Unitary Method — Exercise 8(C)

A can do a piece of work in 6 days and B can do it in 8 days. How long will they take to complete it together?

Answer

Given:

A can do the work in 6 days ⇒ A's 1 day's work = $\dfrac{1}{6}$

B can do the work in 8 days ⇒ B's 1 day's work = $\dfrac{1}{8}$

(A + B)'s 1 day's work

$= \dfrac{1}{6} + \dfrac{1}{8}\\[1em] \text { L.C.M. of 6 and 8 = 24 }\\[1em] = \dfrac{4}{24} + \dfrac{3}{24}\\[1em] = \dfrac{4 + 3}{24}\\[1em] = \dfrac{7}{24}$

Time taken by both together = $\dfrac{24}{7} = 3\dfrac{3}{7}$ days

Hence, A and B together will complete the work in $3\dfrac{3}{7}$ days.

A and B working together can do a piece of work in 10 days. B alone can do the same work in 15 days. How long will A alone take to do the same work?

Answer

Given:

(A + B)'s 1 day's work = $\dfrac{1}{10}$

B's 1 day's work = $\dfrac{1}{15}$

A's 1 day's work

$= \dfrac{1}{10} - \dfrac{1}{15}\\[1em] \text { L.C.M. of 10 and 15 = 30 }\\[1em] = \dfrac{3}{30} - \dfrac{2}{30}\\[1em] = \dfrac{3 - 2}{30}\\[1em] = \dfrac{1}{30}$

Time taken by A alone = 30 days

Hence, A alone will take 30 days to do the work.

A can do a piece of work in 4 days and B can do the same work in 5 days. Find, how much work can be done by them working together in: (i) one day (ii) 2 days.

What part of work will be left, after they have worked together for 2 days?

Answer

Given:

A's 1 day's work = $\dfrac{1}{4}$

B's 1 day's work = $\dfrac{1}{5}$

(i) (A + B)'s 1 day's work

$= \dfrac{1}{4} + \dfrac{1}{5}\\[1em] \text { L.C.M. of 4 and 5 = 20 }\\[1em] = \dfrac{5}{20} + \dfrac{4}{20}\\[1em] = \dfrac{5 + 4}{20}\\[1em] = \dfrac{9}{20}$

Hence, in one day they can do $\dfrac{9}{20}$ of the work.

(ii) Work done in 2 days

$= 2 \times \dfrac{9}{20}\\[1em] = \dfrac{18}{20}\\[1em] = \dfrac{9}{10}$

Hence, in 2 days they can do $\dfrac{9}{10}$ of the work.

Work left after 2 days

$= 1 - \dfrac{9}{10}\\[1em] = \dfrac{10 - 9}{10}\\[1em] = \dfrac{1}{10}$

Hence, $\dfrac{1}{10}$ of the work will be left.

A and B take 6 hours and 9 hours respectively to complete a work. A works for 1 hour and then B works for two hours.

(i) How much work is done in these 3 hours?

(ii) How much work is still left?

Answer

Given:

A's 1 hour's work = $\dfrac{1}{6}$

B's 1 hour's work = $\dfrac{1}{9}$

(i) Work done by A in 1 hour = $\dfrac{1}{6}$

Work done by B in 2 hours = $2 \times \dfrac{1}{9} = \dfrac{2}{9}$

Total work done in these 3 hours

$= \dfrac{1}{6} + \dfrac{2}{9}\\[1em] \text { L.C.M. of 6 and 9 = 18 }\\[1em] = \dfrac{3}{18} + \dfrac{4}{18}\\[1em] = \dfrac{3 + 4}{18}\\[1em] = \dfrac{7}{18}$

Hence, $\dfrac{7}{18}$ of the work is done in these 3 hours.

(ii) Work still left

$= 1 - \dfrac{7}{18}\\[1em] = \dfrac{18 - 7}{18}\\[1em] = \dfrac{11}{18}$

Hence, $\dfrac{11}{18}$ of the work is still left.

A, B and C can do a piece of work in 12, 15 and 20 days respectively. How long will they take to do it working together?

Answer

Given:

A's 1 day's work = $\dfrac{1}{12}$

B's 1 day's work = $\dfrac{1}{15}$

C's 1 day's work = $\dfrac{1}{20}$

(A + B + C)'s 1 day's work

$= \dfrac{1}{12} + \dfrac{1}{15} + \dfrac{1}{20}\\[1em] \text { L.C.M. of 12, 15 and 20 = 60 }\\[1em] = \dfrac{5}{60} + \dfrac{4}{60} + \dfrac{3}{60}\\[1em] = \dfrac{5 + 4 + 3}{60}\\[1em] = \dfrac{12}{60}\\[1em] = \dfrac{1}{5}$

Time taken by all together = 5 days

Hence, A, B and C together will take 5 days to complete the work.

Two taps can fill a cistern in 10 hours and 8 hours respectively. A third tap can empty it in 15 hours. How long will it take to fill the empty cistern, if all of them are opened together?

Answer

Given:

First tap fills in 10 hours ⇒ 1 hour's work = $\dfrac{1}{10}$

Second tap fills in 8 hours ⇒ 1 hour's work = $\dfrac{1}{8}$

Third tap empties in 15 hours ⇒ 1 hour's work = $-\dfrac{1}{15}$ (emptying)

Work done in 1 hour when all are opened together:

$= \dfrac{1}{10} + \dfrac{1}{8} - \dfrac{1}{15}\\[1em] \text { L.C.M. of 10, 8 and 15 = 120 }\\[1em] = \dfrac{12}{120} + \dfrac{15}{120} - \dfrac{8}{120}\\[1em] = \dfrac{12 + 15 - 8}{120}\\[1em] = \dfrac{19}{120}$

Time taken to fill the cistern = $\dfrac{120}{19} = 6\dfrac{6}{19}$ hours

Hence, the cistern will be filled in $6\dfrac{6}{19}$ hours.

Mohit can complete a work in 50 days, whereas Anuj can complete the same work in 40 days.

Find:

(i) work done by Mohit in 20 days.

(ii) work left after Mohit has worked on it for 20 days.

(iii) time taken by Anuj to complete the remaining work.

Answer

Given:

Mohit's 1 day's work = $\dfrac{1}{50}$

Anuj's 1 day's work = $\dfrac{1}{40}$

(i) Work done by Mohit in 20 days

$= 20 \times \dfrac{1}{50}\\[1em] = \dfrac{20}{50}\\[1em] = \dfrac{2}{5}$

Hence, Mohit does $\dfrac{2}{5}$ of the work in 20 days.

(ii) Work left

$= 1 - \dfrac{2}{5}\\[1em] = \dfrac{5 - 2}{5}\\[1em] = \dfrac{3}{5}$

Hence, $\dfrac{3}{5}$ of the work is left.

(iii) Time taken by Anuj to do $\dfrac{3}{5}$ of the work

$= \dfrac{3}{5} \div \dfrac{1}{40}\\[1em] = \dfrac{3}{5} \times 40\\[1em] = 24 \text{ days}$

Hence, Anuj will take 24 days to complete the remaining work.

Joseph and Peter can complete a work in 20 hours and 25 hours respectively.

Find:

(i) work done by both together in 4 hrs.

(ii) work left after both worked together for 4 hrs.

(iii) time taken by Peter to complete the remaining work.

Answer

Given:

Joseph's 1 hour's work = $\dfrac{1}{20}$

Peter's 1 hour's work = $\dfrac{1}{25}$

(Joseph + Peter)'s 1 hour's work

$= \dfrac{1}{20} + \dfrac{1}{25}\\[1em] \text { L.C.M. of 20 and 25 = 100 }\\[1em] = \dfrac{5}{100} + \dfrac{4}{100}\\[1em] = \dfrac{5 + 4}{100}\\[1em] = \dfrac{9}{100}$

The total work done by Joseph and Peter in 1 hour = $\dfrac{9}{100}$.

(i) Work done by both in 4 hours

$= 4 \times \dfrac{9}{100}\\[1em] = \dfrac{36}{100}\\[1em] = \dfrac{9}{25}$

Hence, both together do $\dfrac{9}{25}$ of the work in 4 hours.

(ii) Work left

$= 1 - \dfrac{9}{25}\\[1em] = \dfrac{25 - 9}{25}\\[1em] = \dfrac{16}{25}$

Hence, $\dfrac{16}{25}$ of the work is left.

(iii) Time taken by Peter to do $\dfrac{16}{25}$ of the work

$= \dfrac{16}{25} \div \dfrac{1}{25}\\[1em] = \dfrac{16}{25} \times 25\\[1em] = 16 \text{ hours}$

Hence, Peter will take 16 hours to complete the remaining work.

A is able to complete $\dfrac{1}{3}$ of a certain work in 10 hrs and B is able to complete $\dfrac{2}{5}$ of the same work in 12 hrs.

Find:

(i) how much work can A do in 1 hour?

(ii) how much work can B do in 1 hour?

(iii) in how much time will the work be completed, if both work together?

Answer

Given:

A completes $\dfrac{1}{3}$ of the work in 10 hours.

B completes $\dfrac{2}{5}$ of the work in 12 hours.

(i) A's 1 hour's work

$= \dfrac{1}{3} \div 10\\[1em] = \dfrac{1}{3} \times \dfrac{1}{10}\\[1em] = \dfrac{1}{30}$

Hence, A can do $\dfrac{1}{30}$ of the work in 1 hour.

(ii) B's 1 hour's work

$= \dfrac{2}{5} \div 12\\[1em] = \dfrac{2}{5} \times \dfrac{1}{12}\\[1em] = \dfrac{2}{60}\\[1em] = \dfrac{1}{30}$

Hence, B can do $\dfrac{1}{30}$ of the work in 1 hour.

(iii) (A + B)'s 1 hour's work

$= \dfrac{1}{30} + \dfrac{1}{30}\\[1em] = \dfrac{1 + 1}{30}\\[1em] = \dfrac{2}{30}\\[1em] = \dfrac{1}{15}$

Time taken to complete the work together = 15 hours

Hence, working together they will complete the work in 15 hours.

Shaheed can prepare one wooden chair in 3 days and Shaif can prepare the same chair in 4 days. If they work together, in how many days will they prepare:

(i) one chair?

(ii) 14 chairs of the same kind?

Answer

Given:

Shaheed's 1 day's work = $\dfrac{1}{3}$

Shaif's 1 day's work = $\dfrac{1}{4}$

(Shaheed + Shaif)'s 1 day's work

$= \dfrac{1}{3} + \dfrac{1}{4}\\[1em] \text { L.C.M. of 3 and 4 = 12 }\\[1em] = \dfrac{4}{12} + \dfrac{3}{12}\\[1em] = \dfrac{4 + 3}{12}\\[1em] = \dfrac{7}{12}$

The total work done by Shaheed and Shaif in 1 day = $\dfrac{7}{12}$

(i) Time taken to prepare 1 chair = $\dfrac{12}{7} = 1\dfrac{5}{7}$ days

Hence, working together they will prepare one chair in $1\dfrac{5}{7}$ days.

(ii) Time taken to prepare 14 chairs

$= 14 \times \dfrac{12}{7}\\[1em] = 2 \times 12\\[1em] = 24 \text{ days}$

Hence, they will prepare 14 chairs in 24 days.

A, B and C together finish a work in 4 days. If A alone can finish the same work in 8 days and B in 12 days, find how long will C take to finish the work.

Answer

Given:

(A + B + C)'s 1 day's work = $\dfrac{1}{4}$

A's 1 day's work = $\dfrac{1}{8}$

B's 1 day's work = $\dfrac{1}{12}$

C's 1 day's work

$= \dfrac{1}{4} - \dfrac{1}{8} - \dfrac{1}{12}\\[1em] \text { L.C.M. of 4, 8 and 12 = 24 }\\[1em] = \dfrac{6}{24} - \dfrac{3}{24} - \dfrac{2}{24}\\[1em] = \dfrac{6 - 3 - 2}{24}\\[1em] = \dfrac{1}{24}$

Time taken by C alone = 24 days

Hence, C will take 24 days to finish the work.