Enroll to CMAT crash course<\/a><\/p>\n Take CMAT mock tests here<\/a><\/p>\n Download CMAT previous papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>Pipe ‘P’ can fill a tank in 36 hours and pipe ‘Q’ can fill this tank in 45 hours. If both the pipes are opened simultaneously, then how much time will be taken to fill this tank ?<\/p>\n a)\u00a020 hours<\/p>\n b)\u00a0$40 \\frac{1}{2}$ hours<\/p>\n c)\u00a09 hours<\/p>\n d)\u00a042 hours<\/p>\n Question 2:\u00a0<\/b>A tank has three pipes. The first one fills up the tank in 30 min and the second one can fill it up in 45 min. The third one drains out the tank. With all the pipes open it takes 27 min to fill up the tank. Find out how long it takes the third pipe alone to empty a full tank?<\/p>\n a)\u00a056 min<\/p>\n b)\u00a050 min<\/p>\n c)\u00a052 min<\/p>\n d)\u00a054 min<\/p>\n Question 3:\u00a0<\/b>Pipes A, B and C together can fill a tank in 4 hours. Pipes A and B together can fill the tank in 9 hours. If pipes A, B and C are opened simultaneously, and after 3 hours, pipes A and B are closed, then how many hours will C alone require to fill the tank?<\/p>\n a)\u00a02.25<\/p>\n b)\u00a02<\/p>\n c)\u00a01.5<\/p>\n d)\u00a01.8<\/p>\n Question 4:\u00a0<\/b>A water tank is filled in 5 hours by three pipes X, Y and Z the pipe Z is thrice as fast as Y and Y is twice as fast as X. How much time will pipe X alone take to fill the water tank?<\/p>\n a)\u00a060 hrs<\/p>\n b)\u00a045 hrs<\/p>\n c)\u00a040 hrs<\/p>\n d)\u00a035 hrs<\/p>\n Question 5:\u00a0<\/b>Pipe A is an inlet pipe that can fill an empty cistern in 69 hours. Pipe B can drain the filled cistern in 46 hours. When the cistern was filled the two pipes are opened one at a time for an hour each, starting with Pipe B. how long will it take for the cistern to be empty?<\/p>\n a)\u00a011 days 10 hours<\/p>\n b)\u00a011 days 7 hours<\/p>\n c)\u00a011 days 12 hours<\/p>\n d)\u00a01 days 13 hours<\/p>\n Get 5 CMAT mocks at just Rs.299<\/a><\/p>\n Download CMAT Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>Pipes A and B, when working together, can fill an empty tank in 8 hours. If B is stopped\u00a0after 2 hours, it will take a total of 11 hoursto fill the tank. How long would it take A to fill the empty tank alone?<\/p>\n a)\u00a015 hours<\/p>\n b)\u00a012 hours<\/p>\n c)\u00a013 hours<\/p>\n d)\u00a016 hours<\/p>\n Question 7:\u00a0<\/b>Pipe A could fill an empty cistern in 18 hours while Pipe B can drain filled cistern in 30 hours. When the cistern is empty. Pipe A is turned on for an hour and then turned off. Now Pipe B is allowed to drain out water from the cistern for an hour and then turned off. The pipes were alternately left open for an hour each time till the cistern was full. How much time did it take for the cistern to be full?<\/p>\n a)\u00a090 hours<\/p>\n b)\u00a086 hours 40 minutes<\/p>\n c)\u00a045 hours<\/p>\n d)\u00a086 hours 48 minutes<\/p>\n Question 8:\u00a0<\/b>Two pipes P and Q can fill a cistern in 18 and 24 minutes respectively. Both pipes are opened, but after some time, the slower pipe is closed. The cistern gets filled in 12 minutes from the start. When is the slower pipe closed?<\/p>\n a)\u00a0After 9 minutes<\/p>\n b)\u00a0After 8 minutes<\/p>\n c)\u00a0After 10 minutes<\/p>\n d)\u00a0After 6 minutes<\/p>\n Question 9:\u00a0<\/b>Two taps can separately fill a cistern in 20 minutes and 25 minutes. Both taps are open for 10 minutes after which the slower one is closed. How long will it take to fill the remaining portion by the other tap alone?<\/p>\n a)\u00a02 minutes<\/p>\n b)\u00a04 minutes<\/p>\n c)\u00a010 minutes<\/p>\n d)\u00a05 minutes<\/p>\n Question 10:\u00a0<\/b>Two pipes can fill a tank in 5 and 6 hours respectively. A third pipe will empty it in 12 hours. If all the pipes are\u00a0opened, then find the time taken to either fill or empty the tank.<\/p>\n a)\u00a0Filled in 30\/17 hours<\/p>\n b)\u00a0Filled in 60\/17 hours<\/p>\n c)\u00a0Filled in 30\/11 hours<\/p>\n d)\u00a0Emptied in 9\/2 hours<\/p>\n Take CMAT Mock Tests<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Required answer<\/p>\n = $ \\frac{1}{36} + \\frac{1}{45} = \\frac{81}{36 \\times 45} = \\frac{1}{20}$<\/p>\n If both the pipes are opened simultaneously, then 20 hrs will be taken to fill this tank.<\/p>\n So, the answer would be option a)20 hours<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 1st and 2nd pipes fill the tank so addition occurs but 3rd can drain out,<\/p>\n $\\ \\frac{\\ 1}{30}+\\ \\frac{\\ 1}{45}-\\ \\frac{\\ 1}{x}=\\ \\frac{\\ 1}{27}$<\/p>\n by solving this we get the value of x=54<\/p>\n Hence option ‘D’ is correct.<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n A and B works for 3 hours , so remaining work =36-3$\\times\\ 9$=9<\/p>\n so C alon can fill the tank in\u00a0$\\ \\frac{\\ 9}{5}=1.8\\ hours$,<\/p>\n hence option ‘D’ is correct.<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n A water tank is filled in 5 hours by three pipes X, Y and Z the pipe Z is thrice as fast as Y and Y is twice as fast as X.<\/p>\n $ \\frac{Z}{Y}=\\frac{3}{1}\u00a0 and\u00a0 \\frac{Y}{X}=\\frac{2}{1} $<\/p>\n Now equate both the ratios by making Y equal in both,<\/p>\n $ \\frac{Z}{Y}=\\frac{6}{2} and \\frac{Y}{X}=\\frac{2}{1} $<\/p>\n So, Z : Y : X = 6 : 2 : 1 (this is the ratio of efficiencies of the pipes)<\/p>\n If three pipes fill the tank in 5 hours, there fore the total work done is<\/p>\n $=Time \\times Total\u00a0 Efficiency\u00a0 of\u00a0 pipes$<\/p>\n $=5\\times (6+2+1) = 45$<\/p>\n So X will take time = $\\frac{Work}{X’s\u00a0 efficiency}$<\/p>\n $=\\frac{45}{1} = 45 hours$<\/p>\n Option B is correct.<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Pipe A can fill cistern in 69 hours.<\/p>\n Pipe\u00a0B can empty cistern in 46 hours.<\/p>\n Let’s assume total work = 138<\/p>\n Efficiency of A = 2<\/p>\n Efficiency of B = -3 (Subtraction sign is due to outlet pipe.)<\/p>\n For empty the\u00a0cistern, Alternatively pipe B and A respectively work. So in every 2 hours 1 part of tank is empty out of total 138.<\/p>\n Then we can say that for emptying 135 parts, we need 270\u00a0hours and last 1 hour B will empty 3 parts.<\/p>\n So 270 + 1 = 271 hours<\/p>\n If we convert\u00a0271 hours in days by dividing 24. Then got 11 days 7 hours.<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n A can fill the empty tank alone in ( 12\/1 \u00d7 1 ) = 12 hours.<\/p>\n Step-by-step explanation:<\/p>\n let, ( A + B ) means A and B work together.<\/p>\n ( A + B) can fill an empty tank in 8 hours and after 2 hours B was stopped. That means first 2 hours A and B worked together.<\/p>\n ( A + B) can fill a 2\/8 = 1\/4 part in 2 hours.<\/p>\n after 2 two hours, A fill the remaining part of tank which is ( 1 – 1\/4 ) = 3\/4 in\u00a0 \u00a0 ( 11- 2) = 9 hours<\/p>\n It means A can fill ( 3\/4 \u00d7 1\/9 ) = 1\/12 part in 1 hour.<\/p>\n Therefore, A can fill the empty tank alone in ( 12\/1 \u00d7 1 ) = 12 hours.<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n A can fill the cistern in 18 hours and B can empty in 30 hours 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n P fill the cistern = 18 min<\/p>\n Q fill the cistern = 24 min<\/p>\n Total work done = LCM (18,24)= 72<\/p>\n P’s efficiency = $\\frac{72}{18}$ = 4<\/p>\n Q’s efficiency = $\\frac{72}{24}$ = 3<\/p>\n Tank will be fiiled= $\\frac{72}{7}$hours<\/p>\n Tap P is open for 12 min\u00a0 \u00a0and taps Q closed after some time<\/p>\n so<\/p>\n as we know that ,<\/p>\n efficiency multiplied by time = work<\/p>\n $4\\times 12 + 3\\times T$= 72<\/p>\n T = 8 min<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $ work done = efficiency \\times time $<\/p>\n Two taps P1 and P2\u00a0 can separately fill a cistern in 20 minutes and 25 minutes.<\/p>\n work done = LCM( 20,25 ) =100<\/p>\n efficiency of P1 = $ \\frac{100}{20} = 5 $<\/p>\n efficiency of P2 = $ \\frac{100}{25} = 4 $<\/p>\n Both taps are open for 10 minutes after which the slower one is closed.<\/p>\n $ work done = ( efficiency of P1 + efficiency of P2 ) \\times 10 + efficiency of P1 \\times x $<\/p>\n where x = time required by P1 to fill the remaining portion by the other tap alone<\/p>\n substituting<\/p>\n $ 100 = ( 5 + 4) \\times 10 + 5x $<\/p>\n solving x = 2 minutes<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $ work = efficiency \\times time $<\/p>\n P1,P2 and P3 be the pipes<\/p>\n P1 fill the tank in 5 hrs<\/p>\n P2 fill the tank in 6 hrs<\/p>\n P3 empty the tank in 12 hrs<\/p>\n work done = LCM of P1,P2 and P3 = LCM (5,6,12) = 60<\/p>\n efficiency of P1 = $ \\frac{60}{5} = 12 $<\/p>\n efficiency of P2 = $ \\frac{60}{6} = 10 $<\/p>\n efficiency of P3 = $ \\frac{60}{12} = 5 $<\/p>\n efficiency of P1 + P2 + P3 = 12 + 10 – 5 = 17<\/p>\n $ 60 = 17 \\times time $<\/p>\n $ time = \\frac{60}{17} hrs $<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/sergdfhd.png\")
<\/figure>\n
\nLCM of 18 and 30 is 90 units
\nIn one hour A can fill 5 units
\nIn one hour B can empty 3 units
\nSo in 2 hours 2 units of work is done
\nFor 86 hours 86 units of work is done
\nNow that 4 units is filled by A in 4\/5 hours
\nSo total time 86 hours 48 minutes<\/p>\n