Question

Pat's average weekly income from her part time job was 15% more in September than it was in July. her average weekly income in July was 10% less than it was in August. if her average weekly income in September was $207, what was her average weekly income in August?​

Answer 1

her average weekly income in August was $200 .

Explanation:

Let in July income was x , So in September :

⇒
`x+(15x)/(100)`

According to question , her average weekly income in September was $207 i.e.

⇒
`x+(15x)/(100)=207`

⇒
`(115x)/(100)=207`

⇒
`x=207((100)/(115))`

⇒
`x=180`

Now , average weekly income in July was 10% less than it was in August , let income of august = y, i.e.

Income in august =
`180+(August(10))/(100)`

⇒
`y=180 + (y(10))/(100)`

⇒
`(9)/(10)y=180`

⇒
`y=\$200`

Therefore , her average weekly income in August was $200 .

Answer 2

Her average weekly income in August was $200.

Explanation:

To find the income in july :

Given that,

• Her average weekly income in September was $207.
• Her part time job was 15% more in September than it was in July.

Let,

'x' be the 100% of income she got in July.

Therefore, the income $207 in September is 115% of x.

⇒ 115% of x = 207

⇒ 1.15 × x = 207

⇒ x = 207 / 1.15

⇒ x = 180

∴ Her average weekly income in July was $180.

To find the income in August :

Given that,

• Her average weekly income in July was 10% less than it was in August.
• We found that, $180 is the average weekly income in July which is 10% less than August.

Let,

'y' be the 100% of income she got in August.

Therefore, the income $180 in July is 90% of y.

⇒ 90% of y = 180

⇒ 0.9 × y = 180

⇒ y = 180/0.9

⇒ y = 200

∴ Her average weekly income in August was $200.