Question

Six men working $10$ hours a day can do a piece of work in $24$ days. In how many days will $9$ men working for $8$ hours a day do the same work?

Answer 1

Method 1: The problem involves $3$ sets of variables, namely - Number of men, Working hours per day and Number of days.

Number of Men	Number of hours per day	Number of days
$6$	$10$	$24$
$9$	$8$	$x$

Step 1: Consider the number of men and the number of days. As the number of men increases from $6$ to $9$, the number of days decreases. So it is in Inverse Variation.
Therefore the proportion is $9:6::24:x$ .....(1)
Step 2: Consider the number of hours worked per day and the number of days. As the number of hours working per day decreases from $10$ to $8$, the number of days increases. So it is inverse variation.
Therefore the proportion is $8:10::24:x$ .....(2)
Combining (1) and (2), we can write as
$\begin{array}{c}9:6\\ 8:10\end{array}::24:x$
We know, Product of extremes $=$ Product of Means.
$\begin{array}{ccccc}Extremes& & Means& & Extremes\\ 9& :& 6::24& :& x\\ 8& :& 10& & \end{array}$
So, $9\times 8\times x=6\times 10\times 24$
$x=\frac{6\times 10\times 24}{9\times 8}=20$ days
Method 2: (Using arrow marks)

Number of Men	Number of hours per day	Number of days
$6$	$10$	$24$
$9$	$8$	$x$

Step 1: Consider men and days. As the number of men increases from $6$ to $9$, the number of days decreases. It is in inverse variation.
The multiplying factor $=\frac{6}{9}$
Step 2: Consider the number of hours per day and the number of days. As the number of hours per day decreases from $10$ to $8$, the number of days increases. It is also in inverse variation.
The multiplying factor $=\frac{10}{8}$
$\therefore$ $x=\frac{6}{9}\times \frac{10}{8}\times 24=20$ days.