GMAT Math : Geometry

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Example Questions

Example Question #943 : Problem Solving Questions

Fill in the circle with a number so that the graph of the resulting equation has -intercept (0,5) :

$4 x+ \bigcirc y = 18$

Possible Answers:

$\frac{5}{4}$

$\frac{4}{5}$

$\frac{5}{18}$

$\frac{18 }{5}$

Correct answer:

$\frac{18 }{5}$

Explanation:

Let be the number in the circle. The equation can be written as

4 x+N y = 18

Substitute 0 for and 5 for ; the equation becomes

$4 \cdot 0+N \cdot 5 = 18$

0+5N = 18

5N = 18

$N = \frac{18 }{5}$

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Example Question #11 : Calculating X Or Y Intercept

Fill in the circle with a number so that the graph of the resulting equation has -intercept (6,0) :

$4x+5y = \bigcirc$

Possible Answers:

The graph cannot have (6,0) as its -intercept regardless of the value written in the circle.

$\frac{3}{2}$

$\frac{6}{5}$

Correct answer:

Explanation:

Let be the number in the circle. The equation can be written as

4x+5y = N

Substitute 0 for and 6 for ; the resulting equation is

$4 \cdot 6+5 \cdot 0 = N$

N = 24 + 0 = 24

24 is the correct choice.

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Example Question #951 : Problem Solving Questions

Fill in the circle with a number so that the graph of the resulting equation has -intercept (-3,0) :

$\bigcirc x+ 5 y = 18$

Possible Answers:

$-\frac{6}{5}$

$-\frac{5}{6}$

The graph cannot have (-3,0) as its -intercept regardless of the value written in the circle.

Correct answer:

Explanation:

Let be the number in the circle. The equation can be written as

N x+ 5 y = 18

Substitute 0 for and for ; the resulting equation is

$N (-3)+ 5 \cdot 0= 18$

-3N + 0= 18

-3N = 18

N = -6

is the correct choice.

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Example Question #952 : Problem Solving Questions

Fill in the circle with a number so that the graph of the resulting equation has -intercept (5,0) :

$4 x+ \bigcirc y = 18$

Possible Answers:

$-\frac{18}{5}$

$-\frac{9}{10}$

$-\frac{5}{4}$

The graph cannot have (5,0) as its -intercept regardless of the value written in the circle.

Correct answer:

The graph cannot have (5,0) as its -intercept regardless of the value written in the circle.

Explanation:

Let be the number in the circle. The equation can be written as

4 x+N y = 18

Substitute 0 for ; the resulting equation is

$4 x+N \cdot 0 = 18$

4x= 18

$x = \frac{9}{2}$

The -intercept is $\left ( \frac{9}{2}, 0 \right )$ regardless of what number is written in the circle.

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Example Question #953 : Problem Solving Questions

Fill in the circle with a number so that the graph of the resulting equation has -intercept (0, 7) :

$4x+5y = \bigcirc$

Possible Answers:

$\frac{7}{4}$

$\frac{7}{5}$

Correct answer:

Explanation:

Let be the number in the circle. The equation can be written as

4x+5y = N

Substitute 7 for and 0 for ; the resulting equation is

$4 \cdot 0+5 \cdot 7= N$

N = 0+35

N = 35

35 is the correct choice.

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Example Question #954 : Problem Solving Questions

Fill in the circle so that the graph of the resulting equation has no -intercepts:

$y = 4x^{2}+ 6 x + \bigcirc$

Possible Answers:

The graph will have at least one -intercept regardless of the value written in the circle.

Correct answer:

Explanation:

Let be the number in the circle. Then the equation can be rewritten as

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

Substitute 0 for and the equation becomes

$4x^{2}+ 6 x + C = 0$

Equivalently, we are seeking a value of for which this equation has no real solutions. This happens in a quadratic equation $Ax^{2}+ Bx+ C = 0$ if and only if

$B^{2}-4AC < 0$

Replacing with 4 and with 6, this becomes

$6^{2}-4 \cdot 4 \cdot C < 0$

36-16C < 0

16C > 36

$C > \frac{36}{16} = \frac{9}{4}$

Therefore, must be greater than $\frac{9}{4}$ . The only choice fitting this requirement is 4, so this is correct.

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Example Question #955 : Problem Solving Questions

Fill in the circle so that the graph of the resulting equation has exactly one -intercept:

$y = 4x^{2}+ \bigcirc x + 8$

Possible Answers:

None of the other choices is correct.

Correct answer:

None of the other choices is correct.

Explanation:

Let be the number in the circle. Then the equation can be rewritten as

$y = 4x^{2}+ Bx + 8$

Substitute 0 for and the equation becomes

$4x^{2}+ Bx + 8 = 0$

Equivalently, we are seeking a value of for which this equation has exactly one solution. This happens in a quadratic equation $Ax^{2}+ Bx+ C = 0$ if and only if

$B^{2}-4AC = 0$

Replacing with 4 and with 8, this becomes

$B^{2}-4 \cdot 4 \cdot 8 = 0$

$B^{2}-128= 0$

$B^{2}=128$

$B= \pm \sqrt{128} = \pm \sqrt{64} \cdot \sqrt{2} = \pm 8 \sqrt{2}$

Therefore, either $B=- 8 \sqrt{2}$ or $B= 8 \sqrt{2}$ .

Neither is a choice.

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Example Question #956 : Problem Solving Questions

Find the $y\textup{-intercept}$ for the following equation:

3x-2y=6

Possible Answers:

(0,-3)

(3,0)

(-3,0)

(0,3)

Correct answer:

(0,-3)

Explanation:

To find the $y\textup{-intercept}$ , you must put the equation into slope intercept form:

y=mx+b where (0,b) is the intercept.

Thus,

3x-2y=6

-2y=-3x+6

$y=\frac{3}2{x-3}$

Therefore, your $y\textup{-intercept}$ is (0,-3)

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Example Question #711 : Geometry

Find where g(x) crosses the y-axis.

g(x)=5x^3-6x^2-5x+945

Possible Answers:

945

Correct answer:

945

Explanation:

Find where g(x) crosses the y-axis.

g(x)=5x^3-6x^2-5x+945

A function will cross the y-axis wherever x is equal to 0. This may be easier to see on a graph, but it can be thought of intuitively as well. If x is 0, then we are neither left nor right of the y-axis. This means we must be on the y-axis.

So, find g(0)

g(0)=5(0)^3-6(0)^2-5(0)+945=945

So our answer is 945.

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Example Question #1 : Calculating The Equation Of A Curve

Suppose the points (-3,-2) and (1,3) are plotted to connect a line. What are the -intercept and -intercept, respectively?

Possible Answers:

$-\frac{5}{4},\frac{7}{4}$

$-\frac{4}{5},\frac{7}{4}$

-1,1

$-\frac{7}{5},\frac{7}{4}$

$\frac{7}{5},\frac{7}{4}$

Correct answer:

$-\frac{7}{5},\frac{7}{4}$

Explanation:

First, given the two points, find the equation of the line using the slope formula and the y-intercept equation.

Slope:

$\frac{y_2-y_1}{x_2-x_1} = \frac{3-(-2)}{1-(-3)}=\frac{5}{4}$

Write the slope-intercept formula.

y=mx+b

Substitute a given point and the slope into the equation to find the y-intercept.

$3=\left(\frac{5}{4}\right)(1)+b$

$\frac{12}{4}=\frac{5}{4}+b$

$b=\frac{7}{4}$

The y-intercept is: $b=\frac{7}{4}$ .

Substitiute the slope and the y-intercept into the slope-intercept form.

$y=\frac{5}{4}x+\frac{7}{4}$

To find the x-intercept, substitute y=0 and solve for x.

$0=\frac{5}{4}x+\frac{7}{4}$

$-\frac{7}{4}=\frac{5}{4}x$

$-\frac{7}{4} \cdot (\frac{4}{5})=\frac{5}{4}x \cdot (\frac{4}{5})$

$x=-\frac{7}{5}$

The x-intercept is: $x=-\frac{7}{5}$

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GMAT Math : Geometry

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #943 : Problem Solving Questions

Example Question #11 : Calculating X Or Y Intercept

Example Question #951 : Problem Solving Questions

Example Question #952 : Problem Solving Questions

Example Question #953 : Problem Solving Questions

Example Question #954 : Problem Solving Questions

Example Question #955 : Problem Solving Questions

Example Question #956 : Problem Solving Questions

Example Question #711 : Geometry

Example Question #1 : Calculating The Equation Of A Curve

All GMAT Math Resources

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