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good one

by pradeepkaushal9518 » Thu Apr 15, 2010 7:45 am
At a local appliance manufacturing facility, the workers received a 20% hourly pay raise due to extraordinary performance. If one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged, by approximately what percent would he reduce the number of hours that he worked?
A) 83%
B) 80%
C) 20%
D) 17%
E) 12%
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by itsratul » Thu Apr 15, 2010 8:14 am
17%
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by harshavardhanc » Thu Apr 15, 2010 8:56 am
pradeepkaushal9518 wrote:At a local appliance manufacturing facility, the workers received a 20% hourly pay raise due to extraordinary performance. If one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged, by approximately what percent would he reduce the number of hours that he worked?
A) 83%
B) 80%
C) 20%
D) 17%
E) 12%
product constancy.

A * B = C

if A gets increased by 1/5th (20%), B should be reduced by 1/6th (16.66%) to get C.

{1+(1/5)} A * {1-(1/6)} B = (6/5) A * (5/6) B = A * B = C.
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by eaakbari » Thu Apr 15, 2010 9:04 am
x be salary, h be old no. of hours
H be new no. if hours

xh=1.2xH

H/h= 1/1.2
H/h = 83%

That means new hours are 83 percent of old hours , hence old should be reduced by 17%
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by neoreaves » Thu Apr 15, 2010 9:06 am
we can just use numbers to solve this because we are asked about the increase in percentage which should be true no matter what numbers we take


old rate = $100/hr
new rate = $120/hr

old time = 6 hrs

old wage = 6x100 = $600
new wage = $600

new time = 600/12 = 5 hrs

percent reduction = 1/6 = 17%
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