Question

8 men and 12 boys can finish a piece of work in 5 days while 6 men and 8 boys can finish it in 7 days find the time taken by 1men alone and that by one boy alone to finish the work

Answer 1

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Answer 2

•Given:-

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$\: \: \: \red \bullet \sf8 \: mens \: and \: 12 \: boys \: can \: finish \: a \: work \: in \: 5 \: days$

$\: \: \: \sf \green{ \bullet} \: 6 \: men \: and \: 8 \: boys \: can \: finish \: it \: in \: 7 \: days$

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•To Find out:-

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$\: \: \: \sf \blue \bullet \: time \: taken \: by \: 1 \: man \: alone \: and \: that \: by \: 1 \: boy \: \\ \sf \: to \: finish \: the \: work$

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•Solution:-

•Suppose 1 man can finish the work in x days and 1 boy alone can finish it in y days

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Then,1 man's 1 day's work=1/x

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And 1 boy's 1 day's work=1/y

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8 men and 12 boys can finish the work in 5 days

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$\sf : \implies(8 \: mens \: 1 \: days \: work) + (12 \: boys \: 1days \: work) = \dfrac{1}{5}$

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$: \implies \sf \dfrac{8}{x} + \dfrac{12}{y} = \dfrac{1}{5}$

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Let 1/x=u and 1/y=v

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$: \implies \sf8u + 12v = \dfrac{1}{5} \: \: \: \: ....(1)$

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Again 6 men and 8 boys can finish the work in 7 day

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$\sf : \implies(9 \: mens \: 1 \: days \: work) + (8 \: boys \: 1 \: days \: work) = \dfrac{1}{7}$

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$\sf : \implies \dfrac{6}{x} + \dfrac{8}{y} = \dfrac{1}{7}$

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$: \implies \sf6u + 8v = \dfrac{1}{7} \: \: \: \: ....(2)$

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On multiplying (1) by 3 ,(2) by 4 and subtracting the results, we get:

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$: \implies \sf \: 4v = \Bigg( \dfrac{3}{5} - \dfrac{4}{7} \Bigg) = \dfrac{1}{35}$

$: \implies \sf \: v = \dfrac{1}{140} : \implies \dfrac{1}{y} = \dfrac{1}{140}$

$\implies \sf \: y = 140$

On putting v=1/140 in (2) we get;

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$: \implies \sf \: 6u = \Bigg( \dfrac{1}{7} - \dfrac{8}{140} \Bigg) = \dfrac{12}{140}$

$: \implies \sf \: u = \Bigg( \dfrac{12}{140} \times \dfrac{1}{6} \Bigg) = \dfrac{1}{70}$

$: \implies \sf \dfrac{1}{x} = \dfrac{1}{70}$

$: \implies \sf \: x = 70$

Therefore,

1 man alone can finish the work in 70 days

and 1 boy alone can finish work in 140 days.