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

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Example Questions

Example Question #124 : Properties Of Exponents

x> 0 and y > 0

$x^{2} = 8$

$y^{2} = 18$

Evaluate $(x+ y) ^{2}$ .

Possible Answers:

144

Correct answer:

Explanation:

By the perfect square trinomial pattern,

$(x+ y) ^{2}$

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

$x^{2} = 8$ and $y^{2} = 18$ .

Also, by the Power of a Power Principle,

$(xy) ^{2} = x^{2} y^{2} = 8 \cdot 18 = 144$ ,

so, since and are both positive,

$xy = \sqrt{(xy) ^{2} }= \sqrt{144}= 12$ .

Therefore,

$(x+ y) ^{2}$

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

= 8 +2 (12)+18

= 8 +24 +18

= 50

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Example Question #125 : Properties Of Exponents

$y^{3} = -17$

$y^{9} = ?$

Possible Answers:

4,913

-4,913

Correct answer:

-4,913

Explanation:

By the Power of a Power Principle,

$(y ^{3} ) ^{3} = y ^{3 \cdot 3} = y ^{9}$

Therefore, we substitute, keeping in mind that an odd power of a negative number is also negative:

$y ^{9} = (y ^{3} ) ^{3}$

$= (-17) ^{3}$

$= - 17 ^{3}$

$= - (17 \cdot 17\cdot 17)$

= -4,913

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Example Question #183 : Algebra

x> 0 and y > 0

$x^{2} = 18$

$y^{2} = 8$

Evaluate $(x- y) ^{2}$ .

Possible Answers:

Correct answer:

Explanation:

By the perfect square trinomial pattern,

$(x- y) ^{2}$

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

$x^{2} =1 8$ and $y^{2} = 8$ .

Also, by the Power of a Power Principle,

$(xy) ^{2} = x^{2} y^{2} = 18 \cdot 8 = 144$ ,

so, since and are both positive,

$xy = \sqrt{(xy) ^{2} }= \sqrt{144}= 12$ .

Therefore,

$(x- y) ^{2}$

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

And, substituting:

$x^{2} -2xy+ y ^{2}$

=18 -2 (12)+8

=18 -24 +8

= 2

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Example Question #184 : Algebra

$t^{6} = 144$

$x^{6}= 81$

t > 0, x< 0

Evaluate the expression $(t-x) (t^{2}+tx+x^{2})$ .

Possible Answers:

Correct answer:

Explanation:

Multiply out the expression by using multiple distributions and collecting like terms:

$(t-x) (t^{2}+tx+x^{2})$

$= t (t^{2}+tx+x^{2}) - x (t^{2}+tx+x^{2})$

$= t \cdot t^{2}+ t \cdot tx+ t \cdot x^{2} - x \cdot t^{2}- x \cdot tx- x \cdot x^{2}$

$= t^{3}+ t^{2} x+ t x^{2} - t^{2} x- tx^{2} - x^{3}$

$= t^{3}+ t^{2} x- t^{2} x + t x^{2} - tx^{2} - x^{3}$

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

Since $(t^{3} ) ^{2} = t ^{3 \cdot 2 } = t^{6}$ by the Power of a Power Principle,

$t^{3} = \pm \sqrt{ t^{6}} = \pm \sqrt{144} = \pm 12$ .

However, is positive, so $t^{3}$ is as well, so we choose $t^{3} = 12$ .

Similarly,

$x^{3} = \pm \sqrt{x^{6}} = \pm \sqrt{81} = \pm 9$ .

However, since is negative, as an odd power of a negative number, $x^{3}$ is as well, so we choose $x^{3} = -9$ .

Therefore, substituting:

$(t-x) (t^{2}+tx+x^{2}) = t^{3} - x^{3} = 12 - (-9) = 21$

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Example Question #121 : Properties Of Exponents

and are both positive integers; A is odd. What can you say about the number

$(A + B + 1)^{A+B+1}$ ?

Possible Answers:

$(A + B + 1)^{A+B+1}$ is odd if is odd, and even if is even.

$(A + B + 1)^{A+B+1}$ is even if is even, and can be odd or even if is odd.

$(A + B + 1)^{A+B+1}$ is even if is odd, and odd if is even.

$(A + B + 1)^{A+B+1}$ is even if is odd, and can be odd or even if is even.

$(A + B + 1)^{A+B+1}$ is odd if is odd, and can be odd or even if is even.

Correct answer:

$(A + B + 1)^{A+B+1}$ is odd if is odd, and even if is even.

Explanation:

If is odd, then A + B +1 , the sum of three odd integers, is odd; an odd number taken to any positive integer power is odd.

If is even, then A + B +1 , the sum of two odd integers and an even integer, is even; an even number taken to any positive integer power is even.

Therefore, $(A + B + 1)^{A+B+1}$ always assumes the same odd/even parity as .

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Example Question #1 : Equations Of Lines

Give the equation of a line that passes through the point (4,7) and has slope 1.

Possible Answers:

x = 4

x + y = 11

x-y = 3

x-y = -3

y = 7

Correct answer:

x-y = -3

Explanation:

We can use the point slope form of a line, substituting $m = 1, x_{1} = 4, y _{1} = 7$ .

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

y- 7 = 1 (x-4)

y- 7 = x-4

y- 7 -y +4 = x-4 -y +4

-3 = x-y

x-y = -3

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Example Question #181 : Algebra

A line can be represented by 2x+3y=6 . What is the slope of the line that is perpendicular to it?

Possible Answers:

$\frac{2}{3}$

$\frac{3}{4}$

$\frac{3}{2}$

$-\frac{2}{3}$

Correct answer:

$\frac{3}{2}$

Explanation:

You will first solve for Y, to get the equation in y=mx+b form.

represents the slope of the line, which would be $-\frac{2}{3}$ .

A perpendicular line's slope would be the negative reciprocal of that value, which is $\frac{3}{2}$ .

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Example Question #1 : How To Find The Equation Of A Line

Find the equation the line goes through the points (3,4) and (0,2) .

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

First, find the slope of the line.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{2-4}{0-3}=\frac{-2}{-3}=\frac{2}{3}$

Now, because the problem tells us that the line goes through (0,2) , our y-intercept must be .

Putting the pieces together, we get the following equation:

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

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Example Question #3 : How To Find The Equation Of A Line

A line passes through the points (-4, 6) and (-2, -2) . Find the equation of this line.

Possible Answers:

y=-4x-10

$y=\frac{1}{4}x-8$

$y=-\frac{1}{4}x+10$

y=4x+10

Correct answer:

y=-4x-10

Explanation:

To find the equation of a line, we need to first find the slope.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-6}{-2-(-4)}=\frac{-8}{2}=-4$

Now, our equation for the line looks like the following:

y=-4x+b

To find the y-intercept, plug in one of the given points and solve for . Using (-4, 6) , we get the following equation:

6=-4(-4)+b

Solve for .

6=16+b

b=-10

Now, plug the value for into the equation.

y=-4x-10

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

What is the equation of a line that passes through the points (8,1) and (-2, -7) ?

Possible Answers:

$y=\frac{4}{5}x-\frac{27}{5}$

$y=\frac{27}{5}x-\frac{4}{5}$

$y=\frac{5}{4}x+\frac{32}{5}$

$y=-\frac{4}{5}x+\frac{27}{5}$

Correct answer:

$y=\frac{4}{5}x-\frac{27}{5}$

Explanation:

First, we need to find the slope of the line.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{-7-1}{-2-8}=\frac{-8}{-10}=\frac{4}{5}$

Next, find the -intercept. To find the -intercept, plug in the values of one point into the equation y=mx+b , where is the slope that we just found and is the -intercept.

$1=8(\frac{4}{5})+b$

Solve for .

$1=\frac{32}{5}+b$

$b=-\frac{27}{5}$

Now, put the slope and -intercept together to get $y=\frac{4}{5}x-\frac{27}{5}$

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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 #124 : Properties Of Exponents

Example Question #125 : Properties Of Exponents

Example Question #183 : Algebra

Example Question #184 : Algebra

Example Question #121 : Properties Of Exponents

Example Question #1 : Equations Of Lines

Example Question #181 : Algebra

Example Question #1 : How To Find The Equation Of A Line

Example Question #3 : How To Find The Equation Of A Line

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

All SSAT Upper Level Math Resources

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