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There are 33 balls and 7 boxes. Balls are filled in these boxes such t
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14 Mar 2025, 00:38
14 Mar 2025, 00:38
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There are 33 balls and 7 boxes. Balls are filled in these boxes such that no box is left empty and each ball goes into a box. Which of the following must be true?
I. There are at least 5 balls in at least 1 box.
II. There are at least 4 balls in every box.
III. There are at least 2 boxes with the same number of balls.
(A) Only I
(B) Only II
(C) Only III
(D) Both I and II
(E) None of these
I. There are at least 5 balls in at least 1 box.
II. There are at least 4 balls in every box.
III. There are at least 2 boxes with the same number of balls.
(A) Only I
(B) Only II
(C) Only III
(D) Both I and II
(E) None of these
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Re: There are 33 balls and 7 boxes. Balls are filled in these boxes such t
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18 Mar 2025, 14:17
18 Mar 2025, 14:17
Expert Reply
We know that there are 33 balls filled in 7 boxes such that no box is left empty; we need to check that which of the given options must be true.
I. There are at least 5 balls in at least 1 box - which must be true as even if try to distribute equal number of balls i.e. 4 in each of the 7 boxes, we will still be left with $\(33-(7 \times 4)=33-28=5\)$ balls which will go in one or more of the boxes.
II. There are at least 4 balls in every box - which might not be true as we can have one ball in each of the 7 boxes and the remaining $\(33-(7 \times 1)=33-7=26\)$ balls can go in any one box. So, it's not necessary for each of the 7 boxes to have at least 4 balls.
III. There are at least 2 boxes with the same number of balls - If we put different number of balls in each of the 7 boxes, we can put $\(1,2,3,4,5,6\)$ and 7 balls and we will left with $\(33-28=5\)$ balls and these 5 balls can go all in $\(7^{\text {th \)$ box having 7 balls, so all boxes are filled with different number of balls. Hence it is not necessary to have same number of balls in at least 2 of the boxes.
Hence only statement I must be true, so the answer is $(A)$.
I. There are at least 5 balls in at least 1 box - which must be true as even if try to distribute equal number of balls i.e. 4 in each of the 7 boxes, we will still be left with $\(33-(7 \times 4)=33-28=5\)$ balls which will go in one or more of the boxes.
II. There are at least 4 balls in every box - which might not be true as we can have one ball in each of the 7 boxes and the remaining $\(33-(7 \times 1)=33-7=26\)$ balls can go in any one box. So, it's not necessary for each of the 7 boxes to have at least 4 balls.
III. There are at least 2 boxes with the same number of balls - If we put different number of balls in each of the 7 boxes, we can put $\(1,2,3,4,5,6\)$ and 7 balls and we will left with $\(33-28=5\)$ balls and these 5 balls can go all in $\(7^{\text {th \)$ box having 7 balls, so all boxes are filled with different number of balls. Hence it is not necessary to have same number of balls in at least 2 of the boxes.
Hence only statement I must be true, so the answer is $(A)$.
gmatclubot
Re: There are 33 balls and 7 boxes. Balls are filled in these boxes such t [#permalink]
18 Mar 2025, 14:17
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