# 40 Time and Work Questions for CAT with Answers

Question 36:
Three labourers started working together to complete a task in certain number of days. But in the course of work, they remained absent for few days due to some unavoidable circumstances and therefore the work wascompleted 3 days late. If the first and second labourers(assume that the wage per day are fixed and late completion of work did not affect it.) remained absent for 2 and 4 days more than the third labourer remained, then the third labourer remained absent for
 1 day
 2 days
 3 days
 cannot be determined

Option: 1

Explanation:

Let $\frac{1}{M}$ be the part of the work a labourer can do in a day. Let m be the number of stipulated days in which the work could have been done had none of the worker been absent. Also let n be the number of actual days for which third labourer remained absent. Then

$\frac{{3m}}{M} = 1$ and $\frac{m+3-n}{M}+\frac{m+3-\left( n+2 \right)}{M}+\frac{m+3-\left( n+4 \right)}{M}=1$ $\Rightarrow \frac{3\left( m+1-n \right)}{M}=1\Rightarrow \frac{3\left( m+1-n \right)}{3m}=1\Rightarrow m+1-n=m$ Therefore, n=1

Question 37:
Three labourers started working together to complete a task in certain number of days. But in the course of work, they remained absent for few days due to some unavoidable circumstances and therefore the work wascompleted 3 days late. If the first and second labourers(assume that the wage per day are fixed and late completion of work did not affect it.) got Rs 10 and Rs 20 less than the third labourer got, then the extra money the third labourer earned is
 Rs 10
 Rs 20
 Rs 30
 cannot be determined

Option: 1

Explanation:

Let the money earned by three labourers be x - 10, x - 20 and x respectively. Had they been not absent, then they would have got equal amount of money, i.e.

$\frac{{x - 10 + x - 20 + x}}{3} = x - 10$

the third labourer earned Rs 10 extra.

Question 38:
Three labourers started working together to complete a task in certain number of days. But in the course of work, they remained absent for few days due to some unavoidable circumstances and therefore the work wascompleted 3 days late. If the first and second labourers(assume that the wage per day are fixed and late completion of work did not affect it.) Rs 80 and Rs 100 less than the third labourer got, then the wage per day per labourer is
 Rs 5
 Rs 10
 Rs 20
 cannot be determined

Option: 3

Explanation:

We need to account for 3 x 3 = 9 leaves (or absences). Let the wage per day per labourer be Rs x. Then the number of days for which first and second labourers remained absent will be 80/x and 100/x more than that the third labourer remained, i.e. total of 180/x leaves (or absences). Considering third labourer to be absent for 0, 1, 2 days

$\frac{{180}}{x} = 9,8,7$ respectively,

The only possible values of $\frac{{180}}{x}{\rm{ is }}9$

Question 39:
Three labourers started working together to complete a task in certain number of days. But in the course of work, they remained absent for few days due to some unavoidable circumstances and therefore the work wascompleted 3 days late. If the first and second labourers(assume that the wage per day are fixed and late completion of work did not affect it.) got Rs 15 and Rs 30 less than the third labourer got, then the third labourer remained absent for
 0 days
 1 days
 2 days
 cannot be determined

Option: 1

Explanation:

As done in Q.38 15/x and 30/x should be integers and $\frac{{45}}{x} = 9,8{\rm{ or }}7$

The only possible Value of $\frac{{45}}{x}{\rm{ is }}9$

the third labour remain absent for 0 days.

Question 40:
Three labourers started working together to complete a task in certain number of days. But in the course of work, they remained absent for few days due to some unavoidable circumstances and therefore the work wascompleted 3 days late. If the first and second labourers(assume that the wage per day are fixed and late completion of work did not affect it.) got Rs x and Rs y less then the third labourer got, then which of the following value of (x, y) is feasible?
 (18, 24)
 (18, 30)
 (24, 30)
 all of these

Option: 3

Explanation:

As done in Q.38, for option (1), we have $\frac{{42}}{x} = 9,6{\rm{ or }}3$

$\Rightarrow x = \frac{{14}}{3},7{\rm{ or }}14$

But 18/x is not an integer for any value of x.

this is rejected. For option (2),

 and as 18/x is not an integer for any value of x.

This is also rejected.

Only option (3) remains to be verified us option (4) cannot be valid.