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14 Jan 2021, 01:15
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
47% (01:53) correct
53%
(02:01)
wrong
based on 119
sessions
History
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Time
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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.
Are You Up For the Challenge: 700 Level Questions
(1) Exactly two children have ordered three hamburgers each.
(2) No child has ordered exactly two hamburgers.
Are You Up For the Challenge: 700 Level Questions
14 Jan 2021, 04:29
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(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
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
Updated on: 18 Apr 2024, 06:11
Originally posted by Kvadlamani7 on 14 Jan 2021, 05:24.
Last edited by Bunuel on 18 Apr 2024, 06:11, edited 2 times in total.
Last edited by Bunuel on 18 Apr 2024, 06:11, edited 2 times in total.
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Bunuel
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.
