## IIFT Time and Work Questions PDF

Download important IIFT Time and Work Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Time and Work questions and answers for IIFT exam.

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**Question 1: **A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. Which is the first pair?

a) A and B

b) A and C

c) B and C

d) A and D

**Question 2: **There’s a lot of work in preparing a birthday dinner. Even after the turkey is in the oven, there’s still the potatoes and gravy, yams, salad, and cranberries, not to mention setting the table.

Three friends — Asit, Arnold and Afzal — work together to get all of these chores done. The time it takes them to do the work together is 6 hr less than Asit would have taken working alone, 1 hr less than Arnold would have taken alone, and half the time Afzal would have taken working alone. How long did it take them to do these chores working together?

a) 20 min

b) 30 min

c) 40 min

d) 50 min

**Question 3: **In a watch, the minute hand crosses the hour hand for the third time exactly after every 3 hr 18 min and 15s of normal time. What is the time gained or lost by this watch in one day?

a) 14 min 10 s lost

b) 13 min 48 s lost

c) 13 min 20 s gained

d) 14 min 40 s gained

**Instructions**Boston is 4 hr ahead of Frankfurt and 2 hr behind India. X leaves Frankfurt at 6 p.m. on Friday and reaches Boston the next day. After waiting there for 2 hr, he leaves exactly at noon and reaches India at 1 a.m. On his return journey, he takes the same route as before, but halts at Boston for 1hr less than his previous halt there. He then proceeds to Frankfurt.

**Question 4: **If X had started the return journey from India at 2.55 a.m. on the same day that he reached there, after how much time would he reach Frankfurt?

a) 24 hr

b) 25 hr

c) 26 hr

d) Data insufficient

**Question 5: **Three wheels can complete 60, 36 and 24 revolutions per minute. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?

a) $\frac{5}{2}$ s

b) $\frac{5}{3}$ s

c) $5$ s

d) $7.5$ s

**Question 6: **It takes six technicians a total of 10 hr to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 am, and one technician per hour is added beginning at 5 pm, at what time will the server be completed?

[CAT 2002]

a) 6.40 pm

b) 7 pm

c) 7.20 pm

d) 8 pm

**Question 7: **Three small pumps and a large pump are filling a tank. Each of the three small pump works at 2/3 the rate of the large pump. If all four pumps work at the same time, they should fill the tank in what fraction of the time that it would have taken the large pump alone?

[CAT 2002]

a) 4/7

b) 1/3

c) 2/3

d) 3/4

**Question 8: **A, B and C individually can finish a work in 6, 8 and 15 hours respectively. They started the work together and after completing the work got Rs.94.60 in all. When they divide the money among themselves, A, B and C will respectively get (in Rs.)

a) 44, 33, 17.60

b) 43, 27, 20, 24.40

c) 45, 30, 19.60

d) 42, 28, 24.60

**Instructions**The Weirdo Holiday Resort follows a particular system of holidays for its employees. People are given holidays on the days where the first letter of the day of the week is the same as the first letter of their names. All employees work at the same rate.

**Question 9: **Raja starts working on February 25(Sunday), 1996, and finishes the job on March 2, 1996. How much time would T and J take to finish the same job if both start on the same day as Raja?

a) 4 days

b) 5 days

c) Either a or b

d) Cannot be determined

**Question 10: **A contract is to be completed in 56 days and 104 men are set to work. Each working 8 hours a day, after 30 days, 2/5th of the work is finished. How many additional men may be employed so that work may be completed on time, each man now working 9 hours per day?

a) 56 men

b) 65 men

c) 46 men

d) None of the above

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**Answers & Solutions:**

**1) Answer (D)**

A takes 4 days to complete the work.

So, B takes 8 days to complete the same work.

C takes 16 days to complete the work.

D takes 32 days to complete the same work.

In order to measure, let the total work be of 64 units. Hence, the speed of working of each of the four persons is given below.

A – 16 units/hr

B – 8 units/hr

C – 4 units/hr

D – 2 units/hr

From the given options, we need to find two pairs in such a way that their speeds are in the ratio 3:2. Note that A+D=18 while B+C=12 and the ratio is 3:2

Hence, the first pair is A and D and the second pair is B and C

**2) Answer (C)**

Let the time taken working together be t.

Time taken by Arnold = t+1

Time taken by Asit = t+6

Time taken by Afzal = 2t

1/(t+1) + 1/(t+6) + 1/2t = 1/t

From the options, if time = 40 min, that is, t = 2/3

3/5 + 3/20 + 3/4 = (12+3+15)/20 = 30/20 = 3/2 = 1/t

**3) Answer (B)**

In a normal watch, the minute hand crosses the hour’s hand after every 1 hour 5 minutes and 27 seconds.

So, the third time the hour’s hand crosses the minute’s hand is after 3 hours 16 minutes and 21 seconds.

In this watch, the time taken for this to happen is 3 hours 18 minutes and 15 seconds.

Hence, the watch loses 1 minute and 54 seconds after every 3 hours 18 minutes and 15 seconds.

18 minutes and 15 seconds = 1095 seconds = 1095/3600 $\approx$ .304 hours.

=> 3 hours 18 minutes and 15 seconds = 3.304 hours

So, time lost in a day is $1\frac{54}{60}*\frac{24}{3.304} = \frac{114}{60}*\frac{24}{3.304} \approx 13.8$

So, the time lost by the watch in a day is approximately equal to 13 minutes and 48 seconds.

**4) Answer (A)**

X leaves Frankfurt at 6 PM and reaches Boston at 10 AM, which is 6 AM in Frankfurt => 12-hour journey

Leaves Boston at 12 PM ad reaches India at 1 AM, which is 11 PM in Boston => 11-hour journey

=> Total time = 12 + 2 + 11 = 25 hours

Return journey is 1 hour lesser => 25 – 1 = 24 hours.

**5) Answer (C)**

The first wheel completes a revolution in $\frac{60}{60}=1$ second

The second wheel completes a revolution in $\frac{60}{36}=1\frac{2}{3}$ second

The third wheel completes a revolution in $\frac{60}{24}=2\frac{1}{2}$ second

The three wheels touch the ground simultaneously at time which are multiples of the above three times.

Hence, the required number is $LCM(1,\frac{5}{3},\frac{5}{2}) = 5$ seconds.

So, the correct option is option (c)

**6) Answer (D)**

Let the work done by each technician in one hour be 1 unit.

Therefore, total work to be done = 60 units.

From 11 AM to 5 PM, work done = 6*6 = 36 units.

Work remaining = 60 – 36 = 24 units.

Work done in the next 3 hours = 7 units + 8 units + 9 units = 24 units.

Therefore, the work gets done by 8 PM.

**7) Answer (B)**

Let the work done by the big pump in one hour be 3 units.

Therefore, work done by each of the small pumps in one hour = 2 units.

Let the total work to be done in filling the tank be 9 units.

Therefore, time taken by the big pump if it operates alone = 9/3 = 3 hours.

If all the pumps operate together, the work done in one hour = 3 + 2*3 = 9 units.

Together, all of them can fill the tank in 1 hour.

Required ratio = 1/3

**8) Answer (A)**

Money will be distributed in the ratio of work done in an hour.

i.e. $\frac{1}{6} : \frac{1}{8} : \frac{1}{15}$

or 20:15:8

Hence part of A will be = $\frac{20}{43} \times 94.6 = 44$

part of B will be = $\frac{15}{43} \times 94.6 = 33$

part of C will be = $\frac{8}{43} \times 94.6 = 17.60$

**9) Answer (A)**

Raja takes 7 days to complete the job.

Feb 25, 1996 is a Sunday.

If T and J start working on Sunday, they can complete the work by wednesday because T would have worked for 3 days and J would worked for 4 days, thereby matching the number of days worked by Raja.

Hence, they can complete the job in 4 days.

**10) Answer (A)**

Let ‘W’ be the amount of work. It is given that,

$\dfrac{2}{5}*W = 104*8*30$ … (1)

Let ‘X’ be the number of additional men required to finish the work on time.

$W – \dfrac{2}{5}*W = (104+x)*9*(56-30)$ … (2)

By equation (1) and (2) we can say that,

$\dfrac{2}{3} = \dfrac{104*8*30}{(104+x)*9*(56-30)}$

$\Rightarrow$ $(104+x)*9*26 = 12*104*30$

$\Rightarrow$ $(104+x)=160$

$\Rightarrow$ $x=56$.

Hence, option A is the correct answer.

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