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
I think the correct answer should be 40 which is not one of the choices. Please help.
Set question  4GMAT
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Is the answer B?
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?
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?
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 )
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 )
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.
The new answer would be
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
The new answer would be
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

 Picture 1.png (29.61 KiB) Viewed 150 times
Last edited by chidcguy on Sat May 24, 2008 3:14 pm, edited 1 time in total.
I found the following formulas on Preeti's site  https://gmatmaths.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)
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)
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 refill 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
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 refill 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
I think this is a mistake. Number of persons in at least one set is just #1.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)
Also, notice that there is a space between 2 and P.