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Unit 9: Multi-Step and Mixed Problems

Topic 5

Comprehensive Mixed Practice

This topic is a capstone for Units 1 through 8. Every question draws on at least two distinct skills from different units, mirroring the way the hardest SAT Math questions work. There are no new formulas or techniques here. Instead, the challenge is recognizing which tools to use when a problem does not announce its topic.
On the real Digital SAT, the second math module (the harder one, if you performed well on the first) leans heavily on multi-concept problems. A question might look like a geometry problem at first glance but require setting up a system of equations to solve. Or it might appear to be a straightforward percent question but involve reading data from a table and then applying exponential reasoning. The following strategies will help:
Read the whole problem first. Many students start computing after the first sentence. On multi-step problems, the last sentence often reframes what you need to find, which changes which calculation to do first.
Identify the math domains involved. Ask yourself: is this algebra, geometry, data, or some combination? What formulas or theorems might apply? Narrowing the domain helps you retrieve the right tools.
Work in stages. Find intermediate values one at a time. Label what you find at each step. Many errors on multi-concept problems come from losing track of what a number represents partway through.
Check your answer against the question. After solving, re-read the final question. A common trap is solving for xx when the problem asks for 2x+12x + 1, or finding an area when the problem asks for a perimeter.
The practice questions below are organized by difficulty rather than by topic. Parts A and B are moderate, Parts C and D are challenging, and Part E is SAT-hard.

Core Skills

  • Recognize which math domains and tools a multi-step problem requires, even when the topic is not stated.
  • Chain together results from different units (e.g., use algebra to find a dimension, then use geometry to find an area).
  • Translate complex word problems into equations or expressions before computing.
  • Verify that the final answer addresses what the question actually asks.

Example 1: Algebra + Geometry

A rectangle has a perimeter of 34. Its length is 3 more than its width. What is the area of the rectangle? Step 1 (Algebra): Let the width be ww. Then the length is w+3w + 3.
2w+2(w+3)=34  ⟹  4w+6=34  ⟹  4w=28  ⟹  w=72w + 2(w + 3) = 34 \implies 4w + 6 = 34 \implies 4w = 28 \implies w = 7
So the length is l=10l = 10. Step 2 (Geometry): Area =7×10=70= 7 \times 10 = 70. The area is 70\boxed{70}.

Example 2: Data + Percent

A store tracks daily sales for 5 days: $400\$400, $520\$520, $480\$480, $350\$350, and one missing day. The mean daily sales for the week was $450\$450. If the store's weekly target is 10%10\% more than the actual total sales, what is the weekly target? Step 1 (Data/Mean): Total sales =5×450=2250= 5 \times 450 = 2250. Sum of known days =400+520+480+350=1750= 400 + 520 + 480 + 350 = 1750. Missing day =2250−1750=500= 2250 - 1750 = 500. Step 2 (Percent): The weekly target is 2250×1.10=24752250 \times 1.10 = 2475. The weekly target is $2,475\boxed{\$2{,}475}.

Example 3: Systems + Functions

A company's profit function is P(x)=−x2+14x−40P(x) = -x^2 + 14x - 40, where xx is the price in dollars. At what price does the company break even (profit equals zero), and what price maximizes the profit? Step 1 (Algebra --- solving a quadratic): Set P(x)=0P(x) = 0:
−x2+14x−40=0  ⟹  x2−14x+40=0  ⟹  (x−4)(x−10)=0-x^2 + 14x - 40 = 0 \implies x^2 - 14x + 40 = 0 \implies (x - 4)(x - 10) = 0
Break-even prices: x=4x = 4 and x=10x = 10. Step 2 (Functions --- vertex): The vertex of P(x)=−(x2−14x)−40=−(x−7)2+49−40=−(x−7)2+9P(x) = -(x^2 - 14x) - 40 = -(x - 7)^2 + 49 - 40 = -(x-7)^2 + 9 is at x=7x = 7. Maximum profit is P(7)=9P(7) = 9. Break-even at x=4x = \boxed{4} and x=10x = \boxed{10}; maximum profit at x=7x = \boxed{7}.

Example 4: Trigonometry + Area

In a right triangle, one acute angle is 30∘30^\circ and the hypotenuse is 12. What is the area of the triangle? Step 1 (Trig / Special triangles): The side opposite 30∘30^\circ is 122=6\frac{12}{2} = 6. The side opposite 60∘60^\circ is 636\sqrt{3}. Step 2 (Area): Area =12(6)(63)=183= \frac{1}{2}(6)(6\sqrt{3}) = 18\sqrt{3}. The area is 183\boxed{18\sqrt{3}}.

Key Takeaways

  • Multi-concept problems are not harder because the individual skills are harder. They are harder because you have to identify which skills to use and in what order.
  • Always re-read the final question after solving. Many points are lost by answering a question the problem did not ask.
  • If your answer comes out messy (ugly fractions, large decimals) on a problem that seems like it should be clean, re-check your setup before re-checking your arithmetic. A wrong equation produces a wrong answer no matter how carefully you compute.
  • When in doubt about where to start, assign a variable to the unknown and write an equation. Translating words to algebra is the single most valuable skill on the SAT.