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sets problem

by crazy4gmat » Thu Oct 09, 2008 8:53 am
Club Number of Students
Chess 40
Drama 30
Math 25

Table above shows the number of students in three clubs at McAuliffe School. Although no student is in all three clubs, 10 students are in both chess and drama, 5 students are in both chess and math, and 6 students are in both drama and math. How many different students are in the three clubs?
(A) 68
(B) 69
(C) 74
(D) 79
(E) 84
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by nikhilagrawal » Thu Oct 09, 2008 9:10 am
Is the answer 74
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by rabab » Thu Oct 09, 2008 9:17 am
IOM- C.

Only Chess = 40-10-5 = 25
Only Drama= 30-10-6 = 14
Only Math= 25-5-6= 14
Chess + Drama= 10
Chess + Math= 5
Drama + Math= 6

Totaling 74.

Hope it helps.
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by schumi_gmat » Thu Oct 09, 2008 12:05 pm
Chess + Drama= 10
Chess + Math= 5
Drama + Math= 6

Why are we considering the above 3 combinations?
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by Morgoth » Thu Oct 09, 2008 12:13 pm
We have to find the following

only drama + only chess + only math

use the sets formula

total = drama + chess + math - (drama & chess) - (math & chess) - (drama & math) - 2*(all three) + neither

Since we know that none of students attended all 3, we just have to find the total number of students.

total = 40+30+25-10-5-6 = 95 - 21 = 74

Hope this helps.
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by stop@800 » Thu Oct 09, 2008 10:50 pm
Morgoth wrote:We have to find the following

only drama + only chess + only math
Small correction:
We have to find All students.

only drama + only chess + only math
will be different from total.

In "only drama + only chess + only math"
we will not have any intersection
while in "Total"
we will count each intersection once.

Let me know if you want me to explain with venn diagram.

use the sets formula

total = drama + chess + math - (drama & chess) - (math & chess) - (drama & math) - 2*(all three) + neither

Since we know that none of students attended all 3, we just have to find the total number of students.

total = 40+30+25-10-5-6 = 95 - 21 = 74

Hope this helps.
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by stop@800 » Thu Oct 09, 2008 10:51 pm
Morgoth wrote:We have to find the following

only drama + only chess + only math
Small correction:
We have to find All students.

only drama + only chess + only math
will be different from total.

In "only drama + only chess + only math"
we will not have any intersection
while in "Total"
we will count each intersection once.

Let me know if you want me to explain with venn diagram.

use the sets formula

total = drama + chess + math - (drama & chess) - (math & chess) - (drama & math) - 2*(all three) + neither

Since we know that none of students attended all 3, we just have to find the total number of students.

total = 40+30+25-10-5-6 = 95 - 21 = 74

Hope this helps.
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by vivek.kapoor83 » Thu Oct 09, 2008 10:52 pm
So, stop acc to u ans is not 74 ?
coz they all are adding intersection part also.I got it 53 wihtout intersection.
pls explain by venn diagram also.
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by vivek.kapoor83 » Thu Oct 09, 2008 10:56 pm
or they all be added once. eg. Intersection of C & D = 10. first finding the students who are the part of chess only and drama only and then adding 10 to one of these, as those students will represent in some set.Is it correct.
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by stop@800 » Thu Oct 09, 2008 10:57 pm
Morgoth wrote:We have to find the following

only drama + only chess + only math
Small correction:
We have to find All students.

only drama + only chess + only math
will be different from total.

In "only drama + only chess + only math"
we will not have any intersection
while in "Total"
we will count each intersection once.

Let me know if you want me to explain with venn diagram.

use the sets formula

total = drama + chess + math - (drama & chess) - (math & chess) - (drama & math) - 2*(all three) + neither

Since we know that none of students attended all 3, we just have to find the total number of students.

total = 40+30+25-10-5-6 = 95 - 21 = 74

Hope this helps.
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by Morgoth » Thu Oct 09, 2008 11:10 pm
stop@800 wrote:
Morgoth wrote:We have to find the following

only drama + only chess + only math
Small correction:
We have to find All students.

only drama + only chess + only math
will be different from total.

In "only drama + only chess + only math"
we will not have any intersection
while in "Total"
we will count each intersection once.

Let me know if you want me to explain with venn diagram.

use the sets formula

total = drama + chess + math - (drama & chess) - (math & chess) - (drama & math) - 2*(all three) + neither

Since we know that none of students attended all 3, we just have to find the total number of students.

total = 40+30+25-10-5-6 = 95 - 21 = 74

Hope this helps.

You just repeated what I posted. Check the writing in bold.

If you are trying to say anything apart from that, pls do let me know.
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by stop@800 » Thu Oct 09, 2008 11:46 pm
vivek.kapoor83 wrote:So, stop acc to u ans is not 74 ?
coz they all are adding intersection part also.I got it 53 wihtout intersection.
pls explain by venn diagram also.
The ans is 74.
I just corrected one typo.
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by crazy4gmat » Fri Oct 10, 2008 12:23 am
Thanks for the explanation guys. I thought "How many different students are in the three clubs" meant students who were part of the individual clubs only and so was wondering when 53 was not in the answer
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by vivek.kapoor83 » Fri Oct 10, 2008 2:30 am
I am getting confuse now...
here we have to find the diff students So, we have to take out all the interactions. SO,53 it comes when we take out all intersections. Why you ppl are adding coz by Venn Diagaram. and we reduce the intersection part...then it comes out 53 or we need to add the intersection coz those 10 student have to come under Some group be it be chess or Drama...pls explain..i got it earlier but now confused.
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by raunekk » Fri Oct 10, 2008 2:31 am
ans 74
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