SSAT Upper Level Quantitative (Math)

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SSAT Upper Level Quantitative › SSAT Upper Level Quantitative (Math)

Questions 1 - 10

The area of a right triangle is $50\textup{ in}^2$ . If the base of the triangle is $10\textup{ in}$ , what is the length of the height, in inches?

Explanation

To find the height, plug what is given in the question into the formula used to find the area of a triangle.

$\text{Area}=\frac{base\times height}{2}$

Use the information given in the question:

$\text{Base}=10$

$\text{Area}=50$

$50=\frac{10\times height}{2}$

Now, solve for the height.

$10\times height=100$

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ ; $\angle A \cong \angle D$ and $\angle C \cong \angle F$ .

Which of the following statements would not be enough, along with what is given, to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ ?

The given information is enough to prove the triangles similar.

$\frac{AB}{DE} = \frac{AC}{DF}$

$\frac{AB}{DE} = \frac{BC}{EF}$

$\frac{AC}{DF}= \frac{BC}{EF}$

$\angle B \cong \angle E$

Explanation

Two pairs of corresponding angles are stated to be congruent in the main body of the problem; it follows from the Angle-Angle Similarity Postulate that the triangles are similar. No further information is needed.

Convert into a fraction in simplest form.

$\frac{19}{50}$

$\frac{19}{25}$

$\frac{38}{50}$

$\frac{5}{19}$

Explanation

Think of the decimal as the fraction $\frac{0.38}{1}$ .

Now, multiply the numerator and denominator by for every digit after the decimal. In this case, because there are digits after the decimal, we will be multiplying by twice, or by .

$\frac{0.38}{1}\times 100=\frac{38}{100}$

Now, simplify this fraction.

$\frac{38}{100}=\frac{19}{50}$

Function 2

What equation is graphed in the above figure?

$y = \left \lfloor x\right \rfloor - 4$

$y = \left \lceil x\right \rceil - 4$

$y = \left \lfloor x\right \rfloor - 3$

$y = \left \lceil x\right \rceil - 3$

$y = \left \lfloor x\right \rfloor +3$

Explanation

The greatest integer function, or floor function, $y = \left \lfloor x\right \rfloor$ , pairs each value of with the greatest integer less than or equal to . Its graph is below.

Floor function

The given graph is the above graph shifted downward four units. The graph of any function shifted downward four units is , so the given graph corresponds to equation $y = \left \lfloor x\right \rfloor - 4$ .

There is a line defined by the equation below:

There is a second line that passes through the point and is parallel to the line given above. What is the equation of this second line?

$y=-\frac{3}{4}x+2.75$

$y=-\frac{3}{4}x+1.25$

$y=\frac{3}{4}x+2.75$

$y=\frac{3}{4}x+1.25$

$y=\frac{3}{4}x+2.625$

Explanation

Parallel lines have the same slope. Solve for the slope in the first line by converting the equation to slope-intercept form.

3x + 4y = 12

4y = _–_3x + 12

y = –(3/4)x + 3

slope = _–_3/4

We know that the second line will also have a slope of _–_3/4, and we are given the point (1,2). We can set up an equation in slope-intercept form and use these values to solve for the y-intercept.

y = mx + b

2 = _–_3/4(1) + b

2 = _–_3/4 + b

b = 2 + 3/4 = 2.75

Plug the y-intercept back into the equation to get our final answer.

y = –(3/4)x + 2.75

Find the equation of a line that goes through the point and is parallel to the line with the equation .

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

Explanation

For lines to be parallel, they must have the same slope. The slope of the line we are looking for then must be .

The point that's given in the equation is also the y-intercept.

Using these two pieces of information, we know that the equation for the line must be

Convert ${}9\tfrac{2}{5}$ to an improper fraction.

${}\frac{47}{5}$

$\frac{90}{5}$

$\frac{16}{5}$

$\frac{23}{5}$

$\frac{78}{5}$

Explanation

To convert into an improper fraction, take the whole number and multiply that with the denominator .

$9\cdot 5=45$

Then, we add that to the numerator which is .

Then we take that sum and put it over th denominator which gives us an answer of:

${}\frac{47}{5}$ .

The area of a rectangle is square feet. The width of the rectangle is four-sevenths of its length. Give the length of the rectangle in inches in terms of .

$6\sqrt{ 7K }$

$\frac{ \sqrt{ 7K } }{24}$

$\frac{ \sqrt{ 7K } }{2}$

$\frac{7K} {48}$

Explanation

Let be the length in feet. Then the width of the rectangle in feet is four-sevenths of this, or $\frac{4}{7}L$ . The area is equal to the product of the length and the width, so set up this equation and solve for :