10 children have ordered a total of 17 hamburgers, and each child has

Question

10 children have ordered a total of 17 hamburgers, and each child has ordered at least one hamburger. Has any child ordered more than 3 hamburgers?

(1) Exactly two children have ordered three hamburgers each.
(2) No child has ordered exactly two hamburgers.

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QuantMadeEasy · Answer

(1) Exactly two children have ordered three hamburgers each. 
Taking cases for distribution of 17 burgers among 10 children:
Case 1: 4,3,3,1,1,1,1,1,1,1
Case 2: 3,3,2,2,2,2,1,1,1
Insufficient

(2) No child has ordered exactly two hamburgers. 
Case 1: 4,3,3,1,1,1,1,1,1,1
Case 2: 3,3,3,3,3,1,1
Insufficient

(1)+(2) 
Only possible case: 4,3,3,1,1,1,1,1,1,1 as no child has exactly 2 burger and exactly 2 children got 3 burgers each.

IMO C

Kvadlamani7 · Answer

Question stem says each student ordered at least 1 HB.

S1: Gives 2 possibilities.

a) 1*10 for the first 10 and then the remaining 7 can be distributed as 2,2,1,1,1. So, 5 students get 1 each, 3 students get 2 each and 2 students get 3 each. So no student get >3.

b) 1*10 for the first 10 and then the remaining 7 can be distributed as 2,2,3. So, 7 students get 1 each, 2 students get 3 each and 1 student gets 4. So one student  gets  >3.

S1 is insufficient.

S2: The first 10 can be distributed as 1 each (as per the que stem) and the remaining 7 can be distributed as 2,2,3; 2,5 or 3,4 as no student has ordered exactly 2. So, again we can have 7 students ordering 1 each, 2 ordering 3 each and 1 ordering 4 or 8 students ordering 1 each, 1 ordering 3 and one ordering 6 or 8 students ordering 1 each, 1 student ordering 4 and 1 student ordering 5. These three are the only possible combinations and in each case we have at least one student ordering more than 3.

S2 is sufficient, so B.

Bunuel · Answer

Anyone else want to try this hard and tricky question?

NeoNguyen1989 · Answer

IMO B
(1) it can be 3 students each order 3 hamburgers, 1 order 2 and 6 order 1 => no
OR 
it can be 2 students each order 3 hamburgers, 7 students each order 1 and 1 student order 4 hamburger => yes

Insufficient

(2) 10 students have 1 each so we have 7 hamburgers to be allocated. Since no student has exactly 2 hamburgers so the 7 has be allocated as 2,2 and 3 or 2 and 5. Either case, one student will get 4 hamburgers or more
=> sufficient

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GCMEMBER · Answer

10 children have ordered a total of 17 hamburgers, and each child has ordered at least one hamburger. Has any child ordered more than 3 hamburgers?

(1) Exactly two children have ordered three hamburgers each.

The ten children will get burgers as follows:
3,3,2,2,2,1,1,1,1,1
Or
3,3,4,1,1,1,1,1,1,1

Not sufficient

(2) No child has ordered exactly two hamburgers.

The ten children will get burgers as follows:
4,3,3,1,1,1,1,1,1,1
Or
5,4,1,1,1,1,1,1,1,1
Or
6,3,1,1,1,1,1,1,1,1

Option B

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Pankaj0901 · Answer

IMO ANSWER  B

Given:
10 C ordered 17 HB
Each one ordered at least 1 HB => The 7 HB's could be ordered as the next HB for any C, as every C has ordered 1 already.

S1:  
3+3 HB's by exactly 2 C
=> 10 + 4 = 14 HB's are ordered by 10 C. The rest 3 HB's (17-14) could be ordered by any of the children who had already ordered 1 HB.
 So, not sufficient.

S2: 
No C has ordered exactly 2 HB

Case A: The rest 7 HB's could be ordered by anyone of the 10 C's who had already ordered 1 HB. So, YES, 1 C has definitely ordered more than 3 HB.

Case B: 1 C orders 3 HB's => 12 HB's gone. The rest 5 HB's (17-12) could be bought by 1 C altogether. => YES, 1 C has definitely ordered more than 3 HB's.

Case C: 2 C orders 3 HB's each => 14 HB's gone. The rest 3 HB's (17-14) could be bought by 1 C altogether, that means 1 C has bought total 4 HB's. => YES

Case D: 3 C orders 3 HB's each => 16 HB's gone. The rest 1 HB could not be bought by those C who had bought 1 HB already (as given in statement 2), so it must have been bought by the one of these 3 HB's who had bought 3 HB each. Thus, the total HB that one C would have bought will be equal to 4. => YES

SUFFICIENT

TestPrepUnlimited · Answer

We only need to think about where  the extra 7 hamburgers  went since we know each child must have one.

Statement 1:

Two children took 2 extra hamburgers, so we have 3 extra hamburgers left to allocate. We may give them to a single child so that somebody has 4 hamburgers, or we distribute them among the children with only one hamburger so that nobody has more than 3. Insufficient.

Statement 2:

This means we cannot split the extra 7 hamburgers into any 1's. We must have 7 = 2 + 2 + 3 or 7 = 2 + 5 since we are not allowed to give any child one extra hamburger only. Thus no matter how we split the 7 there was always one number greater than 2, which means somebody will have more than 3 hamburgers total. Sufficient.

Ans: B

GMATGuruNY · Answer

Statement 1:
Two children A and B with 3 hamburgers each = 2*3 = 6 hamburgers
Remaining 8 children C through J with at least one hamburger each = 8*1 = 8 hamburgers
Remaining hamburgers to be distributed among C through J = 17-6-8 = 3

Case 1: C receives the 3 remaining hamburgers 
In this case, C has a total of 4 hamburgers, so the answer to the question stem is YES.

Case 2: C, D and E each receive 1 more hamburger
In this case, C, D, and E each have 2 hamburgers.
Since no child has more than 3 hamburgers, the answer to the question stem is NO.
INSUFFICIENT.

Statement 2:
10 children with at least 1 hamburger each = 10*1 = 10
Remaining hamburgers to be distributed = 17-10 = 7

Since no child receives exactly 2 hamburgers, none of the 10 children may receive exactly 1 more hamburger.
Since 7 is ODD, it is not possible for every child who receives additional hamburgers to receive exactly 2. 
Implication:
At least 1 child must receive 3 OR MORE hamburgers, yielding a total of 4 OR MORE hamburgers for that child.
Thus, the answer to the question stem is YES.
SUFFICIENT.

B

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10 children have ordered a total of 17 hamburgers, and each child has

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