# Practice GMAT Questions

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The picture above, which is not drawn entirely to scale, has four concentric circles. The ratio of the radii of the four circles is 4:3:2:1. Assuming x is the radius of the inner circle, what is the total area of the shaded regions?

**C**

- Understand what information is given, and what is being asked. Given that the four circles are concentric, we know from geometry that concentric circles share a common center.
- The ratio of the radii is our next important clue: since the ratio of the four circles is 4:3:2:1, the length of the radii is in the ratio of 4:3:2:1. This means that if the small radius were 1cm, then the second one will be 2cm (twice), the third one will be 3cm (three times), and the fourth radius will be 4cm (four times the first radius). Since we are not told actual measurements, let us assume the first radius is X. Then, the second radius will be 2X, the third 3X, and the fourth 4X.
- Now that we know the radii, let us look at the picture, and isolate the two colored regions: In order to find the area of the furthest outside region we need to subtract the area of the biggest circle (which has r = 4X) from the area of the third biggest circle (with r = 3X).
- Let us figure out the areas of the two circles. Using the formula for the area of the circle A = πr
^{2}:

Area of the fourth circle A = π[(4X)^{2}] = [16(X)^{2}]π

Area of the third circle A = π[(3X)^{2}] = [9(X)^{2}]π - The area of the furthest out shaded region is the difference between the areas of fourth and third circles:

[16(X)^{2}]π - [9 (X)^{2}]π = [7(X)^{2}]π - Repeat the same step to find the area of the smaller colored region: For this we need to subtract area of the smallest circle, from the second circle.

Area of the second circle A = [(2X)^{2}]π = [4(X)^{2}]π

Area of the first circle A = (X)^{2}π - The area of the small colored region: [4(X)
^{2}]π - (X)^{2}π = [3 (X)^{2}]π - Finally, now that we found the areas of the two colored regions (steps 5, 7) the total area is their Addition:

[7(X)^{2}]π + [3(X)^{2}]π = [10(X)^{2}]π

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