GMAT Math : Geometry

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Example Questions

Example Question #741 : Geometry

Find the equation of circle whose radius is and center is at (-4,2) .

Possible Answers:

(x-4)^2+(y+2)^2=64

(x-4)^2+(y+2)^2=8

(x+4)^2+(y-2)^2=64

(x+4)^2+(y-2)^2=8

Correct answer:

(x+4)^2+(y-2)^2=64

Explanation:

To solve, simply use the formula given center at $\small (h, k)$ and radius of $\small r$ .

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

(x-(-4))^2+(y-(2))^2=8^2

(x+4)^2+(y-2)^2=64

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Example Question #14 : Calculating The Equation Of A Circle

Two circles on the coordinate plane have the origin as their center. The outer circle has twice the area as the inner circle, the equation of which is

$x^{2}+ y^{2} = 5$ .

Give the equation of the outer circle.

Possible Answers:

$x^{2}+ y^{2} = 5 \sqrt{2}$

$x^{2}+ y^{2} = 10$

$x^{2}+ y^{2} = 40$

$x^{2}+ y^{2} = 20$

$x^{2}+ y^{2} = 25$

Correct answer:

$x^{2}+ y^{2} = 10$

Explanation:

The equation of a circle centered at the origin is

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

where is the radius of the circle. Since the equation of the outer circle is

$x^{2}+ y^{2} = 5$ ,

$r^{2} = 5$ ,

and the area is

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

The area of the larger circle is twice this, or $2 \cdot 5 \pi = 10 \pi$ ; that is,

$A_{2} = \pi R^{2} = 10 \pi$

$R^{2} = 10$ ,

and the equation of that outer circle is

$x^{2}+ y^{2} = 10$ .

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Example Question #15 : Calculating The Equation Of A Circle

Two circles on the coordinate plane have the origin as their center. The outer circle has area five times that of the inner circle; the region between them has area $20 \pi$ . Give the equation of the inner circle.

Possible Answers:

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

$x^{2}+ y^2 = 5$

$x^{2}+ y^2 =16$

$x^{2}+ y^2 = 8$

$x^{2}+ y^2 = 25$

Correct answer:

$x^{2}+ y^2 = 5$

Explanation:

Let be the radius of the inner circle. The area of the inner circle is $A = \pi r^{2}$ ; the outer circle has area five times this, or $A '= 5 \pi r^{2}$ ; the region between them has area equal to the difference of these quantities, or

$A'' = A ' - A= 5 \pi r^{2} - \pi r^{2} = 4 \pi r^{2}$

This is equal to $20 \pi$ , so

$4 \pi r^{2} = 20 \pi$

$\frac{4 \pi r^{2} }{4 \pi}= \frac{20 \pi}{4 \pi }$

$r^{2} =5$

A circle with its center at the origin has as its equation

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

so the inner circle has as its equation

$x^{2}+ y^2 = 5$ .

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Example Question #1 : How To Graph An Exponential Function

Give the -intercept(s) of the graph of the equation

$y = 2^{x } - 20$

Possible Answers:

$\left ( 2 + \log_{2} 5, 0 \right )$

(-20,0)

The graph has no -intercept.

(10,0 )

$\left ( 1 + \log_{2} 5, 0 \right )$

Correct answer:

$\left ( 2 + \log_{2} 5, 0 \right )$

Explanation:

Set y = 0 and solve for :

$y = 2^{x } - 20$

$2^{x } - 20 = 0$

$2^{x } - 20+ 20 = 0 + 20$

$2^{x } = 20$

$\log_{2} \left ( 2 ^{x}\right ) = \log_{2} 20$

$x = \log_{2} 20$

$= \log_{2} 4 + \log_{2} 5$

$= 2 + \log_{2} 5$

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Example Question #981 : Gmat Quantitative Reasoning

Define a function as follows:

$f(x) = 3 ^{x-2}$

Give the -intercept of the graph of .

Possible Answers:

(0,0)

$(\ln 3, 0)$

(2, 0)

(3,0)

The graph of has no -intercept.

Correct answer:

The graph of has no -intercept.

Explanation:

Since the -intercept is the point at which the graph of intersects the -axis, the -coordinate is 0, and the -coordinate can be found by setting f(x) equal to 0 and solving for . Therefore, we need to find such that $3 ^{x-2} = 0$ . However, any power of a positive number must be positive, so $3 ^{x-2} > 0$ for all real , and $3 ^{x-2} = 0$ has no real solution. The graph of therefore has no -intercept.

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Example Question #1 : Graphing

Define a function as follows:

$f(x) = 2 ^{x-3}+ 5$

Give the vertical aysmptote of the graph of .

Possible Answers:

x = 5

The graph of does not have a vertical asymptote.

x= 2

x= 3

$x= - \frac{5}{2}$

Correct answer:

The graph of does not have a vertical asymptote.

Explanation:

Since any number, positive or negative, can appear as an exponent, the domain of the function $f(x) = 2 ^{x-3}+ 5$ is the set of all real numbers; in other words, f(x) is defined for all real values of . It is therefore impossible for the graph to have a vertical asymptote.

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Example Question #981 : Gmat Quantitative Reasoning

Define a function as follows:

$f(x) = 3 ^{x-3}- 9$

Give the -intercept of the graph of .

Possible Answers:

(3,0)

(5,0)

The graph of has no -intercept.

(12,0)

(9,0)

Correct answer:

(5,0)

Explanation:

$3 ^{x-3}- 9 = 0$ .

$3 ^{x-3}= 9$

$3 ^{x-3}= 3 ^{2}$

x-3 = 2

x = 5

The -intercept is therefore (5,0) .

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Example Question #742 : Geometry

Define a function as follows:

$f(x) = 2 ^{x-1} + 7$

Give the horizontal aysmptote of the graph of .

Possible Answers:

y = 1

y = 7

$y = -\frac{7}{2}$

y = 0

y = 2

Correct answer:

y = 7

Explanation:

The horizontal asymptote of an exponential function can be found by noting that a positive number raised to any power must be positive. Therefore, $2 ^{x-1} > 0$ and $y= 2 ^{x-1} + 7 > 7$ for all real values of . The graph will never crosst the line of the equatin y = 7 , so this is the horizontal asymptote.

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Example Question #143 : Coordinate Geometry

Define functions and as follows:

$f(x) = 2^{x +2}$

$g(x) = 4^{x-1}$

Give the $y \,$ -coordinate of the point of intersection of their graphs.

Possible Answers:

Correct answer:

Explanation:

First, we rewrite both functions with a common base:

$f(x) = 2^{x +2}$ is left as it is.

$g(x) = 4^{x-1}$ can be rewritten as

$g(x) = \left (2 ^{2} \right )^{x-1}$

$g(x) = 2 ^{2 (x-1)}$

$g(x) = 2 ^{2x-2}$

To find the point of intersection of the graphs of the functions, set

f(x) = g(x)

$2^{x +2}= 2 ^{2x-2}$

The powers are equal and the bases are equal, so we can set the exponents equal to each other and solve:

x+ 2= 2x-2

x+ 2 -x +2= 2x-2 -x +2

x = 4

To find the $y\,$ -coordinate, substitute 4 for in either definition:

$f(x) = 2^{x +2}$

$f(4) = 2^{4+2}= 2^{6} = 64$ , the correct response.

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Example Question #2 : Graphing

Define a function as follows:

$f(x) = 4 ^{x+2} - 3$

Give the -intercept of the graph of .

Possible Answers:

$\left ( 0, \frac{3}{4}\right )$

(0,13)

(0,2)

(0,1)

(0,-3)

Correct answer:

(0,13)

Explanation:

The -coordinate ofthe -intercept of the graph of is 0, and its -coordinate is f(0) :

$f(x) = 4 ^{x+2} - 3$

$f(0) = 4 ^{0+2} - 3 = 4 ^{2} - 3 = 16 - 3 = 13$

The -intercept is the point (0,13) .

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GMAT Math : Geometry

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #741 : Geometry

Example Question #14 : Calculating The Equation Of A Circle

Example Question #15 : Calculating The Equation Of A Circle

Example Question #1 : How To Graph An Exponential Function

Example Question #981 : Gmat Quantitative Reasoning

Example Question #1 : Graphing

Example Question #981 : Gmat Quantitative Reasoning

Example Question #742 : Geometry

Example Question #143 : Coordinate Geometry

Example Question #2 : Graphing

All GMAT Math Resources

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