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# 40 Time and Work Questions for CAT with Answers 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
 12 days
 13 days
 14 days
 15 days
Option:

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

${m^2} + {n^2} + {16^2} = 16\times29$

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

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

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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?
 4
 5
 6
 cannot be determined
Option:

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= $y-\left( x-10 \right)=5$

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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
 15 days
 16 days
 18 days
 none of these
Option:

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

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

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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
 15th day
 16th day
 17th day
 18th day
Option:

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.

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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?
 100%
 120%
 125%
 can not be determined
Option:

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.

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Question 6:
12 men and 16 boys can do a piece of work in 5 days and 13 men and 24 boys can do it in 4 days. Compare the daily work done by a man with that done by a boy.
 3 : 2
 2 : 1
 4 : 7
 3 : 1
Option:

12 men + 16 boys can do the work in 5 days.

5 x (12 men + 16 boys) can do the work in 1 day.

Similarly 4 x (13 men + 24 boys) can do the same work in 1 day.

=> 52 men + 96 boys = 60 men + 80 boys.

=> 8 men = 16 boys or 1 man = 2 boys, i.e. 2 : 1

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Question 7:
If 5 men and 3 boys can reap 23 hectares in 4 days and if 3 men and 2 boys can reap 7 hectares in 2 days, then how many boys must assist 7 men in order that they may reap 45 hectares in 6 days?
 1
 2
 3
 4
Option:

5 men + 3 boys can reap 23 hectares in 4 days.

3 men + 2 boys can reap 7 hectares in 2 days.

14 (5 men + 3 boys) can reap 23 x 14 hectares in 4 days.

23 (3 men + 2 boys) can reap 7 x 2 x 23 hectares in 4 days.

14 (5 men + 3 boys) = 23 (3 men + 2 boys).

1 man = 4 boys. Now 5 men + 3 boys = 23 boys.

23 boys can reap 23 hectares in 4 days.

30 boys can reap 45 hectares in 6 days.

But 30 boys = 28 boys + 2 boys = 7 men + 2 boys.

Hence 2 boys must assist 7 men.

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Question 8:
X can do a job in 10 days, Y in 15 days and Z in 18 days. Y and Z begin the work but have to leave after 3 days. How many days will X take to finish the job?
 57/9 days
 57/11 days
 53/12 days
 6.5 days
Option:

(Y + Z)'s one day's work = 1/15 + 1/18 = 11/90.

(Y + Z)'s 3 day's work = 1/15 + 1/18 = 33/90.

Remaining Work = 1 — 33/90 = 57/90. ,

Time taken by X = 57/90 ÷ 1/10 = 57/9 days.

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Question 9:
10 men and 12 children complete a certain piece of work in 10 days. Each child takes thrice the time taken by a man to complete the work. The time taken by 12 men to finish the same work is
 11.66 days
 10 days
 10.33 days
 12.16 days
Option:

3 Children = 1 Man; 10 Men + 12 Children = 10 Men + 4 Men = 14 Men.

Work = 14 x 10 = 12 x D where D is the required number of days.

D = 14 x 10/12 = 140/12 days = 11.66 days.

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Question 10:
5 men and 8 women can do a job in 8 days, while 4 men and 6 women can do it in 10 days. How many days will 10 women take to finish the job?
 10 days
 40 days
 20 days
 None of these
Option:

Let X = 1 man's one day's job, Let Y = 1 woman's one day's job.

According to the given condition, 5X + 8Y = 1/8 and 4X + 6Y = 1/10.

On solving the two equations, we get : X = 1/40, Y = 0.

Women have not done any amount of work.

Women can never complete the work.

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Question 11:
To do a certain piece of work, B would take three times as long as A and C together and C twice as long as A and B together. The three men working together can complete the work in 10 days. How long would B take by himself to complete the same piece of work?
 24 days
 30 days
 40 days
 36 days
Option:

By the question, 3 times B's daily work = (A + C)'s daily work.

Add B's daily work to both sides.

=> 4 times B's daily work = (A + B + C)'s daily work = 1/10.

=>B's daily work = 1/40 .... (1)

Also, 2 times C's daily work = (A + B)'s daily work.

Add C's daily work to both sides.

=>3 times C's daily work = (A + B + C)'s daily work = 1/10.

=>C's daily work = 1/30 .... (2)

Now A's daily work = 1/10 — (1/40 + 1/30) = 1/24 ...(3)

=>A, B and C can do the work in 24, 40 and 30 days respectively.

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Question 12:
A and B working separately can finish a work in 8 and 12 days respectively. If they work for a day alternately (beginning with A), then in how many days will the work be completed?
 9 days
 9.5 days
 10 days
 9.8 days
Option:

A's one day's work = 1/8.

Then B comes and does 1/12 of the whole work. Hence at the end of two days,

(1/8 + 1/12 = 5/24) of the whole work has been done.

in the next 2 days, 5/24 of the whole work will again be done.

=>In these 4 days, 5/24 + 5/24 = 10/24 of the work has been done.

Again in the next four days, 10/24 of the work will be done.

=>In these 4 + 4 = 8 days, 10/24 + 10/24 = 20/24 of the whole work will be done.

Remaining work = 1 — 20/24 = 4/24 = 1/6th of the work.

Now A does 1/8th of the work in one day.

Hence remaining work = 116 — 1/8 = 1/24.

This will be now completed by B. B does 1/12 of the work in 1 day.

Hence he will complete 1/24 of the work in 1/24 ÷ 1/12 = 1/2 days.

Hence total time taken = 4 + 4 + 1 + 1/2 = 9.5 days.

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Question 13:
Raman and Bose working separately can finish a work in 50 and 40 days respectively. They begin the work together and after 10 days, Bose goes away. In how many days will the work be completed by Raman alone?
 27 days
 28 days
 350/17 days
 55/2 days
Option:

(Raman's + Bose's) one day's work $= \frac{1}{{50}} + \frac{1}{{40}} = \frac{9}{{200}}$

They worked together for 10 days.

Hence their 10 day's work =$\frac{9}{200}\times 10=\frac{9}{20}$

Remaining work $= 1 - \frac{9}{{20}} = \frac{{11}}{{20}}$ This is completed by Raman alone.

Raman will require, $\frac{{11}}{{20}} \div \frac{1}{{50}} = \frac{{55}}{2}$ days to complete the remaining work.

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Question 14:
Two men, A and B, working separately can mow a field in 10 and 12 hours respectively. If they work for an hour alternately, B beginning at 9 am, then at what time will the mowing be finished?
 7:30 pm
 8:00 pm
 8:30 pm
 9:00 pm
Option:

In the first hour 1 , B mows $\frac{1}{{12}}$ of the field.

In the second hour, A mows $\frac{1}{{10}}$ of the field.

In 2 hours, $\frac{{11}}{{60}}$ of the field is mown.

In 10 hours, $\frac{{55}}{{60}}$ of the field is mown.

Now remaining work is $1 - \frac{{55}}{{60}} = \frac{5}{{60}} = \frac{1}{{12}}$field remains to be mown.

In the 11th hour, B mows remaining$\frac{1}{{12}}$ of the field.

the total time required is 11 hours.

The work started at 9 am

:. it would be finished at 8.00 pm.

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Question 15:
25 men can reap a field in 20 days. When should 15 men leave the work, if the whole field is to be reaped in 371/2 days after they leave the work?
 after 4 days
 after. 6 days
 after 5 days
 after 3 days
Option:

=> 25 men can reap the field in 20 days,

=> 10 men can reap the field in 20 x 25/10 or 50 days.

When 15 men leave the work, 10 men remain and these can reap in 37.5 days

37.5/50 or 3/4 of the field.

Hence all men must work till (1 — 3/4) or 1/4 of the field is reaped.

Now 25 men reap 1/4 of the field in 20/4 or 5 days.

Hence 15 men should leave the work after 5 days.

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Question 16:
If A and B can do a piece of work in 8 days, B and C in 12 days and C and A in 16 ‘' days, then in how many days will C finish it when working alone?
 80 days
 36 days
 96 days
 88 days
Option:

(A + B)'s one day's work = 1/8, (B + C)'s one day's work = 1/12.

(A +C)'s one day's work = 1/16. Add up all the three equations, we get

=> 2(A + B + C)'s one day's work = 1/8 + 1/12 + 1/16 = 26/96.

=> (A + B + C)'s one day's work = 26/192.

=>A's one day's work = (A + B + C)'s one day's work — (B + C)'s one day's

work = 26/192 — 1/12 = (26 — 16)/192 = 10/192.

Since A completes 10/192 of the work in 1 day, he will complete the work in 192/10 = 19.2 days. By a similar logic, B will require 13.7 days and C will require 96 days.

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Question 17:
A is twice as good a workman as B and together they finish a piece of work in 20 p days. In how many days can A alone finish the work?
 30 days
 25 days
 20 days
 32 days
Option:

A's one day's work : B's one day's work = 2 : 1.

=>Out of every three parts of work done, 2 will be done by A and 1 by B

Now, (A B)'s one day's work = 1/20.

A's share = 2/3 x 1/20 = 2/60 = 1/30.

=>In 1 day, A does 1/30 of the work. Hence, A will need a total of 30 days to do the job alone.

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Question 18:
The rates of working of A and B are in the ratio 4 : 5. What is the ratio of the number of days taken by them to finish a job when working separately?
 2 : 3
 3 : 2
 5 : 4
 4 : 3
Option:

The Time Taken is always inversely proportional to the Efficiencies (or the Rates of Working) = time taken by A : B = 5 : 4.

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Question 19:
Ram can do a piece of work in 20 days which Shyam can do in 30 days. They begin together with the condition that Ram shall leave the job 3 days before the actual completion of work. What is the total number of days required to complete the work?
 14 days
 19 days
 27 days
 9 days
Option:

Let X be the days required to complete the work $\frac{{x - 3}}{{20}} + \frac{x}{{30}} = 1$

Solving $x\approx 14$

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Question 20:
Rama can do a piece of work in 10 days and Rami alone can do it in 5 days. Rama and Rami undertook to do it for Rs.400. With the help of their neighbour Snooty, they finished it in 2 days. The share of Snooty out of the remuneration is
 Rs.120
 Rs.200
 Rs.160
 Rs.180
Option:

(Rama + Rami + Snooty)'s one day's work = 1/10 + 1/5 + 1/X.

(Rama + Rami + Snooty)'s two day's work = 2 x (1110 + 1/5 + 1/X).

{because they need 2 days to do the job)

=> 2 [ 1/10 + 115 + 1/X ] = 1 = On solving we get, X = 5 days.

This means that Snooty can do the whole work alone in 5 days.

=> In one day, he does 1/5 of the work.

=>In two days, he does (2 x 1/5) of the work.

=> Snooty's share = 1/5 x 2 x 400 = Rs.160.

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Question 21:
Samtaprasad can do a piece of work in 50 days. He worked only for 5 days and was infected with Malaria and had to leave the job. Shantaprasad completed the remaining work in 30 days. The number of days, in which both together can complete the work is
 22 days
 20 days
 25 days
 27 days
Option:

Samtaprasad's 1 day work =, 1/50. 5 days work = (5/50) = 1/10.

=>Remaining work = 1 - (1/10) = 9/10.

As Shantaprasad completes 9/10 work in 30 days, so the work done by him in one day = 9/300. When they work together, the work done by them in one day = (1/50) + (9/300) = 1/20.

=>Number of days required by them, working together = 20.

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Question 22:
25 days of Ram's wages can be paid by a certain sum of money. The same amount of money is sufficient to pay Badriprasad's wages for 20 days. The number of days for which the money will be sufficient to pay the wages of both if they work together is
 10 days
 11 days
 100/9 days
 110/9 days
Option:

Ram's 1 day's wages = 1/25th of the money, Badriprasad's 1 day's wages

= 1/25 + 1120 = 9/100 of the total money. Hence the money will be sufficient

for 100/9 days if both of them work together.

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Question 23:
Gagan is thrice as good a worker as Dilip and takes 8 days less to do a piece of work than Dilip. In how many days can Dilip do the complete work?
 4 days
 24 days
 12 days
 18 days
Option:

Let Dilip's 1 day's work = X Gagan's 1 day's work = 3X.

Total time taken by Dilip to do the whole job = 1/X.

Total time taken by Gagan to do the whole job = 1/3X.

It is given that 1/3X + 8 = 1/X X = 1/12 = Dilip's one day's work.

Time taken by Dilip to complete the whole work alone = 12 days.

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Question 24:
Billy is four times as good a workman as Silly and can build a wall in 45 days less than the number of days required by Silly. What is the time they will take, working together, to build two such walls?
 12 days
 24 days
 9 days
 18 days
Option:

Let Billy's time to complete one work = x days.

Then, Silly's time to complete one work = 4x days.

Since Billy take 45 days less than Silly to build a wall.

i.e., 4x - x = 45 => x = 15 and 4x = 60.

Working together they will take $\frac{1}{{\frac{1}{{15}} + \frac{1}{{60}}}} = 12$ days to build one wall.

So, to build two such walls, 24 days will be required.

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Question 25:
X men can finish a job in 40 days. If 5 extra Men join them, the job takes 10 days less. What is the value of X?
 15
 20
 10
 18
Option:

Work = W. Let the number of men be X.

40X = W and (40 - 10)(X + 5)

= W => 40X = 30(X + 5) => 10X = 150=> X = 15.

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Question 26:
A,B and c together earn Rs. 300 a day, while A and C together earn Rs.188 and B and C together earn Rs. 152 a day .What is the daily earning of C?
 Rs.15
 Rs.40
 Rs.60
 Rs.35
Option:

C's earning for a day = (A + B + C)'s earning for a day - (A + B)'s for a day = 300 - (148 + 112) = Rs.40.

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Question 27:
A and B together can do a piece of work In 6 days and A alone can do It in 9 day,. In how days B alone do it?
 15 day
 18 days
 21 days
 20 days
Option:

A and B can do 1/6 of the work in 1 day.

A alone can do 1/9 of the work in 1 day.

B alone can do (1/6 —1/9) or 1/18 of the work in 1 day.

B can do the whole work in 18 days.

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Question 28:
A alone can finish a job in 12 days and B alone can do It In 20.days. If they work together and finish it, then the share of A In total wages of Rs100 is
 Rs.56.25
 Rs.67.50
 Rs.62.50
 Rs.50
Option:

Ratio of times taken by A and B to do the job = 12 : 20 = 3 : 5.

Ratio of work done by them when they work together = 5 : 3.

Share of A = 5/(5 + 3) x 100 = 5/8 x 100 = Rs.62.5.

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Question 29:
Two pipes, P and Q can fill a cistern in 12 and 15 minutes respectively. Both are opened together, but at the end of 3 minutes the first Is turned off. How much longer will the cistern take to fill completely?
 8.25 min
 10 min
 9 min
 8.5 min
Option:

P and Q fill (1/12 + 1/15) or 3/20 of the cistern in 1 minute.

In 3 minutes, 9/20 of the cistern is filled (1 — 9/20) or 11/20 of the cistern is empty when the first pipe P is closed.

Now Q can fill 1/15 of the cistern in one minute.

Q can fill 11/20 of the cistern in 11/20 x 15/1 or $8_4^1$ minutes.

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Question 30:
A certain number of men can do a work in 60 days. If there were 8 more men, it could be finished in 10 days less. How many men are there?
 30
 50
 40
 42
Option:

The original number of men together with 8 men more can finish the work in (60 — 10) or 50 days.

Now 8 men can do in 50 days what the original number of men can do in 10 days.

The original number of men = 50 x 8/10 = 40.

Short-cut : 8 men's 1 day's work = 1/50 — 1/60 = 1/300.

Now in one day 1/300 of the work is done by 8 men.

In one day 1/60 of the work is done by 8 x 5 or 40 men.

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Question 31:
Two men undertake to do a piece of work for Rs.200. One alone could do it in 6 days, the other In 8 days. With the assistance of a boy they finish it in 3 clays. What Is the share of the boy?
 Rs.25
 Rs.50
 Rs.60
 Rs.30
Option:

1st man's a days work = 3/6, 2nd man's 3 days work = 3/8.

The boy's 3 days' work = 1 — (3/6 + 3/8) = 1/8.

Clearly Rs.200 should be divided amongst them in the proportion of 3/6 : 3/8 : 1/8 or 4 : 3 : 1.

1st man's share = 4/8 of Rs.200 = Rs.100.

2nd man's share = 3/8 of Rs.200 = Rs.75.

The boy's share = 1/8 of Rs.200 = Rs.25.

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Question 32:
A contract Is to be completed In 56 days and 104 men were set to work, each working 8 hours a day. After 30 days, 2/5 of the work Is finished. How many additional men may be employed so that work may be completed on time?
 180 men
 64 men
 76 men
 96 men
Option:

Total number of men =$104 \times \frac{8}{8} \times \frac{{30}}{{26}} \times \frac{{3/5}}{{2/5}} = 180$

Therefore, Additional men required = 180 — 104 = 76.

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Question 33:
Two coal loading machines each working 12 hours per day for 8 days handles 9 tons of coal with an efficiency of 90%. While 3 other coal loading machines at an efficiency of 80% set to handle 12 tons of coal In 6 days. Find how many hours per day each should work?
 12 hrs/day
 16 hrs/day
 20 hrs /day
 18 hrs/day
Option:

$\frac{{{N_1} \times {D_1} \times {H_1} \times {E_1}}}{{{W_1}}} = \frac{{{N_2} \times {D_2} \times {H_2} \times {E_2}}}{{{W_2}}}$ (By chain Rule).

$= \frac{{2 \times 8 \times 12 \times 90}}{9} = \frac{{3 \times 6 \times {H_2} \times 80}}{{12}} \Rightarrow {H_2} = 16{\rm{ hrs}}/{\rm{day}}$

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Question 34:
If the cost of printing a book of 320 leaves with 21 lines On each page and on an average 11 words in each line is Rs.19, then find the cost of printing a book with 297 leaves,28 lines on each page and 10 words In each line.
 Rs $22\frac{3}{8}$
 Rs $20\frac{3}{8}$
 Rs $21\frac{3}{8}$
 Rs $21\frac{3}{4}$
Option:

By chain rule $\frac{{19}}{{320 \times 21 \times 11}} = \frac{x}{{297 \times 28 \times 10}}$

$\Rightarrow x = {\rm{Rs}}.21\frac{3}{8}$

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Question 35:
A can do a job in 10 days. B can do in 12 days and C can do in 15 days. All begin together but A leaves the work after 2 days and B leaves 3 days before the work is finished. How long did the work last?
 5 days
 9 days
 4 days
 7 days
Option:

Let the job be completed in x days.

Only C remains from the start to the finish of the work. C worked for x days.

A worked for 2 days and B worked for (x-3) days.

$\Rightarrow \frac{2}{{10}} + \frac{{x - 3}}{{12}} + \frac{x}{{15}} = 1 \Rightarrow x = 7$

The job is finished in 7 days.

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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:

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

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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:

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.

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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:

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 $\frac{{180}}{x}{\rm{ is }}9$

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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:

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.

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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:

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.

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