A person can do the job as fast as his two sons working together. If one son does the job in $6$ and the other in $12$ days then how many days does it take for father to do the job?
Answer
613.8k+ views
Hint:
Let us assume that the father does the job in $x$ days and it is given that the father does the job as fast as the two sons working together that means
One day work of father $ = $ sum of one day work of both his sons.
Complete step by step solution:
Here we are given that the person can do the job as fast as his two sons can working together that means that one day work if both his sons is equivalent to the work of father alone and we are given that one son completes the work in $6$ days and other in $12$ days and we need to find the days taken by father to complete the job.
As from the first statement we get to know that father do the work as fast as his two sons do working together that means that the father will complete the work in less than $12$ days and also that the father takes less than $6$ days and hence the father must complete the work in less than $6$ days and also we know that
One day work of father$ = $ sum of one day work of both his sons.
We know that one son complete the work in $6$ days
Hence one son’s one day work$ = \dfrac{1}{6}$
And other complete the work in $12$ days
So his one day work$ = \dfrac{1}{{12}}$
Also we have assumed that the father complete the work in $x$ days
So one day work of father$ = \dfrac{1}{x}$
Now we know that
One day work of father $ = $ one day work of one son$ + $one day work of another son
$\dfrac{1}{x} = \dfrac{1}{6} + \dfrac{1}{{12}}$
$\Rightarrow \dfrac{1}{x} = \dfrac{{2 + 1}}{{12}}$
$\Rightarrow \dfrac{1}{x} = \dfrac{3}{{12}} = \dfrac{1}{4}$
Hence we get that $x = 4$
Hence father can complete the work in $4$ days.
Note:
If one person completes the work in $n$ days, then the one day work of that person$ = \dfrac{1}{n}$. So $m$persons with the same efficiency can complete the work in $\dfrac{n}{m}$ days. If efficiency is not the same then we have to calculate by using one day work of both the persons.
Let us assume that the father does the job in $x$ days and it is given that the father does the job as fast as the two sons working together that means
One day work of father $ = $ sum of one day work of both his sons.
Complete step by step solution:
Here we are given that the person can do the job as fast as his two sons can working together that means that one day work if both his sons is equivalent to the work of father alone and we are given that one son completes the work in $6$ days and other in $12$ days and we need to find the days taken by father to complete the job.
As from the first statement we get to know that father do the work as fast as his two sons do working together that means that the father will complete the work in less than $12$ days and also that the father takes less than $6$ days and hence the father must complete the work in less than $6$ days and also we know that
One day work of father$ = $ sum of one day work of both his sons.
We know that one son complete the work in $6$ days
Hence one son’s one day work$ = \dfrac{1}{6}$
And other complete the work in $12$ days
So his one day work$ = \dfrac{1}{{12}}$
Also we have assumed that the father complete the work in $x$ days
So one day work of father$ = \dfrac{1}{x}$
Now we know that
One day work of father $ = $ one day work of one son$ + $one day work of another son
$\dfrac{1}{x} = \dfrac{1}{6} + \dfrac{1}{{12}}$
$\Rightarrow \dfrac{1}{x} = \dfrac{{2 + 1}}{{12}}$
$\Rightarrow \dfrac{1}{x} = \dfrac{3}{{12}} = \dfrac{1}{4}$
Hence we get that $x = 4$
Hence father can complete the work in $4$ days.
Note:
If one person completes the work in $n$ days, then the one day work of that person$ = \dfrac{1}{n}$. So $m$persons with the same efficiency can complete the work in $\dfrac{n}{m}$ days. If efficiency is not the same then we have to calculate by using one day work of both the persons.
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