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# A part time employee whose hourly wage was increased by 25 %

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Senior Manager
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A part time employee whose hourly wage was increased by 25 % [#permalink]

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21 Nov 2003, 12:28
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A part time employee whose hourly wage was increased by 25 % decided to reduce the number of hours workd per week so that the employee's total weekly income whould remain unchgned .By wht percent should the number of hours worked be reudced?
A. 12.5%
B. 20%
C 25%
D. 50%
E. 75%
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shubhangi

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21 Nov 2003, 12:54
yes, I guess so. m I missing something?

again, I tried putting in values..

normal: 1 day - 10 hrs. if $10/hr, then wage =$100. 7 days = \$700

increased: 1 day - (10-X) hrs
wage = (10-X)*12.5 . 7 days = 7*(10-X)*12.5 = 700
gives X = 2. ....decreased 20%

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21 Nov 2003, 12:59
20% is right.. but guess its messed up.. any short or clear.. way?
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shubhangi

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21 Nov 2003, 13:17
consider,
hours = h1 and h2
wage = w1 and w2
x = number of hrs reduced

h1*w1 = h2*w2
= (h1-x) * 1.25 w1
gives,
1.25x = .25h1
x/h1 = 25/125 = 1/5 = .2 -->20%
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21 Nov 2003, 14:25
shubhangi wrote:
20% is right.. but guess its messed up.. any short or clear.. way?

For Rate x Hours = constant means Rate and Hours are inversely proportional.

Hence if rate is multiplied by 5/4 (25% increase) hours must be multiplied by the reciprocal or 4/5 (a 20% decrease) to maintain inverse proportionality.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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21 Nov 2003, 14:31
20% is right.. but guess its messed up.. any short or clear.. way?

Well, yes, sort of. See enough of these problems and you start to learn the "inverses" that return you to your starting point:

20% rise 16.666% drop
25% rise 20% drop
33.3333% rise, 25% drop
50% rise 33.3333% drop
100% rise 50% drop

Make sense?
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# A part time employee whose hourly wage was increased by 25 %

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