sanju09 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
We can set up two equations with 2 unknowns to solve the system.
If 40% of the lights expected to be off were actually only, then 60% of the lights expected to be off were actually off.
If 90% of the lights expected to be on were actually on, then 10% of the lights expected to be on were actually off.
If we let the total number of lights = 100, we know that 80 are on and 20 are off.
Letting:
x = expected on
y = expected off
We then get:
.9x + .4y = 80
and
.1x + .6y = 20
and of course
x + y = 100
Now we can solve for x and y. Let's multiply the second equation by 10:
x + 6y = 200
x = 200 - 6y
Subbing into our 3rd equation:
(200 - 6y) + y = 100
200 - 5y = 100
100 = 5y
20 = y
Therefore, 20 lights were expected to be off and, since x + y = 100, 80 lights were expected to be on.
Back to the original question:
What percent of the lights that are on, are the lights that were not expected to be on?
% = part/whole * 100%
The whole = lights that are on = 80
The part = the lights that are on which weren't expected to be on = 40%(20) = 8
So, % = 8/80 * 100% = 1/10 * 100% = 10%... choose A.
