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.
[spoiler]OA=B[/spoiler].
I don't understand this question. Can any expert explain it to me? Thanks for your help.
10 children have ordered a total of 17 hamburgers, and
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 ceilidh.erickson
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First, assess the given information:
17 hamburgers for 10 children who each eat at least 1 > 7 additional hamburgers. It's only these 7 that we need to worry about.
It helps to think through a few of the ways these could be distributed:
 7 kids get 1 additional burger each (2 each), 3 kids don't (1 each) > 7*2 + 3*1 = 17
 3 kids get 2 additional (3 each), 1 kid gets 1 additional (2 each), 6 kids get 1 each > 3*3 + 1*2 + 6*1 = 17
 1 kid gets all 7 extra (8 total), 9 kids get 1 each > 1*8 + 9*1 = 17
The more burgers any one child gets, the fewer are left > the fewer children get more than 1.
If we want to know if anyone had more than 3, we'd need some information about how many kids had exactly 1, exactly 2, etc.
(1) Exactly two children have ordered three hamburgers each.
Try to come up with different scenarios using this information that could give you a "yes" answer or a "no" answer to the question.
2 kids have 3 each > 2*3 = 6 burgers accounted for, 11 not accounted for, with 8 kids not accounted for. That's 3 bonus hamburgers.
Case1: evenly dividing 3 bonus hamburgers
2 kids have 3, 3 kids have 2, 5 kids have 1 > 2*3 + 3*2 + 5*1 = 17
Does anyone have more than 3? No.
Case2: giving 1 kid all bonus hamburgers
2 kids have 3, 1 kid has 4, 7 kids have 1 > 2*3 + 1*4 + 7*1 = 17
Does anyone have more than 3? Yes. Insufficient.
(2) No child has ordered exactly two hamburgers.
Try to come up with different scenarios using this information that could give you a "yes" answer or a "no" answer to the question.
No one ordered exactly 2 > the 7 bonus hamburgers can't be given away 1 at a time. It has to be 2 or 3 at a time, etc.
You might infer from this that since we have an odd number of bonus hamburgers, we can't just give them away in 2's (giving kids 3 total). We have to give away some odd number greater than 1. Giving any kid 3 bonus burgers would always give us a "yes" answer.
Or test cases:
Case1: maximize one kid's burgers
1 kid gets all 7 extra (8 total), 9 kids get 1 each > 1*8 + 9*1 = 17
Does anyone have more than 3? Yes.
Case2: minimizing any one kid's burgers
2 kids get 2 extra (3 total), 1 kid gets 3 extra (4 total), and 7 kids get 1 each > 2*3 + 1*4 + 7*1 = 17
Does anyone have more than 3? Yes.
We always get a "yes" answer, so this is sufficient.
The answer is B.
17 hamburgers for 10 children who each eat at least 1 > 7 additional hamburgers. It's only these 7 that we need to worry about.
It helps to think through a few of the ways these could be distributed:
 7 kids get 1 additional burger each (2 each), 3 kids don't (1 each) > 7*2 + 3*1 = 17
 3 kids get 2 additional (3 each), 1 kid gets 1 additional (2 each), 6 kids get 1 each > 3*3 + 1*2 + 6*1 = 17
 1 kid gets all 7 extra (8 total), 9 kids get 1 each > 1*8 + 9*1 = 17
The more burgers any one child gets, the fewer are left > the fewer children get more than 1.
If we want to know if anyone had more than 3, we'd need some information about how many kids had exactly 1, exactly 2, etc.
(1) Exactly two children have ordered three hamburgers each.
Try to come up with different scenarios using this information that could give you a "yes" answer or a "no" answer to the question.
2 kids have 3 each > 2*3 = 6 burgers accounted for, 11 not accounted for, with 8 kids not accounted for. That's 3 bonus hamburgers.
Case1: evenly dividing 3 bonus hamburgers
2 kids have 3, 3 kids have 2, 5 kids have 1 > 2*3 + 3*2 + 5*1 = 17
Does anyone have more than 3? No.
Case2: giving 1 kid all bonus hamburgers
2 kids have 3, 1 kid has 4, 7 kids have 1 > 2*3 + 1*4 + 7*1 = 17
Does anyone have more than 3? Yes. Insufficient.
(2) No child has ordered exactly two hamburgers.
Try to come up with different scenarios using this information that could give you a "yes" answer or a "no" answer to the question.
No one ordered exactly 2 > the 7 bonus hamburgers can't be given away 1 at a time. It has to be 2 or 3 at a time, etc.
You might infer from this that since we have an odd number of bonus hamburgers, we can't just give them away in 2's (giving kids 3 total). We have to give away some odd number greater than 1. Giving any kid 3 bonus burgers would always give us a "yes" answer.
Or test cases:
Case1: maximize one kid's burgers
1 kid gets all 7 extra (8 total), 9 kids get 1 each > 1*8 + 9*1 = 17
Does anyone have more than 3? Yes.
Case2: minimizing any one kid's burgers
2 kids get 2 extra (3 total), 1 kid gets 3 extra (4 total), and 7 kids get 1 each > 2*3 + 1*4 + 7*1 = 17
Does anyone have more than 3? Yes.
We always get a "yes" answer, so this is sufficient.
The answer is B.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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Harvard Graduate School of Education
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Statement 1:Gmat_mission wrote: ↑Sat Mar 24, 2018 4:27 am10 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.
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 = 1768 = 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 = 1710 = 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.
The correct answer is B.
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