Difficult Math Problem #123 - Algebra

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Difficult Math Problem #123 - Algebra

by 800guy » Fri Apr 27, 2007 12:46 pm
X percents of the rooms are suits, Y percent of the rooms are painted light blue. Which of the following best represents the least percentage of the light blue painted suits?

1) X-Y
2)Y-X +100
3)100X-Y
4)X+Y-100
e)100-XY


from diff math doc, oa coming when people respond with answers

THIS IS THE LAST QUESTION OF THE DIFFICULT MATH PROBLEMS DOC!!

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800guy wrote:X percents of the rooms are suits, Y percent of the rooms are painted light blue. Which of the following best represents the least percentage of the light blue painted suits?

1) X-Y
2)Y-X +100
3)100X-Y
4)X+Y-100
e)100-XY


from diff math doc, oa coming when people respond with answers

THIS IS THE LAST QUESTION OF THE DIFFICULT MATH PROBLEMS DOC!!
XnY = X + Y - XuY (set theory)

Min value of XnY is when XuY is maximum i.e. 100%

So is the min X + Y - 100 ?

Just scanning the answer choices, X-Y or Y-X would be incorrect
since we want to find rooms that are both "blue" and "suits". That
eliminates A and B.

C doesn't look good either since we are comparing percentages i.e.
100X would be a whole number and you can't technically subtract
a percentage value from this.

That leaves us with D and E :-).

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OA

by 800guy » Mon Apr 30, 2007 11:29 am
OA FOR LAST QUESTION IN DIFF MATH DOC!

Equation from set theory:

n(AUB)=n(A)+n(B)-n(A^B)

where,

A= % of rooms which are suites
B= % of rooms painted blue
A^B means the intersection of the two sets

Now in this case, what we need to find is n(A^B), therefore
n(A^B)=n(A)+n(B)-n(AUB)
= X + Y - n(AUB)

Now this would be least when n(AUB) is maximum, which would happen if these two kinds of rooms are only two kinds available, making n(AUB)=100

Therefore the answer should be X+Y-100