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**Hint:**

If a person does any work (x) in (y) days, then he will do complete work in \[\left( {\dfrac{y}{x}} \right)\]days.

To solve two linear equations \[{a_1}x + {b_1}y + {c_1} = 0;{a_2}x + {b_2}y + {c_2} = 0\] using cross multiplication method \[ = \dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}\]

**Complete step by step solution:**

Let one boy alone can finish the work in x days and one girl alone can finish the work in y days.

In one day a boy can do \[\dfrac{1}{x}th\] of the work, then 8 boys can do \[\dfrac{8}{x}th\] of the work.

In one day a girl can do \[\dfrac{1}{y}th\] of the work, then 12 girls can do \[\dfrac{{12}}{y}th\] of the work.

Given that together they can finish it in 10 days;

\[ \Rightarrow \dfrac{8}{x} + \dfrac{{12}}{y} = \dfrac{1}{{10}}\]

Putting \[\dfrac{1}{x} = a\] and \[\dfrac{1}{y} = b\] in above equation we get;

\[

\Rightarrow 8a + 12b = \dfrac{1}{{10}} \\

\Rightarrow 80a + 120b = 1.......Eq.01 \\

\]

Also, In one day a boy can do \[\dfrac{1}{x}th\] of the work, then 6 boys can do \[\dfrac{6}{x}th\] of the work.

In one day a girl can do \[\dfrac{1}{y}th\] of the work, then 8 girls can do \[\dfrac{8}{y}th\] of the work.

Given that together they can finish it in 14 days;

\[ \Rightarrow \dfrac{6}{x} + \dfrac{8}{y} = \dfrac{1}{{14}}\]

Putting \[\dfrac{1}{x} = a\] and \[\dfrac{1}{y} = b\] in above equation we get;

\[

\Rightarrow 6a + 8b = \dfrac{1}{{14}} \\

\Rightarrow 84a + 112b = 1.......Eq.02 \\

\]

Solving Eq.01 and Eq.02 using cross-multiplication method we get;

\[

80a + 120b = 1 \\

84a + 112b = 1 \\

\Rightarrow \dfrac{a}{{120(1) - 112(1)}} = \dfrac{b}{{84(1) - 80(1)}} = \dfrac{{ - 1}}{{80(112) - 84(120)}} \\

\Rightarrow \dfrac{a}{{120 - 112}} = \dfrac{b}{{84 - 80}} = \dfrac{{ - 1}}{{8960 - 10080}} \\

\Rightarrow \dfrac{a}{8} = \dfrac{b}{4} = \dfrac{{ - 1}}{{ - 1120}} \\

\Rightarrow \dfrac{a}{8} = \dfrac{b}{4} = \dfrac{1}{{1120}} \\

\Rightarrow a = \dfrac{8}{{1120}};b = \dfrac{4}{{1120}} \\

\Rightarrow a = \dfrac{1}{{140}};b = \dfrac{1}{{280}} \\

\]

Putting back the values of a and b in \[\dfrac{1}{x} = a\] and \[\dfrac{1}{y} = b\] respectively we get;

\[

\Rightarrow \dfrac{1}{x} = \dfrac{1}{{140}};\dfrac{1}{y} = \dfrac{1}{{280}} \\

\Rightarrow x = 140;y = 280 \\

\]

**Note:**

This is a question from time and work. Below are some important points to remember while solving such questions.

1) More men can do more work

2) More work means more time required to do work

3) More men can do a piece of work in less time.

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