ACT Math Practice Tests - [ FREE Questions, Answers ... - Free Printable

Let's solve each problem step by step.

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Problem 1:

A car averages 27 miles per gallon. If gas costs $4.04 per gallon, which of the following is closest to how much the gas would cost for this car to travel 2,727 typical miles?

Step 1: Find how many gallons are needed.
$$
\text{Gallons} = \frac{\text{Miles}}{\text{Miles per gallon}} = \frac{2727}{27} = 101 \text{ gallons}
$$

Step 2: Calculate the cost.
$$
\text{Cost} = 101 \times 4.04 = 408.04
$$

So, the answer is:
> D. $408.04

✔ Answer: D

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Problem 2:

When $x = 3$ and $y = 5$, by how much does the value of $3x^2 - 2y$ exceed the value of $2x^2 - 3y$?

Step 1: Plug in values.

First expression:
$$
3x^2 - 2y = 3(3)^2 - 2(5) = 3(9) - 10 = 27 - 10 = 17
$$

Second expression:
$$
2x^2 - 3y = 2(9) - 3(5) = 18 - 15 = 3
$$

Step 2: Find the difference:
$$
17 - 3 = 14
$$

✔ Answer: G. 14

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Problem 3:

What is the value of $x$ when $2x + 3 = 3x - 4$?

Step 1: Solve the equation.
$$
2x + 3 = 3x - 4
$$
Subtract $2x$ from both sides:
$$
3 = x - 4
$$
Add 4 to both sides:
$$
x = 7
$$

✔ Answer: E. 7

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Problem 4:

What is the greatest common factor (GCF) of 42, 126, and 210?

Step 1: Prime factorization:

- $42 = 2 \times 3 \times 7$
- $126 = 2 \times 3^2 \times 7$
- $210 = 2 \times 3 \times 5 \times 7$

Step 2: Identify common prime factors:
All three share: $2, 3, 7$

So,
$$
\text{GCF} = 2 \times 3 \times 7 = 42
$$

✔ Answer: J. 42

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Problem 5:

Sales for a business were 3 million dollars more the second year than the first, and sales for the third year were double the sales for the second year. If sales for the third year were 38 million dollars, what were sales, in millions of dollars, for the first year?

Let:
- First year sales = $x$
- Second year = $x + 3$
- Third year = $2(x + 3)$

Given: Third year = 38 million
$$
2(x + 3) = 38
$$
Divide both sides by 2:
$$
x + 3 = 19
$$
$$
x = 16
$$

✔ Answer: A. 16

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✔ Final Answers:

1. D. $408.04
2. G. 14
3. E. 7
4. J. 42
5. A. 16

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