Jane’s monthly income in the year 2010 was $48,200, where 14% of her salary was withheld for income tax. In the year 2011, her monthly salary was $51,800, where 16% of her income was withheld for income tax. What was her average take home pay during the years 2010-2011?

In 2010: Take home salary = (1-0.14)*(48200*12) = $497,424
In 2011: Take home salary = (1-0.16)*(51800*12) = $522,144

Average take home per month in 2010-11 = (497,424+522,144)/24 = $42,482

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chisnau chisnau · Answer 2 · 2019-06-21T15:51:17+02:00

Answer:

The average take home pay per month during the years 2010-2011 was $42,482.

Step-by-step explanation:

Jane’s monthly income in the year 2010 was $48,200, where 14% of her salary was withheld for income tax.

So, for the year 2010, the amount withheld was = [tex]0.14\times48200=6748[/tex] dollars.

Her take home salary was = [tex]48200-6748=41452[/tex] dollars

This is the monthly pay so her yearly pay is [tex]12\times41452[/tex]=$497424

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In the year 2011, her monthly salary was $51,800, where 16% of her income was withheld for income tax.

So, for the year 2011, the amount withheld was = [tex]0.16\times51800=8288[/tex] dollars.

Her take home salary was = [tex]51800-8288=43512[/tex] dollars

Her yearly pay is [tex]12\times43512[/tex]=$522144

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So, her average take home pay per month during the years 2010-2011 was:

[tex]\frac{497424+522144}{24}[/tex] = $42,482.