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$4$ boys and $5$ girls can do $\frac{1}{2}$ work in $6$ days. after this $1$ boy and $2$ girls are added and $\frac{1}{3}$ work is done in $3$ days. how many boys must be added to complete the remaining work in $1$ day?

My Attempt

$4$ boys and $5$ girls can do $\frac{1}{2}$ work in $6$ days.

$5$ boys and $7$ girls can do $\frac{1}{3}$ work in $3$ days.

$5+x$ boys and $7$ girls can do $\frac{1}{6}$ work in $1$ day.

I could just reach upto here. can anyone help me further?

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  • $\begingroup$ Are we to assume that all girls work exactly equally fast and all boys work exactly equally fast; and (perhaps) boys and girls work at different rates. $\endgroup$ – paw88789 Apr 30 '16 at 0:45
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$1$ boy can do the work in $x$ days and $1$ girl do it in $y$ days.

In $1$ day, $1$ boy can do $1/x$ work.

In $6$ days, $4$ boys can do $24/x$ work.

In $6$ days, $5$ girls can do $30/y$ work.

$24/x + 30/y = 1/2$

One boy and two girls are added

$15/x + 21/y = 1/3$

$\Rightarrow x=y=108$

Boys and girls are equal.

$1/6$th of the work is left.

In $108$ days, $1$ work is completed by $1$ boy.

In $1$ day, $1$ work is completed by $108$ boys.

In $1$ day $1/6$ work is completed by $108/6 = 18$ boys.

$5$ boys and $7$ girls = $12$

Number of boys to be added to complete the remaining work in $1$ day

$= 18 - 12 = 6$

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You will need to figure out the amount of work done by one boy $b$ and one girl $g$. You have: $$ \begin{array}{rcl} (4b+5g)\times 6&=&\frac12,\\ (5b+7g)\times3&=&\frac13. \end{array} $$ Solving this for $b$ gives you $\frac{1}{108}$. The last of your items gives equation $((5+x)b+7g)\times 1 = \frac16$. This equation and the second give: $xb=\frac16-\frac19=\frac{1}{18}$ and, finally $x=\frac{108}{18}=6$.

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Say

$b_w$: The amount of job can be done by a boy in one day

$g_w$: The amount of job can be done by a girl in one day

$(4b_w+5g_w)$ is the job done in one day, hence

All the job is $12(4b_w+5g_w)$ units

$1/3$ of the job can be done by 5 boys and 7 girls so all the job can also be expressed as $9(5b_w+7g_w)$.

$$9(5b_w+7g_w)=12(4b_w+5g_w)$$ $$b_w=g_w$$

(Boys and girls do the same job:)

Let's say $$b_w=g_w=x$$ to make things simpler. All the job is $108x$ units. $1/6$ of the job remained at the end, which is $18x$. You want to finish this amount of job in one day; you have 7 girls who can do $7x$ , and you have 5 boys who can do $5x$, $18x-7x-5x=6x$ of job remains so you need 6 boys.

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Let $b$ be the time (in days) a boy needs alone to complete the entire job and $g$ the time (in days) a girl needs alone to do the same. Then: $$ 4\frac{1}{b}+5\frac{1}{g}=\frac{\frac{1}{2}}{6} $$ and also $$ 5\frac{1}{b}+7\frac{1}{g}=\frac{\frac{1}{3}}{3} $$ which give $b=g=108$. Thus: $$ (5+x)\frac{1}{108}+7\frac{1}{108}=\frac{1-\frac{1}{2}-\frac{1}{3}}{1}=\frac{\frac{1}{6}}{1}\Leftrightarrow 6(x+12)=108\Leftrightarrow x=6 $$

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