ACT Math study guide · Chapter 49
Hard Mixed Practice
The eight questions below are not organized by chapter. Most combine familiar skills from different Parts of this book, while a few present one familiar skill in a less familiar representation. Both forms are useful preparation for mixed ACT Math work because the section does not announce which chapter or method applies before a question begins. Each question below is labeled afterward, in the solution, with the content area or areas it draws from; the question itself gives you no such hint, matching real test conditions.
▸ Questions
1. A rectangle's length is 3 more than twice its width. If its area is 65, find its perimeter.
A) 32 B) 40 C) 26 D) 36
2. The equation x² − 2x + 10 = 0 has a complex solution with a positive imaginary part. Find the sum of its real and imaginary parts.
F) −2 G) −4 H) 4 J) 8
3. In a right triangle, one leg has length 8 and the angle opposite that leg measures 30°. Find the hypotenuse.
A) 16 B) 8√3 C) 4 D) 12
4. In a survey of 200 people, 120 prefer coffee. Of those coffee drinkers, 45 also prefer tea. Find the probability that a randomly chosen coffee drinker also prefers tea.
F) 0.225 G) 0.60 H) 0.15 J) 0.375
5. A car depreciates 15% in value each year. After one year, it is worth $17,000. Find its original value.
A) $19,550 B) $20,000 C) $17,850 D) $14,450
6. A company's profit is modeled by P(x) = −2x² + 80x − 150, where x is hundreds of units sold. Find the maximum profit.
F) 600 G) 650 H) 700 J) 500
7. Find the area of a triangle with vertices at (1, 2), (5, 2), and (3, 8).
A) 24 B) 10 C) 12 D) 16
8. A line y = x + 1 intersects the circle x² + y² = 25 at two points. Find the sum of the x-coordinates of those two points.
F) −1 G) 7 H) −7 J) 1
▸ Answers and Explanations
1: D. Algebra (Part II) + Geometry (Part IV). Let width = w and length = 2w+3. Area: w(2w+3)=65, so 2w²+3w−65=0. The discriminant is 9+520=529=23², so w=(−3+23)/4=5. Length = 2(5)+3=13. Perimeter = 2(5+13)=36. Choice A reports 4w instead of 2(w+length), an incomplete perimeter formula.
2: H. Functions (Part III) + Number & Quantity (Part I). Discriminant: (−2)²−4(1)(10)=4−40=−36. Root: (2±6i)/2=1±3i. The root with positive imaginary part is 1+3i; its real and imaginary parts sum to 1+3=4. Choice F drops the sign on the imaginary part, using 1−3 instead of 1+3.
3: A. Geometry (Part IV). sin(30°)=opposite/hypotenuse=8/hyp, so hyp=8/sin(30°)=8/(1/2)=16. Choice G confuses the 30-60-90 rule, multiplying the leg by √3 instead of dividing by sin(30°).
4: J. Statistics & Probability (Part V). This is a conditional probability restricted to the coffee-drinker group: 45/120=0.375. Choice F divides by the full survey total (200) instead of restricting to the coffee-drinker group (120).
5: B. Integrating Essential Skills (Part VI) + Algebra (Part II). original × (1−0.15) = 17,000, so original = 17,000/0.85 = 20,000. Choice A applies the depreciation rate in the wrong direction, adding 15% to 17,000 instead of dividing by 0.85.
6: G. Functions (Part III) + Integrating Essential Skills (Part VI). The vertex occurs at x=−b/2a=−80/(2·−2)=20. Maximum profit: P(20)=−2(400)+80(20)−150=−800+1600−150=650. Choice F comes from an arithmetic slip while evaluating P(20).
7: C. Geometry (Part IV). The side from (1,2) to (5,2) is horizontal with length 4, serving as the base; the height, the vertical distance to (3,8), is 6. Area=(1/2)(4)(6)=12. Choice A omits the 1/2 factor, computing the area as if for a rectangle instead of a triangle.
8: F. Geometry (Part IV) + Algebra (Part II). Substitute y=x+1 into the circle's equation: x²+(x+1)²=25, giving 2x²+2x−24=0, or x²+x−12=0. Factoring: (x+4)(x−3)=0, so x=−4 or x=3. The sum of the x-coordinates is −4+3=−1. Choice G reverses the sign of one root while combining, computing 4+3 or −4−3 instead of −4+3.
These eight questions are a small, deliberate sample of what synthesis looks like on this section. The dedicated Mixed Challenge Practice that follows extends this same idea to book-wide volume.