A 4×4×4 wooden cube is painted so that one pair of opposite faces is blue, one pair green and one pair red. The cube is now sliced into 64 cubes of side 1 unit each. (i) How many of the smaller cubes have no painted face? (ii) How many of the smaller cubes have exactly one painted face? (iii) How many of the smaller cubes have exactly two painted face? (iv) How many of the smaller cubes have exactly three painted face? (v) How many of the smaller cubes have exactly one face painted blue and one face painted green?


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Question Number 18133 by Tinkutara last updated on 15/Jul/17

$A 4 \times 4 \times 4 wooden cube is painted so$ $that one pair of opposite faces is blue,$ $one pair green and one pair red . The$ $cube is now sliced into 64 cubes of side$ $1 unit each .$ $(i) How many of the smaller cubes have$ $no painted face ?$ $(ii) How many of the smaller cubes have$ $exactly one painted face ?$ $(iii) How many of the smaller cubes have$ $exactly two painted face ?$ $(iv) How many of the smaller cubes have$ $exactly three painted face ?$ $(v) How many of the smaller cubes have$ $exactly one face painted blue and one$ $face painted green ?$
	Answered by sandy_suhendra last updated on 16/Jul/17

	$1) {(4 - 2)}^{3} = 8 (this is like volume)$ $2) 6 (4 - 2) = 12 (this is like surface area)$ $3) 12 (4 - 2) = 24 (this is like the sum of edges)$ $4) 8 (vertices)$ $5) 4 (4 - 2) = 8$
	Commented by Tinkutara last updated on 16/Jul/17

	$Thanks Sir!$
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