## Set question - 4GMAT

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### Set question - 4GMAT

by netigen » Sat May 24, 2008 12:50 pm
In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for none of the three subjects?

A. 19
B. 41
C. 21
D. 57
E. 26

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by chidcguy » Sat May 24, 2008 1:47 pm

Replace & with set intersection symbol and the underline for negation

The question is asking for (P & C & M ) = Total - (P U C U M)

(P U C U M) = P + C + M - (P & C ) - (C & M) - ( P& M) - ( P & C & M )

= 60 + 24 + 17 - 10 -3 -8 -1 =79

120 - 79 = 41 is the number of students who did not take any course. Did you miss the P & C & M, which is the number 70 divisible by 2,5 & 7?

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by netigen » Sat May 24, 2008 2:08 pm
you intersection numbers are off for P & C

P & C = 12

Also, isn't the formula for finding number of students in at least one of the set this -

(P U C U M) = P + C + M - (P & C ) - (C & M) - ( P& M) + 2( P & C & M )

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by chidcguy » Sat May 24, 2008 2:27 pm
I have not only got the number wrong, but also got the formula wrong (see attachment). I am glad that I got it now. I have a huge problem with getting the numbers correct on my scrap sheet. Originally I calculated it to 12 and when the math was being done, it ended up being 10.

60 + 24 + 17 - 12 -3 -8 +1 = 79

120 -79 = 41

If every thing now is squared and B is the correct answer, I hope I get lucky like this on the real exam
Attachments Last edited by chidcguy on Sat May 24, 2008 3:14 pm, edited 1 time in total.

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by netigen » Sat May 24, 2008 2:39 pm
I found the following formulas on Preeti's site - https://gmat-maths.blogspot.com/2008_05_01_archive.html. Help me with the difference between 1 and 6

1. For 3 sets A, B, and C: P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

2. No of persons in exactly one set:
P(A) + P(B) + P(C) – 2P(AnB) – 2P(AnC) – 2P(BnC) + 3P(AnBnC)

3. No of persons in exactly two of the sets: P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC)

4. No of persons in exactly three of the sets: P(AnBnC)

5. No of persons in two or more sets: P(AnB) + P(AnC) + P(BnC) – 2P(AnBnC)

6. No of persons in atleast one set:
P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)

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by luvaduva » Sat May 24, 2008 2:41 pm
I normally draw the old trusty VENN diagrams for these.

Again, hehe, a picture is worth a thousand words so I made a picture. I do these in the following steps:

1) Determine how many students are in Physics, Chemistry, and Math => 60, 24, and 17.

2) Determine the overlap between each of these. In otherwords, how many students are taking two classes => 12, 8, and 3

3) Determine how many students are taking 3 classes => 1

4) Subtract (3) from each item in (2) and re-fill in (2). Students taking 2 classes becomes=> 11, 2, and 7.

5) Subtract the redundancies of (4) and (3) from the values in (1) => 41, 10, and 7.

6) Total (5), (4), and (1) and subtract it from 120 => 120 - 79 = 41.

HTH

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by netigen » Sat May 24, 2008 2:52 pm
I agree Venn diagrams are the best way to go but knowing the right formulas can save me some precious seconds in the test.

Any help with the question above is appreciated.

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by luvaduva » Sat May 24, 2008 3:35 pm
netigen wrote: 6. No of persons in atleast one set:
P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)
I think this is a mistake. Number of persons in at least one set is just #1.

Also, notice that there is a space between 2 and P.

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