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

Question

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. 3
B. 6
C. 9
D. 13
E. 15

prerakgoel03 · Accepted Answer

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ScottTargetTestPrep · Answer

We can use the following formula:

Total = n(L) + n(D) + n(F) - n(exactly two) - 2 * n(all three) + n(none)

68 = 34 + 23 + 32 - x - 2 * (18 - x) + 0

68 = 89 - x - 36 + 2x

68 = 53 + x

x = 15

Answer: E

Chets25 · Answer

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. 3
B. 6
C. 9
D. 13
E. 15

34 - Home lunch
34 - No home lunch

Scenario 1:
Out of 34 without Home lunch, lets say 32 with fruit from cafeteria & remaining 2 out of 23 drink from cafeteria. Therefore, 34 home lunch includes 21 with drink from cafeteria. Therefore, atleast 2 will be 34-21=13.

Scenario 2: 
Out of 34 without Home lunch, lets say 23 with drink from cafeteria & remaining 11 out of 32 fruit from cafeteria. Therefore, 34 home lunch includes 21 with fruit from cafeteria. Therefore, atleast 2 will be 34-21=13.

Hence, answer is 13 (D).

attari92 · Answer

As per the question stem total number of children having only one type of meal and having at least two meals would be equal to 50 and 18 respectively. Now, 50 =(34+23+32)-2(at least two)- all three and that implies all three = 3. Subsequently, only two = 18-3 =15 (AT LEAST TWO = ONLY 2 + ALL 3)

jxav · Answer

34+23+32=89 
89-68=21
21-18=3
So there are 3 children doing all 3 things
Leaves us with 21-(3×2)=15 children doing exactly 2 things

E

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RishiQV · Answer

So, this is a three set venn diagrm question
Let students with lunches = A
Let students with Drinks = B
Let students with Fruit= C

Using a nice formula (Which you can easily derive) :
Total =   -   + all three common + Neither

Here we are told that "18 children did at least 2 of these things" which mean the SUM common area of two and three overlaps is 18. 
This includes AnB, BnC, AnC which is the combination of three 'two-set' overlap area and one 'three-set' overlap area. 
Let 'center(three-set overlap area)' be x
So putting numbers in the formula we get : 68=  - 18 + x + 0
x=3

So, only two set overlap area = number of children did exactly two of these things = 18-3 = 15

nikhilmathew1990 · Answer

Using the above calculations the value of x is x=-3. @bunnel: Are we supposed to take the positive value in this case? Please help.

KarishmaB · Answer

All 68 children had atleast one thing - homelunch, drink, fruit

There are 34 + 23 + 32 = 89 instances of these three things in all. 
18 children had at least two so 18 instances have been double counted at least. Let's remove them. We are left with 89 - 18 = 71
Now we still have 3 instances extra so these 3 must have been counted thrice.

Hence 18 - 3 = 15 children consumed exactly two.

Answer (E)

Oryan · Answer

@bunnel Can you comment to this question... I have the same one like this guy.

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TarunKumar1234 · Answer

Total = n(A) + n(B) + n(C) - n(exactly two) -2* n(A∩B∩C) or, 68 = 34 +23 +32 - x -2*(18-x) or, 68 = 89 -36 +x or, x = 15 So, I think E.

Archit3110 · Answer

total = A+B+C- 2*( atleast) + exactly two
68= 34+23+32 -36+exactly two
exactly two = 68-53 ;15
option E

Fdambro294 · Answer

Let the number of people who took at least 1 item (they took 1 or 2 or 3 items) = I

Let the number of people who took at least 2 items (they took 2 or 3 items) = II

And finally let the number of people who took all 3 items = III

Everyone of the 68 unique people took at least 1 (in other words the Union of the set is defined ——> there is no one in the “neither/nor” group)

Therefore:
(# of ppl who took at least 1) + (# of ppl who took at least 2) + (# of ppl who took 3) = A + B + C

Where A, B, and C = the three tallies ——-> 34, 23, and 32

Note: this can be proven by labeling a Venn diagram with variables and adding up the 3 circles (A + B + C) and then adding up each of I , II, and III

So we have:

I + II + III = 34 + 23 + 32 = 89

Where it must be the case that: I >/= II >/= III

I = Union of the overlapping sets = number of people who bring at least I = all 68 unique people

And

II = # of ppl who bring at least 2 = 18

I + II + III = 89

Becomes

(68) + (18) + III = 89

III = 3 people

And finally, the number of people who bring exactly 2 items =

II - III = 18 - 3 = 15

Answer
15

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Kinshook · Answer

Given: 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.
Asked: If 18 children did at least 2 of these things, how many children did exactly two of these things?

18 = Exactly 2 + Exactly 3
Exactly 3 = 18 - Exactly 2

All = 1 + 2 + 3 - Exactly 2 - 2* Exactly 3
68 = 34 + 23 + 32 - Exactly 2 - 2* Exactly 3 = 89 - Exactly 2 - 2 (18 - Exactly 2) = 89 - 36 + Exactly 2 = 53 + Exactly 2
Exactly 2 = 15

IMO E

Kinshook · Answer

Hi  prerakgoel03 
How did you get lunch value = 34 and not (3/4)*68 = 51

bumpbot · Answer

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