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A certain car depreciates such that its value at the end of

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by BTGmoderatorLU » Wed Aug 14, 2019 3:23 pm

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Source: Veritas Prep

A certain car depreciates such that its value at the end of each year is \(p\)% less than its value at the end of the previous year. If that car was worth \(a\) dollars on December 31,010 and was worth \(b\) dollars on December 31, 2011, what was the car worth on December 31, 2013 in terms of \(a\) and \(b\)?

A. \(\frac{b^3}{a^2}\)

B. \(\frac{b^2}{a}\)

C. \(\frac{b\sqrt{b}}{a}\)

D. \(\frac{b^2\sqrt{b}}{a^2}\)

E. \(\frac{b^2-a}{a}\)

The OA is A
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Source: — Problem Solving |

by Scott@TargetTestPrep » Sun Aug 18, 2019 6:29 pm
BTGmoderatorLU wrote:Source: Veritas Prep

A certain car depreciates such that its value at the end of each year is \(p\)% less than its value at the end of the previous year. If that car was worth \(a\) dollars on December 31,010 and was worth \(b\) dollars on December 31, 2011, what was the car worth on December 31, 2013 in terms of \(a\) and \(b\)?

A. \(\frac{b^3}{a^2}\)

B. \(\frac{b^2}{a}\)

C. \(\frac{b\sqrt{b}}{a}\)

D. \(\frac{b^2\sqrt{b}}{a^2}\)

E. \(\frac{b^2-a}{a}\)

The OA is A
If the car is worth a dollars on December 31, 2010, then by 2013, the value of the car is:

a(1 - p/100)(1 - p/100)(1 - p/100)

However, notice that b = a(1 - p/100), so 1 - p/100 = b/a. Therefore, we can express
a(1 - p/100)(1 - p/100)(1 - p/100) as a(b/a)(b/a)(b/a) = b^3/a^2.

Answer: A

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