**Time and work problems :**

Time and work problems play a major role in quantitative aptitude test. There is no competitive exam without the questions from this topic. In this section, we are going to see, how time and work problems can be solved.

1. If a person can do a piece of work in ‘m’ days, he can do (1/m) part of the work in 1 day.

2. If the number of persons engaged to do a piece of work be increased (or decreased) in a certain ratio the time required to do the same work will be decreased (or increased) in the same ratio.

3. If X is twice as good a workman as Y, then X will take half the time taken by Y to do a certain piece of work.

4. Time and work are always in direct proportion.

**5. If two taps or pipes P and Q take ‘m’ and ‘n’ hours respectively to fill a cistern or tank, then the two pipes together fill (1/m)+(1/n) part of the tank in one hour. Entire tank is filled in mn/(m+n) hours**

**Question 1 : **

A and B can complete a work in 12 days.B and C can complete the same work in 18 days. C and A can complete in 24 days. How many days will take for A, B and C combined together to complete the same amount of work ?

**Solution : **

From the given information, we can have

(A+B) can complete(1/12) part of the work in 1 day

(B+C) can complete (1/18) part of the work in 1 day

(A+C) can complete (1/24) part of the work in 1 day

By adding the three equations, we get,

(A+B) + (B+C) + (A+C) = 1/12 + 1/18 + 1/24

2A + 2B + 2C = (6+4+3) / 72

2(A+B+C) = 13/72

(A+B+C) can complete 13/144 part of the work in 1 day

Therefore (A+B+C) can together complete the work in 144/13 days

That is 11(1/3) days.

**Question 2 : **

A & B can do a work in 15 days, B & C in 30 days and A & C in 18 days. They work together for 9 days and then A left. In how many more days, can B and C finish the remaining work ?

**Solution : **

We can apply L.C.M method to solve this problem

Total work = 90 units (L.C.M of 15,30,18,9)

(A+B) can complete 6 units/day (90/15 = 6)

(B+C) can complete 3 units/day (90/30 = 3)

(A+C) can complete 5 units/day (90/18 = 5)

By adding, we get 2(A+B+C) = 14 units/day

(A+B+C) = 14/2 = 7 units/day

work done by A = (A+B+C) - (B+C) = 7 - 3 = 4 units/day

work done by B = (A+B+C) - (A+C) = 7 - 5 = 2 units/day

work done by C = (A+B+C) - (A+B) = 7 - 6 = 1 unit/day

work done by (A+B+C) in 9 days = 9 x 7 = 63 units.

Balance work = 90 - 63 = 27 units

This 27 units of work to be completed B & C. Because A left after 9 days of work.

No. of days taken by B & C to complete the balance 27 units of work = 27/3 = 9 days. (Because B&C can complete 3 units of work per day)

**Question 3 : **

A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely ?

**Solution : **

From the question, we have to consider an important thing.

That is, pipe B is faster than pipe A.

When two pipes are opened together, the tank will emptied.

So the right choice would be (A) or (B)

Total capacity of the tank = 60 units. (L.C.M of 10,6)

The tank is already two-fifth full.

That is,quantity of water is in the tank

= (2/5)X60 = 24 units

If both the pipes are opened together, this 24 units will be emptied.

work done by pipe A = 60/10 = 6 units/min

work done by pipe B = 60/6 = -12 units/min (emptying the tank)

Adding the above two equations, we get(A+B)=-4 units/min

That is 4 units will be emptied per minute when both the pipes are opened together

Time taken to empty 24 units (2/5 of the tank) is

= 24/4

= 6 min

Time taken to empty the tank is 6 min. Option (A) is correct.

**Question 4 : **

A contractor decided to complete the work in 90 days and employed 50 men at the beginning and 20 men additionally after 20 days and got the work completed as per schedule. If he had not employed the additional men,how many extra days would he have needed to complete the work?

**Solution : **

Given Information:

The work has to completed in 90 days (as per schedule)

Total no. of men appointed initially = 50

50 men worked 20 days and completed a part of the work.

The remaining work is completed by 70 men (50+20 = 70) in 70 days (90-20 = 70)

If the remaining work is completed by 50 men, no. of days taken by them = (70x70)/50 = 98 days.

Hence, extra days needed = 98 - 70 = 28 days.

**Question 5 : **

Three taps A, B and C can fill a tank in 10, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, find the time taken to fill the tank.

**Solution : **

Total work = 60 units. (L.C.M of 10,15,20)

work done by the pipe A = 60/10 = 6 units/hr

work done by the pipe B = 60/15 = 4 units/hr

work done by the pipe C = 60/20 = 3 units/hr

(Given:A is open all the time,B and C are alternately)

1st hour: (A+B) = 10 units/hr

2nd hour: (A+C) = 9 units/hr

3rd hour: (A+B) = 10 units/hr

4th hour: (A+C) = 9 units/hr

5th hour: (A+B) = 10 units/hr

6th hour: (A+C) = 9 units/hr

When we add the above units, we get the total 57 units.

Apart from the 6 hours of operation, to get the total work 60 units, A has to work for half an hour

(Because in one of hour work of A, we will get 6 units)

Hence, time taken to fill the tank is

6 hours and 30 minutes.

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