Question

If 2 men and 3 boys can do a piece of work in $3\frac{3}{7}$ days, while 3 men and 2 boys can do it in days, then in $3\frac{15}{43}$. Then in how many days will 4 boys finish the work?

• A. 5 days
• B. $4\frac{1}{2}days$
  
• C. $5\frac{1}{2}days$
  
• D. 4 days

Answer 1

The correct option is B $4\frac{1}{2}days$


Let a man takes p days to do a piece of work and a boy takes q days to do the same work.
Then,
One day work of a man = $\frac{1}{p}$
One day work of a boy = $\frac{1}{q}$
Since, 2 men and 3 boys can do a piece of work in $3\frac{3}{7}$ days i.e. $\frac{24}{7}$ days,
So, one day work of 2 men and 3 boys will be $\frac{7}{24}.$
$\Rightarrow \frac{2}{p}+\frac{3}{q}=\frac{7}{24}.....\left(i\right)$
Also, as 3 men and 2 boys can do a piece of work in $3\frac{15}{43}$ days i.e. $\frac{144}{43}$
$\Rightarrow \frac{3}{p}+\frac{2}{q}=\frac{43}{144}.....\left(ii\right)$
Now, let $x=\frac{1}{p}andy=\frac{1}{q}$
Substituting $\frac{1}{p}=xand\frac{1}{q}=yin\left(i\right)and\left(ii\right),$
$2x+3y=\frac{7}{24}.....\left(iii\right)$
$3x+2y=\frac{43}{144}.....\left(iv\right)$
Multiplying (iii) by 2 and (iv) by 3, we get
$4x+6y=\frac{7}{12}.....\left(v\right)$
$9x+6y=\frac{43}{48}.....\left(vi\right)$
Subtracting (v) from (vi), we get
$5x=\frac{43-28}{48}=\frac{15}{48}$
$\Rightarrow x=\frac{15}{48\times 5}$
$\Rightarrow x=\frac{1}{16}$

Now, substituting $x=\frac{1}{16}$ in (iii), we get
$2\times \frac{1}{16}+3y=\frac{7}{24}$
$\Rightarrow 3y=\frac{7}{24}-\frac{1}{8}=\frac{7-3}{24}$
$\Rightarrow 3y=\frac{1}{6}$
$\Rightarrow y=\frac{1}{18}$
∴ q = 18 i.e. one boy can do a piece of work in 18 days

So, 4 boys will take $\frac{1}{4}$ times number of days taken by a boy to do the same work.
$As,\frac{q}{4}=\frac{18}{4}=\frac{9}{2}=4\frac{1}{2},$
Therefore, 4 boys will finish the work in $4\frac{1}{2}$ days.
Hence, the correct answer is option (2).