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Intern
Joined: 23 Aug 2008
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23 Aug 2008, 04:29
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Online discussion?
70 students are enrolled in Math, English, or German.
40 students are in Math
35 are in English
30 are in German.
15 are in all three.
How many of the students are enrolled in exactly two of the courses: Math, English and German?

My thinking is that:

Total = 40 + 35 + 30 = 105 students,
If we got 70 students studying only ONE course; and the number of students studying three courses at the same time will be counted 3 times. And the students studying two courses at the same time will be counted two times. Thus, 105 - 2A- 3*15 = 70 given that A as the students studying two courses at the same time; And finally, I cannot solve the puzzle due to negative result. Could you tell me where my wrong point is?
Thank you and looking forward to discussing more. :D
Have a nice weekend!
SVP
Joined: 07 Nov 2007
Posts: 1761
Location: New York

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23 Aug 2008, 04:55
turtle0701 wrote:
Online discussion?
70 students are enrolled in Math, English, or German.
40 students are in Math
35 are in English
30 are in German.
15 are in all three.
How many of the students are enrolled in exactly two of the courses: Math, English and German?

My thinking is that:

Total = 40 + 35 + 30 = 105 students,
If we got 70 students studying only ONE course; and the number of students studying three courses at the same time will be counted 3 times. And the students studying two courses at the same time will be counted two times. Thus, 105 - 2A- 3*15 = 70 given that A as the students studying two courses at the same time; And finally, I cannot solve the puzzle due to negative result. Could you tell me where my wrong point is?
Thank you and looking forward to discussing more. :D
Have a nice weekend!

best way to solve this problem is using venn diagram.

I got 5 as the answer. Is that correct?

Please post Maths questions in Math forum.
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Senior Manager
Joined: 09 Oct 2007
Posts: 459

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23 Aug 2008, 09:04
1
KUDOS
Agree on 5.

Formula is:
Sum of students in different clubs - students in 2 clubs - 2 (Students in all 3 clubs) = Total

You only subtract one time the students who are in 2 clubs because you want to count them once, and you subract twice those who are in all three because you counted them 3 times and want to keep only one.

Therefore:

105 - x - 2(15) = 70
105 - 70 - 30 = 5 = x

Hope that helps.
SVP
Joined: 07 Nov 2007
Posts: 1761
Location: New York

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23 Aug 2008, 11:03
asdert wrote:
Agree on 5.

Formula is:
Sum of students in different clubs - students in 2 clubs - 2 (Students in all 3 clubs) = Total

You only subtract one time the students who are in 2 clubs because you want to count them once, and you subract twice those who are in all three because you counted them 3 times and want to keep only one.

Therefore:

105 - x - 2(15) = 70
105 - 70 - 30 = 5 = x

Hope that helps.

good job.. this is the fastest way..

+1 for you.,
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Manager
Joined: 23 Aug 2008
Posts: 63

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23 Aug 2008, 13:15
I used a venn diagram too, but came to the answer of 20.
For me, total # of students = 105, broken down as
Students enrolled in only one course = 70 (the non-intersecting area of the diagram)
Students enrolled in three courses = 15 (the central area of the diagram)
Students enrolled in two courses = 20 (remainder)

Did I go wrong somewhere too? I think my interpretation of the data is different... Is the correct answer available anywhere? Thanks
Senior Manager
Joined: 09 Oct 2007
Posts: 459

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23 Aug 2008, 21:53
prince13 wrote:
I used a venn diagram too, but came to the answer of 20.
For me, total # of students = 105, broken down as
Students enrolled in only one course = 70 (the non-intersecting area of the diagram)
Students enrolled in three courses = 15 (the central area of the diagram)
Students enrolled in two courses = 20 (remainder)

Did I go wrong somewhere too? I think my interpretation of the data is different... Is the correct answer available anywhere? Thanks

I think your problem is that the total of students is 70 (not 105), including those who attend 2 or 3 clubs.

x2suresh, I thought you were done with your test, weren't you? Thanks for the +1
SVP
Joined: 07 Nov 2007
Posts: 1761
Location: New York

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24 Aug 2008, 08:43
asdert wrote:
prince13 wrote:
I used a venn diagram too, but came to the answer of 20.
For me, total # of students = 105, broken down as
Students enrolled in only one course = 70 (the non-intersecting area of the diagram)
Students enrolled in three courses = 15 (the central area of the diagram)
Students enrolled in two courses = 20 (remainder)

Did I go wrong somewhere too? I think my interpretation of the data is different... Is the correct answer available anywhere? Thanks

I think your problem is that the total of students is 70 (not 105), including those who attend 2 or 3 clubs.

x2suresh, I thought you were done with your test, weren't you? Thanks for the +1

No buddy.. Now trying to improve CR and RC..
I am planning to give in october.
When are you planning?
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Senior Manager
Joined: 09 Oct 2007
Posts: 459

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24 Aug 2008, 13:43
I'm about to schedule for Sept. 12. I feel ready sometimes, and not so much some others :S I'm sure you know how it is. Good luck to us all!
Director
Joined: 01 Aug 2008
Posts: 692

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24 Aug 2008, 15:06
Another way of solving this (still using Venn diagrams only)

lets say,

M' is maths only,
G' is German only,
E' is English only

a is M' ^ E'
b is M' ^ G'
c is E' ^ G'

and we need to find out the value of (a+b+c)

so now
M'+G'+E' + (a+b+c)+15 = 70 (all the students)

M'+G'+E'+ (a+b+c) = 55. ----- (main equation)

M'+a+b+15 = 40 ---> a+b= 25-M' ---- (1)
E'+a+c+15 = 35 ----> a+c= 20-E' ----(2)
G'+b+c+15 = 30 -----> b+c = 15-G' ----(3)

add equations (1), (2) and (3), you get
M'+G'+E' = 60 - 2(a+b+c)

so substitute above M+G+E value in the main equation

you get

60-2(a+b+c)+(a+b+c) = 55.

(a+b+c) = 5. which represents all the students taking exactly two classes.

Hope that helps.

Thanks.
Manager
Joined: 18 Jul 2009
Posts: 166
Location: India
Schools: South Asian B-schools

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08 Aug 2009, 22:22
let
only (math+english) = x;
only (math+german)=y;
only(german+english)=z;

hence set eqn becomes 40+30+35-x-y-z+15=70
40+30+35-(x+y+z)+15=70
(x+y+z)= 50

hence students studying two courses (x+y+z) = 50
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Bhushan S.
If you like my post....Consider it for Kudos

Intern
Joined: 03 Jun 2009
Posts: 49

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08 Aug 2009, 23:41
bhushan252 wrote:
let
only (math+english) = x;
only (math+german)=y;
only(german+english)=z;

hence set eqn becomes 40+30+35-x-y-z+15=70
40+30+35-(x+y+z)+15=70
(x+y+z)= 50

hence students studying two courses (x+y+z) = 50

-----------------
Hi bhushan,

this 50 ie., (x+y+z) consists of the number of students who study only (math+eng+german)3 times more.

Hence the no. of students who study exactly 2 courses is (x+y+z)-3(math intersection english intersection german)

= 50 - 3(15) =5
Intern
Joined: 09 Aug 2009
Posts: 46

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09 Aug 2009, 01:27
hello all,

i just got the equ like this.
40+30+35-x-y-z-15 = 70
105-15-x-y-z=70
x+y+z= 20

Please let me know wat is the problem with this approach

/Prabu
CEO
Joined: 17 Nov 2007
Posts: 3525
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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09 Aug 2009, 01:51
prabu wrote:
hello all,

i just got the equ like this.
40+30+35-x-y-z-15 = 70
105-15-x-y-z=70
x+y+z= 20

Please let me know wat is the problem with this approach

/Prabu

What is reasoning behind the formula? When you added 40+30+35 you added three times students who go to 3 courses. So, you need to subtract two times the number of these students. Use 15*2 instead of 15
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Intern
Joined: 09 Aug 2009
Posts: 46

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09 Aug 2009, 02:16
walker wrote:
prabu wrote:
hello all,

i just got the equ like this.
40+30+35-x-y-z-15 = 70
105-15-x-y-z=70
x+y+z= 20

Please let me know wat is the problem with this approach

/Prabu

What is reasoning behind the formula? When you added 40+30+35 you added three times students who go to 3 courses. So, you need to subtract two times the number of these students. Use 15*2 instead of 15

Hi Walker,

I am assuming that 40 students enrolled in math that includes only math , math+eng, math+german and all the 3.
So i summed up 40+30+35 -x(both math+eng) -y(both math+german) -z(both eng+german) -15(all the 3) = total(70).

please let me know ver i'm going wrong.

/Prabu
CEO
Joined: 17 Nov 2007
Posts: 3525
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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09 Aug 2009, 07:50
prabu wrote:
Hi Walker,

I am assuming that 40 students enrolled in math that includes only math , math+eng, math+german and all the 3.
So i summed up 40+30+35 -x(both math+eng) -y(both math+german) -z(both eng+german) -15(all the 3) = total(70).

please let me know ver i'm going wrong.

/Prabu

In your formula I marked students who enrolled in math+eng+german:

40 (1 time)+30 (1 time) +35 (1 time) -x(both math+eng) -y(both math+german) -z(both eng+german) -15(all the 3 (? times)) = total(70 (1 time)).

to balance your equation you need subtract 2 times from left side:

40 (1 time)+30 (1 time) +35 (1 time) -x(both math+eng) -y(both math+german) -z(both eng+german) -2*15(all the 3 (-2 times)) = total(70 (1 time)).
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