ACT Math Practice: Enhanced Topics, Pacing, and Worked Questions

ACT math gives you 50 minutes for 45 questions, and it is the only section where you may use a calculator. This lesson walks through the format, a pacing plan, and ten original four-choice practice questions. Each solution includes a second independent check and a note on why every tempting wrong choice fails.

Facts verified against ACT’s official publications on October 1–2, 2026.

What's covered

The enhanced ACT math section tests skills from pre-algebra through coordinate and plane geometry, basic trigonometry, and statistics/probability. ACT's Design Framework for the ACT Enhancements (R2519, February 2026) organizes the section around the same content strands the ACT has long published: numbers and quantities, algebra, functions, geometry, and statistics/probability.

Pacing the 45 questions

You have 50 minutes for 45 questions, so the section averages about 67 seconds per question. That is more time per question than the legacy 60-question math section, but the questions still demand a plan.

• First pass: solve the questions you know quickly. Do not let one hard problem eat three minutes.
• Flag and move: mark any question that will take more than 90 seconds and come back later.
• Return pass: use remaining time on flagged questions. A partial educated guess beats a blank.
• Bubble check: leave the last two minutes to make sure every question has an answer recorded.

Worked practice

The ten questions below are original PrepSolution practice items written in the enhanced four-choice format. Each answer was verified with two independent methods.

Question 1: Linear equations

If 3(x − 4) + 2 = 5x − 10, what is the value of x?
A) −2
B) 0
C) 2
D) 4

Solution: Distribute the 3 to get 3x − 12 + 2 = 5x − 10, which simplifies to 3x − 10 = 5x − 10. Subtract 3x from both sides: −10 = 2x − 10. Add 10 to both sides: 0 = 2x, so x = 0.

Check another way: substitute x = 0 into the original equation. Left side: 3(0 − 4) + 2 = −12 + 2 = −10. Right side: 5(0) − 10 = −10. Both sides match.

• A) −2 fails from combining −12 + 2 incorrectly as −14 instead of −10.
• C) 2 fails from a sign slip in the final isolation step.
• D) 4 fails from distributing only the 3 to the x, leaving −4 unchanged.

Question 2: Linear equation word problem

A taxi charges a $5 base fare plus $2 per mile. If a ride costs $23, how many miles was the ride?
A) 9
B) 11
C) 14
D) 18

Solution: let m be the number of miles. The total cost is 5 + 2m = 23. Subtract 5: 2m = 18. Divide by 2: m = 9.

Check another way: work backward from 9 miles. Base fare $5 plus 9 × $2 = $18 gives $23.

• B) 11 fails from computing 23 ÷ 2 and ignoring the base fare.
• C) 14 fails from adding the base fare instead of subtracting: (23 + 5) ÷ 2 = 14.
• D) 18 fails from subtracting the base fare but forgetting to divide by the per-mile rate.

Question 3: Systems of equations

A school sells adult tickets for $8 and student tickets for $5. If 120 tickets are sold for a total of $840, how many student tickets were sold?
A) 40
B) 60
C) 72
D) 80

Solution: let a = adult tickets and s = student tickets. Then a + s = 120 and 8a + 5s = 840. From the first equation, a = 120 − s. Substitute into the second: 8(120 − s) + 5s = 840. That gives 960 − 8s + 5s = 840, so −3s = −120 and s = 40.

Check another way: if 40 student tickets were sold, then 80 adult tickets were sold. Revenue is 80 × $8 + 40 × $5 = $640 + $200 = $840, and 80 + 40 = 120 tickets.

• B) 60 fails from averaging the two ticket counts without using the prices.
• C) 72 fails from dividing the revenue shortfall by the student price instead of the price difference.
• D) 80 fails because 80 is the number of adult tickets, or from swapping the two prices.

Question 4: Functions

For the function f(x) = x² − 4x + 1, what is f(−3)?
A) −20
B) −2
C) 4
D) 22

Solution: substitute x = −3. f(−3) = (−3)² − 4(−3) + 1 = 9 + 12 + 1 = 22.

Check another way: rewrite the function by completing the square: f(x) = (x − 2)² − 3. Then f(−3) = (−5)² − 3 = 25 − 3 = 22.

• A) −20 fails from two sign errors: treating (−3)² as −9 and −4(−3) as −12.
• B) −2 fails from evaluating at x = 3: 9 − 12 + 1 = −2.
• C) 4 fails from computing −3² without parentheses: −9 + 12 + 1 = 4.

Question 5: Percentages

A jacket is on sale for 25% off. If the sale price is $63, what was the original price?
A) $47.25
B) $63.00
C) $84.00
D) $78.75

Solution: a 25% discount means the sale price is 75% of the original. Let P be the original price. Then 0.75P = 63, so P = 63 ÷ 0.75 = 84.

Check another way: 25% of $84 is $21. Subtracting the discount from $84 gives $63, which matches the sale price.

• A) $47.25 fails from taking 25% off the sale price instead of the original price.
• B) $63.00 fails by treating the sale price as the original price.
• D) $78.75 fails from adding 25% of the sale price to the sale price.

Question 6: Ratios

A fruit punch recipe uses orange juice and pineapple juice in a 2:5 ratio. If the pitcher contains 15 cups of pineapple juice, how many cups of orange juice does it contain?
A) 3
B) 5
C) 6
D) 10

Solution: set up the proportion orange/pineapple = 2/5. With 15 cups of pineapple, orange = (2/5) × 15 = 6.

Check another way: 6 cups of orange juice to 15 cups of pineapple juice reduces to 2:5 when both are divided by 3.

• A) 3 fails from dividing 15 by 5 but forgetting to multiply by 2.
• B) 5 fails from treating 5 as the answer or using the wrong part of the ratio.
• D) 10 fails from using 2/3 of 15 instead of 2/5 of 15.

Question 7: Geometry — area and perimeter

A rectangle has a perimeter of 34 units. If the length is 7 units longer than the width, what is the area of the rectangle?
A) 49
B) 60
C) 84
D) 119

Solution: let w be the width. Then the length is w + 7. The perimeter is 2w + 2(w + 7) = 34. Simplify: 4w + 14 = 34, so 4w = 20 and w = 5. The length is 12, and the area is 5 × 12 = 60.

Check another way: a rectangle with width 5 and length 12 has perimeter 2(5) + 2(12) = 34 and length exactly 7 more than width.

• A) 49 fails from assuming a square with side 7, which gives the wrong perimeter.
• C) 84 fails from using width 7 and length 12 without solving the perimeter correctly.
• D) 119 fails from multiplying half the perimeter (17) by 7.

Question 8: Coordinate geometry — slope

What is the slope of the line passing through the points (2, −3) and (6, 5)?
A) −2
B) 1/2
C) 2
D) 4

Solution: slope m = (5 − (−3)) / (6 − 2) = 8/4 = 2.

Check another way: using point-slope form with m = 2 and (2, −3), y + 3 = 2(x − 2), or y = 2x − 7. When x = 6, y = 12 − 7 = 5, matching the second point.

• A) −2 fails from reversing the subtraction in the numerator: (−3 − 5)/(6 − 2) = −8/4.
• B) 1/2 fails from using run/rise instead of rise/run.
• D) 4 fails from dividing the rise by the x-coordinate of the first point instead of the run.

Question 9: Probability

A fair six-sided number cube is rolled twice. What is the probability that the sum of the two rolls is 7?
A) 1/6
B) 1/7
C) 1/12
D) 7/36

Solution: there are 6 × 6 = 36 equally likely outcomes. The pairs that sum to 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1) — six outcomes. The probability is 6/36 = 1/6.

Check another way: for any first roll from 1 to 6, there is exactly one second roll that makes the sum 7. That is 6 favorable ordered pairs out of 36.

• B) 1/7 fails from using the target sum as the denominator.
• C) 1/12 fails from counting only three of the six ordered pairs.
• D) 7/36 fails from using the sum value as the numerator.

Question 10: Statistics — mean

Five test scores have a mean of 84. If a sixth score of 90 is added, what is the new mean?
A) 84
B) 85
C) 87
D) 90

Solution: the original five scores have a total of 84 × 5 = 420. Adding the sixth score gives a total of 420 + 90 = 510. The new mean is 510 ÷ 6 = 85.

Check another way: if the new mean were 85 across six scores, the total would be 85 × 6 = 510. The original total was 420, so the added score would need to be 90 — which matches the problem.

• A) 84 fails from assuming the mean does not change when a new score is added.
• C) 87 fails from averaging the old mean and the new score: (84 + 90)/2.
• D) 90 fails from treating the newly added score as the new mean.

Classify your mistakes

After you work the questions above, sort every miss into one of three buckets. The bucket tells you what to do next.

Choose your next drill

• If algebra questions were your main miss, practice more linear, system, and function questions on the free ACT practice page.
• If geometry or statistics cost you points, work those topics in isolation first, then mix them back into full sections.
• If time was the issue, follow the pacing plan in our ACT study plan and log every pacing slip.
• For the overall enhanced format, section counts, and rollout dates, see the enhanced ACT guide.

Use these ten questions as a diagnostic, not as a score prediction. A short set cannot estimate your ACT scaled score — only a full official form with its own scoring key can do that.