CAT Exam  >  CAT Tests  >  Test Level 3: Time & Work - 2 - CAT MCQ

Test Level 3: Time & Work - 2 - CAT MCQ


Test Description

10 Questions MCQ Test - Test Level 3: Time & Work - 2

Test Level 3: Time & Work - 2 for CAT 2024 is part of CAT preparation. The Test Level 3: Time & Work - 2 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 3: Time & Work - 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 3: Time & Work - 2 below.
Solutions of Test Level 3: Time & Work - 2 questions in English are available as part of our course for CAT & Test Level 3: Time & Work - 2 solutions in Hindi for CAT course. Download more important topics, notes, lectures and mock test series for CAT Exam by signing up for free. Attempt Test Level 3: Time & Work - 2 | 10 questions in 20 minutes | Mock test for CAT preparation | Free important questions MCQ to study for CAT Exam | Download free PDF with solutions
Test Level 3: Time & Work - 2 - Question 1

A flat of 3 rooms is painted by A, B and C in 5 days. A works for the whole time, B only on the first two days and C only on the last three days. This work could have been done by B and C in 6 days (both working full time) without involving A. If B and C working together can do as much work in two days as A can do in 3 days, then how long will it take for A, B and C, respectively, to do this work alone?  

Detailed Solution for Test Level 3: Time & Work - 2 - Question 1

Let A, B and C individually take a, b and c days to complete the same task, respectively.
B and C can finish the task in 6 days working together.
Therefore,

Again, two days of work of B and C equals to 3 days of A's work.
Therefore,

From equation (i),
3/a = 2/6 
a = 9 days
Now, A and B work for first two days and A and C work for last three days to complete the same task.
Therefore,

Substitute values of a and from equation (ii).

c = 9 days
Substitute values of a and c in equation (i).

b = 18 days
Hence, this is the required solution.

Test Level 3: Time & Work - 2 - Question 2

A tap can fill a tank in 24 hours and an outlet can empty the full tank in 30 hours. In how many hours the empty tank will be filled, if both the tap and the outlet are opened simultaneously ?

Detailed Solution for Test Level 3: Time & Work - 2 - Question 2

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test Level 3: Time & Work - 2 - Question 3

Three pipes A, B and C are attached to a tank. A and B can fill it in 20 and 30 minutes, respectively, while C can empty it in 15 minutes. If A, B and C are kept open successively for 1 minute each, how soon will the tank be filled if pipe A is opened first?  

Detailed Solution for Test Level 3: Time & Work - 2 - Question 3

The part of tank filled by Pipe A in 1 minute = 1/20
The part of tank filled by Pipe B in 1 minute = 1/30
The part of tank emptied by Pipe C in 1 minute = 1/15
So, totally in 3 minute (because open for 1 minute each) = 1/20 + 1/30 - 1/15 = 1/60
Consider 55/60 can be filled in 3 × 55 = 165 minute, remaining part is 1/12.
A can fill 1/20 in 1 minute.
So, capacity left is, (1/12) - (1/20) = 1/30
B can fill the remaining 1/30 in 1 minute.
So, total time taken = ((3 × 55) + 1 + 1) = 167 minute
Hence, it will take 167 minutes to fill the tank.

Test Level 3: Time & Work - 2 - Question 4

There are two walls to be built. Area of one wall is twice that of the other. The labourers work on the larger wall for half a day. For the remaining half of the day, half of the labourers work on the larger wall, while the remaining work on the smaller wall. At the end of the day, the larger wall is completely built, while some portion of the smaller wall is left to be built, which is completed by one labourer the next day. How many labourers worked on the first day?  

Detailed Solution for Test Level 3: Time & Work - 2 - Question 4

Let 4x be the total number of labourers.
4x labourers' work in half a day = 2x labourers' work in one day
Half of 4x, i.e. 2x labourers' work in second half of the day = x labourers' work in one day
According to given information, at the end of the day, the larger wall is completed.
Total 3x labourers are required to work full day to complete the work.
For smaller wall:
2x labourers' work in second half of the day = x labourers' work in one day
But, at the end of the first day, it was not completed and needed one labourer's full-day work.
Therefore, total number of labourers required for smaller wall = x + 1
x + 1 is half of the 3x.
x + 1 = 3x/2
x = 2
Total number of labourers who worked on the first day = 4 × 2 = 8

Test Level 3: Time & Work - 2 - Question 5

A and B worked together and completed a piece of work in 24 days. They got some money for their work. If they had distributed that money according to their work, A would have received 50% more money than B. In how many days could B alone complete the same work?  

Detailed Solution for Test Level 3: Time & Work - 2 - Question 5

Let work done by A in 1 day be A, and that done by B in 1 day be B.
Let total work be W.
While working together, A and B took 24 days to complete the work. Therefore, we have
24(A + B) = W ...(i)
Also, as A received 50% more money than B, it is fair enough to assume that A must have done 1.5 times as much work in the same time as B.
Thus, A = 1.5B ...(ii)
Now, plugging the value of A as 1.5B from equation (ii) into equation (i), we have 24(2.5 B) = W or B = W/60 ...(iii)
Suppose, B took d days to do the same work alone.
Thus, dW/60 = W, or d = 60
Thus, working alone, B would take 60 days to do the work.
Hence, answer option (4) is correct.

Test Level 3: Time & Work - 2 - Question 6

Pipe A fills a tank of 700 litres capacity at the rate of 40 litres/minute. Another pipe B fills the same tank at a rate of 30 litres/minute.Pipe C, which is at the bottom of the tank drains the tank at the rate of 20 litres/minute. Pipe A is kept open for a minute and then closed. Pipe B is kept open for a minute and then closed. Pipe C is kept open for a minute and then closed and the whole cycle is repeated. What minimum time will it take for the given tank to overflow?  

Detailed Solution for Test Level 3: Time & Work - 2 - Question 6

After 1 minute, 40 litres is filled.
After 2 minutes, another 30 litres is filled (total 70 litres).
After 3 minutes, 20 litres is emptied.
Therefore, at the end of 3 minutes, 50 litres is filled.
50 litres is filled in 3 minutes.
Pipe A fills the tank at the rate of 40 litres a minute.
Pipe B at the rate of 30 litres a minute and Pipe C drains the tank at the rate of 20 litres a minute.
If each of them is kept open for a minute in the order A - B - C, the tank will have 50 litres of water at the end of 3 minutes. After 13 such cycles, the tank will have 13 × 50 = 650 litres of water.
It will take 13 × 3 = 39 minutes for the 13 cycles to be over. At the end of the 39th minute, Pipe C will be closed and Pipe A will be opened. It will add 40 litres to the tank. Therefore, at the end of the 40th minute, the tank will have 650 + 40 = 690 litres of water.

Test Level 3: Time & Work - 2 - Question 7

Directions: Study the following tables and answers the questions that follow.

Darbar Toy Company has to go through the following stages for the launch of a new toy

The profile of the company’s manpower is

Q. Given this situation, the minimum number of days in which the company can launch a new toy going through all the stages is

Detailed Solution for Test Level 3: Time & Work - 2 - Question 7

Interpretation of the first row of the first table in the question: 
Design and Development requires 30 expert man- days or 60 non-expert man-days.
Hence, work done in 1 expert man-day = 3.33% and work done in 1 non-expert man-day = 1.66%. Further, from the second table, it can be interpreted that: A is an expert at design and development. Hence, his work rate is 3.33% per day and B, D and E are ready to work as non-experts on design and development, hence their work rate is 1.66% per day each.
Thus, in 1 day the total work will be 
A + B + D + E = 3.33 + 1.66 + 1.66 + 1.66 = 8.33% 
work.
Thus, 12 days will be required to finish the design and development phase.

Test Level 3: Time & Work - 2 - Question 8

Directions: Study the following tables and answers the questions that follow.

Darbar Toy Company has to go through the following stages for the launch of a new toy

The profile of the company’s manpower is

Q. If A and C refuse to have anything to do with the manufacturing set up. The number of days by which the project will get delayed will be

Detailed Solution for Test Level 3: Time & Work - 2 - Question 8

Interpretation of the first row of the first table in the question: 
Design and Development requires 30 expert man- days or 60 non-expert man-days.
Hence, work done in 1 expert man-day = 3.33% and work done in 1 non-expert man-day = 1.66%. Further, from the second table, it can be interpreted that: A is an expert at design and development. Hence, his work rate is 3.33% per day and B, D and E are ready to work as non-experts on design and development, hence their work rate is 1.66% per day each.
Thus, in 1 day the total work will be 
A + B + D + E = 3.33 + 1.66 + 1.66 + 1.66 = 8.33% 
work.
Thus, 12 days will be required to finish the design and development phase.
Increase of number of days 

Test Level 3: Time & Work - 2 - Question 9

Directions : Read the following and answer the questions that follow.

A fort contains a granary, that has 1000 tons of grain. The fort is under a siege from an enemy army that has blocked off all the supply routes.
The army in the fort has three kinds of soldiers:
Sepoys → 2,00,000.
Mantris → 1,00,000
Footies → 1,00,000
100 Sepoys can hold 5% of the enemy for one month. 100
Mantris can hold 10% of the enemy for 15 days. 50
Footies can hold 5% of the enemy for one month.
A sepoy eats 1 kg of food per month, a Mantri eats 0.5 kg of food per month and a footie eats 3 kg of food. (Assume 1 ton = 1000 kg).
The king has to make some decisions based on the longest possible resistance that can be offered to the enemy.
If a king selects a soldier, he will have to feed him for the entire period of the resistance. The king is not obliged to feed a soldier not selected for the resistance.
(Assume that the entire food allocated to a particular soldier for the estimated length of the resistance is redistributed into the king’s palace in case a soldier dies and is not available for the other soldiers.)

Q. If the king strategically attacks the feeder line on the first day of the resistance so that the grain is no longer a constraint, the maximum time for which the resistance can last is

Detailed Solution for Test Level 3: Time & Work - 2 - Question 9

For these questions, since food is no longer a constraint, the constraint then becomes the number of lives. Then, the assumption will be that the resistance lasts for one month with a loss of either 2000 sepoys, 2000 mantris or 1000 footies.
Length of resistance 

= 250 months.

Test Level 3: Time & Work - 2 - Question 10

Directions : Read the following and answer the questions that follow.

A fort contains a granary, that has 1000 tons of grain. The fort is under a siege from an enemy army that has blocked off all the supply routes.
The army in the fort has three kinds of soldiers:
Sepoys → 2,00,000.
Mantris → 1,00,000
Footies → 1,00,000
100 Sepoys can hold 5% of the enemy for one month. 100
Mantris can hold 10% of the enemy for 15 days. 50
Footies can hold 5% of the enemy for one month.
A sepoy eats 1 kg of food per month, a Mantri eats 0.5 kg of food per month and a footie eats 3 kg of food. (Assume 1 ton = 1000 kg).
The king has to make some decisions based on the longest possible resistance that can be offered to the enemy.
If a king selects a soldier, he will have to feed him for the entire period of the resistance. The king is not obliged to feed a soldier not selected for the resistance.
(Assume that the entire food allocated to a particular soldier for the estimated length of the resistance is redistributed into the king’s palace in case a soldier dies and is not available for the other soldiers.)

Q. If the feeder line is opened after 6 months and prior to that the king had made decisions based on food availability being a constraint then the number of months (maximum) for which the resistance could last is

Detailed Solution for Test Level 3: Time & Work - 2 - Question 10

For these questions, since food is no longer a constraint, the constraint then becomes the number of lives. Then, the assumption will be that the resistance lasts for one month with a loss of either 2000 sepoys, 2000 mantris or 1000 footies.
In 6 months, the resistance will have lost 12000 mantris. He would also have lost all other soldiers since he has not fed them.

Information about Test Level 3: Time & Work - 2 Page
In this test you can find the Exam questions for Test Level 3: Time & Work - 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Test Level 3: Time & Work - 2, EduRev gives you an ample number of Online tests for practice

Top Courses for CAT

Download as PDF

Top Courses for CAT