"> Word Problem 2 | Job Exam Rare Mathematics (JEMrare)

Word Problem 2

Problem 2: If 3 men or 6 boys can do a piece of work in 20 days, then in how many days 6 men and 8 boys will finish the same work?



Solution:

First Case (work is done by 3 men or 6 boys)

or

3M=6B

\Rightarrow 1M = \frac{6}{3}B

\Rightarrow 1M = 2B----(1)

2nd Case (work is done by 6 men and 8 boys)

or

6M+8B

Putt value of M from eq(1)

\Rightarrow 6M+8B=6\left ( 2B \right )+8B=12B+8B=20B

So, 6 men and 8 boys are equal to 20 boys;

Now if 1st case is taken with boys only, then 6 boys do the work in 20 days. In 2nd case now all men and boys are equal to 20 boys, so we can suppose that 20 boys can do the work in ‘x’ days.

We arrange data as below;

Solving for x;

\frac{x}{20}=\frac{6}{20}

\Rightarrow x = \frac{6\times 20}{20}

\Rightarrow x = 6

 So, 20 boys will finish the same work in 6 days