Yesterday, Manhattan GMAT posted a 700 level GMAT question on our blog. Today, they have followed up with the answer:
First, we figure out the area of the smallest circle. A1 =r2 =
12 =
.
Now, we find the area of the second smallest circle (n = 2). A2 = A1 + (2(2) – 1) =
+ 3
= 4
. This means that the radius of the second smallest circle is 2 (since the area is
r2).
The third smallest circle has area A3 = A2 + (2(3) – 1) = 4
+ 5
= 9
. This means that the radius of this circle is 3.
Finally, the fourth smallest circle (that is, the largest circle) has area A4 = A3 + (2(4) – 1) = 9
+ 7
= 16
. This means that the radius of this circle is 4.
The sum of all the areas is + 4
+ 9
+ 16
= 30
.
The sum of all the circumferences is 2 times the sum of all the radii. The sum of all the radii is 1 + 2 + 3 + 4 = 10, so the circumferences sum up to 20
.
Thus, the sum of all the areas, divided by the sum of all the circumferences, is 30/(20
) = 1½.