A set consists of 15 integers, each of which is multiple of

Question

A set consists of 15 integers, each of which is multiple of 5, or even or both.If 12 numbers are multiple of 5, 7 numbers are even, how many of the numbers are multiple of 10?

A.0 
B.3
C.4
D.5
E.6

Please explain ..

carpeD · Answer

this is set theory in disguise.

multiple of 5 = 12
multiple of 2 = 7
total number = 15

so multiple of both 2 and 5 is (12 + 7 - 15 ) = 4. Answer C

beckee529 · Answer

i get C for the answer:

from the question, we know that

total in set = 15 elements (all of which is a multiple of 5, even or both)
in other words, set consists of numbers odd/even multiple of 5, or some other even

multiple of 5 = 12 elements
multiple of 2 = 7 elements (may also be even multiple of 5 which is also multiple of 10)

Not sure if my logic is correct but -
If we know that 12 out of 15 is a multiple of 5 then we know that the remaining 3 must be even but not multiple of 10. We are told that there are 7 total even numbers so 7 - 3 =  4 multiples of 10  <-- C

(even though the question is not asking) final we have:
8 odd multiples of 5
4 even multiples of 5 (so also mult of 10)
3 even non-multiple of 5

kollisat · Answer

To be multiple of 10, numbers have to be even and multiple of 5. Essentially, the problem is asking for how many numbers belong to "both" category.

Total = A + B + Neither - Both
15 = 12 + 7 + 0 - Both
Both = 19 - 15 = 4
Ans C.

A set consists of 15 integers, each of which is multiple of

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