Every student can learn time and work problems very easily by practicing with the worksheet on the Time required to complete a piece of work. All the important topics are covered in this worksheet. A variety of problems are available with the solutions. Also, an explanation will help the students to learn the way of solving problems. Practicing the worksheet is very important to get good knowledge on the concept of Time required to complete a piece of work.

Also, Read: Worksheet on Work Done in a Given Period of Time

## Solved Examples on Time Required to Complete a Piece of Work

1. John can fence the park in 8 days and Liam can do the same work in 6 days. If they work together, how long will it take to finish the work?

## Solution:

The given details are John can fence the park in 8 days. That is A = 8 days.

John can fence the park in one day is \(\frac{1}{A}\) = \(\frac{1}{8}\).

Liam can complete the same work in 6 days. That is B = 6 days.

Liam can complete work in one day = \(\frac{1}{B}\)= \(\frac{1}{6}\).

If John and Liam working together, then they can finish the work in one day is

\(\frac{1}{A}\) + \(\frac{1}{B}\) =

\(\frac{1}{8}\) + \(\frac{1}{6}\).

= \(\frac{A + B}{AB}\) = \(\frac{6 + 8}{(6)(8)}\).

= \(\frac{14}{48}\).

= \(\frac{7}{24}\).

So, John and Liam can complete the \(\frac{7}{24}\)th part of the work in one day.

Total work is completed by together in \(\frac{24}{7}\) = 3.4 days.

Therefore, John and fence together can fence the park in 3.4 days.

2. Andrew and Marvel together level the ground of a place in 12 hours. If Andrew alone takes 16 hours to do the same work, then how much time will Marvel alone takes to finish the work?

## Solution:

As per the given information, Andrew and Marvel both can take time to level the ground of a place is \(\frac{AB}{A + B}\) = 12 days.

Andrew and Marvel both can level the ground of a place in one day = \(\frac{A + B}{AB}\) = \(\frac{1}{12}\) —-(1).

Andrew alone can finish the same work in 16 days. That is A = 16 days.

One day work by Andrew alone is \(\frac{1}{A}\) = \(\frac{1}{16}\) —–(2).

Let us assume Marvel alone can complete the work in ‘B’ days.

Then, one day work by Marvel alone is equal to \(\frac{1}{B}\)—(3).

If Andrew and Marvel working together, then

equation (1) = equation (2) + equation (3).

\(\frac{1}{12}\) = \(\frac{1}{16}\) + \(\frac{1}{B}\).

\(\frac{1}{B}\) = \(\frac{1}{12}\) – \(\frac{1}{16}\).

\(\frac{1}{B}\) = \(\frac{4 + 3}{48}\).

\(\frac{1}{B}\) = \(\frac{7}{48}\).

Marvel alone can do the work in one day = \(\frac{7}{48}\)th part of the work.

Marvel alone can complete the total work in \(\frac{48}{7}\) = 6.85 = 7 days.

Therefore, Marvel alone can level the ground of a place in 7 days.

3. William can do a piece of work in 15 days and Warner can finish it in 10 days. They work together for 5 days, then William goes away. In how many days will Warner finish the remaining work?

## Solution:

As per the details, William can complete a piece of work in 15 days. That is A = 15 days.

So, William can complete the work in one day = \(\frac{1}{A}\) = \(\frac{1}{15}\).

Warner can finish it in 10 days. That is B = 10 days.

So, Warner can complete the work in one day = \(\frac{1}{B}\) = \(\frac{1}{10}\).

William and Warner together can complete the work in one day is

\(\frac{1}{A}\) + \(\frac{1}{B}\) = \(\frac{1}{15}\) + \(\frac{1}{10}\).

= \(\frac{A + B}{AB}\) = \(\frac{2 + 3}{30}\).

= \(\frac{5}{30}\).

= \(\frac{1}{6}\).

So, William and Warner together can complete the \(\frac{1}{6}\)th part of the work in one day.

William and Warner working together for 5 days, then work done by both in 5 days is 5 × \(\frac{1}{6}\) = \(\frac{5}{6}\).

So, the work completed by William and Warner together in 5 days is \(\frac{5}{6}\)th part of the work.

The remaining work is 1 – \(\frac{5}{6}\).

= \(\frac{6 – 5}{6}\).

= \(\frac{1}{6}\).

The remaining work is completed by Warner.

Warner can complete the work in 10 days. So, \(\frac{1}{6}\)th of the work is completed by warner is

= 10 ×\(\frac{1}{6}\).

= \(\frac{5}{3}\).

= 1.6 days.

Therefore, Warner alone can finish the remaining work in 1.6 days.

4. Olivia reads 38 pages of a book in 50 days. In how many days will he reads 190 pages?

## Solution:

As per the details, Olivia can read the 38 pages of a book in 50 days.

So, Olivia can read one page of a book in \(\frac{50}{38}\) days.

Then, Olivia can read the 190 pages of a book in 190 × \(\frac{50}{38}\).

=\(\frac{500}{2}\).

= 250 days.

Therefore, Olivia can read 190 pages of a book in 250 days.

5. Noah can paint a picture in 15 minutes while Emma can do the same in 20 minutes. Working together how much time will they take to paint 55 pictures?

## Solution:

As per the given information, Noah can take time to paint a picture is A = 15 minutes.

Noah can complete the picture in one minute = \(\frac{1}{A}\) = \(\frac{1}{15}\).

Emma can take time to paint a picture is B = 20 minutes.

Emma can complete the picture in one minute = \( = [latex]\frac{1}{20}\).

By working together, Noah and Emma can complete the picture in one minute is

\(\frac{1}{A}\) + \(\frac{1}{B}\) = \(\frac{1}{15}\) + \(\frac{1}{20}\).

\(\frac{A + B}{AB}\) = \(\frac{4 + 3}{60}\).

= \(\frac{7}{60}\).

Therefore, Noah and Emma can paint a picture in one minute is \(\frac{7}{60}\)th part of the picture.

So, Noah and Emma both can paint a picture in

\(\frac{AB}{A + B}\) = \(\frac{60}{7}\) =8.57 minutes.

Therefore, Noah and Emma together can paint a picture in 8.57 minutes.

So, the time taken by the Noah and Emma together to paint 55 pictures is

= 55 × 8.57

= 471 minutes= 7.8 hours

Finally, Noah and Emma can paint 55 pictures in 7.8 hours.

6. Mihira and Mia can reap a field in 30 days. Mia and Ira can reap the field in 40 days and Ira and Mihira can reap it in 50 days. In how many days will they take to reap

(i) the field together

(ii) the field separately?

## Solution:

As per the information, Mihira and Mia can reap a field in 30 days.

So, Mihira and Mia can reapa field in one day = \(\frac{1}{30}\).

Mia and Ira can reap the field in 40 days.

So, Mia and Ira can reap the field in one day = \(\frac{1}{40}\).

Ira and Mihira can reap the field in 50 days.

So, Ira and Mihira can reap the filed in one day = \(\frac{1}{50}\).

By working together,

(Mihira + Mia) + (Mia + Ira) + (Ira + Mihira)’s one day’s work = \(\frac{1}{30}\) + \(\frac{1}{40}\) + \(\frac{1}{50}\).

= 2(Mihira + Mia + Ira)’s one day’s work = \(\frac{20 + 15 + 12}{600}\).

= (Mihira + Mia + Ira)’s one day’s work = \(\frac{1}{2}\) \(\frac{47}{600}\).

= \(\frac{47}{1200}\).

So, Mihira, Mia, and Ira together can reap the field in one day is equal to the \(\frac{47}{1200}\)th part of the field.

Therefore, Mihira, Mia, and Ira together can reap the complete filed in \(\frac{1200}{47}\) days = 25.5 days.

Mihira alone can reap the field = The Total number of days to reap the field together – the number of days to reap the field by Mia and Ira together.

= 25.5 – 40 = 14 days.

So, Mihira alone can reap the field in 14 days.

Mia alone can reap the field = The Total number of days to reap the field together – the number of days to reap the field by Mihira and Ira together.

= 25.5 – 50 = 24.5 days

So, Mia alone can reap the field in 24.5 days.

Ira alone can reap the field = The Total number of days to reap the field together – the number of days to reap the field by Mihira and Mia together.

= 25.5 – 30 days = 4.5 days.

So, Ira alone can reap the field in 4.5 days.

(i) Mihira, Mia, and Ira together can reap the field in 25.5 days.

(ii) Mihira. Mia and Ira alone can reap the field in 14 days, 24.5 days, and 4.5 days respectively.

7. Sophia and Harper can polish the furniture of a house in 16 days. Sophia alone can do it 1/4th of it in 20 days. In how many days will Harper take to finish the same work alone?

## Solution:

As per the given details, Sophia and Harper can polish the furniture of a house in 16 days.

Sophia and Harper can polish the furniture of a house in one day = \(/frac{A + B}{AB}[\latex] = [latex]/frac{1}{16}[\latex].

Sophia alone can do the [latex]/frac{1}{4}[\latex]th of the work in 20 days.

So, Sophia alone can polish the furniture of a house in 20 × 4 = 80days.

Sophia alone can polish the furniture of a house in one day = [latex]\frac{1}{A}\) = \(\frac{1}{80}\).

Let us assume, Harper alone can polish the furniture of a house in ‘B’ days.

Harper alone can polish the furniture of a house in one day = \(\frac{1}{B}\).

If Sophia and Harper both are working together, then they can complete the work in one day is

\(\frac{A + B}{AB}\) = \(\frac{1}{A}\) + \(\frac{1}{B}\).

Substitute the values in the above equation, then we will get

\(\frac{1}{16}\) = \(\frac{1}{80}\) + \(\frac{1}{B}\).

\(\frac{1}{B}\) = \(\frac{1}{16}\) – \(\frac{1}{80}\).

\(\frac{1}{B}\) = \(\frac{5 – 1}{80}\).

\(\frac{1}{B}\) = \(\frac{4}{80}\).

\(\frac{1}{B}\) = \(\frac{1}{20}\).

So, Harper alone can polish the furniture of a house in 20 days.

8. 10 men can do a piece of work in 12 days. In how many days will it take 8 men to do the same work?

## Solution:

As per the details, 10 men can do a piece of work in 12 days.

So, 1 men can do a piece of work in \(\frac{12}{10}\) = \(\frac{6}{5}\).

Then, 8 men can do the same work in 8 × \(\frac{6}{5}\).

= \(\frac{48}{5}\).

= 9.6 days.

Therefore, 8 men can do a piece of work in 9.6 days.

9. Mason can do work in 10 days. Noah can do it 20 days alone and Mason, Noah, Catherin together in 4 days. At what time can Catherin do it alone?

## Solution:

As per the given information, Mason can do work in 10 days. That is A = 10 days.

Mason can complete a work in one day = \(\frac{1}{A}\) = \(\frac{1}{10}\).

Noah can do work in 20 days. That is B = 20 days.

Noah alone can do the work in one day =\(\frac{1}{B}\) = \(\frac{1}{20}\).

Let us consider, Catherin can do the work in ‘C’ days.

So, Catherin can complete the work in one day = \(\frac{1}{C}\).

Mason, Noah, and Catherin together can complete the work in 4 days. That is

\(\frac{ABC}{AB + BC +CA}\) = 4 days.

So, Work done by together in one day = \(\frac{AB + BC +CA}{ABC} \) = \(\frac{1}{4}\).

If Mason, Noah, and Catherin working together, then they can complete the work in one day is

\(\frac{AB + BC +CA}{ABC} \) = \(\frac{1}{A}\) + \(\frac{1}{B}\) + \(\frac{1}{C}\).

Substitute the values in the above equation. Then, we will get

\(\frac{1}{4}\) = \(\frac{1}{10}\) + \(\frac{1}{20}\) + \(\frac{1}{C}\).

\(\frac{1}{C}\) = \(\frac{1}{4}\) – \(\frac{1}{10}\) – \(\frac{1}{20}\).

=\(\frac{1}{C}\) = \(\frac{50 – 20 -10}{200}\).

= \(\frac{1}{C}\) = \(\frac{20}{200}\).

= \(\frac{1}{C}\) = \(\frac{1}{10}\).

So, Catherin can complete the work in one day is \(\frac{1}{10}\)th part of the work.

Therefore, Catherin alone can complete the total work in 10 days.

10. Ethan can alone do a piece of work in 40 days and Evelyn in 20 days. They work together but 4 days before the completion of work Evelyn leaves. In how many days is the remaining work finished?

## Solution:

As per the information, Ethan can alone do a piece of work in A = 40 days.

Ethan can complete the work in one day \(\frac{1}{A}\) = \(\frac{1}{40}\).

Evelyn can complete the same work in B = 20 days.

Evelyn can complete the work in one day \(\frac{1}{B}\) = \(\frac{1}{20}\).

If they work together,

\(\frac{1}{A}\) + \(\frac{1}{B}\) = \(\frac{1}{40}\) + \(\frac{1}{20}\).

\(\frac{A +B}{AB}\) = \(\frac{1 + 2}{40}\).

\(\frac{A + B}{AB}\) = \(\frac{3}{40}\).

They can complete the work in one day is \(\frac{3}{40}\).

Total work is complete by together in \(\frac{40}{3}\) = 13.3 days.

Evelyn leaves the work 4 days before the completion of the work. 13 – 4 = 9

Only 9 days, Evelyn and Ethan works together. That is

9 × \(\frac{3}{40}\) = \(\frac{27}{40}\).

So, in 9 days, Ethan and Evelyn completed the \(\frac{27}{40}\)th part of the work.

The remaining work is 1 – \(\frac{27}{40}\)

= \(\frac{40 – 27}{40}\).

= \(\frac{13}{40}\).

The remaining work is completed by Ethan.

Ethan alone can complete the total work in 40 days.

So, Ethan can complete the \(\frac{13}{40}\)th part of the work in

\(\frac{13}{40}\) × 40 =13 days.

Therefore, Ethan alone can complete the remaining work in 13 days.

11. Elijah can do work in 24 days and Isabella in 30 days. They work together for 4 days, then Isabella leaves and Elijah alone continues the work. After 1 day, Andrew joins Elijah and the work is completed in 10 more days. In how many days can Andrew do it alone?

## Solution:

As per the given information, Elijah can do work in 24 days. That is A = 24 days.

Elijah can complete the work in one day = \(\frac{1}{A}\) = \(\frac{1}{24}\).

Isabella can do work in 30 days. That is B = 30 days.

Isabella can complete the work in one day = \(\frac{1}{B}\) = \(\frac{1}{30}\).

Elijah and Isabella work together for 4 days. That is

4 × (\(\frac{1}{A}\) + \(\frac{1}{B}\)).

= 4 × (\(\frac{1}{24}\) + \(\frac{1}{30}\)).

= 4 × \(\frac{5 + 4}{120}\).

= 4 × \(\frac{9}{120}\).

= 4 × \(\frac{3}{40}\).

= \(\frac{3}{10}\).

Elijah and Isabella completed the \(\frac{3}{10}\)th part of the work in 4 days.

The remaining work is 1 – \(\frac{3}{10}\) = \(\frac{7}{10}\).

Elijah works alone for one day = The remaining work – Elijah’s one day’s work.

= \(\frac{7}{10}\) – \(\frac{1}{24}\).

= \(\frac{84 – 5}{120}\).

= \(\frac{79}{120}\).

Andrew can do it along in \(\frac{79}{120}\) days.

12. Brusly, Ria, Mythili can finish a work in 20 days. Brusly and Mythilitogether can do thrice as much as Ria does. Also, Brusly and Ria can do twice as much as Mythili does. Find the time taken by each to finish the work.

## Solution:

As per the given information, Brusly, Ria, Mythili can finish a work in 20 days. That is

\(\frac{ABC}{AB + BC + CA}\) = 20 days.

Brusly, Ria, Mythili can finish the work in one day is \(\frac{AB + BC + CA}{ABC}\) = \(\frac{1}{20}\).

Let us assume, Ria can complete the work in ‘x’ days.

Then, Brusly and Mythili both can complete the work in \(\frac{x}{3}\) (Brusly and Mythili can do thrice as much as Ria does).

The Ratio of the work done by Ria and (Brusly + Mythili) is x : \(\frac{x}{3}\).

= 3 : 1.

Ria can complete the work in \(\frac{3}{4}\) × 20 = 15 days.

Ria’s one day work = \(\frac{1}{A}\) = \(\frac{1}{15}\).

(Brusly + Mythili) both can complete the work in \(\frac{1}{4}\) × 20 = 5 days.

If Mythili can complete the work in ‘x’ days.

Then, Brusly and Ria both can complete the work in \(\frac{x}{2}\) (Brusly and Ria can do twice as much as Mythili does).

The Ratio of the work done by Mythili and (Ria + Brusly) is x : \(\frac{x}{2}\).

= 2 : 1.

Mythili can complete the work in \(\frac{2}{3}\) × 20 = 13 days.

Mythili’s one day’s work = \(\frac{1}{B}\) = \(\frac{1}{13}\).

(Ria + Brusly) together can complete the work in \(\frac{1}{3}\) × 20 = 6.6 days.

Brusly, Mythili, and Ria working together, then

\(\frac{1}{A}\) + \(\frac{1}{B}\) + \(\frac{1}{C}\) = \(\frac{1}{15}\) + \(\frac{1}{13}\) + \(\frac{1}{C}\).

= \(\frac{AB + BC + CA}{ABC}\) = \(\frac{1}{15}\) + \(\frac{1}{13}\) + \(\frac{1}{C}\).

= \(\frac{1}{20}\) = \(\frac{1}{15}\) + \(\frac{1}{13}\) + \(\frac{1}{C}\).

= \(\frac{1}{C}\) = \(\frac{1}{20}\) – \(\frac{1}{15}\) – \(\frac{1}{13}\).

\(\frac{1}{C}\) = \(\frac{(15)(13) –(20)(13) – (20)(15)}{(20(15)(13)}\).

= \(\frac{195 –260 – 300}{3900}\).

= \(\frac{365}{3900}\).

= \(\frac{73}{780}\).

Brusly alone can finish the work in \(\frac{780}{73}\) = 10.6 days.

Therefore, Brusly alone can complete the work in 10.6 days.