SAT Math Test 3 (Medium)

Try our third free SAT Math practice test. This one is even more challenging, with a mix of medium and difficult practice questions. This portion of the SAT Math test does allow a calculator, but it won’t be needed for every question. Topics include algebra, data analysis, and advanced math. Continue your math prep right now with our free SAT Math practice questions.

Directions: Solve each problem and select the best of the answer choices provided. The use of a calculator is permitted.

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Question 1
The circumference of a right cylinder is half its height. The radius of the cylinder is x. What is the volume of the cylinder in terms of x?

A
$2 \pi x^3$
B
$3 \pi x^3$
C
$3\pi^2x^3$
D
$4\pi^2x^3$
Question 1 Explanation: 
The correct answer is (D). Begin by noting the relevant formulas:

Circumference: $C=2 \pi r$
Volume: $V = \pi r^2 h$

Because the radius is equal to x, and the circumference is equal to half the height, we can write:

$C = 2 \pi r = \frac{1}{2} h$, or $h = 4 \pi r$

Substitute this expression for h into the volume formula:

$V = \pi x^2 (4 \pi x) = 4 \pi^2x^2$
Question 2
In the given figure, the measure of angle OAC is 60 degrees, and the center of the circle is O. If the circle has a radius of 6, what is the length of segment DB?

A
$3$
B
$3\sqrt{2}$
C
$6$
D
$6\sqrt{2}$
Question 2 Explanation: 
The correct answer is (C).

Since O is the center, AO = DO = CO = BO = 6. This means that both triangles are isosceles.

If triangle AOC is isosceles, and angle ∠OAC = 60°, then angle ∠ACO = 60°. This leaves 180° − 120° = 60° degrees for the third angle ∠AOC.

Triangle AOC = equilateral, and angle ∠DOB is vertical with angle ∠AOC and will also equal 60 degrees.

Because DO = BO, angle ∠OBD = ∠ODB. To solve for angle ∠OBD: 180° - 2x = 60°, so x = 60°. All three angles = 60°, so therefore, triangle DOB will also be equilateral.

Length of DB = 6, which is choice (C).
Question 3
A perfect sphere with a diameter of 5 meters is inscribed in a cube. Which of the following best approximates the volume of the space between the sphere and the cube?

A
15 in3
B
25 in3
C
45 in3
D
60 in3
Question 3 Explanation: 
The correct answer is (D).

It may help you to draw a picture to better visualize the sphere and the cube.

Volume of the space between the figures = total volume of the cube - volume of the sphere

The volume of the cube is: $length * width * height = 5*5*5 = 125$

The volume of the sphere is:
$\frac{4}{3}$ $\pi$ $($$\frac{5}{2}$$)$$^3$ ≈ $65.44$

So the volume of the space between the cube is: $125 - 65.44 ≈ 59.55 ≈ 60$
Question 4
On a coordinate plane, (a, b) and (a + 5, b + c), and (13, 10) are three points on line l. If the x-intercept of line l is −7, what is the value of c?

A
1.5
B
2.0
C
2.5
D
3.0
Question 4 Explanation: 
The correct answer is (C).

Recall that all lines can be written in the form y = mx + b, where m is the slope of the line (rise/run), and b is the y-intercept of the line.

Given that the coordinate plane uses the variables a and b for x and y, we can rewrite the line equation as: b = ma + k, where the variable k is used to replace the original variable b for the y-intercept to avoid confusion.

Using this revised equation in conjunction with the given points, we can first solve for the slope of the line in terms of c. Given that the slope of a line is the change in the y variable divided by the change in the x variable, calculate the line’s slope:

$slope = \frac{y_2 - y_1}{x_2 - x_1} = \frac{(b+c)-b}{(a+5)-a} = \frac{c}{5}$

Substitute this value for the slope into the linear equation:

$b = \frac{c}{5} * a + k$

Notice that the question only asks for the value of c, which is a part of the slope of the line. If we use the given information to determine the actual value of the slope, we can equate the expression containing c with the actual value and solve for c.

Given that the line has an x-intercept of −7, we can deduce that (−7, 0) is a point on the line. Calculate the slope of the line using this point and the point (13,10):

$slope = \frac{10 - 0}{13-(-7)} = \frac{10}{20} = \frac{1}{2}$

Equate this value with the expression containing c:

$\frac{1}{2} = \frac{c}{5}$

$c = 5 * \frac{1}{2} = 2.5$
Question 5
Under which conditions is the expression $\frac{ab}{a-b}$ always less than zero?

A
$a$ $<$ $b$ $<$ $0$
B
$0$ $<$ $b$ $<$ $a$
C
$a$ $<$ $0$ $<$ $b$
D
$b$ $<$ $a$ $<$ $0$
Question 5 Explanation: 
The correct answer is (A).

This is an example of a problem where it may be simplest to work backwards (check each of the answers to see which one works). For each of the answers, choose values for a and b that meet the criteria then plug these values into the original expression to see if they satisfy the requirement.

A) $a < b < 0: Let~a=-2 ~and~b=-1$

$→ \frac{[(-2)(-1)]}{[-2 - (-1)]} = \frac{2}{-1} = -2$ (works)

(Note: this makes sense since if a and b are both negative (less than zero), their product will be positive; and if a is less than b, then ab will be negative; thus, we will have a positive numerator and a negative denominator, resulting in a negative quotient.)

B) $0 < b < a: Let~b=1~and~a=2$

$→ \frac{[(2)(1)]}{[2 - 1]} = \frac{2}{1} = 2$ (does not work)

C) $a < 0 < b: Let~a=-1~and~b=1$

$→ \frac{[(-1)(1)]}{[(-1) - (1)]} = \frac{-1}{-2} = \frac{1}{2}$ (does not work)

D) $b < a < 0: Let~b=-2~and~a=-1$

$→ \frac{[(-1)(-2)]}{[-1 - (-2)]} = \frac{2}{1} = 2$ (does not work)
Question 6

What is the value of a + b?

A
wxxy + z
B
360 − x + y + w + z
C
180 − y + zw
D
w + x + y + z − 360
Question 6 Explanation: 
The correct answer is (D).

Recall that angles forming a straight line and the 3 interior angles of a triangle each sum to 180°. Using these relationships and substituting expressions for the unlabeled angles, we can then solve the equations in terms of a and b. We can then combine these equations to solve for a + b.

For the top triangle:
a + (180 − w) + (180 − x) = 180°
360 + awx = 180
a = −180 + w + x

For the bottom triangle:
b + (180 − y) + (180 − z) = 180°
360 + byz = 180
b = −180 + y + z

Adding the 2 equations together:
a + b = (−180 + w + x) + (−180 + y + z)
a + b = −360 + w + x + y + z
Question 7
A cylindrical birthday cake with a height of 4 inches is cut into two pieces such that each piece is of a different size. If the ratio of the volume of the larger slice to the volume of the smaller slice is 5 to 3, what is the degree measure of the cut made into the cake?

A
115°
B
120°
C
135°
D
145°
Question 7 Explanation: 
The correct answer is (C).

Since the ratio of the larger slice to the smaller slice is 5 to 3, the ratio of the area of the smaller slice to the area of the entire cake must be 3 to 8.

This ratio is the same as the ratio of the interior angle of the smaller slice to 360° (the entire cake).

We can therefore set up a proportion:

$\frac{3}{8} = \frac{x}{360}$

360(3) = 8x
1080 = 8x
135 = x
Question 8
Parallelogram QRST has an area of 120 and its longest side (QT) is 24. The angle opposite the vertical is 30°, and the vertical is from R to point U, which lies along QT. What is the length of the hypotenuse of Triangle QRU, rounded to the nearest whole number?

A
6
B
8
C
10
D
13
Question 8 Explanation: 
The correct answer is (C).

Since the area is 120 and the base is 24, we know that the height (RU) must be 5.

Given that the angle opposite the vertical is 30°, we can observe a 30°, 60°, 90° triangle.

Recall that the ratio of the side lengths of a 30°, 60°, 90° triangle is:
$x:x\sqrt{3}:2x$

In this case, the smallest side length x is 5, so:
$5:5\sqrt{3}:2*5$

The hypotenuse is 2 * 5 = 10.
Question 9
A pool that holds 35,000 cubic feet of water is being filled by a pump at a rate of 200 cubic feet per minute. At the same time, water is draining out through an open valve accidentally left open. If the pool is full in 200 minutes, how many cubic feet of water are draining out per minute?

A
5
B
15
C
25
D
35
Question 9 Explanation: 
The correct answer is (C).

Write an equation to model the situation:

$\frac{200 ft^3}{min}$(# of minutes) $-$ $\frac{x ft^3}{min}$(# of minutes) $= 35,000 $$ft^3$

Substituting 200 for the number of minutes, the unknown rate can be found:

$\frac{200 ft^3}{minutes}$$ * $$200$ $minutes$ $-$$\frac{x ft^3}{minutes}$$*$$200$ $minutes$ $= 35,000 ft^3$

Taking $200$ common:

$200(200-x) = 35,000$

Divide by 200:

$200 - x = 175$

$x = 25$ $\frac{ft^3}{min}$
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