sales problem

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sales problem

by hemant_rajput » Thu Jan 17, 2013 8:01 am
Q16 . The owner of an art shop conducts the business in the following manner. Every once in a while he raises his prices by x%, then a while later he reduces all the new prices by x%. After one such up-down cycle, the price of a painting decreased by 441. after a second up-down cycle, the painting was sold for 1944.81 . What was the original price of the painting?
a. 2756.25
b. 2256.25
c. 2500
d. 2000
[spoiler]oa : a[/spoiler]
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by GMATGuruNY » Thu Jan 17, 2013 8:22 am
hemant_rajput wrote:Q16 . The owner of an art shop conducts the business in the following manner. Every once in a while he raises his prices by x%, then a while later he reduces all the new prices by x%. After one such up-down cycle, the price of a painting decreased by 441. after a second up-down cycle, the painting was sold for 1944.81 . What was the original price of the painting?
a. 2756.25
b. 2256.25
c. 2500
d. 2000
[spoiler]oa : a[/spoiler]
We can plug in the answers, which represent the original price.

In the first cycle, the original price is reduced by 441.
Since the STARTING PRICE in the second cycle is lower, the REDUCTION in the second cycle must be LESS than 441.
Thus, the original price must be somewhat less than 1944 + 2(441) = 2826.
The most likely answer choice is A.

As we evaluate the answers, we must bear in mind the following:
In each cycle, there is a STARTING PRICE and a REDUCTION.
From one cycle to the next, the STARTING PRICE and the REDUCTION must decrease by the SAME PERCENTAGE.
If the STARTING PRICE from the first cycle to the second cycle decreases by 10%, then the REDUCTION from the first cycle to the second cycle must also decrease by 10%.

Answer choice A: 2756.25
Percent decrease in the STARTING PRICE from the first cycle to the second cycle = 441/2756.25 * 100 = 16%.
Starting price in the second cycle = 2756.25 - 441 = 2315.25.
Reduction during the second cycle = 2315.25 - 1944.81 = 370.44.
Percent decrease in the REDUCTION from the first cycle to the second cycle = (441 - 370.44)/441 * 100 = 16%.
Success!

The correct answer is A.
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by Tommy Wallach » Fri Jan 18, 2013 1:14 am
As far as I can see, the method proposed by Guru is the only way to do this question, which makes it unlikely as a real GMAT question. I say this because a real GMAT question would also give you the option to solve it algebraically, like so:

P * x = P - 441

P * x^2 = 1944.81

At first glance, this may seem like an oversimplification, but actually, we can express the fact that the percent goes up and down by the same amount as a single variable, because it would be distinct depending on the value of x (for example, if x = 10, we would multiply by 1.1 * . 9 = .99; if x = 20, we would multiply by 1.2 * .8 = .96).

The two equation I have here would usually solve to a friendly quadratic, except these numbers are so ridiculously terrible, the quadratic is insoluble without the help of a friendly website! (https://www.math.com/students/calculator ... dratic.htm)

Just for fun, here's the math!:

P = 1944.81/x^2

1944.81/x^2 * x = 1944.81/x^2 - 441

1944.81/x = 1944.81/x^2 - 441

1944.81x = 1944.81 - 441x^2

441x^2 + 1944.81x - 1944.81 = 0

If you enter this data into the website posted above, it'll solve you to:

x = .84

And because: 1/.84 = 1.19 (approximate)

All we have to do is one last calculation:

1944.81 * 1.19 * 1.19 = 2754 (approximate)

Phew!

-t
Tommy Wallach, Company Expert
ManhattanGMAT

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