Question

a can do a piece of work in 6 days and b can do it in 10 days. How long will they take to complete it together​

Answer 1

The above Question falls from Time & Work

Given

A can do a piece of work in 6 days

B can do a piece of world in 10 days

Let the total Number of Days be x

Solution :-

A can do piece of Work in 1 days = $\bf \dfrac{1}{6}$

B can do the piece of Work in 1 days = $\bf \dfrac{1}{10}$

Therefore, adding A + B

$\longrightarrow \tt \dfrac{1}{6} + \dfrac{1}{10}$

$\longrightarrow \tt\dfrac{5 + 3}{30}$

$\longrightarrow \tt \dfrac{8}{30}$

So, $\tt \dfrac{8}{30}$ part of work is completed in 1 day.

So, total part of work will be completed in

$\sf \dfrac{1}{ \frac{8}{30}}$

Therefore, $\bf \dfrac{30}{8}$ of a Day.

Additional Information

• If A person do a piece of work in n days, therefore, in one day He will do 1/n Work.

• M men can do a piece of work in T hours, then Total effort or work = MT man hours

• More work means more time required to do work.

• More men can do more work in less time

Answer 2

Let us consider the work of A =$\tt{\dfrac{1}{6}}$

Let us take the work of B => $\tt{\dfrac{1}{10}}$

So , we have been asked to find the total time A and B will take together to complete a specific work.

For this we simply need to find out the sum of these two fractions.

$\longrightarrow{\tt{\dfrac{1}{6} + \dfrac{1}{10}}}$

$\longrightarrow{\tt{LCM \implies}}$

$\begin{array}{c | c} 2 & 6 , 10 \\\cline{1-2} 3 & 3 , 5 \\\cline{1-2} 5 & 1 , 5 \\\cline{1-2} & 1 , 1 \end{array}$

So the LCM : 30

$\longrightarrow{\tt{\dfrac{5 + 3}{30}}}$

$\longrightarrow{\tt{\dfrac{8}{30}}}$

So both of them will complete the work :

8/30 units of 1 day .

And if they need to complete the whole work it will take $\longrightarrow{\tt{\dfrac{1}{\dfrac{8}{30}}}}$

Or simply $\tt{ \dfrac{30}{8}}$