Question

If a's salary is 10% more than that of b,then how much percent is b's salary less than that of a?

Answer 1

let B s salary is X

Salary of A = X+X×10/100 =11X/10

Difference= 11X/10 -X =X/10


Difference% = (X/10 / 11X/10 )× 100= 9.09


So salary of B is 9.09% less than A s.

Answer 2

Given:

The a's salary is 10% more than b's salary.

To find:

The percentage by which b's salary is less than a.

Solution:

Let the salary of b be x units.

So,

the salary of a is

$= x + \frac{10}{100} x$

$= x + \frac{1}{10} x$

$= \frac{11x}{10} \: units$

Now,

the percentage by which b's salary is less than a is

$= \frac{decrease \: in \: salary}{salary \: of \: a} \times 100$

$= \frac{ \frac{11x}{10} - x }{ \frac{11x}{10} } \times 100$

(decrease in salary = salary of a - salary of b)

After solving the above, we get

$= \frac{ \frac{x}{10} }{ \frac{11x}{10} } \times 100$

$= \frac{x}{11x} \times 100$

$= \frac{1}{11} \times 100$

$= 9.09 \: \%$

Hence, b's salary is less than a's salary by 9.09℅.