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.

**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 |

**There are 5 pencil-cases on the desk. Each pencil-case contains at least 10 pencils, but not more than 14 pencils. Which of the following could be the total number of pencils in all five pencil-cases?**

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 |

**A bird traveled 72 miles in 6 hours flying at constant speed. At this rate, how many miles did the bird travel in 5 hours?**

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 |

**Three kids own a total of 96 comic books. If one of the kids owns 16 of the comic books, what is the average (arithmetic mean) number of comic books owned by the other two kids?**

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 |

**If David has twice as many nickels as Tom, and Tom has 15 more nickels than John, what is the value in**

*dollars*of David’s nickels if John has 6 nickels?$2.10 | |

$1.40 | |

$42.00 | |

$21.00 |

If John has 6 nickels, then Tom has 15 + 6 = 21 nickels.

If David has twice as many nickels as Tom, then David has 21 × 2 = 42 nickels.

A nickel is worth 5 cents, so 42 nickels will be worth 42 × 5 = 210 cents. Because every dollar is worth 100 cents, this is equivalent to:

$\dfrac{210}{100}$ = 2.1 dollars.

Question 5 |

**A “Triangle” is any positive integer greater than 1 that has only three positive integer factors: itself, its square root, and 1. Which of the following is a Triangle?**

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 |

**The sum of two positive integers is 13. The difference between these numbers is 7. What is their product?**

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*:

2

*x*= 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 |

**What is the difference between the median and the mode in the following set of data?**72, 44, 58, 32, 34, 68, 94, 22, 67, 45, 58

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 |

**The table below shows the mass, radius, axis period, radius of orbit, and period of revolution of the Sun and the planets in our solar system. Based on the table, if Earth, Mars, or Jupiter was chosen at random, what is the probability that the chosen planet’s mass would be greater than 10 × 10**

^{24}?13% | |

33% | |

66% | |

100% |

First, analyze the table and note each planet’s mass:

Earth: 5.98 × 10

^{24}kg

Mars: 6.37 × 10

^{23}kg

Jupiter: 1.90 × 10

^{27}kg

Only Jupiter is greater than 10 × 10

^{24}kg. One out of the three planets, or $\frac{1}{3}$ = 33.33%.

Question 9 |

**The price of Kryptokoin over an 8 month period is shown in the illustration. Paul takes cash from his shoebox and buys 10 of the coins in April. Three months later he sells 5 of the coins, and in August he sells the remaining 5 coins. He puts the cash from each of these sales back in the shoebox. Assuming no other cash was taken or added, by what amount has the cash in the shoebox increased?**

$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 |

**Every May, Robo Carwash offers a “Buy 4 Get 2 Free” promotion — for every 4 carwash tickets purchased at $8**

*per ticket*, the customer receives 2 additional carwash tickets for free. What is the true cost per carwash ticket for a customer who takes advantage of the promotion?$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 |

**Approximately 4 out of every 25,000 males has a genetic mutation resulting in hemophilia. Approximately how many Americans would be expected to have this genetic mutation if the current male population of America is 163 million?**

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 |

**You've been tasked with creating the**

*rectangular*pen (fenced-in area) shown above using a maximum of 300 m of fence. The pen must have an area of at least 5400 m^{2}. What set of inequalities should be used to best describe the requirements?$2l + 2 w ≤ 300$
$l \cdot w ≥ 5400$ | |

$l + w ≤ 300$
$l \cdot w ≤ 5400$ | |

$2l + 2 w ≤ 300$
$2l \cdot 2w ≥ 5400$ | |

$l + w ≤ 300$
$2l \cdot 2w ≤ 5400$ |

Perimeter is simply the distance around an object. In this case we have:

$P = l + w + l + w$

$P = 2l + 2w$

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 400 m

^{2}:

$A ≥ 5400$

$l \cdot w ≥ 5400$

List |

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