GMAT Math

In a class of 100 students, numbered 1 to 100, all the even numbered students opt physics, all the students whose number is divisible by 3 opt Chemistry and all the students whose number is divisible by 5 opt Mathematics. How many students opt none of these subjects.
A) 3
B) 16
C) 26
D) 50
E) 74

kaarthikvs wrote:In a class of 100 students, numbered 1 to 100, all the even numbered students opt physics, all the students whose number is divisible by 3 opt Chemistry and all the students whose number is divisible by 5 opt Mathematics. How many students opt none of these subjects.
A) 3
B) 16
C) 26
D) 50
E) 74

We have asked to select the number of numbers that are not divisible by 2, 3, or 5

Numbers divisible by 2 ----> 100/2 ------> 50 All even numbers 2,4,......100.

Numbers divisible by 3 ----> (100/3)- even multiples of 3 ------> 17 the pattern of multiples of 3 is 3,6,9,12,....99 i.e. odd, even, odd, even,........odd so their will be 17 odd numbers and 16 even numbers. Here we will only consider odd multiples because even multiples we have already counted as 2's multiples.

Numbers divisible by 5 ----> only 7 cases -----> 5,25,35,55,65,85,95 ------> Rest all multiples are either of 2 or 3 or both and so not to be counted again.

Number of number divisible by 2,3,or 5 between 1 and 100 (both inclusive) = 50 + 17 + 7 = 74
Number of number not divisible by 2,3,or5 between 1 and 100 (both inclusive) = 26 ---Answer

Regards,

Abhijit

Thanks a lot Abhijit.
Numbers divisible by 2 ----> 100/2 ------> 50
Numbers divisible by 3 ----> (100/3)- both even and odd multiples of 3 ------> 33
Numbers divisible by 5 ----> All 20 cases
Is there any way to find the solution using the above three data (by using n(AUBUC) sets formula) ?

kaarthikvs wrote:Thanks a lot Abhijit.
Numbers divisible by 2 ----> 100/2 ------> 50
Numbers divisible by 3 ----> (100/3)- both even and odd multiples of 3 ------> 33
Numbers divisible by 5 ----> All 20 cases
Is there any way to find the solution using the above three data (by using n(AUBUC) sets formula) ?

The Formula is TOTAL = A + B + C -(AΩB)-(BΩC)-(AΩC)+ AUBUC + NONE

Here We have to find NONE ; How many students opt none of these subjects.

A = 50 = number of students opt for physics i.e. number of all multiples of 2
B = 33 = number of students opt for chemistry i.e. number of all multiples of 3
C = 20 = number of students opt for mathematics i.e. number of all multiples of 5
AΩB = 16(Take LCM of 2 and 3 and divide 100 by that) = number of students opt for physics AND Chemistry i.e. number of all multiples of 2 and 3
BΩC = 6(Take LCM of 3 and 5 and divide 100 by that) = number of students opt for chemistry AND maths i.e. number of all multiples of 3 and 5
AΩC = 10(Take LCM of 2 and 5 and divide 100 by that) = number of students opt for physics AND maths i.e. number of all multiples of 2 and 5
AΩBΩC = 3(Take LCM of 2,3,and 5 and divide 100 by that) number of students opt for physics AND maths AND chemistry i.e. number of all multiples of 2 and 5 and 3

so 100 = 50 + 33 + 20 - 16 - 6 - 10 + 3 + NONE
100 = 106 - 32 + NONE ---------> 100 = 74 + NONE -------> NONE = 26

If you need clear understanding of SET concept, Revisit CBSC Class XI Maths Book(Volume I) written by Prof. R.D.Sharma

Regards,

Abhijit.

Great stuff on using sets to solve this one - I love the discussion.

And just for anyone who's stymied by the set theory on this and looking for an alternative, I'd check out the answer choices and what the question is *really* asking: "How many numbers between 1 and 100 are not divisible by 2, 3, or 5?"

Note that half the numbers are even, alone, so the number has to be less than 50, which only leaves you with three choices:

3
16
26

And it's not 3, since you can easily list more than 3 prime numbers (other than 2, 3, 5). So since the only plausible options are 16 and 26, if you can find a 17th number by listing out numbers that fit the bill, you're done since you've eliminated 16. And that's not that hard to do just by thinking of 1 and prime numbers:

1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43...

By this point if you've listed 12 qualifying numbers before you even got halfway to the limit, it's pretty clear that it's more than 16, so it has to be 26 because that's the only remaining possible answer.

I think Bryan's answer is the best so far. But I challenge: give the exact answer, if the numbers go from 1 to 1000, divisibility by 2,3, 5, and 7.
misterholmes
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