Question

A bowl of sweets was placed on a table to be distributed among three brothers rajan, sajal and karan. rajan arrived first and ate what he thought was his share of sweets and left. then, sajal arrived. he thought that he was the first one to arrive and ate the number of sweets, he thought was his share and left. lastly, karan arrived. he again thought he was the first one to arrive and he took what he thought was his share. if $16$ sweets are left in the bowl finally, how many sweets did the bowl contain initially?

Answer 1

Finding the number of sweets:

$3$ boys ate one-third of the sweets in their turn.

Now let us consider the initial number of sweets to be

$(3\times 3\times 3)x=27x$

The sweets taken by Rajan are one third of total:
Therefore $\frac{27x}{3}=9x$
Remaining sweets after taken by Rajan:
$$\begin{gathered} =27x-9x \\ =18x \end{gathered}$$
Now the number of sweets left are $18x$

The sweets taken by Sajal are one third of $18x$ and can be calculated as:
$\frac{18x}{3}=6x$
Remaining sweets:
$$\begin{gathered} =18x-6x \\ =12x \end{gathered}$$
Now the number of sweets left are $12x$.
The sweets taken by Karan are one third of $12x$ which are
$\frac{12x}{3}=4x$
the sweets taken by karan are $4x$.
Now the remaining sweets after karan are:
$12x-4x=8x$
But the final remaining sweets in the bowl are $16$.
So

$$\begin{gathered} 8x=16 \\ \Rightarrow x=2 \end{gathered}$$

The total number of sweets are $=27x=(27\times 2)=54$

Therefore, the total number of sweets in the bowl are$54$