Here you can download a free Time and Work questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the Time and Work of answers for the given questions. These questions will help you to make practice and solve the Time and Work questions in the MAH MBA CET exam. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Time and Work MCQ<\/strong> PDF for MBA-CET 2022 for free.<\/p>\n Download Time and Work Questions for MAH-CET<\/a><\/p>\n Enroll to MAH-CET 2022 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>The speed of a motor boat in still water is 20 km\/h.It travels 150 km downstream and then returns to the starting point. If the round trip takes a total of 16 hours, what is the speed (in km\/h) of the flow of river?<\/p>\n a)\u00a08<\/p>\n b)\u00a06<\/p>\n c)\u00a05<\/p>\n d)\u00a04<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the flow of the river = s<\/p>\n The speed of motor boat in still water(m)\u00a0= 20 km\/h<\/p>\n Downstream speed = 20 + s<\/p>\n Upstream speed = 20 – s<\/p>\n Time taken for boat to travel 150 km downstream = $\\frac{150}{20+s}$<\/p>\n Time taken for boat to travel 150 km upstream = $\\frac{150}{20-s}$<\/p>\n Total time taken = 16 hours<\/p>\n $\\frac{150}{20+s}$ +\u00a0$\\frac{150}{20-s}$ = 16<\/p>\n 150[$\\frac{20-s+20+s}{(20+s)(20-s)}$] = 16<\/p>\n $\\frac{150\\times40}{16}$ =\u00a0(20+s)(20-s)<\/p>\n 375 = 400 – s$^2$<\/p>\n s$^2$ = 25<\/p>\n s = 5 km\/h<\/p>\n The speed of the flow of the river = s = 5 km\/h<\/p>\n Hence, the correct answer is Option C<\/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 Free MAH-CET mock tests here<\/a><\/strong><\/span><\/p>\n Enrol to 5 MAH CET Latest Mocks For Just Rs. 299<\/a><\/strong><\/span><\/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 Question 16:\u00a0<\/b>Kamal and Anil can dig a pond in 8 days and 14 days, respectively. If the total expense of digging is \u20b9 4,400, then how much money will Anil earn?<\/p>\n a)\u00a0\u20b9 1400<\/p>\n b)\u00a0\u20b9 1600<\/p>\n c)\u00a0\u20b9 1500<\/p>\n d)\u00a0\u20b9 1300<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of days required for Kamal to dig the pond = 8 days<\/p>\n $\\Rightarrow$ Workdone by Kamal in 1 day = $\\frac{1}{8}$ of the work<\/p>\n Number of days required for Anil to dig the pond = 14 days<\/p>\n $\\Rightarrow$ Workdone by Anil in 1 day = $\\frac{1}{14}$ of the work<\/p>\n Ratio of their efficiency = $\\frac{1}{8}:\\frac{1}{14}$ = 14 : 8<\/p>\n Total expense of digging =\u00a0\u20b9 4,400<\/p>\n Money that can be earned by Anil = $\\frac{8}{14+8}\\times4400$ = \u20b9 1600<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 17:\u00a0<\/b>A car covered 150 km in 5 hours. If it travels at one-third its usual speed, then how much more time will it take to cover the same distance?<\/p>\n a)\u00a012 hours<\/p>\n b)\u00a014 hours<\/p>\n c)\u00a010 hours<\/p>\n d)\u00a08 hours<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, the car covered\u00a0150 km in 5 hours<\/p>\n Speed of the car =\u00a0$\\frac{150}{5}$ = 30 km\/h<\/p>\n One third of the speed =\u00a0$\\frac{1}{3}\\times30$ = 10 km\/h<\/p>\n Time required for the car to cover 150 km with one third speed =\u00a0$\\frac{\\text{Distance}\\\\ }{\\text{Speed}}$ =\u00a0$\\frac{150}{10}$ = 15 hours<\/p>\n $\\therefore\\ $Extra time required to cover the distance with one third speed = 15 – 5 = 10 hours<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 18:\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 18)\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 Question 19:\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 19)\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\\ $Difference between their speeds = 30 km\/h<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 20:\u00a0<\/b>Mohan travels three equal distances at speeds of 12 km\/h, 18 km\/h and 24 km\/h.If he takes a total of 13 hours, then what is the total distance covered?<\/p>\n a)\u00a0214 km<\/p>\n b)\u00a0212 km<\/p>\n c)\u00a0216 km<\/p>\n d)\u00a0218 km<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total distance be 3d<\/p>\n Time taken to cover the distance at speed 12 km\/h =\u00a0$\\frac{d}{12}$ hour<\/p>\n Time taken to cover the distance at speed 18 km\/h = $\\frac{d}{18}$ hour<\/p>\n Time taken to cover the distance at speed 24 km\/h = $\\frac{d}{24}$ hour<\/p>\n Given, total time = 13 hours<\/p>\n $\\Rightarrow$ \u00a0$\\frac{d}{12}+\\frac{d}{18}+\\frac{d}{24}=13$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{6d+4d+3d}{72}=13$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{13d}{72}=13$<\/p>\n $\\Rightarrow$\u00a0\u00a0 d = 72 km<\/p>\n $\\therefore\\ $Total distance = 3d = 3 x 72 = 216 km<\/p>\n Hence, the correct answer is Option C<\/p>\n