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## Over lapping problem......again, kinda tricky

tagged by: Mr.Hollywood

This topic has 2 expert replies and 1 member reply
Mr.Hollywood Rising GMAT Star
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Over lapping problem......again, kinda tricky Tue Mar 13, 2012 1:48 pm
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
A school has 3 classes, math class has 14 students. English class has 10 students, PE class has 11 students. There are 20 students taking only one class, 3 students are taking all three classes. How many students are taking two classes?

A) 6
B) 3
C) 9
D) 18

I personally think it's A

What do you guys think?

Thanked by: icemanKK
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Mike@Magoosh GMAT Instructor
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Tue Mar 13, 2012 5:17 pm
Hi, there. I'm happy to help with this.

First of all, I'll say: this question is unusual in form, because GMAT PS questions invariably have 5 answer choices. Also, the numbers seem a bit smaller and simpler than I've seen on analogous GMAT questions. Also, on the real GMAT, answer choices are almost always in numerical order. I am suspicious of the source of these questions.

A school has 3 classes, math class has 14 students. English class has 10 students, PE class has 11 students. That's a sum of 14 + 10 + 11 = 35. This sum counts "doublers" twice, and counts "triplers" three times.

There are 20 students taking only one class. Remove the singletons, 35-20 = 15, and that number represents the doublers counted twice and the triplers counted three times.

3 students are taking all three classes. There are three triplers, so they count for 9 in the total. 15 - 9 = 6, which represents the doublers counting twice. That means, there must be 3 doublers.

Does all this make sense? Please let me know if you have any questions on what I've said.

Mike

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http://gmat.magoosh.com/

Thanked by: Mr.Hollywood
Mr.Hollywood Rising GMAT Star
Joined
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Wed Mar 14, 2012 12:02 pm
Mike@Magoosh wrote:
Hi, there. I'm happy to help with this.

First of all, I'll say: this question is unusual in form, because GMAT PS questions invariably have 5 answer choices. Also, the numbers seem a bit smaller and simpler than I've seen on analogous GMAT questions. Also, on the real GMAT, answer choices are almost always in numerical order. I am suspicious of the source of these questions.

A school has 3 classes, math class has 14 students. English class has 10 students, PE class has 11 students. That's a sum of 14 + 10 + 11 = 35. This sum counts "doublers" twice, and counts "triplers" three times.

There are 20 students taking only one class. Remove the singletons, 35-20 = 15, and that number represents the doublers counted twice and the triplers counted three times.

3 students are taking all three classes. There are three triplers, so they count for 9 in the total. 15 - 9 = 6, which represents the doublers counting twice. That means, there must be 3 doublers.

Does all this make sense? Please let me know if you have any questions on what I've said.

Mike
Thank you it's great. Although I'm not so sure about the "This sum counts "doublers" twice" part. Can you demonstrate a little further regarding the doublers? I do understand the triplers.

### GMAT/MBA Expert

Mike@Magoosh GMAT Instructor
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Wed Mar 14, 2012 2:02 pm
Mr.Hollywood wrote:
Thank you it's great. Although I'm not so sure about the "This sum counts "doublers" twice" part. Can you demonstrate a little further regarding the doublers? I do understand the triplers.
OK, let's be concrete.

Let's say:
A is taking Math
B is taking Math
C is taking English
D is taking PE
E is taking PE

F is taking Math and English
G is taking Math and English
H is taking Math and PE
I is taking Math and PE
J is taking English and PE

K is taking Math, English, and PE
L is taking Math, English, and PE

Here, I have marked the "singletons" in purple, the doublers in green, and the triplers in red.

Who is in Math? A, B, F, G, H, I, K, and L ---> 8 people
Who is in English? C, F, G, J, K, and L ---> 6 people
Who is in PE? D, E, H, I, J, K, and L ---> 7 people

8 + 6 + 7 = 21

That sum of 21 counts the two triplers (K & L) three time --- they are included on all three lines of the sums. The sum of 21 counts the 5 doublers (F, G, H, I, and J) each twice --- each one of those is included on two of the three lines of sums.

Thus (5 singletons) + 2*(five doublers) + 3*(two triplers) = 5 + 2*5 + 3*6 = 21

Does this make sense?

Mike

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