GMAT Math : Problem-Solving Questions

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Example Questions

Example Question #31 : Circles

What is the area of a circle which goes through the points (2,4), (14,4), (2,9) ?

Possible Answers:

$169\pi$

$144 \pi$

$36\pi$

$\frac{169\pi }{4}$

$72\pi$

Correct answer:

$\frac{169\pi }{4}$

Explanation:

As can be seen in this diagram, the three points form a right triangle with legs of length 5 and 12.

Right_triangle

A circle through these three points circumscribes this right triangle.

An inscribed right, or $90^{\circ }$ , angle intercepts a $2 \times 90 = 180 ^{\circ }$ arc, or a semicircle, making the hypotenuse a diameter of the circle. The diameter of the circle is therefore the hypotenuse of the right triangle, which we can find via the Pythagorean Theorem:

$d = \sqrt{ 5^{2} + 12 ^{2}} = \sqrt{ 25+ 144} = \sqrt{169} = 13$

The radius of the circle is half this, or $\frac{13}{2}$ .

The area of the circle is therefore:

$A = \pi r^{2} = \pi \cdot \left ( \frac{13}{2} \right) ^{2} =\frac{169\pi }{4}$

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Example Question #5 : Calculating The Area Of A Circle

A circle on the coordinate plane has equation

$x^{2} + y ^{2} = 50$

Which of the following represents its area?

Possible Answers:

$50 \pi$

$10 \pi \sqrt{2}$

$100 \pi$

$25 \pi$

$5 \pi \sqrt{2}$

Correct answer:

$50 \pi$

Explanation:

The equation of a circle centered at the origin is

$x^{2} + y ^{2} = r ^{2}$

where is the radius of the circle.

The area of a circle is $A = \pi r ^{2}$ ; since in the equation of the circle, $r ^{2} = 50$ , we can substitute to get the area $A = \pi \cdot 50 = 50 \pi$ .

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Example Question #2 : Calculating The Area Of A Circle

Refer to the above figure, in which the dot indicates the center of the larger circle. The larger circle is one yard in diameter. Give the area of the gray region in square inches.

Possible Answers:

$405 \pi \textrm{ in}^{2}$

$243 \pi \textrm{ in}^{2}$

$972 \pi \textrm{ in}^{2}$

$108 \pi \textrm{ in}^{2}$

$1,620 \pi \textrm{ in}^{2}$

Correct answer:

$243 \pi \textrm{ in}^{2}$

Explanation:

The larger circle has diameter one yard, or 36 inches; its radius is half that, or 18 inches, so its area is

$A_{1} = \pi r ^{2} = \pi \cdot 18 ^{2} = 324 \pi$ square inches.

The diameter of the smaller circle is equal to the radius of the larger, or 18 inches; its radius is half that, or 9 inches, so its area is:

$A_{2} = \pi r ^{2} = \pi \cdot 9 ^{2} = 81 \pi$ square inches.

The area of the gray region is the difference of the two:

$A= 324 \pi - 81 \pi = 243 \pi$ square inches.

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Example Question #2 : Radius

For $10, Brandon can order either a 12"-diameter pizza, two 6"-diameter pizzas, or three 4"-diameter pizzas. Which option is the best value, assuming all pizzas are the same thickness?

Possible Answers:

three 4" pizzas

one 12" pizza

two 6" pizzas

Cannot be determined

All three pizzas have the same value.

Correct answer:

one 12" pizza

Explanation:

For a circle, $Area = \pi r ^{2}$ .

Therefore, the area of the 12" pizza $=6^{2}\pi =36\pi$ .

The area of the two 6" pizzas $=2\times 3^{2}\pi=2\times 9\pi =18\pi$ .

The area of the three 4" pizzas = $3\times 2^{2}\pi=3\times 4\pi =12\pi$ .

The 12" pizza is the best option.

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Example Question #3 : Radius

Refer to the above circle. The smaller circle has diameter feet. In terms of , give the area of the gray region in square inches.

Possible Answers:

$108 \pi k \textrm{ in}^{2}$

$216 \pi k \textrm{ in}^{2}$

$432 \pi k \textrm{ in}^{2}$

$108\pi k ^{2} \textrm{ in}^{2}$

$432 \pi k ^{2} \textrm{ in}^{2}$

Correct answer:

$108\pi k ^{2} \textrm{ in}^{2}$

Explanation:

Multiply the diameter of the smaller circle by 12 to convert it to inches - this will be 12k , which is also the radius of the larger circle. The radius of the smaller circle will be half this, or .

Using the area formula for a circle $A = \pi r^{2}$ , we can substitute these quantities for and subtract the area of the smaller circle from that of the larger:

$A_{1} = \pi r^{2} = \pi \left (12k \right ) ^{2} = 144 \pi k^{2}$ square inches

$A_{2} = \pi r^{2} = \pi \left (6k \right ) ^{2} = 36 \pi k^{2}$ square inches

$A = A_{1} - A_{2} = 144 \pi k ^{2}- 36 \pi k ^{2} = 108 \pi k ^{2}$ square inches

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Example Question #32 : Geometry

A circle on the coordinate plane has equation

$x^{2} + 4x + y ^{2} + 6y = 12$

What is its area?

Possible Answers:

$5 \pi$

$25 \pi$

$36 \pi$

$12 \pi$

$10 \pi$

Correct answer:

$25 \pi$

Explanation:

The area of a circle is equal to $A = \pi r ^{2}$ , where is the radius.

The standard form of the area of a circle with radius and center (h,k ) is

$(x-h)^{2} + (y-k)^{2} = r^{2}$

Once we get the equation in standard form, we know $r ^{2}$ , which can be multiplied by $\pi$ .

Complete the squares:

$\left (\frac{4}{2} \right )^{2} = 4 ;\left (\frac{6}{2} \right )^{2} = 9$

so $x^{2} + 4x + y ^{2} + 6y = 12$ can be rewritten as follows:

$x^{2} + 4x + 4 + y ^{2} + 6y + 9 = 12 + 4 + 9$

$\left (x + 2 \right ) ^{2} +\left ( y + 3 \right ) ^{2} = 25$

Therefore, $r^{2} = 25$ and $A = \pi r ^{2} = 25 \pi$ .

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Example Question #4 : Radius

A square has the same area as a circle with a radius of 12 inches. What is the sidelength of that square, in terms of $\pi$ ?

Possible Answers:

$144\pi\textrm{ in }$

$12\sqrt{\pi} \textrm{ in }$

$12\pi^{2}\textrm{ in }$

$\sqrt{12\pi} \textrm{ in }$

$12\pi\textrm{ in }$

Correct answer:

$12\sqrt{\pi} \textrm{ in }$

Explanation:

The area of a circle with radius 12 is

$A = \pi r^{2} = \pi \cdot 12^{2} = 144\pi$ .

This is also the area of the square, so the sidelength of that square is the square root of the area:

$s = \sqrt{A} = \sqrt{144\pi} = \sqrt{144} \cdot \sqrt{\pi}= 12\sqrt{\pi}$

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Example Question #271 : Problem Solving Questions

A circle on the coordinate plane has equation $x^{2} + y ^{2} = 49$ .

What is its area?

Possible Answers:

$7 \pi$

$49 \pi$

$14\pi$

$7 \sqrt{\pi }$

$28 \pi$

Correct answer:

$49 \pi$

Explanation:

The equation of a circle centered at the origin is

$x^{2} + y ^{2} = r ^{2}$ ,

where is the radius of the circle.

The area of a circle is $A = \pi r ^{2}$ .

From the equation given in the question stem, we know that $r ^{2} = 49$ , so we can plug this into the area formula:

$A = \pi \cdot 49 = 49 \pi$

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Example Question #271 : Problem Solving Questions

Two circles are constructed; one is inscribed inside a given equilateral triangle, and the other is circumscribed about the same triangle.

The circumscribed circle has circumference $20 \pi$ . Give the area of the inscribed circle.

Possible Answers:

$100 \pi$

$200 \pi$

$25 \pi$

$400 \pi$

$50 \pi$

Correct answer:

$25 \pi$

Explanation:

Examine the diagram below, which shows the triangle, its three altitudes, and the two circles.

Thingy

The three altitudes of an equilateral triangle divide one another into two segments each, the longer of which is twice the length of the shorter. The length of each of the longer segments is the radius of the circumscribed circle, and the length of each of the shorter segments is the radius of the inscribed circle. Therefore, the inscribed circle has half the radius of the circumscribed circle.

The circumscribed circle has circumference $20 \pi$ , so its radius is

$20 \pi \div 2 \pi = 10$

The inscribed circle has radius half this, or 5, so its area is

$A= \pi r^{2} = \pi \cdot 5^{2}= 25 \pi$

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Example Question #41 : Geometry

Two circles are constructed; one is inscribed inside a given equilateral triangle, and the other is circumscribed about the same triangle.

The inscribed circle has circumference $20 \pi$ . Give the area of the circumscribed circle.

Possible Answers:

$100 \pi$

$25 \pi$

$200 \pi$

$400 \pi$

$50 \pi$

Correct answer:

$400 \pi$

Explanation:

Examine the diagram below, which shows the triangle, its three altitudes, and the two circles.

Thingy

The three altitudes of an equilateral triangle divide one another into two segments each, the longer of which is twice the length of the shorter. The length of each of the longer segments is the radius of the circumscribed circle, and the length of each of the shorter segments is the radius of the inscribed circle. Therefore, the circumscribed circle has twice the radius of the inscribed circle.

The inscribed circle has circumference $20 \pi$ , so its radius is

$20 \pi \div 2 \pi = 10$

The circumscribed circle has radius twice this, or 20, so its area is

$A= \pi r^{2} = \pi \cdot 20^{2}= 400 \pi$

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GMAT Math : Problem-Solving Questions

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #31 : Circles

Example Question #5 : Calculating The Area Of A Circle

Example Question #2 : Calculating The Area Of A Circle

Example Question #2 : Radius

Example Question #3 : Radius

Example Question #32 : Geometry

Example Question #4 : Radius

Example Question #271 : Problem Solving Questions

Example Question #271 : Problem Solving Questions

Example Question #41 : Geometry

All GMAT Math Resources

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