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Sixteen rectangles are formed by choosing all possible combinations of [#permalink]
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11 Sep 2015, 03:42
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L = 1 or 2 or 3 or 4 . B = 1 or 2 or 3 or 4.
Remember square is also a kind of rectangle.
Total combinations of area (a) i.e L X B are as following.
1X1 , 1X2 , 1X3 , 1X4..............1 , 2 , 3 , 4 2X1 , 2X2 , 2X3 , 2X4..............2 , 4 , 6 , 8 3X1 , 3X2 , 3X3 , 3X4..............3 , 6 , 9 , 12 4X1 , 4X2 , 4X3 , 4X4..............4 , 8 , 12 , 16
So total 16 combinations for area (a) can be made and 7 out of these combinations satisfy our condition i.e 4<= a <=8
So the prob is 7 / 16 ........Eq 1.
Total combinations of perimeter (p) i.e 2(L + B) are as following.
4 , 6 , 8 , 10................................. 2(1+1) , 2(1+2) , 2(1+3) , 2(1+4) 6 , 8 , 10 , 12............................... 2(2+1) , .... so on...................... 8 , 10 , 12 , 14...............................2(3+1) , 2(3+2) , ....so on.......... 10 , 12 , 14 , 16..............................2 (4+1) , 2 (4+2) , ....so on........
So total 16 combinations of perimeter (p) can be made & 7 out of these combinations satisfy our condition i.e 10<= p <=12
There for prob is 7 / 16 ..................Eq 2.
Multiply Eq 1 & Eq 2.
7 / 16 * 7 / 16 = 49 / 256.
Ans is B.
Hope the answer is right and the approach is correct.
