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# There are 10 people studying A, 11 people studying B and 14

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There are 10 people studying A, 11 people studying B and 14 [#permalink]

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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?
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25 Feb 2006, 21:33
10+11+14=35 total

35-20-3= 12

What am I missing here?
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25 Feb 2006, 21:45
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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25 Feb 2006, 21:49
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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25 Feb 2006, 22:00
There was a similar question posted few days back ...

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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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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25 Feb 2006, 22:28
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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27 Feb 2006, 17:26
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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28 Feb 2006, 00:00
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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28 Feb 2006, 22:37
hmmmmm.......... all for 3 but i still stick with 9. laxi, did you forget to explain this with OA?
Director
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01 Mar 2006, 03:10
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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01 Mar 2006, 03:52
ipc302 wrote:
There was a similar question posted few days back ...

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 .

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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01 Mar 2006, 09:55
laxieqv wrote:
ipc302 wrote:
There was a similar question posted few days back ...

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 .

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.
01 Mar 2006, 09:55
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