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!!
Difficult Math Problem #123 - Algebra
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- jayhawk2001
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XnY = X + Y - XuY (set theory)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!!
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 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
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