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Download Time and Work Questions for SSC MTS<\/a><\/p>\n Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>Ten men or twelve women can finish the same workin 10 days. If 5 men and 2 women undertake the work together, how many days will they take to complete the work?<\/p>\n a)\u00a015<\/p>\n b)\u00a040<\/p>\n c)\u00a020<\/p>\n d)\u00a060<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total work = time $\\times$ (men + women)<\/p>\n Ten men or twelve women can finish the same workin 10 days so,<\/p>\n $10m \\times 10 = 12w \\times 10$<\/p>\n 5m = 6w<\/p>\n Let\u00a0time taken to complete same work by 5 men and 2 women be d days.<\/p>\n Total work done by\u00a05 men and 2 women = (5m + 2w) $\\times$ d<\/p>\n 12w $\\times 10$ =\u00a0(6w + 2w) $\\times$ d<\/p>\n d = 120\/8 = 15 days<\/p>\n $\\therefore$ They will completed the same work in 15 days.<\/p>\n Question 2:\u00a0<\/b>Two cars A and B leave Delhi at 8:30 a.m. and at 9 a.m. for Shimla, respectively. They travel at the speeds of 40km\/h and 50 km\/h respectively. How many kilometres away from Delhi will the two cars be together?<\/p>\n a)\u00a045 km<\/p>\n b)\u00a0100 km<\/p>\n c)\u00a0200 km<\/p>\n d)\u00a05 km<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Cars A and B leave Delhi at 8:30 a.m. and at 9 a.m. for Shimla, respectively so,<\/p>\n Distance traveled by car A in 30 min = speed $\\times times = 40 \\times 1\/2 = 20 km<\/p>\n (30 min = 30\/60 hr = 1\/2 hr)<\/p>\n Relative speed of car B = 50 – 40 = 10 km\/hr<\/p>\n Time taken by car B to cover 20 km\u00a0 = 20\/10 = 2 hr<\/p>\n (Time = distance\/speed)<\/p>\n Distance covered in 2 hr = 50 \\times 2 = 100 km<\/p>\n $\\therefore$ 100\u00a0kilometers away from Delhi will the two cars be together.<\/p>\n Question 3:\u00a0<\/b>A and B, working together, can complete a work in d days. Working alone, A takes (8 + d) days and B takes (18 + d) days to complete the same work. A works for 4 days. The remaining work will be completed by B alone, in:<\/p>\n a)\u00a024 days<\/p>\n b)\u00a016 days<\/p>\n c)\u00a018 days<\/p>\n d)\u00a020 days<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n To find the time when\u00a0A and B, working together = $\\sqrt{more days taken by A \\times more days taken by B}$<\/p>\n d = $\\sqr{8 \\times 18}$ = 12<\/p>\n Time taken by A to complete the work = (8 + d) = 8 + 12 = 20<\/p>\n Time taken by B to complete the work = (18 + d) = 18 + 12 = 30<\/p>\n Let the total work be 60.<\/p>\n (LCM of 20 and 30 is 60.)<\/p>\n Efficiency of A = 60\/20 = 3<\/p>\n Efficiency of B = 60\/30 = 2<\/p>\n Work done by A in 4 days = efficiency $\\times time = 3 \\times 4 = 12$<\/p>\n Remaining work = 60 – 12 = 48<\/p>\n Time taken by B to complete the remaining work = 48\/2 = 24 days<\/p>\n $\\therefore$\u00a0The remaining work will be completed by B alone, in 24 days.<\/p>\n Question 4:\u00a0<\/b>The time taken by a boat to travel 13 km downstream is the same as time taken by it to travel 7 km upstream. If the speed of the stream is 3 km\/h, then how much time (in hours) will it take to travel a distance of 44.8 km in still water?<\/p>\n a)\u00a0$4\\frac{12}{25}$<\/p>\n b)\u00a0$5\\frac{3}{5}$<\/p>\n c)\u00a0$5\\frac{2}{5}$<\/p>\n d)\u00a0$4\\frac{13}{25}$<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of the stream = 3 km\/h<\/p>\n Let the speed of the boat be v.<\/p>\n Speed in downstream = 3 + v<\/p>\n Speed in upstream = v – 3<\/p>\n Time = distance\/speed<\/p>\n According to the question,<\/p>\n The time taken by a boat to travel 13 km downstream = time taken by boat to travel 7 km upstream<\/p>\n 13\/(3 + v) = 7\/(v – 3)<\/p>\n 13v – 39 = 21 + 7v<\/p>\n 6v = 60<\/p>\n v = 10 km\/h<\/p>\n Speed of boat = 10 km\/h<\/p>\n The time taken by a boat to travel\u00a0a distance of 44.8 km in still water = 44.8\/10 = 112\/25 = $ 4 \\frac{12}{25}$ hr<\/p>\n Question 5:\u00a0<\/b>Eight persons can finish a work in 20 days. After 5 days they were requested to complete the work in the next 8 days. How many more persons should join the group to fulfil the requirement?<\/p>\n a)\u00a023<\/p>\n b)\u00a015<\/p>\n c)\u00a07<\/p>\n d)\u00a012<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total work = number of persons $\\times time = 8 \\times 20 = 160$<\/p>\n Work done in 5 days = 8 $times 5 = 40$<\/p>\n Remaining Work = 160 – 40 = 120<\/p>\n According to question,<\/p>\n Number of man $\\times$\u00a0time = 120<\/p>\n Number of man $\\times$ 8 = 120<\/p>\n Number of man = 120\/8 = 15<\/p>\n Number of\u00a0persons should join the group = 15 – 8 = 7<\/p>\n Take a free SSC MTS Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>A can finish a work in 20 days and B can finish the same work in 25 days. They begin together, but B left the work after 5 days. How many more days will A take to finish the remaining work?<\/p>\n a)\u00a011<\/p>\n b)\u00a08<\/p>\n c)\u00a016<\/p>\n d)\u00a021<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work be 100 units.<\/p>\n ($\\because LCM of 20 and 25 is 100.$)<\/p>\n Efficiency of A = work\/time = 100\/20 = 5 unit\/day<\/p>\n Efficiency of B = 100\/25 = 4 unit\/day<\/p>\n Work done by A and B in 5 days = efficiency $\\times times =\u00a0(4 +\u00a0 5) \\times$ 5 = 45 units<\/p>\n Remaining work = 100 – 45 = 55 units<\/p>\n Time taken by A\u00a0to complete 55 units work = 55\/5 = 11 days<\/p>\n Question 7:\u00a0<\/b>A can do a piece of work in 6 days. B can do it in 9 days. With the assistance of C they completed the work in 3 days. In how many days can C alone do the work ?<\/p>\n a)\u00a018<\/p>\n b)\u00a08<\/p>\n c)\u00a012<\/p>\n d)\u00a016<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work be 18 units.<\/p>\n ($\\because$ LCM of 6, 9 and 3 is 18.)<\/p>\n Efficiency of A = work\/time = 18\/6 = 3 units\/day<\/p>\n Efficiency of B = 18\/9 = 2\u00a0units\/day<\/p>\n Efficiency of (A + B + C)\u00a0= 18\/3 = 6\u00a0units\/day<\/p>\n Efficiency of C =\u00a0Efficiency of (A + B + C) –\u00a0Efficiency of A –\u00a0Efficiency of B = 6 – 3 – 2 = 1 unit\/day<\/p>\n Time taken by C to complete work alone = 18\/1 = 18 days<\/p>\n Question 8:\u00a0<\/b>A train crosses a platform 180 m long in 60 sec at a speed of 72 km\/h. The time taken by the train to cross an electric pole is:<\/p>\n a)\u00a051 sec<\/p>\n b)\u00a00.51 min<\/p>\n c)\u00a05.1 sec<\/p>\n d)\u00a05.1 min<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Length of platform = 180 m<\/p>\n Time = 60 sec<\/p>\n Speed = 72 km\/hr\u00a0= 72$\\times$ 5\/18 = 20 m\/sec<\/p>\n Total length =\u00a0length of platform +\u00a0length of train = 180 +\u00a0length of train<\/p>\n Speed = length\/time<\/p>\n 180 + length of train = 20 $\\times$ 60<\/p>\n Length of train =1200 – 180 = 1020 m<\/p>\n The time taken by the train to cross an electric pole = 1020\/20 = 51 sec<\/p>\n Question 9:\u00a0<\/b>The two trains leave Varanasi for Lucknow at 11:00 a.m. and at 11:30 a.m., respectively and travel at a speed of 110 km\/h and 140 km\/h, respectively. How many kilometers from Varanasi will both trains meet?<\/p>\n a)\u00a0$255\\frac{1}{3}$ km<\/p>\n b)\u00a0$238\\frac{2}{3}$ km<\/p>\n c)\u00a0$246\\frac{1}{3}$ km<\/p>\n d)\u00a0$256\\frac{2}{3}$ km<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the train which leaves Varanasi at 11:00 be A and another train be B.<\/p>\n Speed of train A = 110 km\/hr<\/p>\n Speed of train B = 140 km\/hr<\/p>\n Distance covered by train A in 30 min(30\/60 = 1\/2 hr)\u00a0= speed $\\times time = 110 \\times 1\/2 = 55 km$<\/p>\n At 11:30, distance between both trains will be 55 km.<\/p>\n Relative speed of train B = 140 – 110 = 30 km\/hr<\/p>\n Time taken by train B\u00a0to meet = 55\/30 = 11\/6 hr<\/p>\n Distance covered by train B in 11\/6 hr = 140 $\\times 11\/6 = 256 \\frac{2}{3}$<\/p>\n $\\therefore 256 \\frac{2}{3}$\u00a0kilometers from Varanasi will both trains meet.<\/p>\n Question 10:\u00a0<\/b>A contract is to be completed in 75 days and 187 men are to work 15 hours per day. After 65 days, $\\frac{3}{5}$ of the work is completed. How many additional men may be employed, so that the work may be completed in time, each man now working 17 hours per day?<\/p>\n a)\u00a0532<\/p>\n b)\u00a0495<\/p>\n c)\u00a0308<\/p>\n d)\u00a0514<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $10\\times\\ \\ \\ $From the formula,<\/p>\n Total work =\u00a0$\\frac{mdh}{w1}$<\/p>\n here, m = man\/woman<\/p>\n d =days<\/p>\n h = hour<\/p>\n w = work<\/p>\n Total work = $75 \\times\u00a0187 \\times 15$\u00a0=\u00a0210375<\/p>\n Work done in 65 days =\u00a0210375 $\\times \\frac{3}{5}$ = 126255<\/p>\n Remaining work =\u00a0210375 –\u00a0126255 = 84150<\/p>\n $(187 + m) \\times\u00a010 \\times 17$ = 84150<\/p>\n (187 + m) = 495<\/p>\n m = 495 -187 = 308<\/p>\n $\\therefore$ 308\u00a0additional men may be employed, so that the work may be completed in time.<\/p>\n Question 11:\u00a0<\/b>Two cyclists X and Y start at the same time from place A and go towards place B at a speed of 6 km\/h and 8 km\/h,respectively. Despite stopping for 15 minutes during the journey, Y reaches 10 minutes earlier than X. The distance between the places A and B is:<\/p>\n a)\u00a06 km<\/p>\n b)\u00a016.5 km<\/p>\n c)\u00a08 km<\/p>\n d)\u00a010km<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Time = distance\/speed<\/p>\n Time taken by x to reach destination =\u00a0Time taken by y to reach destination + 10 min<\/p>\n Let the Distance be d.<\/p>\n $\\frac{d}{6} = \\frac{d}{8}+ \\frac{15}{60}+\u00a0\\frac{10}{60}$ $\\therefore$\u00a0The distance between the places A and B is 10 km.<\/p>\n Question 12:\u00a0<\/b>Amit and Sunil together can complete a work in 9 days, Sunil and Dinesh together can complete the same work in 12 days, and Amit and Dinesh together can complete the same work in 18 days. In how many days will they complete the work if Amit, Sunil and Dinesh work together?<\/p>\n a)\u00a012 days<\/p>\n b)\u00a08 days<\/p>\n c)\u00a014 days<\/p>\n d)\u00a016 days<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work be 36 units.<\/p>\n ($\\because$ LCM of 9, 12 and 18 is 36.)<\/p>\n Efficiency of\u00a0Amit and Sunil = work\/time = 36\/9 = 4 units\/day<\/p>\n Efficiency of\u00a0Sunil and Dinesh = 36\/12 = 3 units\/day<\/p>\n Efficiency of\u00a0Amit and Dinesh = 36\/18 = 2\u00a0units\/day<\/p>\n Efficiency of Amit,\u00a0Sunil and\u00a0Dinesh = $\\frac{4 + 3 + 2}{2}$ = 9\/2<\/p>\n Time taken by Amit, Sunil and Dinesh to complete the work = $\\frac{36}{9\/2}$ = 8 days<\/p>\n Question 13:\u00a0<\/b>A, B and C can individually complete a task in 24 days,16 days and 32 days respectively. If A and C start the work and worked for 6 days and left, then the number of days required by B to complete the remaining task, is:<\/p>\n a)\u00a0$9$<\/p>\n b)\u00a0$17\\frac{1}{2}$<\/p>\n c)\u00a0$7\\frac{1}{2}$<\/p>\n d)\u00a0$12\\frac{1}{2}$<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work be 96 units.<\/p>\n ($\\because$ LCM of 24, 16 and 32 is 96.)<\/p>\n Efficiency of A = work\/time = 96\/24 = 4<\/p>\n Efficiency of B = 96\/16 = 6<\/p>\n Efficiency of C = 96\/32 = 3<\/p>\n Efficiency of A and C = 4 + 3 = 7<\/p>\n Work done by A and C in 6 days = 7 $\\times$ 6 = 42 units<\/p>\n Remaining work = 96 – 42 = 54 units<\/p>\n Time taken by B to complete 54 units work = 54\/6 = 9 days<\/p>\n Question 14:\u00a0<\/b>Sixteen men can finish a work in 8 days. Eight men and nine women working together can finish the same work in 10 days. In how many days will twenty women finish the same work?<\/p>\n a)\u00a09<\/p>\n b)\u00a012<\/p>\n c)\u00a011<\/p>\n d)\u00a013<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total work = man\/woman $\\times$ time<\/p>\n Work is same so,<\/p>\n 16m $\\times 8 = (8m + 9w) \\times 10<\/p>\n $\\Rightarrow$ 128m = 80m + 90w<\/p>\n $\\Rightarrow$ 48m = 90w<\/p>\n $\\Rightarrow$ 8m = 15w<\/p>\n Let the 20 women complete work in days so,<\/p>\n Total work = 20w $\\times$ d<\/p>\n $\\Rightarrow$ (8m + 9w) \\times 10\u00a0= 20w $\\times$ d<\/p>\n $\\Rightarrow$ (15w + 9w) \\times 10 = 20w $\\times$ d<\/p>\n $\\Rightarrow$ 24w \\times 10 = 20w $\\times$ d<\/p>\n $\\Rightarrow$ d = 12 days<\/p>\n $\\therefore$\u00a0Twenty women finish the same work in 12 days.<\/p>\n Question 15:\u00a0<\/b>Pipes A and B can fill a tank in 10 hours and 40 hours, respectively. C is an outlet pipe attached to the tank. If all the three pipes are opened simultaneously, it takes 80 minutes more time than what A and B together take to fill the tank. A and B are kept opened for 7 hours and then closed and C was opened. C will now empty the tank in:<\/p>\n a)\u00a049 hours<\/p>\n b)\u00a038.5 hours<\/p>\n c)\u00a042 hours<\/p>\n d)\u00a045.5 hours<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work be 40 units.<\/p>\n $(\\because$ L.C.M. of 10 and 40 is 40.)<\/p>\n Efficiency of A = work\/time = 40\/10 = 4 units\/hour<\/p>\n Efficiency of B = 40\/40 = 1\u00a0unit\/hour<\/p>\n Time time taken by pipe A and B = $\\frac{40}{4 +\u00a01}$ = 8 hours<\/p>\n Time time taken by pipe A, B and\u00a0C together = 8 hours + 80\/60 hours = 28\/3 hours<\/p>\n Efficiency of\u00a0 A, B and\u00a0C together = $\\frac{40}{28\/3}$ = 30\/7\u00a0units\/hour Work done by pipe A and B in 7 hours = (1 + 4)$\\times $7 = 35 units<\/p>\n Time taken by pipe C to empty the tank = $\\frac{35}{5\/7}\u00a0=\u00a0\u00a049$<\/p>\n
\n$\\frac{d}{6} = \\frac{d}{8}+ \\frac{1}{4}+ \\frac{1}{6}$
\n$\\frac{d}{6} = \\frac{3d + 6 + 4}{24}$
\nd= $\\frac{3d + 10}{4}$
\n4d = 3d +\u00a010
\nd = 10 km<\/p>\n
\nEfficiency of C alone = 30\/7 – 5 = -5\/7 (- as pipe C is an outlet pipe and does negative work)<\/p>\n