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

Question

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. $\frac{b^2−a}{a}$

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}

ScottTargetTestPrep · Accepted Answer

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

sashiim20 · Answer

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...

amanvermagmat · Answer

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

pushpitkc · Answer

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

premprabhs111 · Answer

Could you please elaborate on how you got this equation:

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

SS.56 · Answer

Let the value of a be 1.
b(31, 2011)= 1*( \frac{9}{10} )
31, 2012= 1* \frac{9}{10} * \frac{9}{10} 
31, 2013= 1* \frac{9}{10} * \frac{9}{10} * \frac{9}{10} =  \frac{729}{1000}

Option A
 \frac{b^3}{a^2} 
= \frac{729}{1000}

Answer A

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

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