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All questions of Time & Work for UPSC CSE Exam

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P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 50 days to complete the same work. then Q alone can do it in
  • a)
    30 days
  • b)
    25 days
  • c)
    20 days
  • d)
    15 days
Correct answer is option 'B'. Can you explain this answer?

Varun Kapoor answered
Work done by P and Q in 1 day = 1/10
Work done by R in 1 day = 1/50
Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50
But Work done by P in 1 day = Work done by Q and R in 1 day . Hence the above equation can be written as
Work done by P in 1 day * 2 = 6/50
=> Work done by P in 1 day = 3/50
=> Work done by Q and R in 1 day = 3/50
Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25
So Q alone can do the work in 25 days

Anup can dig a well in 10 days. but particularly in difficult time the work is such that due to fatigue every subsequent day efficiency of a worker falls by 10%.If Anup is given a task of digging one such well in the difficult time, then in how many days will he finish the work?
  • a)
    12th day
  • b)
    15 th day
  • c)
    11th day
  • d)
    Never
Correct answer is option 'D'. Can you explain this answer?

Shalini Patel answered
Let us assume that digging one well = 40 unit
Efficiency of Anup = 40/10 = 4 unit/day
Now efficiency is falling by 10%
So, this is case of GP
⇒ 4,18/5,81/25,…… where common ratio = 9/10
Now, sum of infinite GP = a / (1 – r)
⇒ 4/ (1 – 9/10) = 40
Hence, it is clear that Anup will takes infinite time, so he will never finish the work.

Sheldon had to cover a distance of 60 km. However, he started 6 minutes later than his scheduled time and raced at a speed 1 km/h higher than his originally planned speed and reached the finish at the time he would reach it if he began to race strictly at the appointed time and raced with the assumed speed. Find the speed at which he travelled during the journey described.
  • a)
    25 km/h 
  • b)
    10 km/h 
  • c)
    6 km/h
  • d)
    15 km/h
Correct answer is option 'A'. Can you explain this answer?

Asf Institute answered
Solve this question through options.
⇒  For instance, if he travelled at 25 km/h, his original speed would have been 24 km/h.
⇒ The time difference can be seen to be 6 minutes in this case = 60 / 24 – 60 / 25 = 0.1 hrs = 6 mins
Thus, 25 km/h is the correct answer. 
So Option A is correct

P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?
  • a)
    8/15
  • b)
    7/15
  • c)
    11/15
  • d)
    2/11
Correct answer is option 'A'. Can you explain this answer?

Varun Kapoor answered
Amount of work P can do in 1 day = 1/15

Amount of work Q can do in 1 day = 1/20

Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60

Amount of work P and Q can together do in 4 days = 4 x (7/60) = 7/15

Fraction of work left = 1 – 7/15= 8/15

A can do a piece of work in 90 days, B in 40 days and C in 12 days. They work for a day each in turn, i.e., first day A does it alone, second day B does it alone and 3rd day C does it alone. After that the cycle is repeated till the work is finished. They get Rs 240 for this job. If the wages are divided in proportion to the work each had done. Find the amount A will get?
  • a)
    14
  • b)
    24
  • c)
    34
  • d)
    36
Correct answer is 'B'. Can you explain this answer?

Ishani Rane answered
Work done by A in 1 day = 1/90
Work done by B in 1 day = 1/40
Work done by C in 1 day = 1/12
Work done in 3 days = 1/90 + 1/40 + 1/12 = 43/360
in 8 * 3 = 24 days , work completed = 8 * 43/360 = 344/360
Remaining work = 1 - 344/360 = 16/360
in 25th day, A works and completes 1/90 work .
Remaining work = 16/360 - 1/90 = 12/360
in 26th day, B works and completes 1/40 work .
Remaining work = 12/360 - 1/40 = 1/120
in 27th day, C works and completes this entire 1/120 work
A worked 9 days by doing 1/90 work each day. Total work done by A = 9 * 1/90 = 1/10
B worked 9 days by doing 1/40 work each day. Total work done by B = 9 * 1/40 = 9/40
C worked 9 days by doing 1/12 work in the initial 8 days and 1/120 work in the 9th day.
Total work done by C = 8 * 1/12 + 1/120 = 81/120
Work done by A : Work done by B : Work done by C
= 1/10 : 9/40 : 81/120 
= 12 : 27 : 81
Total amount that they get = 240
Amount that A get = 240 * 12/(12+27+81) = Rs.24

6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in - days.
  • a)
    4 days
  • b)
    6 days
  • c)
    2 days
  • d)
    8 days
Correct answer is option 'A'. Can you explain this answer?

Vikas Choudhury answered
Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b 
Work done by 6 men and 8 women in 1 day = 1/10 
=> 6m + 8b = 1/10
=> 60m + 80b = 1    (1)
Work done by 26 men and 48 women in 1 day = 1/2 
=> 26m + 48b =1/2
=> 52m + 96b = 1    (2)
Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200
Work done by 15 men and 20 women in 1 day 
= 15/100 + 20/200 =1/4
=> Time taken by 15 men and 20 women in doing the work = 4 days

P can finish a work in 18 days. Q can finish the same work in 15 days. Q worked for 10 days and left the job. how many days does P alone need to finish the remaining work?
  • a)
    8
  • b)
    5
  • c)
    4
  • d)
    6
Correct answer is option 'D'. Can you explain this answer?

Aruna Singh answered
Work done by P in 1 day = 1/18

Work done by Q in 1 day = 1/15

Work done by Q in 10 days = 10/15 = 2/3

Remaining work = 1 – 2/3 = 1/3

Number of days in which P can finish the remaining work = (1/3) / (1/18) = 6

In what time would a cistern be filled by three pipes of diameter of 1 cm, 2 cm and 3 cm if the largest pipe alone can fill the cistern in 49 minutes, the amount of water flowing through each pipe being proportional to the square of its diameter?
  • a)
    31.5 minutes
  • b)
    63 minutes
  • c)
    126 minutes
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?

Amita Das answered
Since the amount of water flowing through each pipe is proportional to the square of its diameter , if efficiency of longest pipe (3 cm) = 1/49
Then efficiency of pipe (2 cm) =  4 / (49 x 9)
and efficiency of pipe (1 cm)   =  1 / (49 x 9)
Now , let the cistern is filled by all the three pipes in x minutes.
So , 1 / 49 + 4 / 49 x 9 + 1 / 49 x 9 = 1 / x 
= x = 31.5 minutes. 

Mayank can do 50% more work than Shishu in the same time. Shishu alone can do a piece of work in 30 hours. Shishu starts working and he had already worked for 12 hours when Mayank joins him. How many hours should Shishu and Mayank work together to complete the remaining work?
  • a)
    6
  • b)
    12
  • c)
    4.8
  • d)
    7.2
Correct answer is option 'D'. Can you explain this answer?

Ishani Rane answered
Shishu alone does the work in 30 hours 

So in 1 hour he does 1/30 of the work 

Mayank in 1 hour does 1/30 + 1/2*1/30= 1/30 +1/60 = 3/60 = 1/20 of the work 

Together in 1 hour they do 1/30 +1/20 = 5/60 = 1/12 of the work 

Together they can finish the work in 12 hours 

Shishu in 12 hours does 12/ 30 = 2/5 

Remaining work = 3/5 

3/5 X 12 = 36/5 = 7.2 hours

Chetan is thrice as efficient as Mamta and together they can finish a piece of work in 60 days. Mamta will take how many days to finish this work alone?
  • a)
    80
  • b)
    160
  • c)
    240
  • d)
    320
Correct answer is option 'C'. Can you explain this answer?

Ritika Choudhury answered
  • Chetan is thrice as efficient as Mamta.
  • Let, Mamta takes 3x days and Chetan takes x days to complete the work.
  • ∴ 1/x + 1/3x = 1/60 ⇒ x = 80.
  • ∴ Mamta will take 80 × 3 = 240 days to complete the work.

Read the passage below and solve the questions based on it.
The tank at a water supply station is filled with water by several pumps. At first three pumps of Ihe same capacity are turned on: 2.5 hours later, two more pumps (both the same) of a different capacity are set into operation. After 1 hour, the additional pumps were set into operation; the tank was almost filled to its capacity (15 m3 were still lacking): in another hour the tank was full. One of the two additional pumps could have filled the tank in 40 hours
Q. What is the volume of the tank?
  • a)
    60 m3
  • b)
    80 m3
  • c)
    75 m3
  • d)
    90 m3
Correct answer is option 'A'. Can you explain this answer?

Bhavya Saha answered
Let us assume that, first three pumps fills the tank in x hours .
so,
→ Efficiency of each pump = (1/x) m³ / hour .
then,
→ Efficiency of three pump = (3/x) m³ / hour .
 
now,
→ First three pumps works for = 2.5h + 1h + 1h = 4.5 hours.
 
so,
→ Water filled by 3 pumps in 4.5 hours = 4.5 * (3/x) = (13.5/x) m³ .
 
now, given that,
→ Time taken by additional pump to fill the tank = 40 hours.
so,
→ Efficiency of 2 additional tanks = 2 * (1/40) = (1/20) m³ / h .
 
and,
→ Additional pumps work for = 1 + 1 = 2 hours.
 
so,
→ Water filled by additional pumps in 2 hours = 2 * (1/20) = (1/10) m³ .
 
therefore,
→ (13.5/x) + (1/10) = 1
→ (13.5/x) = 1 - (1/10)
→ (13.5/x) = (9/10)
→ x = 135/9 = 15 hours.
 
since given that, in last 1 hour they filled 15 m³ .
 
hence,
→ 3 * (1/15) + (1/20) = 15 m³
→ (1/5) + (1/20) = 15
→ (4 + 1)/20 = 15
→ (5/20) = 15
→ (1/4) = 15
→ 1 = 60 m³ (Ans.) (Option A)

Refer to the data below and answer the questions that follow.
Anoop was writing the reading comprehension sections in Lhe DOG entrance examinations, There were four passages of exactly equal length in terms of number of words and die four passages had 5, 8, 8 and 6 questions following each of them respectively. It is known that Anoop can answer exactly 12 questions in the time he takes to read any one of the four passages. Assume that his rate of reading and answering questions remains the same throughout the section.
Q.
By what per cent should Anoop increase his reading speed if he has to cut down on his total time spent on the section by 20%? Assume that the time spent on answering the questions is constant and as given in the directions.
  • a)
    36.36%
  • b)
    54.54%
  • c)
    50.50%
  • d)
    45.45%
Correct answer is option 'D'. Can you explain this answer?

Maulik Rane answered
To solve this problem, let's first find out the total time Anoop takes to read all four passages and answer all the questions.

Let the time he takes to read one passage be T. Since there are four passages, he takes 4T time to read all the passages. It is given that he can answer 12 questions in the time he takes to read one passage. So, the time he takes to answer one question is T/12.

There are a total of 5+8+8+6 = 27 questions. The time he takes to answer all the questions is 27 * (T/12) = 27T/12 = 9T/4.

Now, the total time spent on the section is the sum of the time spent on reading all the passages and answering all the questions: 4T + 9T/4 = 25T/4.

To cut down on his total time spent on the section by 20%, the new total time should be 80% of the original time, which is 0.8 * (25T/4) = 5T.

Since the time spent on answering the questions remains constant, the time spent on reading should reduce to 5T - 9T/4 = 11T/4. The new time he takes to read one passage is (11T/4) / 4 = 11T/16.

Now, let's find out the percentage increase in reading speed. The original time to read one passage is T, and the new time is 11T/16. Since speed is inversely proportional to time, the new speed will be 16/11 times the original speed.

The percentage increase in speed is [(16/11 - 1) * 100] = [(5/11) * 100] = 45.45%.

So, Anoop should increase his reading speed by 45.45% to cut down on his total time spent on the section by 20%.

A can work twice as fast as B. A and C together can work thrice as fast as B. If A, B and C complete a job in 30 days working together, in how many days can each of them complete the work?
  • a)
    60, 120, 120
  • b)
    50, 100, 120
  • c)
    60, 100, 80
  • d)
    40, 80, 100
Correct answer is option 'A'. Can you explain this answer?

Shilpa Choudhury answered
A, B and C complete a job in 30 days working together,
⇒ 1/A + 1/B + 1/C = 1/30
A can work twice as fast as B,
⇒ 1/B = 1/2A
A and C together can work thrice as fast as B,
⇒ 1/B = 1/3(1/A + 1/C)
Solving,
⇒ 3/B = 1/A + 1/C
⇒ 3/B = 1/30 – 1/B
⇒ 1/B = 120
Then,
⇒ 1/A = 1/60
⇒ 1/C = 1/120
∴ A, B and C can complete work in 60, 120 and 120 days respectively. 

X can do a piece of work in 20 days. He worked at it for 5 days and then Y finished it in 15 days. In how many days can X and Y together finish the work?
  • a)
    12 days
  • b)
    15 days
  • c)
    10 days
  • d)
    5 days
Correct answer is option 'C'. Can you explain this answer?

Rahul Mehta answered
  • X’s five day work = 5/20 = 1/4. Remaining work = 1 – 1/4 = 3/4.
  • This work was done by Y in 15 days. Y does 3/4th of the work in 15 days, he will finish the work in 15 × 4/3 = 20 days.  
  • X & Y together would take 1/20 + 1/20 = 2/20 = 1/10 i.e. 10 days to complete the work.

Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is 3/4th full, a leak develops in, through which one-fourth of water supplied by both the pipes goes out. What is the total time taken to fill the tank?
  • a)
    18 hours
  • b)
    14 hours
  • c)
    15 hours
  • d)
    13 hours
Correct answer is option 'D'. Can you explain this answer?

Tanvi nair answered
Understanding the problem:
- Pipe 1 can fill the tank in 20 hours.
- Pipe 2 can fill the tank in 30 hours.
- When the tank is 3/4th full, a leak develops and one-fourth of the water supplied by both pipes goes out.

Solution:
1. Let's calculate the filling rate of each pipe:
- Pipe 1 can fill 1/20 of the tank in 1 hour.
- Pipe 2 can fill 1/30 of the tank in 1 hour.
2. When the tank is 3/4th full, the total amount of water in the tank is 3/4 of the tank's capacity.
3. At this point, the leak starts and one-fourth of the water supplied by both pipes goes out. This means only 3/4 of the water supplied by both pipes remains in the tank.
4. Let the total time taken to fill the tank be x hours. In x hours, the combined filling rate of both pipes is 1/x of the tank's capacity.
5. The effective filling rate after the leak starts is (3/4) * (1/20 + 1/30) = (3/4) * (1/20 + 1/30) = (3/4) * (1/12) = 1/16 of the tank's capacity.
6. Equating the effective filling rate to the combined filling rate:
1/x = 1/16
x = 16 hours
Therefore, the total time taken to fill the tank is 16 hours, which corresponds to option 'D'.

There are three taps A, B and C in a tank. They can fill the tank in 25 hrs, 20 hrs and 10 hrs respectively. At first all of them are opened simultaneously. Then after 1 hrs, tap C is closed and tap A and B are kept running. After the 4th hour, tap B is also closed. The remaining work is done by tap A alone.
Find the percentage of work done by tap A itself?
  • a)
    72%
  • b)
    70%
  • c)
    71%
  • d)
    65%
Correct answer is option 'B'. Can you explain this answer?

Akshita Kaur answered
Given Information:
- Tap A can fill the tank in 25 hours.
- Tap B can fill the tank in 20 hours.
- Tap C can fill the tank in 10 hours.
- All taps are opened simultaneously for the first hour.
- After the first hour, tap C is closed and only taps A and B are kept running.
- After the fourth hour, tap B is also closed.
- The remaining work is done by tap A alone.

Calculation:

Let's assume the capacity of the tank is 100 units.

Work done by each tap in 1 hour:
- Tap A can fill 100/25 = 4 units of the tank in 1 hour.
- Tap B can fill 100/20 = 5 units of the tank in 1 hour.
- Tap C can fill 100/10 = 10 units of the tank in 1 hour.

Work done in the first hour:
- Since all taps are opened simultaneously, the total work done in the first hour is 4 + 5 + 10 = 19 units.

Work done in the second and third hour:
- After the first hour, only taps A and B are running.
- So, the work done in the second and third hour is (4 + 5) * 2 = 18 units.

Work done in the fourth hour:
- After the fourth hour, tap B is closed and only tap A is running.
- So, the work done in the fourth hour is 4 units.

Remaining work:
- The total capacity of the tank is 100 units.
- The work done in the first four hours is 19 + 18 + 4 = 41 units.
- Therefore, the remaining work is 100 - 41 = 59 units.

Work done by tap A alone:
- The remaining work of 59 units is done by tap A alone.
- Since tap A can fill 4 units of the tank in 1 hour, it will take 59/4 = 14.75 hours for tap A to complete the remaining work.

Percentage of work done by tap A:
- The total time taken to complete the remaining work is 14.75 hours.
- The total time taken to fill the tank by all taps is 4 hours (first four hours) + 14.75 hours (remaining work by tap A) = 18.75 hours.
- The percentage of work done by tap A is (14.75 / 18.75) * 100 = 78.67%.

Therefore, the percentage of work done by tap A itself is approximately 78.67%, which is closest to option 'b' (70%).

In a military camp there is sufficient food supply for 200 soldiers for 40 days. After 10 days, certain number of men join and everyone eats 50% more than earlier. Now the food lasts for another 10 days. How many soldiers joined the camp after 10 days?
  • a)
    150
  • b)
    300
  • c)
    250
  • d)
    200
Correct answer is option 'D'. Can you explain this answer?

EduRev CLAT answered
Let eating capacity of each soldier initially = 1 unit/day.
Initial food supply = 200 × 1 × 40 = 8000 units.
Supply left after 10 days = 200 × 1 × 30 units.
Let ‘s’ numbers of soldiers joined the camp.
New capacity of each soldier = 1.5 units/day
∴ The remaining supply is consumed by (200 + s) solders in another 10 days.
∴ 200 × 1 × 30 = (200 + s) × 1.5 × 10
⇒ 400 = 200 + s
⇒ s = 200
Hence, option D.

A tank can be filled in 30 minutes by 20 pumps. If five pumps go out of order, what time will be taken by the remaining pumps?
  • a)
    40 mins
  • b)
    38 mins
  • c)
    32 mins
  • d)
    30 mins
Correct answer is option 'A'. Can you explain this answer?

Glance Learning Institute answered
Answer: Option A
Explanation :We know, 
Time ∝ 1/no. of pumps
∴ Time × (Number of pumps) = constant.
⇒ 20 × 30 = 15 × x
⇒ x = 20 × 30/15 = 40 mins
Hence, option (a).

Two pipes, when working one at a time, can fill a cistern in 3 hours and 4 hours, respectively while a third pipe can drain the cistern empty in 8 hours. All the three pipes were opened together when the cistern was 1/12 full. How long did it take for the cistern to be completely full?
  • a)
    2 hours
  • b)
    1 hour 45 minutes
  • c)
    2 hour 11 minutes
  • d)
    2 hour 10 minutes
Correct answer is option 'A'. Can you explain this answer?

EduRev CLAT answered
Let the total amount of work in filling a cistern be 24 units. (LCM of 3, 4 and 8)
Work done by pipe 1 in 1 hour = 24/3 = 8 units.
Work done by pipe 2 in 1 hour = 24/4 = 6 units.
Work done by pipe 3 in 1 hour = 24/ (-8) = -3 units
Total work done in 1 hour = 8 + 6 – 3 = 11 units
The time required to complete 11/12th of the work = 11/12 × 24/ 11 = 2 hours
∴ The correct answer is 2 hours.

A man, a woman or a boy can do a job in 20 days, 30 days or 60 days respectively. How many boys must assist 4 men and 5 women to do the work in 2 days?
  • a)
    16
  • b)
    14
  • c)
    15
  • d)
    None of these
Correct answer is option 'D'. Can you explain this answer?

Avantika verma answered
Given Data:
- A man can do the job in 20 days
- A woman can do the job in 30 days
- A boy can do the job in 60 days

Calculating Efficiency:
- Efficiency of a man = 1/20
- Efficiency of a woman = 1/30
- Efficiency of a boy = 1/60

Efficiency of 4 men, 5 women, and x boys working together:
- (4 * 1/20) + (5 * 1/30) + (x * 1/60) = 1/2
- Simplifying, we get: 1/5 + 1/6 + x/60 = 1/2
- LCM of 5, 6, and 60 is 60
- Multiplying throughout by 60, we get: 12 + 10 + x = 30
- Solving for x, we get: x = 8

Number of Boys required:
- Therefore, 8 boys must assist 4 men and 5 women to do the work in 2 days.
Therefore, the correct answer is option D - None of these.

A factory manufactures dyes. It has 12 men and two machines which can be operated by all of its men. It takes 4 hours to manufacture one dye on the machine with the operator. The machines can work continuously without a break. Without the machine each of the men can manufacture a dye in 8 hours. The policy is such that the production is maximized and the men are ready to work in three shifts of 8 hours per day. What will be the average cost incurred per dye if 1 man hour costs Rs 20 and 1 machine hour costs Rs 15?
  • a)
    Rs 140
  • b)
    Rs 160
  • c)
    Rs 147
  • d)
    Cannot be determined
Correct answer is option 'C'. Can you explain this answer?

Madhavan Kulkarni answered
To maximize the production, four men will work in each shift. 2 men will work with machines and 2 men work alone.
Total cost incurred in one hour

If A and B together can complete a piece of work in 15 days and B alone in 20 days, in how many days can A alone complete the work?
  • a)
    60
  • b)
    45
  • c)
    40
  • d)
    30
Correct answer is option 'A'. Can you explain this answer?

Rahul Mehta answered
1st method:
A and B complete a work in = 15 days
One day's work of (A + B) = 1/ 15
B complete the work in = 20 days;
One day's work of B = 1/20
Then, A's one day's work
=1/15−1/20
=(4−3)/6
=1/60
Thus, A can complete the work in = 60 days.

A and B together can complete a work in 3 days. They start together but after 2 days, B left the work. If the work is completed after two more days, B alone could do the work in
  • a)
    5 days
  • b)
    9 days
  • c)
    10 days
  • d)
    6 days
Correct answer is option 'D'. Can you explain this answer?

Ashwini Chatterjee answered
(A+B)'s one day's work = 1/3 part
(A+B) works 2 days together = 2/3 part
Remaining work = 1−2/3 = 1/3 part
1/3 part of work is completed by A in two days
Hence, one day's work of A = 1/6
Then, one day's work of B = 1/3−1/6 = 1/6
So, B alone can complete the whole work in 6 days.

Two sprinters run the same race of100 m One runs at a speed of 10 m/s and the other runs at 8 m/s. By what time will the first sprinter beat the other sprinter?
  • a)
    1.5 sec
  • b)
    2 sec
  • c)
    2.5sec
  • d)
    3 sec
Correct answer is option 'C'. Can you explain this answer?

Navya Chavan answered
To determine the time it takes for the first sprinter to beat the second sprinter, we need to compare their speeds and calculate the time it takes for the first sprinter to cover a distance of 100m.

Let's start by calculating the time it takes for each sprinter to cover 100m:

- Sprinter 1 runs at a speed of 10m/s, so the time it takes for Sprinter 1 to cover 100m is:
Time = Distance / Speed = 100m / 10m/s = 10 seconds

- Sprinter 2 runs at a speed of 8m/s, so the time it takes for Sprinter 2 to cover 100m is:
Time = Distance / Speed = 100m / 8m/s = 12.5 seconds

From the above calculations, we can see that Sprinter 1 takes 10 seconds to cover 100m, while Sprinter 2 takes 12.5 seconds to cover the same distance.

Now, let's find out the time difference between the two sprinters:

- Time difference = Time taken by Sprinter 2 - Time taken by Sprinter 1
Time difference = 12.5 seconds - 10 seconds = 2.5 seconds

Therefore, the first sprinter beats the second sprinter by a time difference of 2.5 seconds. Hence, the correct answer is option 'C' - 2.5 seconds.

480 persons working 10 hours per day complete one-fourth of a work in 10 days. How many additional persons are to be employed in order to complete the remaining work in 20 days, working 8 hours per day?
  • a)
    400
  • b)
    420
  • c)
    480
  • d)
    500
Correct answer is option 'B'. Can you explain this answer?

Siddharth Yadav answered
Let's break down the given information and solve the problem step by step:

Given:
- 480 persons are working for 10 hours per day.
- They complete one-fourth of the work in 10 days.

Step 1: Calculate the work completed by 480 persons in 10 days.
The work completed by 480 persons in 10 days can be calculated using the formula:
Total Work = Persons × Time
= 480 × 10
= 4800 person-days

Step 2: Calculate the total work required to be completed.
Since they have completed one-fourth of the work in 10 days, the total work required can be calculated as:
Total Work = 4 × Work Completed
= 4 × 4800
= 19,200 person-days

Step 3: Calculate the work done by 480 persons in 20 days.
The work done by 480 persons in 20 days can be calculated using the formula:
Work Done = Persons × Time
= 480 × 20
= 9600 person-days

Step 4: Calculate the remaining work.
The remaining work can be calculated as:
Remaining Work = Total Work - Work Done
= 19,200 - 9600
= 9600 person-days

Step 5: Calculate the additional persons required to complete the remaining work in 20 days.
Let's assume the number of additional persons required is 'x'.
The work done by 'x' additional persons in 20 days can be calculated using the formula:
Work Done = Persons × Time
= x × 20
= 20x person-days

Step 6: Set up an equation to solve for 'x'.
From step 4, we know that the remaining work is 9600 person-days. So, we can set up the equation:
9600 = 20x

Step 7: Solve the equation.
Dividing both sides of the equation by 20, we get:
x = 9600/20
x = 480

Therefore, 480 additional persons need to be employed in order to complete the remaining work in 20 days, working 8 hours per day.

Hence, the correct answer is option B) 480.

Working alone, the times taken by Anu, Tanu and Manu to complete any job are in the ratio 5 : 8 : 10. They accept a job which they can finish in 4 days if they all work together for 8 hours per day. However, Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day. Then, the number of hours that Manu will take to complete the remaining job working alone is
  • a)
    8
  • b)
    9
  • c)
    6
  • d)
    4
Correct answer is option 'C'. Can you explain this answer?

Diksha menon answered
Given Information:
- Ratio of times taken by Anu, Tanu, and Manu to complete a job: 5 : 8 : 10
- They can finish a job in 4 days working together for 8 hours per day
- Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day

Calculating Individual Work Rates:
- Let the individual work rates of Anu, Tanu, and Manu be A, T, and M respectively
- According to the given ratio, A : T : M = 5 : 8 : 10
- Let's assume they work at a speed of 5x, 8x, and 10x units per hour respectively

Calculating Work Done in 4 Days:
- Together they can complete the job in 4 days working 8 hours per day
- Total work done in 4 days = (5x + 8x + 10x) * 4 * 8

Calculating Work Done in the First 6 Days:
- Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day
- Work done by Anu and Tanu in the first 6 days = (5x + 8x) * 6 * 6.67

Calculating Remaining Work:
- Remaining work to be done by Manu = Total work - Work done in the first 6 days

Calculating Time taken by Manu to complete the Remaining Work:
- Let M hours be the time taken by Manu to complete the remaining work alone
- Work done by Manu in M hours = 10x * M
- Equating the remaining work to the work done by Manu, we can find the value of M
Therefore, the number of hours that Manu will take to complete the remaining job working alone is 6 hours.

Anuj can do a piece of work in a certain number of days. To do the same piece of work, Bhanu takes thrice the number of days as Anuj takes whereas Chandu takes thrice as many days as Bhanu does and Dodo takes thrice as many days as Chandu does. Now, they are paired and two groups are formed. The first pair takes one-third the time taken by the second pair to complete the work, which is the first pair?
  • a)
    Anuj and Dodo
  • b)
    Anuj and Chandu
  • c)
    Chandu and Bhanu
  • d)
    Bhanu and Dodo
Correct answer is option 'B'. Can you explain this answer?

Madhavan Kulkarni answered
Let Anuj do the work in x days.
 So, Bhanu = 3x days, Chandu = 9x days and Dodo = 27x days
Now, using options,
Anuj + Chandu = 1/x + l/9x = 10/9x,
So, they take 9x/10 days
Bhanu + Dodo = l/3x + 1/27x = 10/27x, So, they take 27x/10 days
Now, 1/3 of 27x/ 10 = 9x /10.
 So, Anuj and Chandu is first pair.

A and B together can do a piece of work in 50 days. If A is 40% less efficient than B, in how many days can A working alone complete 60% of the work?
  • a)
    70
  • b)
    110
  • c)
    80
  • d)
    105
Correct answer is option 'C'. Can you explain this answer?

Shilpa Choudhury answered
Given:
A and B together can do a piece of work in 50 days.
A is 40% less efficient than B
Concept used:
Total work = Efficiency of the workers × time taken by them
Calculation:
Let the efficiency of B be 5a
So, efficiency of A = 5a × 60%
⇒ 3a
So, total efficiency of them = 8a
Total work = 8a × 50
⇒ 400a
Now,
60% of the work = 400a × 60%
⇒ 240a
Now,
Required time = 240a/3a
⇒ 80 days
∴ A can complete 60% of the work working alone in 80 days.

Anil, Bhuvan and Chandan take 10,20 and 25 days to complete a job. What is the minimum time required to finish the job if not more than 2 of them work together on a single day and no two consecutive days have the same pair of people working?
  • a)
    6.5 days
  • b)
  • c)
    7.5 days
  • d)
Correct answer is option 'B'. Can you explain this answer?

Pooja Sen answered
Assume total amount of work =100 units
A does = 10 units/day,
B does = 5 units/day and
C does = 4 units/day
Possible Pairs:
A + B = 15 units/day,
A + C = 14 units/day,
B + C = 9 units/day
To minimize time, we will use the first two pairs.
So, 15 + 14 + 15 + 14 + 15 + 14 + 15 = 102 Units
So, 7 days are required.

Chapter doubts & questions for Time & Work - CSAT Preparation 2025 is part of UPSC CSE exam preparation. The chapters have been prepared according to the UPSC CSE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for UPSC CSE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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