## Wipro Sample Tough Work Time Problems

Dear Reader, Below are three bit tough (may not be tough to everyone) problems on work and time. Some of the below questions can be bit tricky while others can take bit longer to solve than usual questions. If you find the below questions on the easier side, then you can be confident that you are on track with your preparation.

Question 1

Adhvaith can do a certain work in 30 days. Kashyap can do same work in 25 days. Adhvaith started the work and worked for 9 days. Kashyap came and joined to do the work from the 10th day. How many more days would they have taken together to complete the work?

a)10 3/11 days b)11 2/11 days c) 9 6/11 days d) 8 2/11 days

Answer : c) 9 6/11 days.

Solution:

Adhvaith can do 1/30 of the work in one day
In 9 days he would have completed - 9 x 1/30 = 3/10 of the work
Balance work = 1 - 3/10 = 7/10
Kashyap can do 1/25 of the work in one day
Work that can be done by Adhvaith and Kashyap in one day = (1/30 + 1/25) = 11/150 of the work.
So Adhvaith and Kashyap can complete 7/10 of the work in
7/10 x 150/11 = 105/11 = 9 6/11 days.

Question 2

A private limited company entrusts works to 20 men, working 12 hours a day. This group can complete the work in 24 days. The company now wants to entrust twice the work to 60 men working 4 hours a day. Assume that 2 men of the first group do as much work in one hour as 3 men of second group do in 1 ½ hours. How many number of days will the second group of men take to complete this work?

a) 108 days b)120 days c)124 days d)81 days

Solution :

Let efficiency of men in I group be E1 and that of second group be E2.

Ratios of efficiency of men in I group to that of II group can be found by using the formula,

E1/E2 = Time taken by men in II group to do certain amount of work / Time taken by men in I group to do the same amount of work as that of men in II goup

= (3 x 1.5) : (2 x 1)
= 4.5 : 2

Now, M1D1T1E1W2 = M2D2T2E2W1 --> 1

(where M1 = number of men in I group, M2 = number of men in II group. D1 = number of days required to complete work by group I, D2 = number of days required to complete work by group II. T1 = working hours per day by group I. T2 = working hours per day by group II. w2 = amount of work by group II, w1 = amount of work by group I.)

Since we are to calculate the time taken by group II to complete twice the amount of work as that of group I, W2 = 2 x W1.
We had earlier calculated E1/E2 = 4.5/2.
Also from the question we can infer that,
M1 = 20, M2 = 60.
T1 = 12, T2 = 4.
D1 = 24 and D2 is what we need to find.

Substituting all the values in eq 1, we can find D2 as follows.

D2 = (20 X 24 X 12 X 4.5 X 2)/(60 X 4 X 2 X 1)
= 108 days.

Question 3

Three persons Manmohan, Anna And Sushma working together, can do a job in X hours. When working alone, Manmohan needs an additional six hours to do the job; Anna, working alone needs an additional hour and Sushma working along needs X additional hours. What is the value of X?

a)2/3 b)3/2 c)11/12 d)2

Solution :

In this type of problems where answers cannot be easily found out using equations, it is advisable to go from the given answer choices.
Based on information given one hour work done by all the three together =
1/ X+6 + 1/ X+1 + 1/2X = 1/X
X is not known.
Using the data given
1/ (2/3) +6 + 1/(2/3) +1 + 1/(4/3) = 1/ ( 2/3)
This comes out correctly. Whereas other values given in b), c) and d) do not get the result properly. Hence a) is correct.

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