Question

A earns 10% more than B. By how much percentage is B's earnings less than A's earnings?

Answer 1

B's earning are 9% less than A's.

Explanation:

Let A and B represent the earning for A and B, respectively.

We learn that A = 1.1B ["A earns 10% more than B"]

Divide both sides by 1.1:

(A/1.1) = B

B = 0.909090909 or 0.91 to 2 sig figs

B's earning are (1-0.91 = 0.09) or 9% less than A's.

Answer 2

9.09%

Explanation:

Let B's earnings be represented by the variable 'x'.

Since A earns 10% more than B, A's earnings would be
`\sf 1.1x`.

The percentage by which B's earnings are less than A's earnings is calculated using the formula:

`\sf \textsf{Percentage difference} = \left( \frac{\textsf{Difference in earnings}}{\textsf{A's earnings}} \right) * 100\%`

Substituting the values:

`\sf \textsf{Percentage difference} = \left( (1.1x - x)/(1.1x) \right) * 100\%`

`\sf \textsf{Percentage difference} = \left( (0.1x)/(1.1x) \right) * 100\%`

`\sf \textsf{Percentage difference} = \left( (0.1)/(1.1) \right) * 100\%`

`\sf \textsf{Percentage difference} \approx 9.09\%`

Therefore, B's earnings are approximately 9.09% less than A's earnings.