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There are 10 people studying A, 11 people studying B and 14 [#permalink ]
25 Feb 2006, 21:06

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There are 10 people studying A, 11 people studying B and 14 people studying C. The number of people learning just one is 20. The number of people learning all three is 3. How many people only learn two of three?

Current Student

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10+11+14=35 total
35-20-3= 12
What am I missing here?

Current Student

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Professor wrote:

GMATT73 wrote:

10+11+14=35 total 35-20-3= 12 What am I missing here?

deduct 3 twice.

Why do we need to make the second deduction?

VP

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GMATT73 wrote:

Professor wrote:

GMATT73 wrote:

10+11+14=35 total 35-20-3= 12 What am I missing here?

deduct 3 twice.

Why do we need to make the second deduction?

when u add a, b and c, you counted abc (=3) three times but you should count only once. so you need to deduct 2 (3).

Manager

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There was a similar question posted few days back ...
The answer is 3 .
say
x study both A &B
y study both A &C
z study both B&C
so the people ONLY studyA is 10 -( 3 +x+y)
the people ONLY study B is 11 -(3 +x +z)
the people ONLYstudy C is 14 -(3 +y +z)
But according to question only 20 people study only one subject so
if we add all the 3 above we get 26 -2 *(x+y+z) = 20
it means x + y+ z = 3
so 3 people study 2 kinds .

CEO

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x - Only A
y - Only B
z- Only C
a - A and B
b - B and C
c - C and A
We have to find (a + b + c) = W
We get (x+y+z) + (a + b + c) + 3 = Total
20 + (a + b + c) + 3 = T
T = 23 + (a + b + c) i.e T = 23+W........EQ1
Also
10+11+14 -2 * 3 - (a+b+c) = T i.e T = 29 - W.... EQ2
Solve these two equations and we get W = 3

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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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Represent intersect using '^'
A + (A^B^C) + (A^B) + (A^C) = 10
A + (A^B) + (A^C) = 7 ----(1)
B + (A^B^C) + (A^B) + (B^C) = 11
B + (A^B) + (B^C) = 8 ----(2)
C + (A^B^C) + (B^C) + (C^A) = 14
C + (B^C) + (C^A) = 11 ----(3)
(1) + (2) + (3)
(A + B + C) + 2(A^B) + 2(A^C) + 2(B^C) = 26
(A^B) + (A^C) + (B^C) = 3
Number of people learning two of the three = 3

VP

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3.
Venn diagram where a,b and c represent people learning two things.
10-(3+a+c) + 11 -(3+c+b) + 14-(3+a+b) = 20.
26-20 = 2(a+b+c)
a+b+c = 3 = Total number of people learning two of three things.

VP

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hmmmmm.......... all for 3 but i still stick with 9. laxi, did you forget to explain this with OA?

Director

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I think the answer shoud be 9....
You guys who calculated a+b+c=3 got what the statement was telling, i.e. that 3 people learn all three disciplines...

SVP

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ipc302 wrote:

There was a similar question posted few days back ... The answer is 3 . sayx study both A &B y study both A &C z study both B&C so the people ONLY studyA is 10 -( 3 +x+y) the people ONLY study B is 11 -(3 +x +z) the people ONLYstudy C is 14 -(3 +y +z) But according to question only 20 people study only one subject so if we add all the 3 above we get 26 -2 *(x+y+z) = 20 it means x + y+ z = 3 so 3 people study 2 kinds .

this OE is the clearest one. Thus OA is 3

I just want to complement a little to the bold part:

x= study both A and B without C

y= study both A and C without B

z= study both B and C without A

SVP

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laxieqv wrote:

ipc302 wrote:

There was a similar question posted few days back ... The answer is 3 . sayx study both A &B y study both A &C z study both B&C so the people ONLY studyA is 10 -( 3 +x+y) the people ONLY study B is 11 -(3 +x +z) the people ONLYstudy C is 14 -(3 +y +z) But according to question only 20 people study only one subject so if we add all the 3 above we get 26 -2 *(x+y+z) = 20 it means x + y+ z = 3 so 3 people study 2 kinds .

this OE is the clearest one. Thus OA is 3

I just want to complement a little to the bold part:

x= study both A and B without C

y= study both A and C without B

z= study both B and C without A

laxi, good to see you after a long time. and i am still waiting for your words.

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