eitijan wrote:A small table has a length of 12 inches and a breadth of b inches. Cubes are placed on the surface of the table so as to cover the entire surface. The maximum side of such cubes is found to be 4 inches. Also, a few such tables are arranged to form a square. The minimum length of side possible for such a square is 48 inches. Find b.
A. 8
B. 16
C. 24
D. 32
E. 48
You could test the answer choices to get to the correct answer and to see why GCF and LCM make sense.
C and E are easy to eliminate.
A 12 x 24 table could accommodate cubes with edges of length 12. Two such cubes would fit on the table, as 24/12 = 2.
A 12 x 48 table could also accommodate cubes with edges of length 12. Four such cubes would fit on the table, as 48/12 = 4.
So, since 12 divides evenly into 24 and 48, cubes with edges of a length greater than 4 would fit on the table.
We are left with A, B and D.
A can be eliminated in the following way.
2 x 12 = 24 and 3 x 8 = 24. So 12 x 8 tables could be arranged into a 24 x 24 square. 24 < 48.
D can be eliminated in the following way.
12 x 32 is not a square. So to make a square we need more than one table.
(12 + 12 + 12 + 12) x 32 is not a square and we cannot get 48 by adding 32's.
So D is out too.
To confirm B, you could do the following.
Cubes with sides length 4 will fit on a 12 x 16 table because both 12 and 16 are divisible by 4.
Also, 48 is divisible by 12 and 16, but no number smaller than 48 is divisible by both 12 and 16. So you could make a 48 x 48 square but no square smaller than 48 x 48.
So the correct answer is
B.
