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A shopkeeper increases the price of an article by X%

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by guerrero » Sat Jun 22, 2013 9:32 am
A shopkeeper increases the price of an article by X% and then decreases it by X%. As a result the price of the article is reduced by $180. After one more such change the price is further reduced by $153 Find the original price of the article.

A. $1200
B. $1100
C. $1400
D. $1000
E. $1500

OAA
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by snigdha1605 » Sat Jun 22, 2013 10:13 am
Let the original price be P

New price P1 = P (1+x/100)(1-x/100)

Given p-p1= 180 -------eq a

Similarly,

p2 = p1(1+x/100)(1-x/100)

and p1-p2= 153----------eq b


From eq a and b we get,

p1= (17/20)*p

Replace this in eq a

p-p1 = 180
p - (17*p)/20 = 180

p= 1200
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by GMATGuruNY » Sat Jun 22, 2013 10:32 am
guerrero wrote:A shopkeeper increases the price of an article by X% and then decreases it by X%. As a result the price of the article is reduced by $180. After one more such change the price is further reduced by $153 Find the original price of the article.

A. $1200
B. $1100
C. $1400
D. $1000
E. $1500

OAA
Let F = the first price and S = the second price.

Both prices are reduced by the SAME FRACTION: up X% then down X%.
Let's call this fraction x/y.

When F is reduced by x/y, the decrease is equal to 180:
(x/y)F = 180
x/y = 180/F.

When S is reduced by x/y, the decrease is equal to 153:
(x/y)S = 153
x/y = 153/S.

Since x/y = 180/F and x/y = 153/S, we get:
180/F = 153/S
153F = 180S
17F = 20S
F/S = 20/17.

Since the ratio of F to S = 20x : 17x, and the difference between F and S is equal to 180, we get:
20x - 17x = 180
x = 60.

Thus:
F = 20x = 20*60 = 1200.

The correct answer is A.
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