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**Hint:**Let us assume that $A$ alone can do the work in $x$ days and similarly $B$ alone can do the same work in $y$ days.

Then one day work of $A$$ = \dfrac{1}{x}$

One day work of $B$$ = \dfrac{1}{y}$

Now we have to do the rest calculation.

**Complete step-by-step answer:**

In this question it is given that $A$ and $B$ can do a piece of work in $15$ days. It is also given that $A$’s one day work is $1\dfrac{1}{2}$ times the one day work of $B$.

Now we have to calculate the total time taken by $A$ alone and $B$ alone to do this work.

So first of all let us assume that:

$A$ alone can do the work in $x$ days.

$B$ alone can do the same work in $y$ days.

So if $A$ alone can do the work in $x$ days, then one day work of $A$$ = \dfrac{1}{x}$

And if $B$ alone can do the same work in $y$ days, then one day work of $B$$ = \dfrac{1}{y}$

And in this question we know the relation between both one day’s work.

$A$’s one day work is $1\dfrac{1}{2}$ times the one day work of $B$.

Here we know that

One day work of $A$$ = \dfrac{1}{x}$

One day work of $B$$ = \dfrac{1}{y}$

And $\dfrac{1}{x} = 1\dfrac{1}{2}.\dfrac{1}{y}$

$\dfrac{1}{x} = \dfrac{3}{{2y}}$

$y = \dfrac{{3x}}{2}$ $ - - - - - \left( 1 \right)$

Now it is also given that they can complete work together in $15$ days. So one day work of together is:

$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{{15}}$ $ - - - - - \left( 2 \right)$

Now putting the value of $y = \dfrac{{3x}}{2}$ in equation $\left( 2 \right)$

$\dfrac{1}{x} + \dfrac{2}{{3x}} = \dfrac{1}{{15}}$

$\dfrac{{3 + 2}}{{3x}} = \dfrac{1}{{15}}$

$\dfrac{5}{{3x}} = \dfrac{1}{{15}}$

$x = 25$days

Similarly $y = \dfrac{{3x}}{2}$$ = \dfrac{{75}}{2}$ days

So $A$ alone can do the work in $25$ days.

$B$ alone can do it in $\dfrac{{75}}{2}$ days.

**Note:**In this type of question you need to find the two questions. First equation you will get from the statement that they can complete work together in $15$ days. And the second equation, you will get from the statement that $A$’s one day work is $1\dfrac{1}{2}$ times the one day work of $B$.

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