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20 Apr 2020, 05:50
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Difficulty:
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Arun has 13 boxes of chocolates with him, with an average of 17 chocolates per box. If each box has at least 11 chocolates and no two boxes have equal number of chocolates, then what can be the maximum possible number of chocolates in any box?
A. 23
B. 24
C. 25
D. 29
E. 30
Are You Up For the Challenge: 700 Level Questions
A. 23
B. 24
C. 25
D. 29
E. 30
Are You Up For the Challenge: 700 Level Questions
22 Apr 2020, 03:57
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(Highest + least)/2= avg.
so (x+11)/2=17
X =23
Posted from my mobile device
so (x+11)/2=17
X =23
Posted from my mobile device
21 Jun 2021, 22:35
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Bunuel
Point to note here is that the mean lies at 17. The first box with minimum number of chocolates will have 11 chocolates. If we keep adding 1 extra chocolate in each subsequent box (so that they have distinct number of chocolates), we will get 17 chocolates in the 7th box, the middle box.
Since 17 is the mean, we must have an arithmetic progression such that the last box i.e. the 13th box will have 17 + 6 = 23 chocolates.
Answer (A)
Generic method if mean were not the middle box:
Total chocolates we have = 13 *17
How do we distribute them? We first 11 to each box so we give away 13 *11.
We are left with 13*17 - 13*11 = 13*6 = 78 chocolates.
Each box has 11 chocolates right now and we have 78 extra chocolates left. We must distribute them in a way such that no two boxes have the same number of chocolates.
So we must give at least 0 + 1 + 2 + 3 + 4... + 11 to the 12 boxes so that whatever is left can be given to the 13th box.
0 + 1 + 2 + 3 + 4... + 11 = n(n+1)/2 = 66
We are left with 78 - 66 = 12 chocolates for the last box bringing up its total to 11 + 12 = 23 chocolates.
