Question

A and B can do a piece of work in 18 days ; B and C can do a piece of work in 24 days and C and A in 36 days in what time can they do it , all together​

Answer 1

A and B do a piece of work in 18 days.

Work done by A and B in one day = 1/18 th part of the work

Similarly, B and C do a piece of work in 24 and C and A 36 days.

So, Work done by B and C in one day = 1/24 th part of the work

And work done by C and A in one day = 1/36 th part of the work

We have to find the work done by A, B and C in one day.

What we have to do is, simply add their work which they do in one day.

According to question,

→ 1/(A + B + B + C + C + A) = 1/18 + 1/24 + 1/36

→ 1/(2A + 2B + 2C) = (4 + 3 + 2)/72

→ 1/[2(A + B + C)] = 9/72

→ 1/[2(A + B + C)] = 1/8

→ 1/(A + B + C) = 1/8*2

→ 1/(A + B + C) = 1/16

→ A + B + C = 16 days

Therefore,

All of them complete the work; together in 16 days.

Answer 2

$\bold{\huge{\underline{\underline{\rm{ Answer }}}}}$

$⟼ \: \: 16 \: \text{days}$

$\bold{\huge{\underline{\underline{\rm{ Solution :}}}}} \:$

Work done by A and B in 18 days

$\text{work done by A and B in one day}$

$= \frac{{{1}}}{{{18} }}$

Work done by B and C in 24 days

$\text{Work done by B and C in one day}$

$= \frac{{{1}}}{{{24} }}$

Work done by C and A in 36 days

$\text{Work done by C and A in one days}$

$= \frac{{{1}}}{{{36} }}$

On Adding

Work done by ( A + B + C ) together in 1 day

$= (\frac{ \: 1 \: }{ \: 18 } + \frac{ \: 1 \: }{ \: 24 \: } \: + \frac{ \: 1 \: }{ \: 36 \: } )$

$= \frac{ \: 4 + 3 + 2 \: }{ \: 72 \: }$

$= \frac{ \: 1 \: }{ \: 16 \: }$

Hence, Work done by ( A + B + C) together in 16 days

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