Question

$3$ men and $4$ boys can do a piece of work in $14$ days, while $4$ men and $6$ boys do the same work in $10$ days. How long would it take one man to finish the same work ?

• A. $40$ days
• B. $20$ days
• C. $35$ days
• D. $70$ days

Answer 1

The correct option is D $70$ days

Let one man's one day's work be $x$

Let one boy's one day's work be $y$

According to first condition,

$3x+4y=\frac{1}{14}$ .......Equation $1$

According to second condition,

$4x+6Y=\frac{1}{10}$ .......Equation $2$

Multiplying Equation $1$ by $4$ and Equation $2$ by $3$,

$12x+16y=\frac{4}{14}$ .....Equation $3$

$12x+18y=\frac{3}{10}$ .....Equation $4$

Now solving Equation $3$ and $4$ simultaneously by substracting Equation $3$ by Equation $4$,

we get,

$\therefore -2y=\frac{4}{14}-\frac{3}{10}$

$\therefore -2y=\frac{40-42}{140}$

$\therefore y=\frac{1}{140}$

Putting the value of y in equation $1$, we get,

$3x+\frac{4}{140}=\frac{1}{14}$

$\therefore x=\frac{1}{70}$

Hence one man's one day work is $\frac{1}{70}$

$\therefore$ one man completes the whole work in $70$ days.