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cubes are glued together

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by sanju09 » Wed May 02, 2012 3:24 am
A cube has a different letter on each face. The letters are A, B, C, D, E, and F. Another cube is similar except that the letters on the faces are A, B, C, G, H, and I. Both cubes are opaque. The cubes are glued together face-to-face and set down on an opaque table. If the cubes are not moved, what is the minimum number of different letters that must be visible to a person who walks around the table? (The person can see all of the exposed faces.)
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9



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by mathbyvemuri » Wed May 02, 2012 11:38 pm
7 faces. Answer option "C"
To get minimum number of viewable faces, maximum distinct (and non-common) faces shall be glued ie one- differently lettered face from each cube must be glued. For example D-face from one cube and G-face from other.
Then the viewable faces are: A,B,C,E,F from one cube and A,B,C,H,I from the other. The distinct number of faces viewable are A,B,C,E,F,H,I.
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by klmehta03 » Thu May 03, 2012 5:24 am
why not 8 faces?

CDEF on one cube and ABGH on second cube?
IMO D
Last edited by klmehta03 on Thu May 03, 2012 8:59 am, edited 1 time in total.
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by mathbyvemuri » Thu May 03, 2012 6:44 am
CDEF on one cube is not possible as only one face of a cube is glued and not visible. Similarly ACGH on the other cube, which means B and D are not visible, is also not a possible assumption.
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