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.

Congratulations - you have completed .

You scored %%SCORE%% out of %%TOTAL%%.

Your performance has been rated as %%RATING%%

Your answers are highlighted below.

Question 1 |

**The circumference of a right circular cylinder is half its height. The radius of the cylinder is**

*x*. What is the volume of the cylinder in terms of*x*?$2 \pi x^3$ | |

$3 \pi x^3$ | |

$3\pi^2x^3$ | |

$4\pi^2x^3$ |

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

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

Substitute this expression for

$V = \pi x^2 (4 \pi x) = 4 \pi^2x^2$

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?**

$3$ | |

$3\sqrt{2}$ | |

$6$ | |

$6\sqrt{2}$ |

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).

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?**

15 in^{3} | |

25 in^{3} | |

45 in^{3} | |

60 in^{3} |

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: $125 - 65.44 ≈ 59.55 ≈ 60$

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*?1.5 | |

2.0 | |

2.5 | |

3.0 |

Question 4 Explanation:

The correct answer is (C).

Recall that all lines can be written in the form

Given that the coordinate plane uses the variables

Using this revised equation in conjunction with the given points, we can first solve for the slope of the line in terms of

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

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

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

$c = 5 * \frac{1}{2} = 2.5$

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 |

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

$2 \pi x^3$ | |

$3 \pi x^3$ | |

$3$$\pi$$^2$$x^3$ | |

$4$$\pi$$^2$$x^3$ |

Question 5 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

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

Substitute this expression for

$V$ $=$ $\pi x^2 (4 \pi x)$ $=$ $4$$\pi$$^2$$x^2$

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$$\pi$$^2$$x^2$

Question 6 |

**Under which conditions is the expression**$\frac{ab}{a-b}$

**always less than zero?**

$a$ $<$ $b$ $<$ $0$ | |

$0$ $<$ $b$ $<$ $a$ | |

$a$ $<$ $0$ $<$ $b$ | |

$b$ $<$ $a$ $<$ $0$ |

Question 6 Explanation:

The correct answer is (A).

If the expression $\frac{ab}{a-b}$ $<$ $0$, then $ab$ $<$ $0$ or $a$ $-$ $b$ $<$ $0$ → $a$ $<$ $b.$

The only answer that guarantees a negative answer is $a$ $<$ $b$ $<$ $0.$

Consider $a$ $=$ $-2$ and $b$ $=$ $-1$

$\frac{(-2)*(-1)}{-2-(-1)}$ $=$ $\frac{2}{-1}$ $=$ $-2.$

On the other hand, if $a$ $=$ $-1$ and $b$ $=$ $1$, then:

$\frac{(-1)*(1)}{-1-1}$ $=$ $\frac{-1}{-2}$ $=$ $\frac{1}{2}$

If the expression $\frac{ab}{a-b}$ $<$ $0$, then $ab$ $<$ $0$ or $a$ $-$ $b$ $<$ $0$ → $a$ $<$ $b.$

The only answer that guarantees a negative answer is $a$ $<$ $b$ $<$ $0.$

Consider $a$ $=$ $-2$ and $b$ $=$ $-1$

$\frac{(-2)*(-1)}{-2-(-1)}$ $=$ $\frac{2}{-1}$ $=$ $-2.$

On the other hand, if $a$ $=$ $-1$ and $b$ $=$ $1$, then:

$\frac{(-1)*(1)}{-1-1}$ $=$ $\frac{-1}{-2}$ $=$ $\frac{1}{2}$

Question 7 |

**What is the value of**

*a*+*b*?w − x − xy + z | |

360 − x + y + w + z | |

180 − y + z − w | |

w + x + y + z − 360 |

Question 7 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

For the top triangle:

360 +

For the bottom triangle:

360 +

Adding the 2 equations together:

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 +

*a*−*w*−*x*= 180*a*= −180 +*w*+*x*For the bottom triangle:

*b*+ (180 −*y*) + (180 −*z*) = 180°360 +

*b*−*y*−*z*= 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 8 |

**Circle A is inside Circle B, and the two circles share the same center O. If the circumference of B is four times the circumference of A, and the radius of Circle A is three, what is the difference between Circle B’s diameter and Circle A’s diameter?**

6 | |

9 | |

12 | |

18 |

Question 8 Explanation:

The correct response is (D).

Start by drawing the figure.

If the radius of A is 3, then its diameter is 6. Its circumference is 2

B’s circumference is four times A’s circumference.

B’s circumference = 4(6

Now work backwards to find B's radius from its circumference:

24

The difference between the diameters is then: 24 – 6 = 18.

Start by drawing the figure.

If the radius of A is 3, then its diameter is 6. Its circumference is 2

*πr*= 2(3)*π*= 6*π*.B’s circumference is four times A’s circumference.

B’s circumference = 4(6

*π*) = 24*π.*Now work backwards to find B's radius from its circumference:

24

*π*= 2*πr*, so B’s radius must be 12. Its diameter is then 12 * 2 = 24.The difference between the diameters is then: 24 – 6 = 18.

Question 9 |

**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 the triangle formed from segments*RU*,*QU*, and*QR*, rounded to the nearest whole number?5 | |

8 | |

10 | |

13 |

Question 9 Explanation:

The correct answer is (C).

Since the area is 120 and the base is 24, we know that the height (

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.

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.

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 9 questions to complete.

List |

**Next Practice Test:**

**SAT Math Practice Test 4 >>**

SAT Main Menu >>