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Maria pays \(\dfrac18\) of her monthly income for food, \(\dfrac18\) for utilities, \(\dfrac18\) for student loans, and

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by Vincen » Wed Jul 22, 2020 12:09 pm

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Maria pays \(\dfrac18\) of her monthly income for food, \(\dfrac18\) for utilities, \(\dfrac18\) for student loans, and \(\dfrac45\) of the remainder for rent. If at the end of each month Maria puts \(\dfrac12\) of her remaining income into a CD account, what portion of her monthly income does she puts into the account?


A. \(\dfrac18\)

B. \(\dfrac1{10}\)

C. \(\dfrac7{80}\)

D. \(\dfrac1{16}\)

E. \(\dfrac1{20}\)

[spoiler]OA=D[/spoiler]

Source: Princeton Review
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Re: Maria pays \(\dfrac18\) of her monthly income for food, \(\dfrac18\) for utilities, \(\dfrac18\) for student loans,

by Scott@TargetTestPrep » Sun Jul 26, 2020 9:09 am
Vincen wrote: ↑
Wed Jul 22, 2020 12:09 pm
 Maria pays \(\dfrac18\) of her monthly income for food, \(\dfrac18\) for utilities, \(\dfrac18\) for student loans, and \(\dfrac45\) of the remainder for rent. If at the end of each month Maria puts \(\dfrac12\) of her remaining income into a CD account, what portion of her monthly income does she puts into the account?


A. \(\dfrac18\)

B. \(\dfrac1{10}\)

C. \(\dfrac7{80}\)

D. \(\dfrac1{16}\)

E. \(\dfrac1{20}\)

[spoiler]OA=D[/spoiler]

Solution:

We can let n = monthly income, and we can calculate what she spends on food, utilities, loans, and rent as:

(1/8)n + (1/8)n + (1/8)n + 4/5(5/8)n

(3/8)n + (1/2)n = (3/8)n + (4/8)n = (7/8)n

Thus, we see that (1/8)n is left over, and since half of that is put into a CD account, (1/16)n is put into a CD.

Alternate Solution:

Let’s pick a convenient number for Maria’s income, such as 800. Then, she spends 100 each on food, utilities, and student loans, totaling 300. She has 800 - 300 = 500 left. Since 4/5 of this money is spent on rent, she spends 500(4/5) = 400 on rent and has 500 - 400 = 100 left. Since she saves half of this in a CD account, she saves 100 / 2 = 50. Thus, she saves 50/800 = 1/16 of her income each month.

Answer: D

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