If a sportsperson runs \({1 \over {6}}th\) of the distance in a minute, then in how many minutes can he complete the distance?

Distance covered in a minute = \({1 \over {6}}\)

Number of minutes required to cover the distance = 1/Distance covered in a minute = 6 minutes . Hence (a).

15 women can do a piece of work in 5 days working 12 hours a day. In how many days will 20women complete the same work working 10hours a day?

\({W_{1} \over {M_{1}*D_{1}}}\) = \({W_{2} \over {M_{2}*D_{2}}}\)

\({W \over {M_{1}*D_{1}}}\) = \({W \over {M_{2}*D_{2}}}\)

M _{1} * D _{1 } *H _{1 } = M _{2} * D _{2 } *H _{2 }

15 * 5 * 12 = 20 * D _{2} * 10

D _{2} = 4.5 days. Hence (b).

A can do a piece of work in 6 days and B in 13 days. In how many days can they complete the

Work if they work together?

Total work = LCM of 6 and 13 = 78 units. A takes 6 days to complete the workà 13 units/day B takes 13 days to complete the work à 6 units/day A and B working together = 19 units/day To complete 78 units A and B will take, =>\({78 units \over {19units/day}}\) = 4\({2 \over {19}}\) days. Hence (d).

12 men can complete a piece of work in 11 days. After 5 days, if 6 more men join them, then how many days will it take to complete remaining work?

Total work = Number of days * Number of men

Total work = 11 * 12 = 132 units.

Work completed in 5 days = 5 * 12 = 60 units.

Remaining units = 132 – 60 = 72 units.

6 more men joined, then time taken to complete the remaining work

- 18 * x = 72

- x = 4 days. Hence (c).

A pipe can fill a tank in 12 hours and another pipe fill the tank in 16 hours. How many hours will both the pipes take to fill the tank operating alternately?

Total work = LCM of 12 and 16 = 48 units. First pipe takes 12 hours to fill the tank à 4 units/hour Second pipe takes 16 hours to fill the tank à 3 units/hour First and Second pipes operating alternately = 7 units/2 hours To complete 48 units First and Second pipe will take, => = 13\({2 \over {3}}\) hours.Hence (b).

A can do a work in 30 days, B in 35 days and C in 42 days. A and C start the work initially, if B helps them on every 3^{rd} day, then how many days will be taken by three of them to complete the work?

Total work = LCM of 30, 35 and 42 = 210 units A takes 30 days to complete the work à 7 units/day B takes 35 days to complete the work à 6 units/day C takes 42 days to complete the work à 5 units/day A and C work for two days together à24 units/2days On every third day B will work with A and C, then 42 units will be completed in 3 days To complete 210 units A, B and C will take, =>\({210 units \over {42units/ 3 days}}\) = 15 days.Hence (c).

A can do a job in 18 days and B in 24 days. They start working together but B leaves before 4 days before completion of the work. Find the number of days worked by B.

Total work = LCM of 18 and 24 = 72 units A takes 18 days to complete the jobà 4 units/day B takes 24 days to complete the jobà 3 units/day Work done by A alone in last 4 days = 4* 4 = 16 units Remaining units = 72 – 16 = 56 units Efficiency of A and B = 7 units/day. Time taken by A and B to complete 56 units = 8 days. Therefore, the number of days B worked is 8 days.Hence (b).

Pipe P fills a tank in 7 hours and pipe Q empties the same tank in 8 hours. How long it will take to fill the tank if both the pipes are opened simultaneously?

Total work = LCM of7 and 8 = 56 units. Pipe P takes 7 hours à 8 units/hour Pipe Q takes 7 hours à-7 units/hour If both the pipes are opened simultaneously, 1 unit will be completed in 1 hour. Number of days to complete 56 units by pipe P and Q = 56 hours. Hence (c).

If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same work will be:

Total work = LCM of 10 and 2 = 10 units 6m + 8b = 10 days à 1 unit/day …..(1) 26m + 48b = 2 days à5 units/day …..(2) On solving both equations will get efficiency of men = \({1 \over {10}}\) units/day, and for boys = \({1 \over {20}}\)units/day 15m + 20b = \({5 \over {2}}\) units/ day. Time taken to complete 10 units by 15 mens and 20 boys = \({10 \over {5/2}}\) = 4 days.Hence (d).

Two pipes A and B can fill a tank in 36 hours and 32 hours respectively. Pipe C can empty the tank in 24 hours. If all the three pipes are opened together, in what time will the \({5 \over {8}}\)^{th} of tank be filled?

Total work = LCM of 36, 32 and 24 = 288 units Pipe A takes 36 hours à 8 units/hour Pipe B takes 32 hours à 9 units/hour Pipe Q takes 24 hours à -12 units/hour If all the three pipes are opened together, 5 units will be completed in 1 hour \({5 \over {8}}th\)of total unit = 180 units. To complete 180 units it will take = \({180 \over {5}}\) = 36 hours. Hence (a).

Three people can complete a piece of work in 8, 12 and 15 days respectively. Rs.2145 is the total money allocated to complete that work. What amount will each person get if all three are working together?

Total work = LCM of 8, 12 and 15 = 120 units A takes 8 days to complete the work à 15 units/day B takes 12 days to complete the work à 10 units/day C takes 15 days to complete the work à 8 units/day A, B and C working together = 33 units/day Amount of money A will get = \({15 \over {33}}\) * 2145 = Rs.975 Amount of money B will get = \({10 \over {33}}\) * 2145 = Rs.650 Amount of money C will get = \({8 \over {33}}\) * 2145 = Rs.520. Hence (d).

Elan and John can do a piece of work in 9 days. After working for 6 days together, their friend Ram breaks 66.67% of the work done and leaves. Now Elan and John take 7 days more to complete the remaining work. Elan alone can complete the job in 25 days. Find the time taken by John to complete the work individually._{.}

Time for Elan and John to do the work together = 9 days

Time for Elan to do the work alone = 25 days

Total work = LCM of 9 and 25 , total wor k to be done = 225 units

Work done by the three = \({225 \over {9}}\) = 25 units per day and by Elan = \({225 \over {25}}\) = 9 units per day

Work done by the three in 6 days = 25 * 6 = 150 units , work broken by their friend Ram = \({2 \over {3}}\) * 150 = 100 units

Rema ining work to be completed = 225 – 50 = 175 units

Time taken by Elan and John to do the remaining work of 175 units = 7 days

Efficiency of Elan + John = = 25 units per day

Efficiency of John = (25 – 9) = 16 units per day.

Therefore, time take n by John to complete the whole work alone = \({225 \over {16}}\) = 14 \({1 \over {16}}\) days . Hence (c).

A leak at the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 litres a minute. When both are opened, the leak takes 24 hours to empty the full tank. Find the capacity of the tank?

Time for leak to empty the full tank = 6 hours

Time for leak to empty along with inlet pipe = 24 hours

Total work = LCM of 6 and 24 = 2 4 units

Efficiency of leak alone = 4 units/hour

Efficiency of leak + pipe = -1 units/hour

Time for filling the tank = \({24 \over {3}}\) = 8 units/hour

Filling rate is 4 litres/minute

Therefore, Capacity of the tank = 4 * 60 * 8 = 1920 litres . Hence (a).

A tank has 28 filling and emptying pipes. Each filling pipe can fill the tank in 24 hours and each emptying pipe can empty the tank in 16 hours. If all the pipes are kept open, then the tank is filled in 8 hours. How many of these are emptying pipes?

Let the number of filling and emptying pipes be n and (28-n) respectively Time taken by filling pipe to fill = 24 hours Time taken by emptying pipe to empty = 16 hours Total work = LCM of 24 and 16 = 48 units Efficiency of filling pipe = 2 units/hour Efficiency of emptying pipe = 3 units/hour If all the pipes open, time to fill the tank = 8 hours Efficiency of filling = Total unit/8 = 6 units/hour Total units filled in 1 hour – total units emptied in hour = 6 units 2n – 3(28 – n) = 6 à n = 18 Number of emptying pipes = 28 – n = 10 pipes. Hence (b).

B is 40% more efficient than A and he can complete thrice the work in 90 days. How many days would A take to complete 9 times the same work?

Let the efficiency of A and B be 100% and 14 0%

Ratio of A and B efficiency = 5: 7

Ratio of A and B time taken = 7: 5

Time taken by B to complete one full work = \({90 \over {3}}\) = 30 days.

Time taken by A to complete one full work = 30 * \({7 \over {5}}\) = 42 days

Therefore, time taken by A to complete 9 times the work = 42 * 9 = 378 days . Hence (d).