$A$ can do $\dfrac{2}{3}$ of a certain work in $16$ days and $B$ can do $\dfrac{1}{4}$ of the same work in $3$ days. In how many days can both finish the work, working together?
Answer
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Hint:Since they are individually doing the same work, we can find their capacities. We need to find the work done by them in a single day. Then we can calculate the number of days needed to finish the work.
Formula used:If $x$ is the number of days and w is the work done per day, then total work done is $wx$.
Complete step-by-step answer:
Given that $A$ can complete $\dfrac{2}{3}$of the given work in $16$ days.
So $A$ can complete the total work (that is $\dfrac{3}{3} = 1$ of the work) in $16 \times \dfrac{3}{2} = 24$ days.
Also given $B$ can complete $\dfrac{1}{4}$ of the same work in $3$ days.
So $B$ can complete the total work in $3 \times 4 = 12$days.
Therefore, the work done by $A,B$ in a single day is $\dfrac{1}{{24}},\dfrac{1}{{12}}$ respectively.
$ \Rightarrow {w_A} = \dfrac{1}{{24}},{w_B} = \dfrac{1}{{12}}$
So the total work done by them in a single day is $w = {w_A} + {w_B} = \dfrac{1}{{24}} + \dfrac{1}{{12}} = \dfrac{1}{{24}} + \dfrac{2}{{24}} = \dfrac{3}{{24}} = \dfrac{1}{8}$
Let they take $x$ days to complete the whole work, working together.
We have to find $x$.
If $x$ is the number of days and $w$ is the work done per day, then total work done is $wx$.
Here, $w = \dfrac{1}{8}$ and total work can be considered as $1$.
$ \Rightarrow \dfrac{1}{8}x = 1$
$ \Rightarrow x = 1 \times 8 = 8$
So $A$ and $B$ need $8$ days to finish the work, working together.
Note:The point should be noted is that work done is proportional to the time taken. The more is the time, the more is the work. But the more is the number of workers, the less is the time taken.
Formula used:If $x$ is the number of days and w is the work done per day, then total work done is $wx$.
Complete step-by-step answer:
Given that $A$ can complete $\dfrac{2}{3}$of the given work in $16$ days.
So $A$ can complete the total work (that is $\dfrac{3}{3} = 1$ of the work) in $16 \times \dfrac{3}{2} = 24$ days.
Also given $B$ can complete $\dfrac{1}{4}$ of the same work in $3$ days.
So $B$ can complete the total work in $3 \times 4 = 12$days.
Therefore, the work done by $A,B$ in a single day is $\dfrac{1}{{24}},\dfrac{1}{{12}}$ respectively.
$ \Rightarrow {w_A} = \dfrac{1}{{24}},{w_B} = \dfrac{1}{{12}}$
So the total work done by them in a single day is $w = {w_A} + {w_B} = \dfrac{1}{{24}} + \dfrac{1}{{12}} = \dfrac{1}{{24}} + \dfrac{2}{{24}} = \dfrac{3}{{24}} = \dfrac{1}{8}$
Let they take $x$ days to complete the whole work, working together.
We have to find $x$.
If $x$ is the number of days and $w$ is the work done per day, then total work done is $wx$.
Here, $w = \dfrac{1}{8}$ and total work can be considered as $1$.
$ \Rightarrow \dfrac{1}{8}x = 1$
$ \Rightarrow x = 1 \times 8 = 8$
So $A$ and $B$ need $8$ days to finish the work, working together.
Note:The point should be noted is that work done is proportional to the time taken. The more is the time, the more is the work. But the more is the number of workers, the less is the time taken.
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