The GRE maths sample question given below is a Quantitative Comparison Question in ratio and proportion. This GRE Math question tests the concept of Area of a Circle, concentric circles, and area of rings.

Question 4: Three concentric circles of radii 2 cm, 10 cm, and 15 cm are drawn.

Quantity A | Quantity B |
---|---|

Ratio of area enclosed between the outer circle and middle circle to the area enclosed between the middle circle and inner circle. | Ratio of area enclosed between the middle circle and inner circle and the area of the innermost circle. |

- Quantity A is greater
- Quantity B is greater
- The two quantities are equal
- Cannot be determined

From INR 2999

**Given Data**

Radius of Inner circle r_{1} = 2 cm

Radius of Middle circle r_{2} = 10 cm

Radius of Outer circle r_{3} = 15 cm

**Formula Used**

Area of circle = π × r^{2}

where r is the radius of the circle.

Area of Inner circle A_{1} = π × r_{1}^{2} = π × 2^{2} = 4π cm^{2}

Area of Middle circle A_{2} = π × r_{2}^{2} = π × 10^{2} = 100π cm^{2}

Area of Outer circle A_{3} = π × r_{3}^{2} = π × 15^{2} = 225π cm^{2}

**Quantity A**: Ratio of area enclosed between the outer circle and middle circle to the area enclosed between the middle circle and inner circle.

\\frac{\text{Area enclosed between the outer circle and middle circle}}{\text{Area enclosed between the middle circle and inner circle}}) = \\frac{\text{225π - 100π}}{\text{100π - 4π}})

→ \\frac{\text{125}}{\text{96}}) ∼ **1.3**

**Quantity B**: Ratio of area enclosed between the middle circle and inner circle and the area of the innermost circle.

\\frac{\text{Area enclosed between the middle circle and inner circle}}{\text{Area of the inner circle}}) = \\frac{\text{100π - 4π}}{\text{4π}})

→ \\frac{\text{96}}{\text{4}}) = **24**

**Quantity A** = 1.302

**Quantity B** = 24

**Quantity B > Quantity A**

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