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

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A certain car depreciates such that its value at the end of each year  [#permalink]

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New post 15 Jun 2017, 05:10
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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. \(\frac{b^2−a}{a}\)

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A certain car depreciates such that its value at the end of each year  [#permalink]

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New post Updated on: 15 Jun 2017, 06:40
Bunuel wrote:
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. \(\frac{b^2−a}{a}\)


Let a be =$100
Let p% be = 10%

Therefore b = 90% of 100 = 90/100 * 100 = $90

2012 = 90% of 90 = 90/100 * 90 = $81

Car worth on December 31, 2013 = 90% of 81 = $72.9

Lets put the values of a and b in the options and check which gives the value of $72.9.

A. \(\frac{b^3}{a^2}\) = 90 * 90 * 90/100 * 100 = $72.9 ----------- Answer

B. \(\frac{b^2}{a}\) = 90 * 90/100 = $81.

C. \(\frac{b\sqrt{b}}{a}\) = 90 \(\sqrt{90}\) / 100 = 0.9\(\sqrt{90}\) = 0.9 * 3 \(\sqrt{10}\)

D. \(\frac{b^2\sqrt{b}}{a^2}\) = 90 * 90 * 3 \(\sqrt{10}\) / 100 * 100 = 24.3 \(\sqrt{10}\)

E. \(\frac{b^2−a}{a}\) = (90 * 90) - 100/ 100 = 8100 - 100/100 = 8000/100 = $80

Answer A...

Originally posted by sashiim20 on 15 Jun 2017, 06:30.
Last edited by sashiim20 on 15 Jun 2017, 06:40, edited 2 times in total.
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Re: A certain car depreciates such that its value at the end of each year  [#permalink]

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New post 15 Jun 2017, 06:36
Let a = 10,000 and let p = 10. So here the value is depreciating by 10% every year, or we can say that value at the end of a certain year is 90% of its value at the end of previous year.

So, value on 31 Dec 2010 = 10000
value on 31 Dec 2011 = 90% of 10000 = 9000 (this is 'b')
value on 31 Dec 2012 = 90% of 9000 = 8100
and value on 31 Dec 2013 = 90% of 8100 = 7290.

So we are looking for an option where if we put a=10000 and b=9000, we should get '7290'.

A) b^3/a^2 = 9000^3/10000^2 = 7290

No other option gives this value. Hence A answer
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A certain car depreciates such that its value at the end of each year  [#permalink]

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New post 17 Jun 2017, 05:22
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Bunuel wrote:
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. \(\frac{b^2−a}{a}\)


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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A certain car depreciates such that its value at the end of each year  [#permalink]

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New post 17 Jun 2017, 05:36
If price of the car on December 31st,2010 is 1000(a=10)

Since the car depreciates at p%(assume p=10),
On 31st December, 2011 the car becomes 900(loss of 10%)
Therefore, b =900.

On 31st December 2012, the car will be worth 810(another 10% less)
This can be got by formule \(\frac{b^2}{a}\) because \(810 = \frac{900*900}{1000}\)

One year later,
On 31st December 2013 the car depriciates by another 10%.
This time the cost can be given by \(\frac{b^3}{a^2}\) because \(729 = \frac{900*900*900}{1000*1000}\)

Hence Option A is our answer
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Re: A certain car depreciates such that its value at the end of each year  [#permalink]

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New post 21 Apr 2018, 08:51
ScottTargetTestPrep wrote:
Bunuel wrote:
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. \(\frac{b^2−a}{a}\)


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



Could you please elaborate on how you got this equation:

a(1 - p/100)(1 - p/100)(1 - p/100)
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Re: A certain car depreciates such that its value at the end of each year &nbs [#permalink] 21 Apr 2018, 08:51
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