Question

Take a two digit number 'ab' where a > b. When the difference between this number and the number obtained by reversing its digits is divided by 9, quotient will be equal to

• A. $a+b$
• B. $a-b$
• C. $a^2-b^2$
• D. $a^2+b^2$

Answer 1

The correct option is B

$a-b$

$\text{Given: 2-digit number ab and a > b.}$

$\text{Expanded form of this 2-digit number}ab=10a+b$

$\text{Number obtained by reversing the}\text{digits of}ab=ba$

$\text{Expanded form of the 2-digit number}ba=10b+a$

$\text{Difference between ab and ba is:}=(10a+b)-(10b+a)=10a+b-10b-a=9a-9b=9(a-b)$

$\text{Now,}\frac{9(a-b)}{9}=(a-b)$
$\text{Thus, the quotient will be (a - b).}$