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 $81$ items more than it actually was. If the faulty calculations show 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

The correct option is A $89$

Solution:

Let the number toys be $xy.$ which can be written as $10x+y.$

where, $y$ is the digit of units place and $x$ is the digit of tens place.

And price of each toy $=ab$ which can be written as $10a+b.$

where, $b$ is the digit of units place and $a$ is the digit of tens place.

Now,

A/q , the shopkeeper has $81$ items more than actual.

$10x+y=10y+x+81$

or, $9x-9y=81$

or, $x-y=9$

The only possible solution of above equation is $x=9$ and $y=0.$

So, actual number of toys sold $=90$

The wrongly recorded number of toys sold $=9$

$\therefore$ Wrongly recorded selling price of a toy $=\frac{Rs.882}{9}=Rs.98$

So, the correct selling price of a toy $=Rs.89$

Hence, A is the correct answer.