In CAT, Time and Work questions have become important, as a result of the rise in the number of questions from arithmetic. There are many shortcut formulas and tricks, which are frequently used to solve CAT Time and Work questions in quick time. Alternatively, Time and Work problems can easily be solved by the concept of the LCM method.

Questions from Pipes and Cisterns also fall under Time and Work as the concept requires to solve such questions is the same as the concepts used to solve Time and Work Questions.

To apply the concepts to solve CAT problems on Time and Work, an aspirant must be good at Ratio & Proportion and Percentages. As the level of difficulty of the questions varies a lot, it is advisable to follow the below tips to construct a proper approach to Time and Work questions for CAT exam.

## Time and Work Tips to solve CAT questions:

- Assume the work to be the LCM of the numerical values given in the question.
- Get the one unit work for each of the participants who are mentioned in the question and are supposed to complete the work.
- Often the concepts of ratio and proportion are helpful in simplifying the complex calculations.
- If the options are provided, try testing the options backwards. The option which agrees with the conditions given in the question should be the right answer.

To give you ample practice, we have provided around 40 questions on Time and Work. These questions are of varied difficulty level. These Time and Work questions are divided into 8 practice exercises. The links of all these exercises are given at the bottom.

**Question 1:**

A work was completed by three persons of equal ability, first one doing m hours for m days, second one doing n hours for n days (m and n being integers) and third one doing 16 hours for 16 days. The work could have been completed in 29 days by third person alone with his respective working hours. If all of them do the work together with their respective working hours, then they can complete it in about

[1] 12 days

[2] 13 days

[3] 14 days

[4] 15 days

**Answer:**

**Explanation: **

\(\left( {{m^2} + {n^2} + {{16}^2}} \right)k = 1\;{\rm{and}}\;16 \times 29k = 1\)

\( {m^2} + {n^2} + {16^2} = 16\times29\)

\( \Rightarrow {m^2} + {n^2} = 16\left( {29 - 16} \right) = 16 \times 13 = 208\)

Now the last digits of m, n cannot be (0, 8), (1, 7), (2, 6), (3, 5). Therefore it can only be (4, 4) or (9, 9). On checking, we find \({{m}^{2}}+{{n}^{2}}={{12}^{2}}+{{8}^{2}}\)

Therefore, they can together do the work in

\(\begin{array}{*{20}{c}}{\frac{1}{{\left( {m + n + 16} \right)k}} = \frac{{16.29}}{{\left( {12 + 8 + 16} \right)}} = \frac{{16.29}}{{36}}}\\{ = \frac{{4.29}}{9} = 4\left( {3 + .22} \right) = 12.88 \cong 13.{\rm{ }}}\end{array}\)

**Question 2:**

Three labourers worked together for 30 days, in the course of work, all of them remained absent for few days. One of them was absent for 10 days more than the second labourer and the third labourer did one-third of the total work. How many days more than the third labourer was the first one absent?

[1] 4

[2] 5

[3] 6

[4] cannot be determined

**Answer:**

**Explanation: **

Let k be the part of work labourers can do in one day and x -10, x, y be the number of days for which they remained present. Then

\(\begin{matrix}\left( x-10 \right)k+xk=\frac{2}{3}\text{ }\text{ and }\text{ }yk=\frac{1}{3} \\ {} \\ \left( 2x-10 \right)\frac{1}{3y}=\frac{2}{3}\Rightarrow 2x-10=2y\Rightarrow x-5=y \\\end{matrix}\)The required number of days= \(y-\left( x-10 \right)=5\)

**Question 3:**

A and B do a work in exactly 16 days, B and C do the same work in exactly 12 days while C and A do the same work in about 10 days. If A, B and C can together do the work in integral number of days, then C does the work alone in

[1] 15 days

[2] 16 days

[3] 18 days

[4] none of these

**Answer:**

**Explanation: **

Let a, b, c be the number of days in which A, B, C, can do the work alone. Then

\(\begin{array}{*{20}{c}}{\frac{1}{a} + \frac{1}{b} = \frac{1}{{16}},\frac{1}{b} + \frac{1}{c} = \frac{1}{{12}},\frac{1}{c} + \frac{1}{a} = \frac{1}{{10}}}\\{{\rm{ Let }}\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{n}{\rm{ where \rm\; }}n{\rm{\; is \;\rm an \;\rm integer }}}\end{array}\)

Then

\(\begin{array}{*{20}{c}}{ \Rightarrow \frac{2}{n} = \frac{7}{{48}} + \frac{1}{{10}} \cong \frac{{35 + 24}}{{240}}}\\{ \Rightarrow \frac{n}{2} \cong \frac{{240}}{{59}} \Rightarrow n \cong \frac{{480}}{{59}} \cong 8}\\{{\rm{ and }}\frac{1}{c} = \frac{1}{8} - \frac{1}{{16}} = \frac{1}{{16}} \Rightarrow c = 16}\end{array}\)

**Question 4:**

Two persons A and B can do a work alone in 29 days. A takes the rest of one day after every 4 days and B takes the rest of one day after every 5 days. If A and B starts working together, then the work will be completed on

[1] 15th day

[2] 16th day

[3] 17th day

[4] 18th day

**Answer:**

**Explanation: **

In 18 days, both take the rest of 3 days each and the work, done is \(\left( {\frac{{36 - 6}}{{29}}} \right)\) part of the work.

By 18th day \(\frac{1}{{29}}\) part of the work is done more,So it implies that the work was completed on 17th day.

**Question 5:**

Works W1 and W2 are done by two persons A and B. A takes 80% more time to do the work W1 alone than he takes to do it together with B. How much percent more time B will take to do the work W2 alone than he takes to do it together with A?

[1] 100%

[2] 120%

[3] 125%

[4] can not be determined

**Answer:**

**Explanation: **

Let a, b be the amounts of work done per day. by A-and B-respectively.

Let A and B complete W, in n days. Then

\(\begin{array}{*{20}{c}}{\left( {a + b} \right)n = W,{\rm{ and }}1.8{\rm{ n a }} = {W_1}}\\{ \left( {a + b} \right) = 1.8a \Rightarrow b = 0.8a \Rightarrow a = 1.25b}\end{array}\)

Now let A and B complete \({{W}_{2}}\) in m days. Then B will take \(\left( {\frac{{2.25 - 1}}{1}} \right) \times 100 = 125{\rm{ }}\% {\rm{ more}}\)time to complete \({W_2}\) alone than with A.