Time and Work Problems For Bank Exam

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Time and Work Problems For Bank Exam

Quantitative Aptitude Study Notes for Bank Exam
You know that quantitative aptitude section is most important in bank exams in PO and Clerk and for other competitive exams because if you want good score in bank exam then you have to score good in maths. In competitive exams the most important thing is time management, if you know how to manage your time then you can do well in Bank Exams as well as in other competitive exams. That’s where maths shortcut tricks and formula are comes into action. So continuously we are providing shortcut tricks on different maths topics. Today’s topic is time and work problems for bank exam. This is the one of the most important topic in quantitative aptitude section in bank and SSC exam. You should know how to solve time and work questions in very short time for bank exam. From this chapter around 1-2 questions are given in the SBI and IBPS exams. For this here we are providing shortcut tricks and quicker method to solve time and work problems in very short time. Before going to solve questions for time and work, we study little bit about basics and formulas for time and work. If ‘M1’ persons can do ‘W1’ work in ‘D1’ days and ‘M2’ persons can do ‘W2’ work in ‘D2’ days then we have a very special general formula in the relationship of M1 D1 W2 = M2 D2 W1. The above relationship can be taken as a very basic and all-in-one formula. We also derive:

1.      More men less days and conversely more days less men.
2.      More men more work and conversely more work more men.
3.      More days more work and conversely more work more days.
If we include the working hours (say T1 and T2) for the two groups then the relationship is
M1 D1 T1 W2 = M2 D2 T2 W1.
Again, if the efficiency (say E1 and E2) of the persons in two groups is different than the relationship is
M1 D1 T1 E1 W2 = M2 D2 T2 E2 W1.
Time and work questions with answers:
Now, we should go ahead starting with simpler to difficult and more difficult questions.
Ex. 1: ‘A’ can do a piece of work in 5 days. How many days will he take to complete 3 works of same type?
Solution:
we recall the statement: “ More work more days” It simply means’ that we will get the answer by multiplication.
Thus, our answer = 5*3 = 15 days.
This  way of solving the question is very simple, but you should know how the “basic formula” could be used in this question.
Recall the basic formula: M1 D1 W2 = M2 D2 W1.
As ‘A’ is the only persn to do the work in both the cases, so
M1 = M2 = 1 (useless to carry it)
D1 = 5 days, W1 = 1, D2 = ? and W2 = 3
Putting the values in the formula we have,
5 * 3 = D1 * 1
Or, D2 = 15 days.
Ex. 2: 16 men can do a piece of work in 10 days, how many men ae needed to complete the work in 40 days?
Solution:
To do a work in 10 days, 16 men are needed.
Or, to do the work in 1 day, 16 * 10 men are needed.
So, to do the work in 40 days, 16*10 / 40 = 4 men are needed.
This was the method used for non-objective exams.
We should see how the ‘basic formula’ works here.
M1 = 16, D1 = 10, W1 = 1
And
M2 = 7, D2 = 40, W2 = 1
Thus, form M1 D1 W2 = M2 D2 W1.
16 * 10 = M2 * 40
Or, M2 = 16*10 / 40 = 4 men
By rule of fractions: to do the work in 40 days we need less number of men than 10.  So we should multiply 10 with a fraction which is less than 1. And that fraction is 10/40. Therefore, required number of men
= 16 * 10 / 40 = 4
Ex. 3: A can do a piece of work in 5 days, and B can do it in 6 days. How long will they take if both work together?
Solution:
‘A’ can do 1/5 work in 1 day.
‘B’ can do 1/6 work in 1 day.
Thus, ‘A’ and ‘B’ can do (1/5 + 1/6) work in 1 day.
‘A’ and ‘B’ can do the work in 1/ 1/5+1/6 days = 30/11 = 28/11 days
By the Theorem: A + B can do the work in 5*6 / 5+6 days = 30/11 = 28/11 days
Theorem: if A, B and C can do a work in x, y and z days respectively then all of them working together can finish the work in xyz / xy +yz + xz days.
Ex. 4: In the above question, if C who can do the work in 12 days, joins them, how long will they take to complete the work?
Solution: by the theorem: ‘A’, ‘B’ and ‘C’ can do the work in
5 * 6 * 12 / 5*6 +6*12 + 5*12
= 360/162
= 22/9 days
Theorem and shortcut trick for time and work question: if A and B together can do a piece of work in x das and A alone can do it in y days. Then B alone can do the work in xy / x-y days.
Ex. 5: A and B together can do a piece of work in 6 days and A alone can do it in 9 days. In how many days can B alone do it?
Solution:
A and B can do 1/6th of the work in 1 day.
A alone can do 1/9th of the work in 1 day.
B alone can do the whole work in 18 days.
By the theorem or shortcut trick:
B alone can do the whole work in
= 6*9 / 9-6
54 / 3 = 18 days
Ex. 6: if 3 men or 4 women can reap a field in 43 days, how long will 7 men and 5 women take to reap it?
Solution:
3 men reap 1/43rd of the field in 1 day.
1 men reaps 1 / 43*3 rd  of the field in 1 day.
4 men reap 1/43 rd  of the field in 1 day.
1 woman reaps  1 / 43*4 of the field in 1 day.
7 men and 5 women reap (7 / 43*3 + 5 / 43*4) = 1/12th of the field in 1 day.
7 men and 5 women will reap the whole field in 12 days.
Shortcut method for time and work:
Required number of days = 1 / (7/43*3 + 5/43*4)
43*3*4 / 7*4 + 5*3
= 12 days
Note: the above formula is very easy to remember.
If we divide the question in two arts and call the first part as OR – part and the second part as AND – part then
7 / 43*3 = number of men in AND part / Number of days * Number of men in OR part