This is the first in our series of free SAT Math practice tests. It is based on the fourth section of the SAT, in which you are allowed to use a calculator. The SAT Math test went through a major revision in 2016, changing the focus to real world problems. Questions now include problems you might encounter at work, in your daily life, or in your college math and science courses. Try our SAT Math multiple choice questions to see if you are fully prepared for this section of the test.
The use of a calculator is allowed. Solve each problem and select the best of the answer choices provided.
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Question 1 |
35 | |
45 | |
65 | |
75 |
If all pencil-cases have 10 pencils (the minimum number of pencils every pencil-case can hold), then the total minimum number of pencils is 10 × 5 = 50 pencils. Likewise, if all pencil-cases have 14 pencils (the maximum number of pencils every pencil-case can hold), then the total maximum number of pencils is 14 × 5 = 70 pencils. Thus, the answer must be between 50 and 70. Out of all the possible answer choices, only (C), 65 pencils, matches these criteria.
Question 2 |
12 | |
30 | |
60 | |
14.4 |
First calculate the rate that the bird is traveling at:
$\frac{72~miles}{6~hour} = 12~miles~per~hour$
The distance traveled in 5 hours will be:
$12~miles~per~hour * 5~hours = 60~miles$
Question 3 |
40 | |
42 | |
44 | |
46 |
If one kid owns 16 of the comic books, then the number of comic books owned by the other two kids is: 96 − 16 = 80.
Therefore, the average is: 80 ÷ 2 = 40.
Question 4 |
$2.10 | |
$1.40 | |
$42.00 | |
$21.00 |
Question 5 |
169 | |
100 | |
36 | |
81 |
The only factors of 169 are: 169, 13, and 1. (The square root of 169 is 13.) All of the other options have more than 3 factors; or, put another way, the square roots of the other answers (10, 6, and 9, respectively) are not prime numbers.
Question 6 |
12 | |
22 | |
30 | |
40 |
Use the given information to create a system of equations in two variables and then solve with the method of combination. Translating the problem statement into algebra:
x + y = 13
x − y = 7
Add the 2 equations together:
x + x = 13 + 7;
Now solve for x:
2x = 20; x = 10
The value of y can then be found by plugging x back into either of the two original equations: 10 + y = 13; y = 3
Once the variable values are found, their product can be computed:
x * y = 10 * 3 = 30
Question 7 |
0 | |
2 | |
4 | |
6 |
Start by organizing the data numerically from least to greatest:
22, 32, 34, 44, 45, 58, 58, 67, 68, 72, 94
The mode, or data value that occurs most often, is 58.
The median, or data value in the middle of the data set, is also 58.
The difference between these values is 0.
Question 8 |

13% | |
33% | |
66% | |
100% |
First, analyze the table and note each planet’s mass:
Earth: 5.98 × 1024 kg
Mars: 6.37 × 1023 kg
Jupiter: 1.90 × 1027 kg
Only Jupiter is greater than 10 × 1024 kg. One out of the three planets, or $\frac{1}{3}$ = 33.33%.
Question 9 |

$2250 | |
$750 | |
$1250 | |
$450 |
Bought 10 coins in April:
Price in April = \$150/coin
Cash spent = (10 coins)(\$150/coin) = \$1500
Sold 5 coins in July (3 Months Later):
Price in July = \$250/coin
Cash received = (5 coins)(\$250/coin) = \$1250
Sold 5 coins in August:
Price in August = \$200/coin
Cash received = (5 coins)(\$200/coin) = \$1000
Assuming he had \$T of cash in the shoebox to start with:
\$T − \$1500 + \$1250 + \$1000
\$T + \$750
Cash has increased by \$750.
Question 10 |
$6.00 | |
$5.33 | |
$4.00 | |
$6.67 |
True Cost per carwash ticket = $\frac{Total~Cost}{Number~of~Carwash~Tickets}$
Total cost = (\$8 per ticket)(4 tickets)
= \$32
Number of Carwash Tickets = 4 (paid for) + 2 (free)
=6 tickets
True Cost per carwash = $\frac{\$8}{6~Tickets}$
= \$5.33 per carwash ticket
Question 11 |
41,000 | |
1,019,000 | |
65,000 | |
26,000 |
$\dfrac{x~have~mutation}{163,000,000~males} = \dfrac{4~have~mutation}{25,000~males}$
Solve for x:
$(25,000)(x) = (4)(163,000,000)$
$x = \frac{(4)(163,000,000)}{25,000}$
$x = 26,080 \approx 26,000$
Question 12 |

$2l + 2 w ≤ 300$ $l \cdot w ≥ 5400$ | |
$l + w ≤ 300$ $l \cdot w ≤ 5400$ | |
$2l + 2 w ≤ 300$ $2l \cdot 2w ≥ 5400$ | |
$l + w ≤ 300$ $2l \cdot 2w ≥ 5400$ |
Perimeter is simply the distance around an object. In this case:
$P = l + w + l + w$
$P = 2l + 2w$
P must be a maximum of 300 m, meaning it must be less than or equal to 300 m:
$P ≤ 300$
$2l + 2 w ≤ 300$
The Area of a rectangle is:
$A = l \cdot w$
The area must be at least 5400 m2:
$A ≥ 5400$
$l \cdot w ≥ 5400$
Question 13 |
x = 12 | |
x = 3 | |
x = 0 | |
x = 9 |
We have been given the vertex and one of the zeros of the parabolic function (i.e. a parabola that either opens up or down)

If one of the zeros lies 2 units (horizontally) to the right of the vertex, the other zero will lie 2 units (horizontally) to the left. This means the other zero will be located at x = 3 (3, 0)
Question 14 |

20 | |
8 | |
0 | |
2 |
Looking at Graph B we see that the maximum efficiency of 60% occurs when Speed = 2 cm/sec. We need to determine the “Input Setting” that results in a speed of 2 cm/sec. Using Table A we can determine that an input setting of 8v will give us the desired speed.
Question 15 |

2700 | |
2800 | |
2900 | |
3000 |
Because the relationship is linear, we know that the rate of change is constant. The easiest way to make use of this knowledge is to notice that 21 is halfway between 18 and 24. This means that the number, c, we are looking for will be halfway between 2400 and 3200.
“Halfway” will be the mean (“average”) of these numbers:
$\frac{2400 + 3200}{2} = 2800~mAh$
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