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Download Time and Work Questions for SSC CHSL and MTS<\/a><\/p>\n Enroll to 15 SSC CHSL 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\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 1)\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 2:\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 2)\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 3:\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 3)\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 4:\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 4)\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 5:\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 5)\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 Take a free SSC CHSL Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\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 6)\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 7:\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 7)\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 8:\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 8)\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 9:\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 9)\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 10:\u00a0<\/b>A and B together can complete a piece of work in 15 days. B and C together can do it in 24 days. If A is twice as good a workman as C, then in how many days can B alone complete the work?<\/p>\n a)\u00a060 days<\/p>\n b)\u00a040 days<\/p>\n c)\u00a052 days<\/p>\n d)\u00a045 days<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work = W<\/p>\n Given,\u00a0A is twice as good a workman as C<\/p>\n Let the number of days required for A alone to complete the work = a<\/p>\n $\\Rightarrow$ Number of days required for C alone to complete the work = 2a<\/p>\n Let the\u00a0number of days required for B alone to complete the work = b<\/p>\n Work done by B in 1 day =\u00a0$\\frac{W}{b}$<\/p>\n Work done by A in 1 day = $\\frac{W}{a}$<\/p>\n Work done by C in 1 day = $\\frac{W}{2a}$<\/p>\n A and B together can complete a piece of work in 15 days<\/p>\n $\\Rightarrow$\u00a0 Work done by A and B together in 1 day =\u00a0$\\frac{W}{15}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{W}{a}$ + $\\frac{W}{b}$ = $\\frac{W}{15}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{1}{a}$ =\u00a0$\\frac{1}{15}$ –\u00a0$\\frac{1}{b}$ …………(1)<\/p>\n B and C together can complete the work in 24 days<\/p>\n $\\Rightarrow$\u00a0 Work done by B and C together in 1 day =\u00a0$\\frac{W}{24}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{W}{b}$ +\u00a0$\\frac{W}{2a}$ =\u00a0$\\frac{W}{24}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{1}{b}$ +\u00a0$\\frac{1}{2a}$ =\u00a0$\\frac{1}{24}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{1}{b}$ +\u00a0$\\frac{1}{2}\\left[\\frac{1}{15}-\\frac{1}{b}\\right]$ =\u00a0$\\frac{1}{24}$\u00a0\u00a0 [From (1)]<\/p>\n $\\Rightarrow$ \u00a0$\\frac{1}{2b}$ =\u00a0$\\frac{1}{24}$ –\u00a0$\\frac{1}{30}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{1}{2b}$ =\u00a0$\\frac{5-4}{120}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{1}{b}$ =\u00a0$\\frac{1}{60}$<\/p>\n $\\Rightarrow$\u00a0 b = 60<\/p>\n $\\therefore\\ $Number of days required for B alone to complete the work = 60 days<\/p>\n Hence, the correct answer is Option A<\/p>\n Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 11:\u00a0<\/b>Rahul and Mithun travel a distance of 30 km. The sum of their speeds is 70 km\/h and the total time taken by both to travel the distance is 2 hours 6 minutes. The difference between their speeds is:<\/p>\n a)\u00a035 km\/h<\/p>\n b)\u00a020 km\/h<\/p>\n c)\u00a025 km\/h<\/p>\n d)\u00a030 km\/h<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of Rahul = s<\/p>\n $\\Rightarrow$ Speed of Mithun = 70 – s<\/p>\n Time taken by Rahul to cover 30 km distance =\u00a0$\\frac{30}{s}$<\/p>\n Time taken by Mithun to cover 30 km distance = $\\frac{30}{70-s}$<\/p>\n Given, total time = 2 hours 6 minutes = 2 + $\\frac{6}{60}$ hours = 2 + $\\frac{1}{10}$ hours =\u00a0$\\frac{21}{10}$ hours<\/p>\n $\\Rightarrow$ \u00a0$\\frac{30}{s}+\\frac{30}{70-s}=\\frac{21}{10}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{1}{s}+\\frac{1}{70-s}=\\frac{7}{100}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{70-s+s}{s\\left(70-s\\right)}=\\frac{7}{100}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{70}{70s-s^2}=\\frac{7}{100}$<\/p>\n $\\Rightarrow$ \u00a0$70s-s^2=1000$<\/p>\n $\\Rightarrow$ \u00a0$s^2-70s+1000=0$<\/p>\n $\\Rightarrow$ \u00a0$s^2-50s-20s+1000=0$<\/p>\n $\\Rightarrow$ \u00a0$s\\left(s-50\\right)-20\\left(s-50\\right)=0$<\/p>\n $\\Rightarrow$ \u00a0$\\left(s-50\\right)\\left(s-20\\right)=0$<\/p>\n $\\Rightarrow$ \u00a0$s-50=0$ \u00a0 or \u00a0 $s-20=0$<\/p>\n $\\Rightarrow$\u00a0 s = 50 km\/h or \u00a0 s = 20 km\/h<\/p>\n When speed of Rahul = 50 km\/h, speed of mithun = 20 km\/h<\/p>\n When speed of Rahul = 20 km\/h, speed of mithun = 50 km\/h<\/p>\n $\\therefore\\ $Difference between their speeds = 30 km\/h<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 12:\u00a0<\/b>10 men working 5 hours\/day earn \u20b9300. How much money will 15 men working 10 hours\/day earn?<\/p>\n a)\u00a0\u20b9 800<\/p>\n b)\u00a0\u20b9 600<\/p>\n c)\u00a0\u20b9 650<\/p>\n d)\u00a0\u20b9 900<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a010 men working 5 hours\/day earn \u20b9300<\/p>\n 1 man working 5 hours\/day earn \u20b930<\/p>\n 15 men working 5 hours\/day\u00a0 earn 30 x 15 = \u20b9450<\/p>\n $\\Rightarrow$\u00a0 15 men working 10 hours\/day earn 450 x 2 =\u00a0\u20b9900<\/p>\n Question 13:\u00a0<\/b>P and Q can finish a work in 10 days and 5 days, respectively. Q worked for 2 days and left the job. In how many days can P alone finish the remaining work?<\/p>\n a)\u00a06 days<\/p>\n b)\u00a04 days<\/p>\n c)\u00a010 days<\/p>\n d)\u00a08 days<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work be W<\/p>\n Number of required for P to finish the work = 10 days<\/p>\n $=$>\u00a0 Work done by P in 1 day = $\\frac{W}{10}$<\/p>\n Number of required for Q to finish the work = 5 days<\/p>\n $=$> Work done by Q in 1 day = $\\frac{W}{5}$<\/p>\n $=$> Work done by Q in 2 days = $\\frac{2W}{5}$<\/p>\n Remaining work = $\\text{W}-\\frac{2W}{5}$ =\u00a0$\\frac{3W}{5}$<\/p>\n $\\therefore\\ $Number of days required for P to finish the remaining work = $\\frac{\\frac{3W}{5}}{\\frac{W}{10}}$ = $\\frac{3W}{5}\\times\\frac{10}{W}$ = 6 days<\/p>\n Hence, the correct answer is Option A<\/p>\n Free SSC Preparation Videos<\/a><\/p>\n Download SSC CHSL\u00a0 Previous Papers PDF<\/a><\/p>\n Question 14:\u00a0<\/b>Raju can finish a piece of work in 20 days. He worked at it for 5 days and then Jakob alone finished the remaining work in 15 days. In how many days can both finish it together?<\/p>\n a)\u00a020 days<\/p>\n b)\u00a012 days<\/p>\n c)\u00a010 days<\/p>\n d)\u00a016 days<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total work = W<\/p>\n Number of days required for Raju to complete the work = 20 days<\/p>\n $=$>\u00a0 Work done by Raju in 1 day = $\\frac{W}{20}$<\/p>\n $=$>\u00a0 Work done by Raju in 5 days =\u00a0$\\frac{5W}{20}=\\frac{W}{4}$<\/p>\n Remaining work =\u00a0$W-\\frac{W}{4}=\\frac{3W}{4}$<\/p>\n $\\therefore\\ $Work done by Jakob in 15 days = $\\frac{3W}{4}$<\/p>\n $=$>\u00a0 Work done by Jakob in 1 day\u00a0= $\\frac{3W}{60}=\\frac{W}{20}$<\/p>\n $=$> \u00a0Work done by Raju and Jakob in 1 day =\u00a0$\\frac{W}{20}+\\frac{W}{20}=\\frac{2W}{20}=\\frac{W}{10}$<\/p>\n $=$> \u00a0Number of days required for both Raju and Jakob to complete the work =\u00a0$\\frac{W}{\\frac{W}{10}}=10$ days<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 15:\u00a0<\/b>Ramu works 4 times as fast as Somu. If Somu can complete a work in 20 days independently, then the number of days in which Ramu and Somu together can complete the work is:<\/p>\n a)\u00a04 days<\/p>\n b)\u00a05 days<\/p>\n c)\u00a03 days<\/p>\n d)\u00a06 days<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of days required for Somu to complete work = 20 days<\/p>\n Ramu works 4 times as fast as Somu<\/p>\n $=$> Number of days required for Ramu to complete work = $\\frac{20}{4}=5$ days<\/p>\n Let the Total Work = W<\/p>\n Work done by Ramu in 1 day =\u00a0$\\frac{W}{5}$<\/p>\n Work done by Somu in 1 day =\u00a0$\\frac{W}{20}$<\/p>\n Work done by Ramu and Somu in 1 day =\u00a0$\\frac{W}{5}+\\frac{W}{20}=\\frac{5W}{20}=\\frac{W}{4}$<\/p>\n Number of days required for both Ramu and Somu together to complete work =\u00a0$\\frac{W}{\\frac{W}{4}}=4$ days<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 16:\u00a0<\/b>Water flows into a tank 180 m $\\times$ 140 m through a rectangular pipe of 1.2m $\\times$ 0.75 m at a rate of 15 km\/h. In what time will the water rise by 4 m?<\/p>\n a)\u00a07 hours 28 minutes<\/p>\n b)\u00a06 hours 42 minutes<\/p>\n c)\u00a08 hours 12 minutes<\/p>\n d)\u00a05 hours 46 minutes<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Dimensions of the tank = 180 m $\\times$ 140 m<\/p>\n Dimensions of the rectangular pipe =\u00a01.2m $\\times$ 0.75 m<\/p>\n Rate of water flow = 15km\/h =\u00a0$15\\times\\frac{5}{18}$ m\/s =\u00a0$\\frac{25}{6}$ m\/s<\/p>\n $\\therefore\\ $Time required to rise the water by 4 m = $\\frac{180\\times140\\times4}{1.2\\times0.75\\times\\frac{25}{6}}$<\/p>\n $=\\frac{180\\times140\\times4\\times6\\times10\\times100}{12\\times75\\times25}$<\/p>\n $=26880$ sec<\/p>\n $=\\frac{26880}{3600}$ hr<\/p>\n $=\\frac{112}{15}$ hr<\/p>\n $=7\\frac{7}{15}$ hr<\/p>\n $=$ 7 hr +\u00a0$\\frac{7}{15}\\times60$ min<\/p>\n $=$ 7 hr 28 min<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 17:\u00a0<\/b>A can finish a piece of work in a certain number of days. B takes 45% more number of days to finish the same work independently. They worked together for 58 days and then the remaining work was done by B alone in 29 days. In how many days could A have completed the work, had he worked alone?<\/p>\n a)\u00a0110 days<\/p>\n b)\u00a0118 days<\/p>\n c)\u00a098 days<\/p>\n d)\u00a0120 days<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the number of days required for A alone to complete the work = 100a<\/p>\n $=$>\u00a0 Number of days required for B alone to complete the work = 145a<\/p>\n Let the total work = W<\/p>\n Work done by A in 1 day =\u00a0$\\frac{W}{100a}$<\/p>\n Work done by B in 1 day = $\\frac{W}{145a}$<\/p>\n Work done by A and B in 1 day = $\\frac{W}{100a}+\\frac{W}{145a}=\\frac{245W}{14500a}=\\frac{49W}{2900a}$<\/p>\n Work done by A and B in 58 days$=58\\times\\frac{49W}{2900a}=\\frac{49W}{50a}$<\/p>\n Remaining work $=W-\\frac{49W}{50a}$<\/p>\n Remaining work is completed by B in 29 days<\/p>\n $=$>\u00a0 $W-\\frac{49W}{50a}=\\frac{29W}{145a}$<\/p>\n $=$> \u00a0$W=\\frac{29W}{145a}+\\frac{49W}{50a}$<\/p>\n $=$> \u00a0$W=\\frac{1450W+7105W}{145\\times50\\times a}$<\/p>\n $=$> \u00a0$W=\\frac{8555W}{145\\times50\\times a}$<\/p>\n $=$> \u00a0$W=\\frac{59W}{50a}$<\/p>\n $=$> \u00a0$50a=59$<\/p>\n $=$> \u00a0$100a=118$<\/p>\n $\\therefore\\ $Number of days required for A alone to complete the work = 100a = 118 days<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 18:\u00a0<\/b>Shyam can complete a task in 12 days by working 10 hours a day. How many hours a day should he work to complete the task in 8 days?<\/p>\n a)\u00a012<\/p>\n b)\u00a015<\/p>\n c)\u00a016<\/p>\n d)\u00a014<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, Shyam can complete the task in 12 days working 10 hours a day<\/p>\n $=$> Total time required for Shyam to complete the task = 12 x 10 = 120 hours<\/p>\n $\\therefore\\ $ Number of hours Shyam should work in a day to complete task in 8 days = $\\frac{120}{8}$ = 15 hours<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 19:\u00a0<\/b>A boat can go 3.6 km upstream and 5.4 km downstream in 54 minutes, while it can go 5.4 km upstream and 3.6 km downstream in 58.5 minutes. The time (in minutes) taken by the boat in going 10 km downstream is:<\/p>\n a)\u00a048<\/p>\n b)\u00a050<\/p>\n c)\u00a045<\/p>\n d)\u00a054<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of speed of stream be u and speed of boat in still water be\u00a0v.<\/p>\n Speed of boat in upstream = u – v<\/p>\n Speed of boat in downstream = u + v<\/p>\n A boat can go 3.6 km upstream and 5.4 km downstream in 54 minutes so,<\/p>\n Time = distance\/speed<\/p>\n $\\frac{3.6}{u – v} + \\frac{5.4}{u + v} = \\frac{54}{60}$<\/p>\n $\\frac{3.6}{u – v} + \\frac{5.4}{u + v} = \\frac{9}{10}$<\/p>\n $\\frac{36}{u – v} + \\frac{54}{u + v} = 9$<\/p>\n 4(u +\u00a0v) + 6(u – v) = $u^2 – v^2$<\/p>\n 10u – 2v =\u00a0$u^2 – v^2$ —(1)<\/p>\n Boat\u00a0can go 5.4 km upstream and 3.6 km downstream in 58.5 minutes so,<\/p>\n $\\frac{5.4}{u – v} + \\frac{3.6}{u + v} = \\frac{58.5}{60}$<\/p>\n $\\frac{54}{u – v} + \\frac{36}{u + v} = \\frac{585}{60}$<\/p>\n $\\frac{6}{u – v} + \\frac{4}{u + v} = \\frac{13}{12}$<\/p>\n 72(u + v) + 48(u – v) = 13($u^2 – v^2$)<\/p>\n 120u + 24v =\u00a013($u^2 – v^2$)<\/p>\n Put the value of eq(1),<\/p>\n 120u + 24v = 13 $\\times\u00a0(10u – 2v)$<\/p>\n 120u + 24v = 130u – 26v<\/p>\n 10u = 50v<\/p>\n u = 5v —(2)<\/p>\n put the value of u in eq(1),<\/p>\n 50v – 2v = $25v^2 – v^2$<\/p>\n 48v = 24v^2<\/p>\n v = 2<\/p>\n put the value of v in eq(2),<\/p>\n u = 5$\\times$ 2 = 10<\/p>\n Speed of boat in downstream = u + v = 10 + 2 = 12<\/p>\n The time taken by the boat in going 10 km downstream = distance\/speed = 10\/12 hours = 60 $\\times$ 10\/12 = 50 min<\/p>\n Question 20:\u00a0<\/b>A, B and C can individually complete a task in 24 days, 20 days and 18 days,\u00a0respectively. B and C start the task, and they work for 6 days and leave. The number of days\u00a0required by A alone to finish the remaining task, is:<\/p>\n a)\u00a0$15\\frac{2}{3}$ days<\/p>\n b)\u00a0$12\\frac{1}{2}$ days<\/p>\n c)\u00a0$8\\frac{4}{5}$ days<\/p>\n d)\u00a0$10$ days<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total taskj be 360.<\/p>\n ($\\because$ LCM of 24, 20 and 18 is 360.)<\/p>\n Efficiency of A = 360\/24 = 15<\/p>\n Efficiency of B = 360\/20 = 18<\/p>\n Efficiency of C = 360\/18 = 20<\/p>\n Task done by B and C in 6 days = total efficiency $\\times$ time = (18 + 20) $\\times$ 6 = 228<\/p>\n Remaining Task = 360 – 228 = 132<\/p>\n Time taken to complete\u00a0remaining task by A = work\/efficiency = 132\/15 = 8 $\\frac{12}{15}$\u00a0=\u00a08 $\\frac{4}{5}$<\/p>\n