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All questions of Work & Time for Mechanical Engineering Exam

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15 men could finish a piece of work in 210 days. But at the end of 100 days, 15 additional menare employed. In how many more days will the work be complete?
  • a)
    80 days
  • b)
    60 days
  • c)
    55 days
  • d)
    50 days
Correct answer is option 'C'. Can you explain this answer?

Ishani Rane answered
15 men took 210 days. 
Therefore,
1 man requires 15 * 210 = 3150 days to complete the same work 
So, in 1 day 1 man complete 1/3150 of the total work 
Thus,
In 100 days 15 men completes (15 * 1/3150 * 100) = 10/21 of the total work 
After 100 days there are 30 men,
They will complete the remaining task i.e. 1 - 10/21 of the total work in 1/30 * 11/21 * 3150 = 55 days.

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?

Future Foundation Institute answered
Let distance between the two places = d km
Let total time taken by faster horse = t hr
⇒ Total time taken by slower horse = (t + 5) hr,
Therefore,
speed of the faster horse = d/t km/hr
speed of the slower horse = d/(t + 5) km/hr 
The two horses meet each other in 3 hour 20 min i.e. in 3(1/3) hr = 10/3 hr
In this time, total distance travelled by both the horses together is d. 
∴ d/(t+5) * 10/3 + d/t * 10/3 = d
⇒ 10/(3(t+5)) + 10/3t = 1
⇒ 10t + 10(t+5) = 3t(t+5)
⇒ 20t + 50 = 3t2 + 15t
⇒ 3t2 − 5t − 50 = 0
⇒ 3t2 + 10t − 15t − 50 = 0
⇒ t(3t + 10) − 5(3t + 10) = 0
⇒ (3t + 10)(t − 5) = 0
⇒ t = 5 (ignoring -ve value) 
Thus, Total time taken by slower horse = 5 + 5 = 10 hr
So Option B is correct

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?

Wizius Careers answered
Correct Answer :- d
Explanation : The total no. of days in which Anoop can dig the well is 10 days.
Anoop's one day efficiency is 10%.
On day one Anoop performs 10% of his work efficiency, then the next day he won't be able to perform because as per the question the efficiency of a worker falls by 10%.
thus, 10%-10% = 0.

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?

Academic Studio answered
Since P to R is double the distance of P to Q,
Therefore, it is evident that the time taken from P to R and back would be double the time taken from P to Q and back (i.e. double of 6.5 hours = 13 hours).
Since going from P to R takes 9 hours, coming back from R to P would take 4 hours i.e. 13- 9 = 4
So Option A is correct

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

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?

Spectrum Coaching Institute answered
Initial distance = 25 dog leaps
Per-minute dog makes 5 dog leaps and cat makes 6 cat leaps = 3 dog leaps
⇒  Relative speed = 2 dog leaps / minutes
⇒  An initial distance of 25 dog leaps would get covered in 12.5 minutes.
So Option D is correct

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?

Naroj Boda answered
Since the amount of water flowing through each pipe is proportional to square of its diameter so 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 cistern is filled by all three pipes in x minutes.

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.

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

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

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)

A can do a piece of work in 20 days. He works at it for 5 days and then B finishes it in 10 moredays. In how many days will A and B together finish the work?
  • a)
    8 days
  • b)
    10 days
  • c)
    12 days
  • d)
    6 days
Correct answer is option 'A'. Can you explain this answer?

Gowri Chakraborty answered
Work done by A in 5 days = 1/20*5
=1/4
remaining work=(1-1/4)
=3/4
now,3/4 work is done by B in 10 days
whole work will be done by B in 10*4/3= 40/3 days
as 1 day's work of A = 1/20 
1 day's work of B =3/40
(A+B)'s 1 day's work= 1/20+3/40
= 1/8
so,both finished the work in 8 day's.

Sashi can do a piece of work in 25 days and Rishi can do it in 20 days. They work for 5 days andthen Sashi goes away. In how many more days will Rishi finish the work?
  • a)
    10 days
  • b)
    12 days
  • c)
    14 days
  • d)
    None of these
Correct answer is option 'D'. Can you explain this answer?

Aarav Sharma answered
Given:
Sashi can do the work in 25 days
Rishi can do the work in 20 days
They worked together for 5 days

To find:
In how many more days will Rishi finish the work?

Solution:
Let's assume that the total work is 100 units.
So, Sashi can do 4 units of work in a day, and Rishi can do 5 units of work in a day.

In the first 5 days, the total work done by Sashi and Rishi together is:
Sashi's work = 4 units/day * 5 days = 20 units
Rishi's work = 5 units/day * 5 days = 25 units
Total work done = 20+25 = 45 units

Remaining work = 100 - 45 = 55 units

Now, only Rishi is working on the remaining work. So, the time taken by Rishi to complete the remaining work will be:
Time = Remaining work / Rishi's efficiency
Time = 55 units / 5 units/day = 11 days

Therefore, the total time taken by Rishi to complete the work is:
Time taken = 5 days (when Sashi and Rishi worked together) + 11 days (when only Rishi worked)
= 16 days

Hence, the correct answer is option D, none of these.

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%.

Pipe A can fill the tank in 4 hours, while pipe B can fill it in 6 hours working separately. Pipe C can empty the whole tank in 4 hours. Ramesh opened the pipe A and B simultaneously to fill the empty tank. He wanted to adjust his alarm so that he could open the pipe C when it was half-filled, but he mistakenly adjusted his alarm at a time when his tank would be 3/4th filled. What is the time difference between both the cases, to fill the tank fully.
  • a)
    48 min.
  • b)
    54 min.
  • c)
    30 min.
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Stuti Choudhury answered
In ideal case :
Time taken to fill the half tank by A and B = 50 / 41.66 = 6 / 5 hours
Time taken by A,B and C to fill rest half of the tank = 50 / 16.66 = 3 hours
Total time = 
In second case :
Time taken to fill 3 / 4 tank by A and B = 75 / 41.66 = 9 / 5 hours
Time taken by A,B and C to fill rest 1 / 4 tank = 25 / 16.66 = 3 / 2 hours
Total time = 9 / 5 + 3 / 2 = 3 hours 18 minutes
Therefore, difference in time = 54 minutes

In a fort there was sufficient food for 200 soldiers for 31 days. After 27 days 120 soldiers left thefort. For how many extra days will the rest of the food last for the remaining soldiers?
  • a)
    12 days
  • b)
    10 days
  • c)
    8 days
  • d)
    6 days
Correct answer is option 'D'. Can you explain this answer?

Prerna Choudhary answered
After 27 days, food left = 4 × 200 = 800 soldier days worth of food. Since, now there are only 80
soldiers, this food would last for 800/80 = 10 days. Number of extra days for which the food lasts
= 10 – 4 = 6 days.

If 10 persons can do a job in 20 days, then 20 person with twice the efficiency can do the same job in:
  • a)
    5 days
  • b)
    40 days
  • c)
    10 days
  • d)
    20 days
Correct answer is option 'A'. Can you explain this answer?

Tejas Das answered
Given:
- 10 persons can do a job in 20 days.

To find:
- How long will it take for 20 persons with twice the efficiency to do the same job?

Solution:

Let's assume that the amount of work to be done is W.

If 10 persons can do the job in 20 days, then the work done by each person in one day is W/10.

Now, let's consider the work done by 20 persons with twice the efficiency in one day. Each person will now do the work at a rate of 2(W/10), which can be simplified as W/5.

Therefore, the work done by 20 persons in one day is W/5.

Since the amount of work to be done remains the same, we can set up the following equation:

Work done by 10 persons in 20 days = Work done by 20 persons in x days

(W/10) * 20 = (W/5) * x

Simplifying the equation, we get:

20 = 2x

x = 10 days

Conclusion:
- If 10 persons can do a job in 20 days, then 20 persons with twice the efficiency can do the same job in 10 days. Therefore, the correct answer is option 'C' - 10 days.

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'.

The number of days required by A,B and C to work individually is 6, 12 and 8 respectively. They started a work doing it alternatively. If A has started then followed by B and so on, how many days are needed to complete the whole work?
  • a)
    8
  • b)
    7.5
  • c)
    8.5
  • d)
    9.5
Correct answer is option 'A'. Can you explain this answer?

Jay Kapoor answered

   ► 6 / 8 = 3 / 4
In 3 days A,B,C do 3 / 8 work,
In 6 days A,B,C do 3 / 4 work
Rest work = 1 / 4, which is less than 3 / 8, On the 7th day, 1 / 6 more work will be done by A
Now rest work = 
Now, this rest work (1 / 12) will be done by B in 1 complete day.
Thus, total number of days = 6 + 1 + 1 = 8 days.
Alternatively : Efficiency of A = 16.66% , Efficiency of B = 8.33% , Efficiency of C = 12.5%
Efficiency of A + B = 25% , Efficiency of A + B + C = 37.5%
In 3 days A,B,C completes 37.5% work, In 6 days A,B,C completes 75% work
Rest work = 25%
This 25% work will be completed by A and B in next 2 days, Thus total 6 + 2 = 8 days are needed.

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.

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.

Ajay and Vijay together can do a piece of work in 6 days. Ajay alone does it in 10 days. Whattime does Vijay require to do it alone?
  • a)
    20 days
  • b)
    15 days
  • c)
    25 days
  • d)
    30 days
Correct answer is option 'B'. Can you explain this answer?

Aarav Sharma answered
Given:
Ajay and Vijay together can do a piece of work in 6 days.
Ajay alone does it in 10 days.

To find:
Time taken by Vijay to do the work alone.

Solution:
Let the work be 1 unit.

Let's assume that Vijay takes x days to finish the work alone.

According to the given information,

Ajay and Vijay together can do the work in 6 days.

So, the work done by Ajay and Vijay in one day is 1/6.

Ajay alone does the work in 10 days.

So, the work done by Ajay in one day is 1/10.

Let's assume that the work done by Vijay in one day is v.

So, the work done by Ajay and Vijay together in one day is (1/6) = (1/10) + v.

Simplifying the above equation, we get:

v = 1/6 - 1/10
v = 1/15

So, Vijay can do the work alone in 15 days.

Therefore, the correct option is (b) 15 days.

HTML representation of solution:

Given:

Ajay and Vijay together can do a piece of work in 6 days.

Ajay alone does it in 10 days.


To find:

Time taken by Vijay to do the work alone.


Solution:

Let the work be 1 unit.

Let's assume that Vijay takes x days to finish the work alone.


According to the given information,

Ajay and Vijay together can do the work in 6 days.

So, the work done by Ajay and Vijay in one day is 1/6.

Ajay alone does the work in 10 days.

So, the work done by Ajay in one day is 1/10.


Let's assume that the work done by Vijay in one day is v.

So, the work done by Ajay and Vijay together in one day is (1/6) = (1/10) + v.


Simplifying the above equation, we get:

v = 1/6 - 1/10

v = 1/15


So, Vijay can do the work alone in 15 days.


Therefore, the correct option is (b) 15 days.

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

Two typists of varying skills can do a job in 6 minutes if they work together. If the first typisttyped alone for 4 minutes and then the second typist typed alone for 6 minutes, they would be left with 1/5 of the whole work. How many minutes would it take the slower typist to complete the typing job working alone?
  • a) 
    10 minutes
  • b) 
    15 minutes
  • c) 
    12 minutes
  • d) 
    20 minutes
Correct answer is option 'B'. Can you explain this answer?

Ishani Rane answered
Working efficiency of both typist together,
= 100/6 = 16.66% per minute 
Now, let work efficiency of first typist be x and then second typist will be (16.66 - x)
First typist typed alone for 4 minutes and second typed alone for 6 minutes and they left with 1/5 (i.e 20%) of job, means they have completed 80% job.
Now,
First Typist typed in 4 minute + Second typed in 6 minutes = 80%
4 *x + 6 *(16.66 - x) = 80% 
4x + 100% - 6x = 80%
x = 10%
First Typist typed 10% per minutes. Then second typed (16.66 - 10) = 6.66% per minute
Then, Second typist complete the whole job in 100/ 6.66 = 15.01 = 15 minutes.

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%).

6 children and 2 men complete a certain piece of work in 6 days. Each child takes twice the time taken by a man to finish the work. In how many days will 5 men finish the same work?
  • a)
    6
  • b)
    8
  • c)
    9
  • d)
    15
Correct answer is option 'A'. Can you explain this answer?

Shruti Chakraborty answered
Given: 6 children and 2 men complete a certain piece of work in 6 days.

Let's assume that a man can complete the work in 'm' days. Therefore, a child can complete the same work in '2m' days.

Total work done by 2 men in 1 day = 1/m
Total work done by 6 children in 1 day = 6/2m = 3/m
Total work done by 6 children and 2 men in 1 day = 1/m + 3/m = 4/m

As per the given question, 6 children and 2 men complete the work in 6 days. Therefore, total work done in 1 day = 1/6.

Equating the above two equations, we get:

4/m = 1/6
m = 24

Therefore, a man can complete the work in 24 days and a child can complete the work in 48 days.

Let's assume that 5 men can complete the work in 'd' days.

Total work done by 5 men in 1 day = 1/d
Total work done by 6 children in 1 day = 1/48
Total work done by 5 men and 6 children in 1 day = 1/d + 1/48

As per the given question, 6 children and 2 men complete the work in 6 days. Therefore, total work done in 1 day = 1/6.

Equating the above two equations, we get:

1/d + 1/48 = 1/6
1/d = 1/6 - 1/48
1/d = 7/48
d = 48/7 ≈ 6.86

Therefore, 5 men can complete the work in approximately 6.86 days, which can be rounded off to 6 days (Option A).

It takes six days for three women and two men working together to complete a work. Three men would do the same work five days sooner than nine women. How many times does the output of a man exceed that of a woman?
  • a) 
    3 times
  • b) 
    4 times
  • c) 
    5 times
  • d) 
    6 times
Correct answer is option 'D'. Can you explain this answer?

Raghavendra Sharma answered
If 9 women,1 day work= 1/x ---(i) 
So 3 men, 1 day work=1/(x-5) ---(ii)
Also given 3 women and 2 men 1 day work= 1/6 ---(iii)
From (i),3 women 1 day work=1/3x
From (ii,2 men 1 day work=2/[3(x-5)]
So (iii) can be represented as, 1/3x + 2/[3(x-5)]= 1/6
Solving x=10
So 1 woman 1 day work=1/(9*10)= 1/90 and
1 man 1 day work=1/3(10-5)= 1/15 
Output of a men exceed that of a women = (1/15)/(1/90)= 6 times.

If 12 men and 16 boys can do a piece of work in 5 days and 13 men and 24 boys can do it in 4days, compare the daily work done by a man with that done by a boy?
  • a)
    1 : 2
  • b)
    1 : 3
  • c)
    2 : 1
  • d)
    3 : 1
Correct answer is option 'C'. Can you explain this answer?

Arpita Pillai answered
12 × 5 man days + 16 × 5 Boy days
= 13 × 4 man days + 24 × 4 Boy days
Æ 8 man days = 16 Boy days
1 man day = 2 Boy days.
Required ratio of man’s work to boy’s work = 2 : 1.

30 workers can finish a work in 20 days. After how many days should 9 workers leave the job so that the work is completed in total 26 days :
  • a)
    12
  • b)
    10
  • c)
    6
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Moumita Pillai answered
Go through options. Consider option (c)
 ► 30 x 20 = 30 x 6 + 21 x 20
 ► 600 = 600,
Hence presumed option is correct.
Alternatively :
 ► ​30 x 20 = 30 x x + 21 x (26 - x )
 ► ⇒ x = 6

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

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