconds.<\/p>\n $\\frac{L}{\\frac{45}{4}}$ = 24<\/p>\n $\\frac{4L}{45}$ = 24<\/p>\n L = 270 m<\/p>\n Length of the train = L = 270 m<\/p>\n Time required for train to pass\u00a0a 450 m long bridge = $\\frac{L+450}{\\frac{45}{4}}$<\/p>\n =\u00a0$\\frac{270+450}{\\frac{45}{4}}$<\/p>\n =\u00a0$\\frac{720\\times4}{45}$<\/p>\n = 64 sec<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 7:\u00a0<\/b>A and B can do certain work in 18 days and 30 days,respectively. They work together for 5 days. C alone completes the remaining work in 15 days. A and C together can complete $\\frac{5}{6}$th part of the same work in:<\/p>\n a)\u00a06 days<\/p>\n b)\u00a08 days<\/p>\n c)\u00a09 days<\/p>\n d)\u00a05 days<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work = 360 units<\/p>\n Efficiency of A = $\\frac{360}{18}$ = 20 units\/day<\/p>\n Efficiency of B = $\\frac{360}{30}$ = 12 units\/day<\/p>\n A and B worked together for 5 days.<\/p>\n Work done by A and B together in 5 days = 5 x (20 + 12) = 160 units<\/p>\n Remaining work = 360 – 160 = 200 units<\/p>\n C alone completes the remaining work in 15 days.<\/p>\n Efficiency of C = $\\frac{200}{15}$ =\u00a0$\\frac{40}{3}$ units\/day<\/p>\n Efficiency of A and C together = 20 +\u00a0$\\frac{40}{3}$ =\u00a0$\\frac{100}{3}$ units\/day<\/p>\n $\\frac{5}{6}$th of the total work =\u00a0$\\frac{5}{6}\\times$360 = 300 units<\/p>\n Number of days required for A and C together to complete\u00a0$\\frac{5}{6}$th of work = $\\frac{300}{\\frac{100}{3}}$<\/p>\n = $\\frac{300\\times3}{100}$<\/p>\n = 9 days<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 8:\u00a0<\/b>A can complete a work in $11\\frac{1}{2}$ days. B is 25% more efficient than A and C is 50% less efficient than B. Working together A, B and C will complete the same work<\/p>\n a)\u00a05 days<\/p>\n b)\u00a04 days<\/p>\n c)\u00a03 days<\/p>\n d)\u00a08 days<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work = 460 units<\/p>\n A can complete a work in $11\\frac{1}{2}$\u00a0days.<\/p>\n Efficiency of A =\u00a0$\\frac{460}{\\frac{23}{2}}$ = 40 units\/day<\/p>\n B is 25% more efficient than A.<\/p>\n Efficiency of B =\u00a0$\\frac{125}{100}\\times40$ = 50 units\/day<\/p>\n C is 50% less efficient than B.<\/p>\n Efficiency of C =\u00a0$\\frac{50}{100}\\times50$ = 25 units\/day<\/p>\n Efficiency of A, B and C together = 40 + 50 + 25 = 115 units\/day<\/p>\n Number of days required for A, B and C together to complete the work =\u00a0$\\frac{460}{115}$\u00a0= 4 days<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 9:\u00a0<\/b>A train covers 450 km at a uniform speed. If the speed had been 5 km\/h more,\u00a0it would have taken 1 hour less to cover the same distance. How much time will it take to cover 315 km at its usual speed?<\/p>\n a)\u00a07h 52m<\/p>\n b)\u00a06h 30m<\/p>\n c)\u00a06h 18m<\/p>\n d)\u00a07h<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the uniform speed of train = s<\/p>\n According to the problem,<\/p>\n $\\frac{450}{s+5}=\\frac{450}{s}-1$<\/p>\n $\\frac{450}{s}-\\frac{450}{s+5}=1$<\/p>\n $450\\left(\\frac{s+5-s}{s\\left(s+5\\right)}\\right)=1$<\/p>\n $450\\left(\\frac{5}{s^2+5s}\\right)=1$<\/p>\n $s^2+5s-2250=0$<\/p>\n $\\left(s+50\\right)\\left(s-45\\right)=0$<\/p>\n s = -50 or s = 45<\/p>\n s cannot be negative, so s = 45 km\/h<\/p>\n The uniform speed of train =\u00a045 km\/h<\/p>\n Time taken by train to cover 315 km at its usual speed =\u00a0$\\frac{315}{45}$ = 7 hours<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 10:\u00a0<\/b>A train running at 48 km\/h crosses a man going with the speed of 12 km\/h, in the same direction, in 18 seconds and passes a woman coming from the opposite direction in 12 seconds. The speed (in km\/h) of the woman is :<\/p>\n a)\u00a06<\/p>\n b)\u00a08<\/p>\n c)\u00a09<\/p>\n d)\u00a010<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the length of the train = L<\/p>\n Relative speed between train and man = 48 – 12 = 36 km\/h<\/p>\n = 36 x $\\frac{5}{18}$ m\/sec<\/p>\n = 10 m\/sec<\/p>\n Time taken by train to cross the man = 18 seconds<\/p>\n $\\frac{L}{10}$ = 18<\/p>\n L = 180 m<\/p>\n Length of the train = 180 m<\/p>\n Let the speed of the woman = s km\/h<\/p>\n Relative\u00a0speed between train and woman = (48 + s) km\/h<\/p>\n =\u00a0(48 + s)\u00a0x $\\frac{5}{18}$ m\/sec<\/p>\n Time taken by train to cross the woman = 12 seconds<\/p>\n $\\frac{L}{\\left(48+s\\right)\\times\\frac{5}{18}}=12$<\/p>\n $\\frac{180}{\\left(48+s\\right)\\times\\frac{5}{18}}=12$<\/p>\n $\\frac{180\\times18}{\\left(48+s\\right)5}=12$<\/p>\n 48 + s = 54<\/p>\n s = 6<\/p>\n Speed of the woman = 6 km\/h<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 11:\u00a0<\/b>A car can cover a distance of 144 km in 1.8 hours. In what time(in hours) will it cover double the distance when its speed is increased by 20% ?<\/p>\n a)\u00a03<\/p>\n b)\u00a02.5<\/p>\n c)\u00a02<\/p>\n d)\u00a03.2<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of the car = $\\frac{144}{1.8}$ = 80 km\/hr<\/p>\n Speed of the car when increased by 20% = $\\frac{120}{100}\\times$80 = 96 km\/hr<\/p>\n Required time = $\\frac{288}{96}$<\/p>\n = 3 hours<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 12:\u00a0<\/b>To do a certain work, efficiencies of A and B are in the ratio 7:5. Working together, they can complete the work in $17\\frac{1}{2}$ days. In how many days, will B alone complete 50% of the same work?<\/p>\n a)\u00a042<\/p>\n b)\u00a021<\/p>\n c)\u00a015<\/p>\n d)\u00a030<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work = 700 units<\/p>\n Efficiencies of A and B are in the ratio 7:5.<\/p>\n Let the efficiency of A and B are 7p and 5p respectively.<\/p>\n Efficiencies of A and B together = 7p + 5p = 12p units\/day<\/p>\n Working together, they can complete the work in $17\\frac{1}{2}$ days.<\/p>\n $\\frac{700}{12p}$ =\u00a0$17\\frac{1}{2}$<\/p>\n $\\frac{700}{12p}$ =\u00a0$\\frac{35}{2}$<\/p>\n p =\u00a0$\\frac{10}{3}$<\/p>\n Efficiency of B = 5p =\u00a0$\\frac{50}{3}$ units\/day<\/p>\n Number of days required for\u00a0B alone complete 50% of the same work = $\\frac{350}{\\frac{50}{3}}$<\/p>\n =\u00a0$\\frac{350\\times3}{50}$<\/p>\n = 21 days<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 13:\u00a0<\/b>A boat goes 30 km upstream in 3 hours and downstream in 1 hour. How much time (in hours) will this boat take to cover 60 km in still water?<\/p>\n a)\u00a06<\/p>\n b)\u00a03<\/p>\n c)\u00a02<\/p>\n d)\u00a05<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the boat in still water = m<\/p>\n Speed of the stream = s<\/p>\n Upstream speed = m – s<\/p>\n $\\frac{30}{3}$ = m – s<\/p>\n m – s = 10…………(1)<\/p>\n Downstream speed = m + s<\/p>\n $\\frac{30}{1}$ = m + s<\/p>\n m + s = 30………..(2)<\/p>\n Adding (1) and (2),<\/p>\n 2m = 40<\/p>\n m = 20<\/p>\n Speed of the boat in still water = 20 km\/h<\/p>\n Time required for the\u00a0boat to cover 60 km in still water = $\\frac{60}{20}$ = 3 hours<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 14:\u00a0<\/b>A is twice as good a workman as B and together they finish a piece of work in 22 days. In how many days will A alone finish the same work?<\/p>\n a)\u00a030 days<\/p>\n b)\u00a044 days<\/p>\n c)\u00a033 days<\/p>\n d)\u00a011 days<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work = W<\/p>\n A is twice as good a workman as B<\/p>\n Let the number of days required for A alone to complete the work = a<\/p>\n $\\Rightarrow$ Number of days required for B alone to complete the work = 2a<\/p>\n Work done by A in 1 day = $\\frac{W}{a}$<\/p>\n Work done by B in 1 day =\u00a0$\\frac{W}{2a}$<\/p>\n Given, A and B together finish the work in 22 days<\/p>\n $\\Rightarrow$\u00a0 Work done by A and B together in 1 day =\u00a0$\\frac{W}{22}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{W}{a}$ +\u00a0$\\frac{W}{2a}$ =\u00a0$\\frac{W}{22}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{3}{2a}$ = $\\frac{1}{22}$<\/p>\n $\\Rightarrow$\u00a0\u00a0 a = 33<\/p>\n $\\therefore\\ $Number of days required for A alone to complete the work = 33 days<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 15:\u00a0<\/b>The speed of a train is 220% of the speed of a car. The car covers a distance of 950 km in 19 hours. How much distance will the train cover in $3 \\frac{1}{2}$ hours?<\/p>\n a)\u00a0380 km<\/p>\n b)\u00a0385 km<\/p>\n c)\u00a0375 km<\/p>\n d)\u00a0285 km<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of the car =\u00a0$\\frac{950}{19}$ = 50 km\/h<\/p>\n Given, speed of the train is 220% of the speed of the car<\/p>\n $\\Rightarrow$\u00a0 Speed of the train =\u00a0$\\frac{220}{100}\\times50$ =\u00a0 110 km\/h<\/p>\n $\\therefore\\ $Distance covered by the train in\u00a0$3 \\frac{1}{2}$ hours =\u00a0$110\\times3\\frac{1}{2}$ =\u00a0$110\\times\\frac{7}{2}$ = 385 km<\/p>\n Hence, the correct answer is Option B<\/p>\n