Question

A shopkeeper sold a certain number of toys. The number of toys, as well as the price of each toy (in Rs.), was a two-digit number. By mistake, he reversed the digits of both the number of toys he sold and the price of each toy. As a result, he found that his stock account at the end of the day showed 81 times more than it actually was. If the faulty calculation shows a total sale of Rs. 882, find the actual selling price of each toy (in Rs.)
(a) 89
(b) 98
(c) 97
(d) 79

Answer 1

Hint: In order to solve this problem, we need to assume the digits of tens-place and units place separately and then reverse the digits. Further, we need to apply the first condition, we get the values of the two-digit number. According to the second condition we need to proceed and find the selling price of the toy.

Complete step by step answer:
There are two two-digit quantities: the number of toys and the price of the toy.
Both the quantities are two-digit quantities.
Let the number of toys be $xy$ . The actual value of $xy$ is $10x+y$ because the $x$ is in the tens place and $y$ is in the units place.
Similarly, let the price of each toy be $ab$. The actual value of $ab$ is $10a+b$ because the $a$ is in the tens place and $b$ is in the units place.
Now, according to the condition, the digits are reversed.
The faulty number of toys are $yx$. The actual value of a faulty number of toys is $10y+x$ .
The condition says that as the digits are reversed by new calculation there are 81 items left in the stock.
This says that we need to add 81 to the faulty count of the number of toys to equate it with the actual number of toys.
By implementing we get,
$10x-x=10y+x+81..............(i)$
Simplifying further we get,
$\begin{align}
  & 9x-9y=81 \\
 & \Rightarrow x-y=9..................(ii) \\
\end{align}$
All the possible values of $x$ and $y$ are single value and positive.
Therefore the only possible solution for (ii) is $x=9$ and $y=0$ .
So, the actual number of toys is 90.
And the faulty calculation gives the count of toys to be 9.
According to the second condition, it is said that the wrong calculation is carried forward and the total sale is Rs. 882.
The total selling price is given by = number of items x selling price of each toy
By substituting the values, we get,
882 = 9 x selling price of each toy.
The selling price of each toy $=\dfrac{882}{9}=98........(iii)$
The selling price we get is according to the wrong calculation as the selling price is from the wrong calculation and the number of toys is also faulty.
Comparing with our assumption, we get,
$b=9$ and $a=8$ .
The actual price of each toy is $ab$ , that is $89$ .
Hence, the correct option is (a).

Note: There are multiple points that needed to be taken care of. We need to understand that the faulty calculation of the number of toys is less than that of the actual number of toys, therefore, we need to add 81 to the wrong calculation of the number of toys. For the second condition, we need to use the value of the wrong calculation of the number of toys as the selling price of the wrong calculation.
Also, we are asked to find the selling price of each toy and so in the end, we need to reverse the digits after the equation (iii).