Time and Work for Bank PO and SSC

Time and Work is a very important topic in Bank and SSC  Exams and even DI Caselets are asked on this topic for Bank Exams. Hence it is very important to master this topic. The question types which usually come in this topic are given below

1. 2 or 3 persons working and days required to complete the work
2. WAGES Problem
3. Multiple Person of Different genders working and days required to complete the Work
4. Pipes and Cisterns

 CONCEPT

A can do a work in 12 days. hence if he works an equal amount on each day we can say that he does 1/12 the part of the work each day. If he increases his efficiency by 50%, he does 1/12 ×( 1.5) or 1.5/12 or 1/8 the part of the work each day

If A and B can do a work in 12 and 15 days then they will do (1/12 +1/15) or 9/60 or 3/20 part of the work in 1 day. Hence they will do the work in 1/(3/20) or 20/3 days. Instead of approaching like this we can do this by shortcut where we take the LCM of days to be the amount of Work. Hence Assume the Work to be the LCM of 12,15 or 60 units. Hence A does 60/12 or 5 units per day and B does 60/15 or 4 units per day. Hence they do (5+4) or 9 units per day.. and hence to do 60 units they will take 60/9 or 20/3 days.

You will notice that by doing the LCM method, the question becomes much easy to solve.

If you closely see you will find that the ratio of efficiencies is 5units: 4 units or 5: 4. Hence after doing the work if their contractor gives them Rs 900 to distribute, they will do so in the ratio of 5:4. Hence A gets 500  and B gets Rs 400. In this way we can solve the Problems of Wages.

Suppose A does the work in 12 days and B destroys the work in 15 days and if it is asked to calculate the days to complete the work,

Take the LCM of 12 and 15 as 60 units of work then we can say that A does +5 and B does -4 units(destroys). Hence they do 5-4 or 1 units per day. Hence they will do the work in 60/1 or 60 days.

If we correlate this problem with another problem

Pipe A can fill up a cistern in 12 hrs and Pipe B can empty a cistern in 15 hrs. How many hrs will the pipes take up to fill an empty cistern?

Here there is no difference between this question and the Previous example. Negative work done By B is equivalent to emptying the cistern by Pipe B. Here Assume the capacity of the cistern to be 60 liters and proceed accordingly.

For Multiple Persons

If M1 men can do W1 work in D1 days working H1 hours per day with an efficiency of E1 and M2 men can do W2 work in D2 days working H2 hours per day with an efficiency of E2, then

M1 D1 H1 E1 / W1 = M2 D2 H2 E2 / W2

Some Questions:

P, Q, and R can completely solve a problem together in 4 hrs. P and R together take 15 hrs less than Q working alone. Q works on the problem for the first 2 hrs and then P and R join him. After another two hours, Q quits. In how many hours is the problem actually solved?

Solution: Let Q alone can answer the problem in x hours
Then P&R together needs (x-15) hours to solve the problem

In 1 hr, Q can complete 1/x of the question, 
In 1 hr, P&R can complete 1/(x-15) of the question 
In 1 hr, P, Q, and R can complete 1/4th of the question

1/x + 1/(x-15) = 1/4
4(x-15)+4x =x(x-15)
4x - 60 + 4x = x^2 - 15x
x^2 - 23x + 60 = 0
(x-20)(x-3)=0
x = 20 or 3
since (x-15) is positive, x = 20

Hence Q does the problem in 20 hrs and P+R does the problem in 5 hrs. Hence assume the problem as 20 units.

The efficiency of Q and ( P+R) is 20/20 and 20/5  or 1 and 4 units per hr respectively. Hence

1. For the 1^st 2 hrs, Q does 2 units
2. For the next 2 hrs, All 3 do (1+4) ×2  or 10 units.
3. Remaining (20-10-2) or 8 units will be done by (P+R) in 8/4 or 2 hrs. Hence Total time is 2+2+2= 6 hrs.

25 men can reap a field in 20 days. When should 15 men leave the work if the remaining field is to be reaped in 40 days after they leave the work?.

Solution: We Know that M × D= Total work which is Constant

If 15 men leave the field after x days, then equating the total work we can say that

25×20 = 25 × x + 10 × 40  or x= 4 days

A cistern filled in 20 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is thrice as fast as A. How much time will pipe A alone take to fill the tank?

Solution: Assume the Capacity of the tank to be 20 liters. Hence Efficiency of A+B+C will be 20/20 or 1 unit per hr. Hence if A has an efficiency of x units, then B has 3x and C has 2 × 3x or 6x. Hence Combined efficiency is x+3x+6x   or 10 x =1 . Hence x=1/10. Hence A alone will fill the tank in 20/(1/10) or 200 hrs

More techniques in our next post.