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There are 68 children in the cafeteria of a school and all of the chil

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Manager
Joined: 10 Oct 2005
Posts: 51
There are 68 children in the cafeteria of a school and all of the chil [#permalink]

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04 May 2006, 18:34
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95% (hard)

Question Stats:

41% (05:29) correct 59% (02:19) wrong based on 34 sessions

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There are 68 children in the cafeteria of a school and all of the children have something for lunch. Thirty-four of the children brought lunches from home, 23 of the children bought a drink from the cafeteria beverage machine, and 32 of the children bought fruit in the cafeteria. If 18 children did at least 2 of these things, how many children did exactly two of these things?

(A) 3
(B) 6
(C) 9
(D) 13
(E) 15
[Reveal] Spoiler: OA
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore
Re: There are 68 children in the cafeteria of a school and all of the chil [#permalink]

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04 May 2006, 19:33
Draw a venn diagram with the following:

(# All 3) = x
(# Drink + Home) = w
(# Drink + Fruit) = y
(# Home + Fruit) = z
(# Drink only) = 23-w-x-y
(# Home only) = 34-w-x-z
(# Fruit only) = 32-x-y-z

So:

34-x-w-z + 32-x-y-z + 23-w-x-y + x + y + z + w = 68

89 - w - y - z - 2x = 68

w+y+z = 21-2x ----(1)

We also know:
w+y+z+x = 18
w+y+z = 18-x ---(2)

Equate (1) and (2) to get:

21-2x = 18-x
3 = x

Substitute this to (2)

w+y+z = 15.

So Ans = E
Director
Joined: 01 Jan 2008
Posts: 622
Re: There are 68 children in the cafeteria of a school and all of the chil [#permalink]

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23 Jan 2008, 19:39
Ravshonbek wrote:
There are 68 children in the cafeteria of a school and all of the children have something for lunch. Thirty-four of the children brought lunches from home, 23 of the children bought a drink from the cafeteria beverage machine, and 32 of the children bought fruit in the cafeteria. If 18 children did at least 2 of these things, how many children did exactly two of these things?

(A) 3
(B) 6
(C) 9
(D) 13
(E) 15

A - brought lunch from home, P(A) = 34
B - bought a drink, P(B) = 23
C - bought fruit, P(C) = 32

P(A) = P(A1)+P(A2)+P(A3) -- A1 those who only brought lunch from home, A2 - those who brought lunch from home and bought one item, A3 - brought lunch from home and bought two items
P(B) = P(B1)+P(B2)+P(B3)
P(C) = P(C1)+P(C2)+P(C3)

when we add them together P(A2) + P(B2) + P(C2) = 2*P2 //where P2 is the number of kids who did exactly two things, each of those kids is counted twice because if a kid brought lunch and bought a drink he belongs to A2 and B2. Consequenty, P(A3)=P(B3)=P(C3) = P3

P(A)+P(B)+P(C) = P1 + 2*P2+3*P3 = P(A)+P(B)+P(C) = 34+23+32 = 89

P2+P3 = 18 -> since 18 kids did at least two things
P1+P2+P3 = 68 -> because all 68 kids ate
P1+2*P2+3*P3 = 89

if we subtract second equation from the third one we get P2 + 2*P3 = 21, then multiply the first one by 2 and subtract the second one P2 = 18*2-21=36-21=15

CEO
Joined: 29 Mar 2007
Posts: 2560
Re: There are 68 children in the cafeteria of a school and all of the chil [#permalink]

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23 Jan 2008, 20:52
Ravshonbek wrote:
There are 68 children in the cafeteria of a school and all of the children have something for lunch. Thirty-four of the children brought lunches from home, 23 of the children bought a drink from the cafeteria beverage machine, and 32 of the children bought fruit in the cafeteria. If 18 children did at least 2 of these things, how many children did exactly two of these things?

(A) 3
(B) 6
(C) 9
(D) 13
(E) 15

This problem... sucks. took me a whopping 5minutes.

Drawing a venn. Lets say X is the number for all 3.

Food from home and Drink is: A-x
Home and fruit is: B-x
Drink and fruit is: C-x

So at least 2 is A-x+B-x+C-x+x = 18 --> A+B+C-2x=18

Now we have to caculate Home, Drink and Fruit: So

Home: 34-(A-x+B-x+x)
Drink: 23-(A-x+x+c-x)
Fruit: 32-(B-x+x+c-x)

Now make everything (add everything) equal to 68

+34+x-A-B
+23+x-A-C
+32+x-B-C
+A-x
+B-x
+C-x
+x
=68

We get A+B+C+x=21

Finished yet? Nope lol... nwow we can add our two equations together to find x.
A+B+C-2x=18
A+B+C+x=21

we get x=3

Now we have to calculate just 2 groups so... we need A-x+B-x+C-x --> A+B+C-3x=? well since A+B+C=24 3x=9

24-9=15

E.

Man that was just too much!
CEO
Joined: 17 Nov 2007
Posts: 3584
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: There are 68 children in the cafeteria of a school and all of the chil [#permalink]

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23 Jan 2008, 23:31
1
KUDOS
Expert's post
E

68=34+23+32-18-(18-x)
18-x=3
x=15
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Manager
Joined: 02 Jun 2015
Posts: 178
Location: Ghana
There are 68 children in the cafeteria of a school and all of the chil [#permalink]

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01 Feb 2017, 08:12
There are 68 children in the cafeteria of a school and all of the children have something for lunch. Thirty-four of the children brought lunches from home, 23 of the children bought a drink from the cafeteria beverage machine, and 32 of the children bought fruit in the cafeteria. If 18 children did at least 2 of these things, how many children did exactly two of these things?

(A) 3
(B) 6
(C) 9
(D) 13
(E) 15
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Math Expert
Joined: 02 Sep 2009
Posts: 39622
Re: There are 68 children in the cafeteria of a school and all of the chil [#permalink]

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01 Feb 2017, 08:19
duahsolo wrote:
There are 68 children in the cafeteria of a school and all of the children have something for lunch. Thirty-four of the children brought lunches from home, 23 of the children bought a drink from the cafeteria beverage machine, and 32 of the children bought fruit in the cafeteria. If 18 children did at least 2 of these things, how many children did exactly two of these things?

(A) 3
(B) 6
(C) 9
(D) 13
(E) 15

Merging topics. Please refer to the discussion above.
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Re: There are 68 children in the cafeteria of a school and all of the chil   [#permalink] 01 Feb 2017, 08:19
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