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# The price of a car was m dollars. It then depreciated by x%. Later, it

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Math Expert
Joined: 02 Sep 2009
Posts: 54493
The price of a car was m dollars. It then depreciated by x%. Later, it  [#permalink]

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10 Apr 2019, 23:42
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Difficulty:

55% (hard)

Question Stats:

65% (03:03) correct 35% (03:54) wrong based on 17 sessions

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The price of a car was m dollars. It then depreciated by x%. Later, it appreciated by y% to n dollars. If there are no other changes in the price and if $$y = \frac{x}{1 - \frac{x}{100}}$$, then which one of the following must n equal?

(A) 3m/4
(B) m
(C) 4m/3
(D) 3m/2
(E) 2m

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Joined: 01 May 2017
Posts: 82
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Re: The price of a car was m dollars. It then depreciated by x%. Later, it  [#permalink]

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11 Apr 2019, 01:38
The price of a car was m dollars. It then depreciated by x%. Later, it appreciated by y% to n dollars. If there are no other changes in the price and if $$y=x/(1−x/100)$$, then which one of the following must n equal?

(A) 3m/4
(B) m
(C) 4m/3
(D) 3m/2
(E) 2m

depreciated by x% implies, $$m*(1-x/100)$$
Appreciated by y% and final value id n
$$m*(1-x/100)*(1+y/100)$$ = n
$$y = x/(1-x/100)$$

$$m*(1-x/100)(100+(x/(1-x/100))$$ = n
m = n

Option B is correct
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Joined: 15 Aug 2018
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The price of a car was m dollars. It then depreciated by x%. Later, it  [#permalink]

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13 Apr 2019, 10:16
Bunuel wrote:
The price of a car was m dollars. It then depreciated by x%. Later, it appreciated by y% to n dollars. If there are no other changes in the price and if $$y = \frac{x}{1 - \frac{x}{100}}$$, then which one of the following must n equal?

(A) 3m/4
(B) m
(C) 4m/3
(D) 3m/2
(E) 2m

Here's my approach

Setting up the problem algebraically:
$$m * ( 1- \frac{x}{100})*( 1+ \frac{y}{100}) = n$$

We know:
$$y= \frac {x}{1- \frac {x}{100}}$$

So we can substitute for $$y$$:
$$m * ( 1- \frac{x}{100})*( 1+ \frac{\frac {x}{1- \frac {x}{100}}}{100}) = n$$

$$m * ( 1- \frac{x}{100})+ ( 1- \frac{x}{100})(\frac {x}{1- \frac {x}{100}}*\frac {1}{100}) = n$$

$$m * ( 1- \frac{x}{100})+ (x*\frac {1}{100}) = n$$

$$m * 1- \frac{x}{100}+ \frac {x}{100} = n$$

Some maths later:

$$m = n$$
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Re: The price of a car was m dollars. It then depreciated by x%. Later, it  [#permalink]

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13 Apr 2019, 19:11
Bunuel wrote:
The price of a car was m dollars. It then depreciated by x%. Later, it appreciated by y% to n dollars. If there are no other changes in the price and if $$y = \frac{x}{1 - \frac{x}{100}}$$, then which one of the following must n equal?

(A) 3m/4
(B) m
(C) 4m/3
(D) 3m/2
(E) 2m

Let’s simplify y first: Multiply x/(1 - x/100) by 100/100, and we have: y = 100x/(100 - x).

We can create the equation:

m * (1 - x/100) * ( 1 + y/100) = n

m * (100 - x)/100 * (1 + [100x/(100 - x)]/100) = n

m * (100 - x)/100 * (1 + x/(100 - x)) = n

m * (100 - x)/100 * (100 - x + x)/(100 - x) = n

m * (100 - x)/100 * 100/(100 - x) = n

m = n

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Re: The price of a car was m dollars. It then depreciated by x%. Later, it   [#permalink] 13 Apr 2019, 19:11
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# The price of a car was m dollars. It then depreciated by x%. Later, it

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