Problem Solving

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 is A

Chaitanya_1986 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 is A

let lights that were suppose to be on be represented by a and lights that were suppose to be off be represented by b, also assume total no. of lights to be x;

now as per the given condition; 80% of the lights are ON; which comprises of 90% of a and 40% of b;
.9a+.4b=.8x;....1)
as 80% of the lights are ON; therefore remaining 20% are off; which comprises of 10% of a and 60% of b;
hence .1a+.6b=.2x;---2)

multiply 2 with 9 we have;
.9a+5.4b=1.8x;
now subtract 1 from it we have;
5b=x;
put x=5b in 1 we have;
.9a=3.6b;
a=4b;
therefore; required percentage is
(.4b/4b)*100=10; hence A

Chaitanya_1986 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 is A

Like many GMAT problems, the one above can be solved with clever reasoning and perhaps some trial and error.

Let total bulbs = 100.
Number of bulbs on = .8*100 = 80.
Since 80 bulbs are on, and 90% of the bulbs expected to be on are in fact on, the number of bulbs expected to be on must be very high.
90 bulbs expected to be on would be too high, because with 10% of these switched off, the number left on would be 90-9 = 81, which is too many.

Let bulbs expected to be on = 80.
10% of these off = 8 off, 72 on.

Bulbs expected to be off = 100-80 = 20.
40% of these on = 8 on, 12 off.

Total on = 72+8 = 80. This works.

Thus, of the 80 bulbs that are on, 8 were expected to be off.
8/80 = 10%.

The correct answer is A.

Hi manpsingh87,

You wrote: now as per the given condition; 80% of the lights are ON; which comprises of 90% of a and 40% of b. Can you tell me where are we getting the 90% and the 40% from?

Thanks,
B

boazkhan wrote:Hi manpsingh87,

You wrote: now as per the given condition; 80% of the lights are ON; which comprises of 90% of a and 40% of b. Can you tell me where are we getting the 90% and the 40% from?

Thanks,
B

read the stem of the question carefully.. it states..!!!

If 40% of lights that were expected to be off, were in fact on, and 10% of lights that were expected to be on

which means, lights that are on are 40% of b; and 90% of a (as 10% are off)...!!!

boazkhan wrote:Hi manpsingh87,

You wrote: now as per the given condition; 80% of the lights are ON; which comprises of 90% of a and 40% of b. Can you tell me where are we getting the 90% and the 40% from?

Thanks,
B

To solve algebraically, I would use the following approach:

Assume 100 bulbs.

Let x = bulbs expected to be off.
Since 40% of these bulbs are on, the number on = .4x.

Let 100-x = bulbs expected to be on.
Since 10% of these bulbs are off, the number on = .9(100-x).

Since 80 bulbs are on, we get:
.4x + .9(100-x) = 80
4x + 900 - 9x = 800
-5x = -100
x = 20.

Thus, 20 bulbs were expected to be off.
Of these 20 bulbs, the number on = .4*20 = 8.
Thus, of the 80 bulbs that are on, the percentage that were expected to be off = 8/80 = 10%.

can we solve it using any other method..

what about table method?

Yes, it can be solved by the table method

Expected On Expected Off
Light On .9(y) .4(x) 80
Light Off .1(y) .6(x) 20
------------ ------------- ---
y x 100

.9y+.4x = 80
.1y+.6x = 20

Solve for x. You get x=20. Therefore (.4(20)/80)*100 = 10