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Expert replies
by eshwarjayanth » Thu Aug 18, 2011 2:33 am
80% of the lights in a hotel were on at 8.00 pm on some evening. If 40% of lights that were expected to be off, were in fact on, and 10% of lights that were expected to be on, were in fact off; what percent of the lights that are on, are the lights that were not expected to be on?


A. 10
B. 12
C. 100/9
D. 8
E. 18

OA - A
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Source: — Problem Solving |

by GMATGuruNY » Thu Aug 18, 2011 3:25 am
I posted two solutions -- the first non-algebraic, the second algebraic -- here:

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by saketk » Thu Aug 18, 2011 8:47 am
eshwarjayanth wrote:80% of the lights in a hotel were on at 8.00 pm on some evening. If 40% of lights that were expected to be off, were in fact on, and 10% of lights that were expected to be on, were in fact off; what percent of the lights that are on, are the lights that were not expected to be on?


A. 10
B. 12
C. 100/9
D. 8
E. 18

OA - A


100 bulbs
80 are ON
20 are OFF
if x lights were expected to be OFF and out of X, 0.4x are ON
So total OFF = 0.6x
So 0.4X are the lights that are ON and were not expected to be ON. -- WE HAVE TO FIND THIS

also,
if Y lights were expected to be ON and out of Y, 0.1Y are OFF
so total ON = 0.9Y

Total ON = 0.4X+0.9Y =80, 4X+9Y =800
and total OFF = 0.6x+0.1Y = 20, 6x+y= 200

or,
54x+9y =1800 (multiply 9 both sides)
4x+9y = 800
Subtracting the equations we will get
50x = 1000
x= 20
0.4X = 8

8/80 = 10% :)
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by force5 » Thu Aug 18, 2011 10:04 am
explained well +1 A
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