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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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15 Jun 2017, 04:10
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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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Updated on: 15 Jun 2017, 05:40
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

Originally posted by sashiim20 on 15 Jun 2017, 05:30.
Last edited by sashiim20 on 15 Jun 2017, 05:40, 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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15 Jun 2017, 05:36
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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17 Jun 2017, 04:22
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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.

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

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17 Jun 2017, 04:36
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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21 Apr 2018, 07:51
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.

Could you please elaborate on how you got this equation:

a(1 - p/100)(1 - p/100)(1 - p/100)
Re: A certain car depreciates such that its value at the end of each year &nbs [#permalink] 21 Apr 2018, 07:51
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