Question

A and B together can do a piece of work in 20 days ; B and C together can do it in 15 days , C and A together can do it in 12 days . How long will they take to finish the work , working all together ? How long would each take to do the same work ?

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A and B together can do a piece of work in 20 days.

So, it means

$\rm \: {(A + B)'}^{s} \: 1 \: day \: work \: = \: \dfrac{1}{20} - - - (1)$

Further given that,

B and C together can do it in 15 days.

So, it means

$\rm \: {(B + C)'}^{s} \: 1 \: day \: work \: = \: \dfrac{1}{15} - - - (2)$

Also, given that

C and A together can do it in 12 days.

So, it means

$\rm \: {(A + C)'}^{s} \: 1 \: day \: work \: = \: \dfrac{1}{12} - - - (3)$

On adding equation (1), (2) and (3), we get

$\rm \: 2{(A + B + C)'}^{s} \: 1 \: day \: work \: = \: \dfrac{1}{20} + \dfrac{1}{15} + \dfrac{1}{12}$

$\rm \: 2{(A + B + C)'}^{s} \: 1 \: day \: work \: = \: \dfrac{3 + 4 + 5}{60}$

$\rm \: 2{(A + B + C)'}^{s} \: 1 \: day \: work \: = \: \dfrac{12}{60}$

$\rm \: 2{(A + B + C)'}^{s} \: 1 \: day \: work \: = \: \dfrac{1}{5}$

$\rm\implies \:\rm \: {(A + B + C)'}^{s} \: 1 \: day \: work \: = \: \dfrac{1}{10} - - - (4)$

$\rm\implies \:\boxed{\sf{A + B + C \: together \: can \: finish \: a \: work \: in \: 10 \: days }}$

Now, on Subtracting equation (1) from equation (4), we get

$\rm \: {C'}^{s} \: 1 \: day \: work \: = \: \dfrac{1}{10} - \dfrac{1}{20}$

$\rm \: {C'}^{s} \: 1 \: day \: work \: = \: \dfrac{2 - 1}{20}$

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

$\rm\implies \:C \: can \: alone \: finish \: the \: work \: in \: 20 \: days.$

Now, On Subtracting equation (2) from (4), we get

$\rm \: {A'}^{s} \: 1 \: day \: work \: = \: \dfrac{1}{10} - \dfrac{1}{15}$

$\rm \: {A'}^{s} \: 1 \: day \: work \: = \: \dfrac{3 - 2}{30}$

$\rm \: {A'}^{s} \: 1 \: day \: work \: = \: \dfrac{1}{30}$

$\rm\implies \:A \: can \: alone \: finish \: the \: work \: in \: 30 \: days.$

On Subtracting equation (3) from (4), we get

$\rm \: {B'}^{s} \: 1 \: day \: work \: = \: \dfrac{1}{10} - \dfrac{1}{12}$

$\rm \: {B'}^{s} \: 1 \: day \: work \: = \: \dfrac{6 - 5}{60}$

$\rm \: {B'}^{s} \: 1 \: day \: work \: = \: \dfrac{1}{60}$

$\rm\implies \:B \: can \: alone \: finish \: the \: work \: in \: 60 \: days.$

Answer 2

We have been given that, A and B together can do a piece of work in 20 days; B and C together can do it in 15 days, C and A together can do it in 12 days.

Firstly Let us assume,

A + B = 20 days And

$\implies$ (A + B)'s 1 day work $\sf{=\:{\dfrac{1}{20}}}\:---(1)$

B + C = 15 days And

$\implies$ (B + C)'s 1 day work $\sf{=\:{\dfrac{1}{15}}}\:---(2)$

C + A = 12 days And

$\implies$ (C + A)'s 1 day work $\sf{=\:{\dfrac{1}{12}}}\:---(3)$

According to the given condition,

Total time would be, A + B + C = (A + B) + (B + C) + (C + A), And so,

$\implies$ (A + B + C)'s 1 day work = (A + B)'s 1 day work + (B + C)'s 1 day work + (C + A)'s 1 day

Substituting (1), (2) & (3) in (4), we get,

$\sf\implies{2(A\:+\:B\:+\:C)\:=\:{\dfrac{1}{20}}\:+\:{\dfrac{1}{15}}\:+\:{\dfrac{1}{12}}}$

$\sf\implies{2(A\:+\:B\:+\:C)\:=\:{\dfrac{3\:+\:4\:+\:5}{60}}}$

$\sf\implies{2(A\:+\:B\:+\:C)\:=\:{\dfrac{12}{60}}}$

$\sf\implies{(A\:+\:B\:+\:C)\:=\:{\dfrac{12}{60}}\:\times\:2}$

$\sf\implies{(A\:+\:B\:+\:C)\:=\:{\dfrac{12}{60}}\:\times\:2}$

$\sf\implies{\dfrac{1}{10}}$

∴ A + B + C together can complete the work in 10 days.

We know that,

• (A + B + C) - (A + B) = C's work
• (A + B + C) - (B + C) = A's work
• (A + B + C) - (C + A) = B's work

Substituting values, we get,

$\implies$ (A + B + C)'s - (A + B)'s 1 day work = $\sf{{\dfrac{1}{10}}\:-\:{\dfrac{1}{20}}}$

$\implies$ C's 1 day work = $\sf{{\dfrac{2\:-\:1}{20}}}$

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

∴ C can complete the work in 20 days.

$\implies$ (A + B + C)'s - (B + C)'s 1 day work = $\sf{{\dfrac{1}{10}}\:-\:{\dfrac{1}{15}}}$

$\implies$ A's 1 day work = $\sf{{\dfrac{3\:-\:2}{30}}}$

$\implies$ A's 1 day work = $\sf{{\dfrac{1}{30}}}$

∴ A can complete the work in 30 days.

$\implies$ (A + B + C)'s - (C + A)'s 1 day work = $\sf{{\dfrac{1}{10}}\:-\:{\dfrac{1}{12}}}$

$\implies$ B's 1 day work = $\sf{{\dfrac{6\:-\:5}{60}}}$

$\implies$ B's 1 day work = $\sf{{\dfrac{1}{60}}}$

∴ B can complete the work in 60 days.

Hence, A + B + C together will finish the work in 10 days. A alone in 30 days, B alone in 60 days and C alone in 20 days.