Question

If 12 men and 16 boys can do a piece of work in 5 days, 13 men and 24 boys can do it in 4 days, then the ratio of the daily work done by a man to that of a boy is:
A. 2:1
B. 3:1
C. 3:2
D. 5:4

Answer 1

Hint: Assume a variable (say x) for 1 day’s work done by a man and a variable (say y) for 1 day’s work done by a boy. Construct two equations. Using the two information given in the equation and solve to get x and y and then find the ratio of x and y.

Complete step by step solution:
According to the question, we have to find the ratio of work done by a man to that of a boy.
So, let us assume work done daily by a man is x and work done by a woman is y.
According to the question, 12 men and 16 boys can do a piece of work in 5 days.
So, in one day 12 men and 16 days can do $\dfrac{1}{5}$ work.
According to our assumption, 12 men can do 12x work in one day and 16 boys can do 16y work in one day.
So, according to our assumption 12 men and 16 boys can do (12x + 16y) work in one day and according to question 12 men and 16 boys can do $\dfrac{1}{5}$ work in one day.
Therefore, $12x+16y=\dfrac{1}{5}............\left( 1 \right)$
One more information in the question is that 13 men and 24 boys can do the work in 4 days.
Which means in one day 13 men and 24 boys can do $\dfrac{1}{4}$ work.
Similarly, according to our assumption 13 men and 24 boys can do (13x +24y) work in one day.
So, $13x+24y=\dfrac{1}{4}............\left( 2 \right)$
Now, let us solve equation (1) and (2),
$\begin{align}
  & 12x+16y=\dfrac{1}{5}............\left( 1 \right) \\
 & 13x+24y=\dfrac{1}{4}............\left( 2 \right) \\
\end{align}$
Dividing both sides of equation (1) by 4, we will get equation (1) as,
$3x+4y=\dfrac{1}{20}$
Now, multiply this equation with 6 both sides,
$18x+24y=\dfrac{6}{20}$
Now subtracting equation (2) from this equation,
$\begin{align}
  & \dfrac{\begin{align}
  & 18x+24y=\dfrac{6}{20} \\
 & 13x+24y=\dfrac{1}{4} \\
 & -\ \ \ \ -\ \ \ \ \ \ \ \ - \\
\end{align}}{5x=\dfrac{6}{20}-\dfrac{1}{4}} \\
 & 5x=\dfrac{6-5}{20}=\dfrac{1}{20} \\
 & \Rightarrow 5x=\dfrac{1}{20} \\
\end{align}$
Dividing both sides of equation by 5, we will get,
$\begin{align}
  & \Rightarrow x=\dfrac{1}{20\times 5}=\dfrac{1}{100} \\
 & \therefore x=\dfrac{1}{100} \\
\end{align}$
Now let us put $x=\dfrac{1}{100}$ in equation (1), we will get,
$12\left( \dfrac{1}{100} \right)+16y=\dfrac{1}{5}$
Taking all the constant terms to RHS, we will get,
$\begin{align}
  & \Rightarrow 16y=\dfrac{1}{5}-\dfrac{12}{100} \\
 & \Rightarrow 16y=\dfrac{20-12}{100} \\
 & \therefore 16y=\dfrac{8}{100} \\
\end{align}$
Dividing both sides by 16, we will get,
$\begin{align}
  & y=\dfrac{8}{100\times 16} \\
 & \Rightarrow y=\dfrac{1}{200} \\
\end{align}$
We have found the value of x and y,
$\text{Ratio of x and y } =\dfrac{\dfrac{1}{100}}{\dfrac{1}{200}} \\
 =\dfrac{1}{100}\times \dfrac{200}{1}=\dfrac{2}{1} \\
$
Thus, the ratio of work done by a man to a boy daily will be 2:1 and option (A) is the correct answer.

Note:
Shortcut method to solve this question,
If ${{a}_{1}}\ men\ and\ {{b}_{1}}\ boys$can do a piece of work in x days and ${{a}_{2}}\ men\ and\ {{b}_{2}}\ boys$ can do it in y days, then the following relationship is obtained,
$\begin{align}
  & \dfrac{1\ man's\ work}{1\ boy's\ work}=\left[ \dfrac{y{{b}_{2}}-x{{b}_{1}}}{x{{a}_{1}}-y{{a}_{2}}} \right] \\
 & Here, \\
 & {{a}_{1}}=12,\ {{b}_{1}}=6,\ x=5,\ {{a}_{2}}=13,\ {{b}_{2}}=24,\ y=4 \\
\end{align}$
Applying above formula, we will get,
$\begin{align}
  & \dfrac{1\ man's\ work}{1\ boy's\ work}=\left[ \dfrac{4\times 24-5\times 16}{5\times 12-4\times 13} \right] \\
 & \Rightarrow \dfrac{1\ man's\ work}{1\ boy's\ work}=\left[ \dfrac{96-80}{60-52} \right] \\
 & \Rightarrow \dfrac{1\ man's\ work}{1\ boy's\ work}=\dfrac{16}{8} \\
 & \dfrac{1\ man's\ work}{1\ boy's\ work}=\dfrac{2}{1} \\
\end{align}$