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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

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Example Questions

Example Question #111 : Coordinate Geometry

Give the length, in terms of $\displaystyle A$, of a segment on the coordinate plane whose endpoints are $\displaystyle (A+2, A-2)$ and the origin.

Possible Answers:

$2 \sqrt{2A}$ $\displaystyle 2 \sqrt{2A}$

$\sqrt{ 2A^{2}+ 8 }$ $\displaystyle \sqrt{ 2A^{2}+ 8 }$

$\sqrt{ 2A^{2}- 8A+8 }$ $\displaystyle \sqrt{ 2A^{2}- 8A+8 }$

$\sqrt{ 2A^{2}- 8 }$ $\displaystyle \sqrt{ 2A^{2}- 8 }$

$\sqrt{ 2A^{2}+ 8A+8 }$ $\displaystyle \sqrt{ 2A^{2}+ 8A+8 }$

Correct answer:

$\sqrt{ 2A^{2}+ 8 }$ $\displaystyle \sqrt{ 2A^{2}+ 8 }$

Explanation:

The distance between two points $(x_{1}, x_{2})$ $\displaystyle (x_{1}, x_{2})$ and $(y_{1}, y_{2})$ $\displaystyle (y_{1}, y_{2})$ can be obtained using the formula

$d = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$ $\displaystyle d = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$

Setting $x_{1}= y_{1} = 0, x_{2}=A+2, y_{2} = A-2$ $\displaystyle x_{1}= y_{1} = 0, x_{2}=A+2, y_{2} = A-2$, we can find our expression:

$d = \sqrt{\left ( \left [A+2 \right )-0\right ]^{2}+\left [\left (A-2 \right )-0 \right ] ^{2}}$ $\displaystyle d = \sqrt{\left ( \left [A+2 \right )-0\right ]^{2}+\left [\left (A-2 \right )-0 \right ] ^{2}}$ $= \sqrt{\left ( A+2 \right ) ^{2}+ \left (A-2 \right ) ^{2}}$ $\displaystyle = \sqrt{\left ( A+2 \right ) ^{2}+ \left (A-2 \right ) ^{2}}$

$= \sqrt{\left ( A^{2}+4A+4 \right )+\left ( A^{2}-4A+4 \right ) }$ $\displaystyle = \sqrt{\left ( A^{2}+4A+4 \right )+\left ( A^{2}-4A+4 \right ) }$

$= \sqrt{ 2A^{2}+ 8 }$ $\displaystyle = \sqrt{ 2A^{2}+ 8 }$

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Example Question #112 : Coordinate Geometry

Give the length, in terms of $\displaystyle A$, of a segment on the coordinate plane whose endpoints are (A,A) $\displaystyle (A,A)$ and (3,3) $\displaystyle (3,3)$

Possible Answers:

$\sqrt{2A^{2} -18 }$ $\displaystyle \sqrt{2A^{2} -18 }$

A - 3 $\displaystyle A - 3$

$\sqrt{2A^{2}+12A+18 }$ $\displaystyle \sqrt{2A^{2}+12A+18 }$

$\sqrt{2A^{2} +18 }$ $\displaystyle \sqrt{2A^{2} +18 }$

$\sqrt{2A^{2}-12A+18 }$ $\displaystyle \sqrt{2A^{2}-12A+18 }$

Correct answer:

$\sqrt{2A^{2}-12A+18 }$ $\displaystyle \sqrt{2A^{2}-12A+18 }$

Explanation:

The distance between two points $(x_{1}, x_{2})$ $\displaystyle (x_{1}, x_{2})$ and $(y_{1}, y_{2})$ $\displaystyle (y_{1}, y_{2})$ can be obtained using the formula

$d = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$ $\displaystyle d = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$

Setting $x_{1}= y_{1} =3, x_{2}= y_{2} = A$ $\displaystyle x_{1}= y_{1} =3, x_{2}= y_{2} = A$, we can find our expression:

$d = \sqrt{\left (A-3 \right )^{2}+\left (A-3\right )^{2}}$ $\displaystyle d = \sqrt{\left (A-3 \right )^{2}+\left (A-3\right )^{2}}$

$= \sqrt{\left (A^{2}-6A+9 \right ) + \left (A^{2}-6A+9 \right ) }$ $\displaystyle = \sqrt{\left (A^{2}-6A+9 \right ) + \left (A^{2}-6A+9 \right ) }$

$= \sqrt{2A^{2}-12A+18 }$ $\displaystyle = \sqrt{2A^{2}-12A+18 }$

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Example Question #111 : Lines

A line segment on the coordinate plane has endpoints $\displaystyle (a-5,-b)$ and $\displaystyle (b+8, a-4)$. In terms of $\displaystyle a$ and $\displaystyle b$, as applicable, give the $\displaystyle x$-coordinate of its midpoint.

Possible Answers:

$\displaystyle a + b + 3$

$\displaystyle a - b - 4$

$\frac{a - b - 13}{2}$ $\displaystyle \frac{a - b - 13}{2}$

$\frac{a + b + 3}{2}$ $\displaystyle \frac{a + b + 3}{2}$

$\frac{a - b - 4}{2}$ $\displaystyle \frac{a - b - 4}{2}$

Correct answer:

$\frac{a + b + 3}{2}$ $\displaystyle \frac{a + b + 3}{2}$

Explanation:

The $\displaystyle x$-coordinate of the midpoint of a line segment is the mean of the $\displaystyle x$-coordinates of its endpoints. Therefore, the $\displaystyle x$-coordinate is

$x= \frac{(a-5)+(b+8)}{2} = \frac{a+b-5+8}{2} = \frac{a+b+3}{2}$ $\displaystyle x= \frac{(a-5)+(b+8)}{2} = \frac{a+b-5+8}{2} = \frac{a+b+3}{2}$.

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Example Question #391 : Ssat Upper Level Quantitative (Math)

Possible Answers:

$\frac{a + b + 3}{2}$ $\displaystyle \frac{a + b + 3}{2}$

$\displaystyle a + b + 3$

$\frac{a - b - 4}{2}$ $\displaystyle \frac{a - b - 4}{2}$

$\frac{a + b - 4}{2}$ $\displaystyle \frac{a + b - 4}{2}$

$\displaystyle a - b - 4$

Correct answer:

$\frac{a - b - 4}{2}$ $\displaystyle \frac{a - b - 4}{2}$

Explanation:

The $\displaystyle y$-coordinate of the midpoint of a line segment is the mean of the $\displaystyle y$-coordinates of its endpoints. Therefore, the $\displaystyle y$-coordinate is

$y= \frac{(a-4)+(-b)}{2} = \frac{a-b-4}{2}$ $\displaystyle y= \frac{(a-4)+(-b)}{2} = \frac{a-b-4}{2}$.

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Example Question #392 : Ssat Upper Level Quantitative (Math)

A line segment has the endpoints (6,10) $\displaystyle (6,10)$ and $\displaystyle (-2, -4)$. What is the midpoint of the line?

Possible Answers:

(4,6) $\displaystyle (4,6)$

(8,14) $\displaystyle (8,14)$

(2, 3) $\displaystyle (2, 3)$

(4,7) $\displaystyle (4,7)$

Correct answer:

(2, 3) $\displaystyle (2, 3)$

Explanation:

To find the midpoint of a line, take the averages of the x-coordinates and the average of the y-coordinates.

$\text{Midpoint}=(\frac{6+(-2)}{2}, \frac{10+(-4)}{2})=(\frac{4}{2}, \frac{6}{2})=(2,3)$ $\displaystyle \text{Midpoint}=(\frac{6+(-2)}{2}, \frac{10+(-4)}{2})=(\frac{4}{2}, \frac{6}{2})=(2,3)$

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Example Question #4 : How To Find The Midpoint Of A Line Segment

What is the midpoint of a line segment with endpoints (12,2) $\displaystyle (12,2)$ and (-2, 1) $\displaystyle (-2, 1)$?

Possible Answers:

(10, 3) $\displaystyle (10, 3)$

(14,1) $\displaystyle (14,1)$

$(5, \frac{3}{2})$ $\displaystyle (5, \frac{3}{2})$

$(\frac{13}{2}, 0)$ $\displaystyle (\frac{13}{2}, 0)$

Correct answer:

$(5, \frac{3}{2})$ $\displaystyle (5, \frac{3}{2})$

Explanation:

To find the midpoint of a line, just take the average of the $\displaystyle x$-coordinates and the average of the $\displaystyle y$-coordinates.

$\text{Midpoint}=(\frac{12+(-2)}{2}, \frac{2+1}{2})=(\frac{10}{2}, \frac{3}{2})=(5, \frac{3}{2})$ $\displaystyle \text{Midpoint}=(\frac{12+(-2)}{2}, \frac{2+1}{2})=(\frac{10}{2}, \frac{3}{2})=(5, \frac{3}{2})$

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Example Question #2 : How To Find The Midpoint Of A Line Segment

What is the midpoint of a line segments with endpoints $\displaystyle (2a, -b)$ and $\displaystyle (4a, 5b)$?

Possible Answers:

$\displaystyle (6a, 4b)$

(3, 4) $\displaystyle (3, 4)$

$\displaystyle (-a, -2b)$

$\displaystyle (3a, 2b)$

Correct answer:

$\displaystyle (3a, 2b)$

Explanation:

To find the midpoint of a line segment, take the average of the $\displaystyle x$-coordinates, then take the average of the $\displaystyle y$-coordinates.

$\text{Midpoint}=(\frac{2a+4a}{2}, \frac{-b+5b}{2})$ $\displaystyle \text{Midpoint}=(\frac{2a+4a}{2}, \frac{-b+5b}{2})$

$\text{Midpoint}=(\frac{6a}{2}, \frac{4b}{2})=(3a, 2b)$ $\displaystyle \text{Midpoint}=(\frac{6a}{2}, \frac{4b}{2})=(3a, 2b)$

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Example Question #3 : Midpoint Formula

A line has the endpoints $(8,8) \text{ and }(2, -1)$ $\displaystyle (8,8) \text{ and }(2, -1)$. What is its midpoint?

Possible Answers:

(5, 7) $\displaystyle (5, 7)$

$\left(5, \frac{7}{2}\right)$ $\displaystyle \left(5, \frac{7}{2}\right)$

(10, 7) $\displaystyle (10, 7)$

$\left(\frac{7}{2}, 5\right)$ $\displaystyle \left(\frac{7}{2}, 5\right)$

Correct answer:

$\left(5, \frac{7}{2}\right)$ $\displaystyle \left(5, \frac{7}{2}\right)$

Explanation:

The coordinates of the midpoint of a line are just the averages of the coordinates of the endpoints.

$\text{Midpoint}=\left(\frac{8+2}{2}, \frac{8-1}{2}\right)=\left(5, \frac{7}{2}\right)$ $\displaystyle \text{Midpoint}=\left(\frac{8+2}{2}, \frac{8-1}{2}\right)=\left(5, \frac{7}{2}\right)$

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Example Question #4 : Midpoint Formula

A line segment has endpoints $(-2, 0) \text{ and }(1, 5)$ $\displaystyle (-2, 0) \text{ and }(1, 5)$. What is the midpoint of this line segment?

Possible Answers:

(-1, 5) $\displaystyle (-1, 5)$

$\left(-\frac{1}{2}, 5\right)$ $\displaystyle \left(-\frac{1}{2}, 5\right)$

$\left(5, -\frac{1}{2}\right)$ $\displaystyle \left(5, -\frac{1}{2}\right)$

$\left(-\frac{1}{2}, \frac{5}{2}\right)$ $\displaystyle \left(-\frac{1}{2}, \frac{5}{2}\right)$

Correct answer:

$\left(-\frac{1}{2}, \frac{5}{2}\right)$ $\displaystyle \left(-\frac{1}{2}, \frac{5}{2}\right)$

Explanation:

To find the coordinates of the midpoint of a line, take the averages of the x and y coordinates of the endpoints.

$\text{Midpoint}=\left(\frac{-2+1}{2}, \frac{5+0}{2}\right)=\left(-\frac{1}{2}, \frac{5}{2}\right)$ $\displaystyle \text{Midpoint}=\left(\frac{-2+1}{2}, \frac{5+0}{2}\right)=\left(-\frac{1}{2}, \frac{5}{2}\right)$

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Example Question #8 : How To Find The Midpoint Of A Line Segment

What is the midpoint of a line if the endpoints of the line are: $(6,9) \textup{ and }(3,-2)$ $\displaystyle (6,9) \textup{ and }(3,-2)$?

Possible Answers:

$\left(-\frac{3}{2},\frac{1}{2}\right)$ $\displaystyle \left(-\frac{3}{2},\frac{1}{2}\right)$

$\left(\frac{3}{2},\frac{11}{2}\right)$ $\displaystyle \left(\frac{3}{2},\frac{11}{2}\right)$

$\left(\frac{9}{2},\frac{7}{2}\right)$ $\displaystyle \left(\frac{9}{2},\frac{7}{2}\right)$

$\left(-\frac{3}{2},-\frac{11}{2}\right)$ $\displaystyle \left(-\frac{3}{2},-\frac{11}{2}\right)$

$\left(\frac{7}{2},\frac{9}{2}\right)$ $\displaystyle \left(\frac{7}{2},\frac{9}{2}\right)$

Correct answer:

$\left(\frac{9}{2},\frac{7}{2}\right)$ $\displaystyle \left(\frac{9}{2},\frac{7}{2}\right)$

Explanation:

Write the midpoint formula.

$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ $\displaystyle M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Substitute the values.

$M=\left(\frac{6+3}{2},\frac{9-2}{2}\right)= \left(\frac{9}{2},\frac{7}{2}\right)$ $\displaystyle M=\left(\frac{6+3}{2},\frac{9-2}{2}\right)= \left(\frac{9}{2},\frac{7}{2}\right)$

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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #111 : Coordinate Geometry

Example Question #112 : Coordinate Geometry

Example Question #111 : Lines

Example Question #391 : Ssat Upper Level Quantitative (Math)

Example Question #392 : Ssat Upper Level Quantitative (Math)

Example Question #4 : How To Find The Midpoint Of A Line Segment

Example Question #2 : How To Find The Midpoint Of A Line Segment

Example Question #3 : Midpoint Formula

Example Question #4 : Midpoint Formula

Example Question #8 : How To Find The Midpoint Of A Line Segment

All SSAT Upper Level Math Resources

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