GMAT Math : GMAT Quantitative Reasoning

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Example Questions

Example Question #3 : Calculating Whether Point Is On A Line With An Equation

Consider segment $\overline{JK}$ which passes through the points $\left ( 4,5\right )$ and $\left ( 144,75\right )$ .

If the point $\left ( 16,y\right )$ is on $\overline{JK}$ , what is the value of y?

Possible Answers:

Correct answer:

Explanation:

First, use the points to find the equation of JK:

Given that JK passes through (4,5) and (144,75) we can find the slope as follows:

Slope is found via:

$m=\frac{y'-y}{x'-x}$

Plug in and calculate:

$\small \small m=\frac{75-5}{144-4}=\frac{70}{140}=\frac{1}{2}$

Next, we need to use one of our points and the slope to find our y-intercept. I'll use (4,5).

$\small 5=\frac{1}{2}*4+b$

$\small b=5-2=3$

So our answer is:

$\small y=\frac{x}{2}+3$

To find y, we need to plug in 16 for x and solve:

$\small \small y=\frac{16}{2}+3=8+3=11$

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Example Question #672 : Geometry

If f(x) is defined as follows, is the point $\left ( -2,5\right )$ on f(x) ?

$\small f(x)=4x+13$

Possible Answers:

f(x) is undefined at (-2,5)

Cannot be calculated from the provided information

Yes

Correct answer:

Yes

Explanation:

To find out if (-2,5) is on f(x), simply plug the point into f(x)

$\small f(x)=4x+13$

Becomes,

$\small \small 5=4*-2+13=-8+13=5$

So yes, it does!

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Example Question #5 : Calculating Whether Point Is On A Line With An Equation

Which of the following are points along g(x) if

${}g(x)=4x^2-8$ .

Possible Answers:

(3,28)

(-3,-28)

(4,64)

(3,-28)

(2,4)

Correct answer:

(3,28)

Explanation:

One way to solve this one is by plugging in each of the answer choices and eliminating any that don't work out. Begin with our original g(x)

${}g(x)=4x^2-8$

If we plug in 3 we get

${}g(x)=4(3)^2-8=4(9)-8=36-8=28$ ,

So our point is (3,28).

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Example Question #673 : Geometry

What is the equation of a line with slope and a point (1,4) ?

Possible Answers:

y=x+3

y=x-4

y=x+1

y=x+4

Correct answer:

y=x+3

Explanation:

Since the slope and a point on the line are given, we can use the point-slope formula:

$y-y_{1}=m(x-x_{1})$

$m=1\ and\ (1,4)$

y-4=1(x-1)

y-4=x-1

y=x-1+4

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Example Question #674 : Geometry

What is the equation of a line with slope and point (5,2) ?

Possible Answers:

y=5

y=x+2

y=2

y=x+5

Correct answer:

y=2

Explanation:

Since the slope and a point on the line are given, we can use the point-slope formula:

$y-y_{1}=m(x-x_{1})$

$m=0\ and\ (5,2)$

y-2=0(x-5)

y-2=0

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Example Question #3 : Calculating The Equation Of A Line

What is the equation of a line with slope $-\frac{4}{3}$ and a point (-8,7) ?

Possible Answers:

$y=-\frac{4}{3}x-\frac{3}{11}$

$y=\frac{4}{3}x+\frac{11}{3}$

$y=-\frac{4}{3}x-\frac{11}{3}$

$y=\frac{4}{3}x+\frac{32}{3}$

Correct answer:

$y=-\frac{4}{3}x-\frac{11}{3}$

Explanation:

Since the slope and a point on the line are given, we can use the point-slope formula:

$y-y_{1}=m(x-x_{1})$

slope: $-\frac{4}{3}$ and point: (-8,7)

$y-7=-\frac{4}{3}(x-(-8))$

$y-7=-\frac{4}{3}(x+8)$

$y-7=-\frac{4}{3}x-\frac{32}{3}$

$y=-\frac{4}{3}x-\frac{32}{3}+7$

$y=-\frac{4}{3}x-\frac{32}{3}+\frac{21}{3}$

$y=-\frac{4}{3}x-\frac{11}{3}$

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Example Question #911 : Gmat Quantitative Reasoning

Find the equation of the line through the points (4, -2) and (1, 7) .

Possible Answers:

y=-3

y=2x+9

y=3x+3

y=2x-2

y=-3x+10

Correct answer:

y=-3x+10

Explanation:

First find the slope of the equation.

$m =\frac{rise}{run} =\frac{7 + 2}{1-4} = \frac{9}{-3}=-3$

Now plug in one of the two points to form an equation. Here we use (4, -2), but either point will produce the same answer.

$y-(-2)=-3(x-4)$

$y + 2=-3x + 12$

$y=-3x+10$

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Example Question #5 : Calculating The Equation Of A Line

Consider segment $\overline{JK}$ which passes through the points $\left ( 4,5\right )$ and $\left ( 144,75\right )$ .

Find the equation of $\overline{JK}$ in the form y=mx+b .

Possible Answers:

$\small \small y=\frac{x}{2}+6$

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

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

$\small y=\frac{x}{2}+3$

$\small \small y=2x+3$

Correct answer:

$\small y=\frac{x}{2}+3$

Explanation:

Given that JK passes through (4,5) and (144,75) we can find the slope as follows:

Slope is found via:

$m=\frac{y'-y}{x'-x}$

Plug in and calculate:

$\small \small m=\frac{75-5}{144-4}=\frac{70}{140}=\frac{1}{2}$

Next, we need to use one of our points and the slope to find our y-intercept. I'll use (4,5).

$\small 5=\frac{1}{2}*4+b$

$\small b=5-2=3$

So our answer is:

$\small y=\frac{x}{2}+3$

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Example Question #2 : Calculating The Equation Of A Line

Determine the equation of a line that has the points (-2,3) and (6,7) ?

Possible Answers:

y=-2x-2

$y=\frac{1}{2}x+4$

y=2x+2

$y=-\frac{1}{2}x-4$

$y=-\frac{1}{4}x+2$

Correct answer:

$y=\frac{1}{2}x+4$

Explanation:

The equation for a line in standard form is written as follows:

y=mx+b

Where is the slope and is the y intercept. We start by calculating the slope between the two given points using the following formula:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{7-3}{6-(-2)}=\frac{4}{8}=\frac{1}{2}$

Now we can plug either of the given points into the formula for a line with the calculated slope and solve for the y intercept:

y=mx+b

$3=(\frac{1}{2})(-2)+b$

b=4

We now have the slope and the y intercept of the line, which is all we need to write its equation in standard form:

$y=\frac{1}{2}x+4$

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Example Question #7 : Calculating The Equation Of A Line

Give the equation of the line that passes through the -intercept and the vertex of the parabola of the equation

$y = x^{2} + 5x - 6$ .

Possible Answers:

y=6x-6

$y=\frac{5}{2 } x - 6$

y = -6x+6

$y=-\frac{5}{2 } x + 6$

y=-x-6

Correct answer:

$y=\frac{5}{2 } x - 6$

Explanation:

The -intercept of the parabola of the equation can be found by substituting 0 for :

$y = x^{2} + 5x - 6$

$y = 0^{2} + 5 (0) - 6$

y = -6

This point is (0, -6) .

The vertex of the parabola of the equation $y = a x^{2} + bx+c$ has -coordinate $- \frac{b}{2a}$ , and its -coordinate can be found using substitution for . Setting a = 1 and b= 5 :

$x = - \frac{b}{2a} = - \frac{5}{2 \cdot 1} = - \frac{5}{2 }$

$y = \left ( - \frac{5}{2 } \right )^{2} + 5 \left ( - \frac{5}{2 } \right ) - 6$

$y = \frac{25}{4 } - \frac{25}{2 } - 6$

$y = \frac{25}{4 } - \frac{50}{4 } - \frac{24}{4}$

$y = - \frac{49}{4 }$

The vertex is $\left (- \frac{5}{2 },- \frac{49}{4 } \right )$

The line connects the points (0, -6) and $\left (- \frac{5}{2 },- \frac{49}{4 } \right )$ . Its slope is

$m= \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

$= \frac{-6-\left ( - \frac{49}{4 } \right )}{0 - \left (- \frac{5}{2 } \right )}$

$= \frac{-\frac{24 }{4 } + \frac{49}{4 } }{ \frac{5}{2 } }$

$= \frac{ \frac{25}{4 } }{ \frac{5}{2 } }$

$= \frac{25}{4 } \cdot \frac{2 } {5}$

$=\frac{5}{2 }$

Since the line has -intercept (0, -6) and slope $m =\frac{5}{2 }$ , the equation of the line is $y=\frac{5}{2 } x + (- 6)$ , or $y=\frac{5}{2 } x - 6$ .

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GMAT Math : GMAT Quantitative Reasoning

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #3 : Calculating Whether Point Is On A Line With An Equation

Example Question #672 : Geometry

Example Question #5 : Calculating Whether Point Is On A Line With An Equation

Example Question #673 : Geometry

Example Question #674 : Geometry

Example Question #3 : Calculating The Equation Of A Line

Example Question #911 : Gmat Quantitative Reasoning

Example Question #5 : Calculating The Equation Of A Line

Example Question #2 : Calculating The Equation Of A Line

Example Question #7 : Calculating The Equation Of A Line

All GMAT Math Resources

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