Worksheet on Work done in a given period of time is here to explain more about a Time and Work concept to the students. We are providing a variety of problems with simple solutions on this page. Practice more problems on work done in a given period of time and get a complete grip on this concept. Before going to practice the problems, know the basic rule of the work done in a given period of time. We already know the relation between Work was done and Time, it is directly proportional. If the time taken by the person is more, then the work done by the person is also more.
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Solved Problems on Work Done in a Given Period of Time
1. John can decorate a pot in 10 days. How much work can be done in 5 days?
Solution:
As per the information, John can take 10 days of time to decorate a pot.
Let us consider the time is x.
Therefore, that is x = 10 days.
Now, find out the John can complete the work in one day.
So, John can complete the work in one day = \(\frac { 1 }{ x } \) = \(\frac { 1 }{ 10 } \).
Now, to find the work done by John in 5 days, multiply the 5 with the work in one day.
The work done by John in 5 days = 5 × \(\frac { 1 }{ x } \).
= 5 × \(\frac { 1 }{ 10 } \).
Simply the above fraction to get the final answer.
= \(\frac { 1 }{ 2 } \).
Therefore, \(\frac { 1 }{ 2 } \)nd part of the work done by John in 5 days.
2. Brusly can paint a picture in 10 minutes while Liam can do the same in 15 minutes. Working together how many pictures will be painted in 60 minutes?
Solution:
As per the given details, Brusly can take 10 minutes of time to paint a picture.
Let us consider the time is x.
Therefore, that is X = 10 minutes.
Now, find out the John can complete the work in one day.
So, Brusly can paint a picture in 1 minute = \(\frac { 1 }{ X } \) = \(\frac { 1 }{ 10 } \).
Given Liam can complete the same work in 15 minutes.
Let us consider the time is x.
Therefore, that is Y = 15 minutes.
Now, find out the John can complete the work in one day.
So, Liam can complete the paint in one minute = \(\frac { 1 }{ Y } \) = \(\frac { 1 }{ 15 } \).
Brusly and Liam both can complete the paint in one minute = \(\frac { 1 }{ X } \) + \(\frac { 1 }{ Y } \) = \(\frac { 1 }{ 10 } \) + \(\frac { 1 }{ 15 } \).
= \(\frac { X + Y }{ XY } \) = \(\frac { 10 + 15 }{ 150 } \).
= \(\frac { 25 }{ 150 } \).
= \(\frac { 1 }{ 6 } \).
Therefore, Liam and Brusly together can complete the \(\frac { 1 }{ 6 } \)th part of the paint in one minute.
The work done by together in 60 minutes = 60 × \(\frac { 1 }{ 6 } \) = 10.
Therefore, Brusly and Liam together can paint 10 pictures in 60 minutes.
3. William can do a piece of work in 40 days and Warner can do the same work in 60 days. How long will they take together to do it?
Solution:
As per the given details, William can take time to complete a piece of work in 40 days.
Let the work done by William be x.
Therefore, x = 40 days.
Now, find out the William can complete the work in one day.
William can complete the work in one day = \(\frac { 1 }{ x } \) = \(\frac { 1 }{ 40 } \).
Warner can take time to complete the same work in 60 days.
Let the work done by Warner be y.
Therefore, y = 60 days.
Now, find out the Warner can complete the work in one day.
Warner can complete the work in one day = \(\frac { 1 }{ y } \) = \(\frac { 1 }{ 60 } \).
If warner and William working together, then the work done in one day is
\(\frac { 1 }{ x } \) + \(\frac { 1 }{ y } \) = \(\frac { 1 }{ 40 } \) + \(\frac { 1 }{ 60 } \).
= \(\frac { x + y }{ xy } \) = \(\frac { 40 + 60 }{ (40)(60) } \).
= \(\frac { 100 }{ 2400 } \).
= \(\frac { 1 }{ 24 } \).
Total work done by together is \(\frac { xy }{ x+ y } \) = 24 days.
Therefore, William and Warner both can complete the work in 24 days.
4. Andrew, Catherin, Donald can do a piece of work in 2 days, 4 days, and 6 days respectively. In how many days will the work finish if Andrew, Catherin, Donald work together?
Solution:
The given details are Andrew can complete the work in 2 days.
Let the work done by Andrew be x.
Therefore, x = 2 days.
Now, find out the Andrew can complete the work in one day.
Andrew can complete the work in 1 day = \(\frac { 1 }{ x } \) = \(\frac { 1 }{ 2 } \).
Catherin can complete the work in 4 days.
Let the work done by Catherin be y.
Therefore, y = 4 days.
Now, find out the Catherin can complete the work in one day.
Catherin can complete the work in 1 day = \(\frac { 1 }{ y } \) = \(\frac { 1 }{ 4 } \).
Donald can complete the work in 6 days.
Let the work done by Donald be z.
Therefore, z = 6 days.
Now, find out the Donald can complete the work in one day.
Donald can complete the work in 1 day = \(\frac { 1 }{ z } \) = \(\frac { 1 }{ 6 } \).
Andrew, Catherin, and Donald working together, then they can complete the wok in one day = \(\frac { 1 }{ x } \) + \(\frac { 1 }{ y } \) + \(\frac { 1 }{ z } \).
= \(\frac { 1 }{ 2 } \) + \(\frac { 1 }{ 4 } \) + \(\frac { 1 }{ 6 } \).
= \(\frac { xy + yz + zx }{ xyz } \) = \(\frac { (4)(6) + (6)(2) + (2)(4) }{ (2)(4)(6) } \).
= \(\frac { 24 + 12 + 8 }{ 48 } \).
= \(\frac { 44 }{ 48 } \).
= \(\frac { 11 }{ 12 } \).
So, by working together Andrew, Catherin, and Donald can complete the \(\frac { 11 }{ 12 } \)th part of the work in one day.
Total work done by Andrew, Catherin, and Donald together is equal to \(\frac { xyz }{ xy + yz + zx } \) = \(\frac { 12 }{ 11 } \) = 1 day.
Therefore, Andrew, Catherin, and Donald together can complete the total work in 1 day.
5. Emma and Charlotte can do a job in 20 days. Charlotte alone can finish it in 25 days. In how many days can Emma alone finish the job?
Solution:
As per the given information, Emma and Charlotte both can do a job in 20 days.
Let the work done by Emma and Charlotte be x and y.
Therefore, \(\frac { xy }{ x + y } \) = 20 days.
Charlotte alone can finish the work in 25 days. That is y = 25 days.
Now, find out the Charlotte can complete the work in one day.
Charlotte alone can finish the work in one day = \(\frac { 1 }{ y } \) = \(\frac { 1 }{ 25 } \).
Emma alone can finish the work in one day = \(\frac { 1 }{ x } \).
Emma and charlotte both can do the work in one day = \(\frac { 1 }{ x } \) + \(\frac { 1 }{ y } \) =
= \(\frac { 1 }{ x } \) + \(\frac { 1 }{ 25 } \).
= \(\frac { x + y }{ xy } \) = \(\frac { x + 25 }{ 25x } \)
Total work done by Emma and Charlotte together is \(\frac { xy }{ x + y } \) = \(\frac { 25x }{ 25 + x } \) = 20 days.
25x = (25 + x) 20.
5x = (25 + x) 4.
5x = 100 + 4x.
x =100.
Therefore, Emma alone can finish the work in 100 days.
6. Sophia and Olivia can do a piece of work in 8 days. Olivia and Noah can do a piece of work in 12 days and Noah and Sophia can do the same piece of work in 8 days. In how many days will the work be over if Sophia, Olivia, Noah work together?
Solution:
As per the information, Sophia and Olivia can do a piece of work in 8 days.
(Sophia and Olivia)’s one day’s work = \(\frac { 1 }{ 8 } \).
Olivia and Noah can do a piece of work in 12 days.
(Olivia and Noah)’s one day’s work = \(\frac { 1 }{ 12 } \).
Noah and Sophia can do the same piece of work in 8 days.
(Noah and Sophia)’s one day’s work =\(\frac { 1 }{ 8 } \).
(Sophia + Olivia + Noah)’s one day’s work = [(Sophia and Olivia) + (Olivia and Noah) + (Noah and Sophia)]’s one day’s work.
=2 (Sophia + Olivia + Noah)’s one day’s work = \(\frac { 1 }{ 8 } \) + \(\frac { 1 }{ 12 } \) + \(\frac { 1 }{ 8 } \).
2 (Sophia + Olivia + Noah)’s one day’s work = \(\frac { 12 + 8 + 12 }{ 968 } \).
2 (Sophia + Olivia + Noah)’s one day’s work = \(\frac { 32 }{ 96 } \).
(Sophia + Olivia + Noah)’s one day’s work = \(\frac { 1 }{ 16 } \)
Total work done by Sophia, Olivia, and Noah together is 16 days.
7. Miais two times faster than Isabella, and together they finish the work in 27 days. If Isabella and Mia work alone, how many days will they finish the work?
Solution:
The given details are Mia is two times faster than Isabella.
Let us consider the Isabella can complete the work in ‘x’ days.
Then the Mia can complete the work in \(/frac{x}{2}\[latex]days.
The ratio of the work done by Mia and Isabella = 2 : 1.
Mia and Isabella together can finish the work in 27 days.
Mia alone can do the work in [latex]\frac { 2 }{ 3 } \) × 27.
Therefore, Mia alone can finish the work in 18 days.
Isabella alone can do the work in \(\frac { 1 }{ 3 } \) × 27.
Therefore, Isabella can finish the work in 9 days.
8. Lucas can do a piece of work in 10 days and Harper in 25 days. They work together for 6 days, Harper goes away. In how many days will Lucas finish the remaining work?
Solution:
As per the information, Lucas can do a piece of work in 10 days. That is x = 10 days.
Lucas’s one day’s work is \(\frac { 1 }{ x } \) = \(\frac { 1 }{ 10 } \).
Harper can complete the work in 25 days. That is y = 25 days.
Harper’s one day’s work is \(\frac { 1 }{ y } \) = \(\frac { 1 }{ 25 } \).
(Lucas + Harper)’s one day’s work is \(\frac { 1 }{ x } \) + \(\frac { 1 }{ y } \) = \(\frac { 1 }{ 10 } \) + \(\frac { 1 }{ 25 } \).
= \(\frac { 10 + 25 }{ (10)(25) } \)
= \(\frac { 35 }{ 250 } \).
= \(\frac { 7 }{ 50 } \).
So, Harper and Lucas together can do work in one day is \(\frac { 7 }{ 50 } \).
Harper and Lucas can work together for 6 days = 6 × \(\frac { 7 }{ 50 } \).
= \(\frac { 21 }{ 25 } \).
The work done by Harper and Lucas in 6 days is \(\frac { 21 }{ 25 } \).
After 6 days, the Harper stop working.
The remaining work is 1 – \(\frac { 21 }{ 25 } \) is done by Lucas.
That is \(/frac{4}{25}\[latex]th part of the work is done by Lucas.
Lucas can finish the work in 10 days.
The remaining work [latex]\frac { 4 }{ 25 } \) 10= \(\frac { 8 }{ 5 } \) = 1.6 days
Therefore, The remaining work is done by Lucas in 1.6 days.
9. Ethan can do 1/4 of the work in 10 days and Evelyn can do the rest of the work in 10 days. In how many days both Ethan and Evelyn can together do the same work?
Solution:
As per the details, Ethan can do \(\frac { 1 }{ 4 } \) of the work in 10 days.
So, Ethan can complete the total work in 10 × 4 = 40 days.
Ethan can do the work in one day = \(\frac { 1 }{ 40 } \).
The remaining work is 1 – \(\frac { 1 }{ 4 } \) = \(\frac { 3 }{ 4 } \).
So, \(/frac{3}{4}\[latex]th of the work is done by Evelyn in 10 days.
Then, Evelyn can complete the total work in 10×[latex]\frac { 4 }{ 3 } \) = \(\frac { 40 }{ 3 } \) = 13 days.
Evelyn can do the work in one day = \(\frac { 1 }{ 13 } \).
If Ethan and Evelyn working together, then the work done in one day = \(\frac { 1 }{ 40 } \) + \(\frac { 1 }{ 13 } \).
= \(\frac { 13 + 40 }{ (13)(40) } \).
= \(\frac { 53 }{ 520 } \)
Evelyn and Ethan together can complete the work in one day is \(\frac { 53 }{ 520 } \)th part of the work.
Total work done by together is \(\frac { 520 }{ 53 } \) = 9.8 days =10 days.
Therefore, Ethan and Evelyn together can finish the work in 10 days.
10. James can do a piece of work in 30 days and Benjamin in 35 days. They began to work together. But James leaves after some days. Benjamin finishes the remaining work in 15 days. After how many days did James leave?
Solution:
As per the information, James can do a piece of work in 30 days. That is x = 30 days.
James can finish the work in one day = \(\frac { 1 }{ x } \) = \(\frac { 1 }{ 30 } \).
Benjamin can do the same work in 35 days. That is y = 35 days.
Benjamin can complete the work in one day = \(\frac { 1 }{ y } \) = \(\frac { 1 }{ 35 } \).
If James and Benjamin working together, then the work done in one day is
\(\frac { 1 }{ x } \) + \(\frac { 1 }{ y } \) = \(\frac { 1 }{ 30 } \) + \(\frac { 1 }{ 35 } \).
= \(\frac { x + y }{ xy } \) = \(\frac { 7 + 6 }{ 210 } \).
= \(\frac { 13 }{ 210 } \).
James and Benjamin works together for some days. That is
x\(\frac { 13 }{ 210 } \).
The remaining work is 1 – x\(\frac { 13 }{ 210 } \).
= \(\frac { 210 – 13x }{ 210 } \)
Benjamin finish the remaining work in 15 days. That is
\(\frac { 210 – 13x }{ 210 } \) = \(\frac { 1 }{ 15 } \).
15 ×(210 – 13x) = 210.
3150 – 195x = 210.
3150 – 210 = 195x.
2940 = 195x.
X = 15.
Therefore, after 15 days James leaves work.
11. If Amelia, Warner, Catherin can finish the piece of work in 12 days, 16 days, and 20 days respectively, and if all the three work at it together and they are paid $1880 for the work, how should the money be divided among them?
Solution:
As per the information, Amelia can complete the piece of work in 12 days. So, one day’s work by Amelia is \(\frac { 1 }{ 12 } \).
Warner can finish the piece of work in 16 days. Then work done by warner in one day is \(\frac { 1 }{ 16 } \).
Catherin can complete the work in 20 days. So, one day’s work by Catherin is \(\frac { 1 }{ 20 } \).
The ratio of the work done by Amelia, Warner, and Catherin is
\(\frac { 1 }{ 12 } \) : \(\frac { 1 }{ 16 } \) : \(\frac { 1 }{ 20 } \).
= 20 : 18 : 12.
= 10 : 9 : 6.
If three members are working together, then they will get the total payment for the work is $1880.
Amelia payment is \(\frac { 10 }{ 25 } \)× 1880 =$752.
Warner payment is \(\frac { 9 }{ 25 } \) × 1880 = $676.8.
Catherin payment is \(\frac { 6 }{ 25 } \) × 1880 = $451.
12. After working for 6 days, Ria finds that 3/4 of the work is left. He employs James who is 20 % efficient as Ria. How many more days will Ria take to complete the work?
Solution:
As per the given details, Ria worked for 6 days and finds ¾ of the work is left. That means,
¼ of the work is completed in 6 days.
So, the total work is completed in 24 days.
James is 20 % efficient than Ria.
Let us assume Ria can complete the work in ‘x’ days.
Then James can complete the work in \(\frac { 100x }{ 20 } \).
The ratio of the working days of James and Ria is
\(\frac { 100x }{ 20 } \). : x.
100 : 20.
5: 1.
James can complete the work in \(\frac { 5 }{ 6 } \)×24.
So, James can complete the work in 20 days.
Ria can complete the work in \(\frac { 5 }{ 6 } \) ×24.
So, Ria can complete the work in 4 days.
Therefore, Ria takes 4 days of time to complete the work.