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A puzzle that I cannnot solve, please help!

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A puzzle that I cannnot solve, please help! [#permalink]

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New post 23 Aug 2008, 04:29
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I got trouble with this question, do you know about this?
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!

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Re: A puzzle that I cannnot solve, please help! [#permalink]

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New post 23 Aug 2008, 04:55
turtle0701 wrote:
Online discussion?
I got trouble with this question, do you know about this?
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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Re: A puzzle that I cannnot solve, please help! [#permalink]

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New post 23 Aug 2008, 09:04
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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.

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Re: A puzzle that I cannnot solve, please help! [#permalink]

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New post 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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New post 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

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New post 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 :)

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New post 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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New post 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!

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Re: A puzzle that I cannnot solve, please help! [#permalink]

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New post 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.

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Re: A puzzle that I cannnot solve, please help! [#permalink]

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New post 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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New post 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

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New post 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

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Re: A puzzle that I cannnot solve, please help! [#permalink]

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New post 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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New post 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

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Re: A puzzle that I cannnot solve, please help! [#permalink]

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New post 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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Re: A puzzle that I cannnot solve, please help!   [#permalink] 09 Aug 2009, 07:50
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