For a school outing, the school has 20 small ice chests, each of which

Question

For a school outing, the school has 20 small ice chests, each of which can hold up to six cans of soda. At the beginning of the day, 10 of the ice chests are full, with six cans each, and ten are empty. The teachers are going to rearrange where the cans of soda are stored. If every can of soda must be in an ice chest, and each ice chest must have at least one can and not more than six, then what is the largest number of ice chests that can have exactly four cans?

(A) 10
(B) 12
(C) 13
(D) 14
(E) 15

(A) 10
(B) 12
(C) 13
(D) 14
(E) 15

jbrandl1 · Answer

Since we have 10 full ice chests with 6 cans each, that means we have:
10 • 6 = 60 cans to distribute

Since 60 / 4 = 15, we could fill 15 ice boxes exactly.

However, since every ice box must have at least 1 can, we need to redistribute:
15 - 1(1) = 14(4) + 4(1) = 18 boxes
13 - 8(1) = 13(4) + 6(1) + 1(2) = 20 boxes

Therefore,  C .

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abhishekdadarwal2009 · Answer

IMO : C

For a school outing, the school has 20 small ice chests, each of which can hold up to six cans of soda. At the beginning of the day, 10 of the ice chests are full, with six cans each, and ten are empty. The teachers are going to rearrange where the cans of soda are stored. If every can of soda must be in an ice chest, and each ice chest must have at least one can and not more than six, then what is the largest number of ice chests that can have exactly four cans?

(A) 10
(B) 12
(C) 13
(D) 14
(E) 15

(A) 10
(B) 12
(C) 13
(D) 14
(E) 15

sol:

Total number of ice chests:20 ;total number of cans=60,

A:if 10 had 4 cans then left cans be 20 in 10 chests, this can be done but the question asks the greatest number .
B:12 *4=48 ,cans left=12 then 
 C:13 best answer in 13*4=52, left=8 then each of the 6 chests have 1 can left and one chest will have 2 cans. 
D:14*4=16 but we are left with 6 chests and 4 cans so not going to work.
E:15*4=60 5 chests left and no cans left.

Archit3110 · Answer

really a good question
total bottles which can be accomodated; 20*6; 120
10 chests are full so we have 60 bottles
now we need to fill all the 20 chests with atleast 1 can and no more than 6
so 4 cans in a box can be accomodated in 60/4 ; 15 boxes
or say 13*4; +1*6 +2*1 ; 52+6+2 ; 60
so we can have 13 chests with 4 cans each
IMO C

Archit3110 · Answer

abhishekdadarwal2009 according to your solution total chests would be 21

but there are 20 chests ..

abhishekdadarwal2009 · Answer

Hi Thanks for noticing,
Updated the post.

C:13 best answer in 13*4=52, left=8 then each of the 6 chests have 1 can left and one chest will have 2 cans.

CrackverbalGMAT · Answer

Each of the small ice chests can hold a maximum of six cans of soda. There are 20 such chests. Therefore, the maximum number of soda cans that can be stored in these chests is 120.

10 of the ice chests are full, which means that there are 60 cans. We need to rearrange these cans such that each chest must have at least one soda can and every soda can should be there in a chest.

To maximize the number of chests that have exactly four cans, let’s start with the biggest number given in the options, which is option E.

If 15 chests have four cans each, it means that 5 cans do not have any cans at all. This violates the condition given in the question. So, option E is not the right answer.

Let’s look at option D. If 14 chests have four cans each, 56 cans out of 60 are in these 14 chests. The remaining 4 cans can be accommodated in 4 chests, but 2 chests will still be empty. Option D cannot be the right answer.

If 13 cans have four cans each, that’s 52 cans. We are left with 8 cans and 7 chests. Clearly, we can arrange so that every chest has at least one can. Therefore, option C has to be the right answer.

The most important point that needs to be realized here, is that the 60 cans are going to be rearranged among all the 20 cans, satisfying certain conditions. The moment you understand this part, is when you start getting closer to the right answer.

Hope this helps!

ScottTargetTestPrep · Answer

We see that there are a total of 6 x 10 = 60 cans. If 15 ice chests have 4 cans each, then we have 5 empty ice chests left. However, since each ice chest must have at least one can, we see that we can’t have 15 ice chests with 4 cans each.

Let’s say 14 ice chests have 4 cans each; then we have a total of 56 cans in these 14 ice chests. We see that we have 4 cans left but 6 ice chests left. So we would still have at least 2 empty ice chests. So it’s impossible to have 14 ice chests with 4 cans each.

Let’s say 13 ice chests have 4 cans each, so we have a total of 52 cans in these 13 ice chests. We see that we have 8 cans left but 7 ice chests left. So we can have 6 ice chests with 1 can each and the last ice chest with 2 cans. We see that it’s possible to have 13 ice chests with 4 cans each. Therefore, the largest number of ice chests that can have exactly four cans is 13.

Answer: C

luisdicampo · Answer

Deconstructing the Question  
There are 20 ice chests. Initially, 10 are full with 6 cans each, so total cans are  10\cdot6=60 .
After rearranging, every chest must contain between  1 and 6  cans.
We want the  maximum  number of chests that can have exactly  4 cans .

Key idea: let  k  be the number of 4-can chests, then check feasibility for the remaining cans using min/max bounds.

Step-by-step  
Let  k  chests have 4 cans, using  4k  cans.
Remaining cans:  60-4k 
Remaining chests:  20-k

Feasibility requires:
 (20-k)\le 60-4k \le 6(20-k)

Use the left inequality:
 60-4k \ge 20-k 
 40 \ge 3k 
 k \le 13

Check  k=13 :
Remaining cans  =60-52=8  for  20-13=7  chests, possible as  1,1,1,1,1,1,2 .
So the maximum is  13 .

Answer: 13

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For a school outing, the school has 20 small ice chests, each of which

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