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10 Jun 2022, 17:45
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Difficulty:
95%
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Question Stats:
22% (02:07) correct
78%
(02:54)
wrong
based on 23
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The figure given above shows a square sheet of paper, which is divided into nine congruent small squares. An open cube (a 5-faced cube with open top) is to be formed using this sheet of paper such that:
(i) Each face of the cube is one of the small squares.
(ii) The square that form the cube must not be cut out separately; instead they must be folded along the dotted lines to form the cube.
(iii) There should be no overlap of any two squares. The squares that are not required to form the cube maybe cut out from the sheet of paper.
If only square 1 is colored yellow, then in how many ways can a cube with one yellow face be formed?
(A) 11
(B) 8
(C) 6
(D) 4
(E) 2
Are You Up For the Challenge: 700 Level Questions
Attachment:
2021-03-05_22-23-03.png [ 12.24 KiB | Viewed 3044 times ]
ashishpathak
Joined: 01 Jul 2021
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GMAT 1: 620 Q42 V34
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11 Jun 2022, 01:57
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how to solve this, please?
12 Jun 2022, 22:33
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Bunuel
The figure given above shows a square sheet of paper, which is divided into nine congruent small squares. An open cube (a 5-faced cube with open top) is to be formed using this sheet of paper such that:
(i) Each face of the cube is one of the small squares.
(ii) The square that form the cube must not be cut out separately; instead they must be folded along the dotted lines to form the cube.
(iii) There should be no overlap of any two squares. The squares that are not required to form the cube maybe cut out from the sheet of paper.
If only square 1 is colored yellow, then in how many ways can a cube with one yellow face be formed?
(A) 11
(B) 8
(C) 6
(D) 4
(E) 2
Are You Up For the Challenge: 700 Level Questions
Attachment:
2021-03-05_22-23-03.png
ashishpathak
Fun one!
5 must be the bottom of the box.
The only way for 1 to be in is if it is attached to either 2 or 4.
Let's look at when it is attached to 4.
- 4 will be the left side of the box and 1 will be the back.
- That means that 2 is not included.
- If 2 is not included, then the only way for 3 to be included is if it is attached to 6 with 6 as the right side and 3 as the back. But we already have 1 in the back, so that means 3 is not included.
- Now, we have two options for the right side: 6 and 9. If we make the right side 6, we could make the front 7, 8, or 9 (that's three ways to have completed the box). And if we make the right side 9, the only way to do that would be to have 9 attached to 8, which means 8 is the front (that's one more way to have completed the box).
- Those are all the ways to complete the box when the 1 is attached to the 4.
That was four ways. By symmetry, we will also have four options if the 1 were attached to the 2.
Total of 8 ways.
Answer choice B.
