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Guide to 7 Hard Digital SAT Math Problems

By Mr. Osama Ahmad July 9, 2025 6 min read
Guide to 7 Hard Digital SAT Math Problems | The Test Advantage

DSAT Masterclass: A Full Walkthrough of 7 Vicious Math Problems

The Digital SAT math section is known for multi-layered problems that test not just what you know, but how you apply it under pressure. To conquer them, you need a solid strategy. This guide will serve as your text-based companion to one of our expert walkthroughs, breaking down seven high-level problems step-by-step. We'll unpack the core strategies and demonstrate how to leverage Desmos to maximize speed and accuracy.

Problem Walkthroughs

Question 1: The Circle and Inscribed Square

Problem: Given 5x² - 40px + 5y² + 70py = 325p² represents a circle with a center at (-8, 14), find the area of a square inscribed within it.

  1. Standardize the Circle Equation: The standard form is (x-h)² + (y-k)² = r². To achieve this, divide the entire given equation by 5:
    x² - 8px + y² + 14py = 65p²
  2. Complete the Square: Create perfect square trinomials for both x and y. Add (b/2)² to both sides for each variable.
    • For x: (-8p/2)² = 16p²
    • For y: (14p/2)² = 49p²
    The equation becomes: (x² - 8px + 16p²) + (y² + 14py + 49p²) = 65p² + 16p² + 49p²
  3. Factor and Solve for 'p' and 'r²': Factoring yields (x - 4p)² + (y + 7p)² = 130p². Since the center is (-8, 14), we solve -4p = -8 to find p = 2. Now find r²:
    r² = 130(2)² = 130 * 4 = 520.
  4. Find the Square's Area: For an inscribed square, the circle's diameter is the square's diagonal (d = 2r = 2√520). The side of the square (s) is d/√2, so s = (2√520)/√2 = 2√260. The area is s², so Area = (2√260)² = 4 * 260 = 1040.

Question 2: Unit Conversions with Squared Units

Problem: An object's speed increases at 33 inches per second squared. What is this rate in miles per hour squared?

The key is recognizing that because the time unit is squared (seconds²), the time conversion factor must also be squared. There are 3600 seconds in 1 hour, so we'll use 3600².

Setup: [33 in/s²] * [1 ft / 12 in] * [1 mi / 5280 ft] * [3600² s² / 1 hr²]

Plugging 33 * (1/12) * (1/5280) * 3600^2 into Desmos gives 6750.

Question 3: Linear Systems with No Solution

Problem: A system of two linear equations has constants g and h, and has no solution. The ratio g:h is equivalent to 15:k. Find k.

  1. Understand "No Solution": Parallel lines have the same slope but different y-intercepts. For Ax+By=C form, this means A₁/A₂ = B₁/B₂.
  2. Simplify & Standardize: After multiplying to clear fractions, the equations are arranged into Ax+By=C form:
    • Eq 1: gx - hy = 9
    • Eq 2: -27x + 42y = -7
  3. Apply No-Solution Formula: Set the ratios of coefficients equal: g / -27 = -h / 42.
  4. Solve for g/h: Cross-multiply to get 42g = 27h. Divide to find g/h = 27/42, which simplifies to 9/14.
  5. Solve for k: We are given g/h = 15/k. So, 9/14 = 15/k. Cross-multiply again: 9k = 14 * 15 = 210. Thus, k = 210/9 = 70/3.

Question 4: Factoring with Unknowns

Problem: Which quadratic expression has 6x - 5b as a factor, where b is a positive integer?

Instead of finding a root, we use logic to construct the other factor.

  1. Set up Factors: One factor is `(6x - 5b)`. The other must be `(Ax + C)`. All answer choices start with 24x², so `6x * Ax = 24x²` means `A=4`. The other factor starts with `4x`.
  2. Find Last Term: The last terms must multiply to -15b. `-5b * C = -15b` means `C=3`.
  3. Full Factored Form: `(6x - 5b)(4x + 3)`
  4. Expand and Compare: FOIL the expression: 24x² + 18x - 20bx - 15b. The middle term is `(18 - 20b)x`.
  5. Test Choices: Find which choice's middle term yields a positive integer for b. For choice A, -22x, we set -22 = 18 - 20b, which gives -40 = -20b, so b = 2. This is a positive integer, making it the correct answer.

Question 5: Analyzing Radical Functions

Problem: A function `f(x) = a√(-x + b)` passes through (5, 0), and `f(-12) > f(-7)`. Which statement must be true?

  1. Use Point (5, 0): Plugging this in gives 0 = a√(-5 + b). For this to be true, either `a=0` or the term in the square root is 0, which means `-5 + b = 0`, so `b = 5`.
  2. Use the Inequality: The condition is `f(-12) > f(-7)`. We know `b=5`. Let's test `a` in Desmos with the function `f(x) = a√(-x + 5)`.
    • If we let `a` be negative (e.g., -1), the graph is below the x-axis, and `f(-12) < f(-7)`. This violates the condition.
    • If we let `a=0`, then `f(x)=0`, so `f(-12) = f(-7)`. This also violates the condition.
    • Only when `a > 0` is the condition `f(-12) > f(-7)` met.
  3. Evaluate Choices: We know `b=5` and `a` is a positive number. Choices comparing `a` and `b` fail because `a` could be 1 (less than b) or 10 (greater than b). The only choice that must be true, which can be confirmed by looking at the graph for any positive `a`, is that the value of the function at x=3 is less than the value at x=-10 (or `f(3) < f(-2b)` since `-2b = -10`).

Question 6: Consecutive Integers & Percentages

Problem: A set of five consecutive even integers is ordered least to greatest. The sum of the first and fourth is 5,860% more than the quotient of the fifth and 30. If the first integer is p% less than the fifth, find p.

  1. Define Integers: Let the first even integer be `x`. The set is: `x, x+2, x+4, x+6, x+8`.
  2. Translate First Sentence: "Sum of first and fourth" is `x + (x+6)`. "5,860% more than" is a multiplier of `1 + 58.60 = 59.6`. "Quotient of fifth and 30" is `(x+8)/30`. The equation is `x + x+6 = 59.6 * ((x+8)/30)`.
  3. Solve for x: Using Desmos to find the intersection of both sides of the equation yields `x = 742`. The integers are 742, 744, 746, 748, 750.
  4. Solve for p: Translate the second sentence. "First integer" (742) "is" (=) "p% less than" (multiplier of `1 - p/100`) "the fifth integer" (750). The equation is `742 = (1 - p/100) * 750`. Solving this for `p` (as `x` in Desmos) gives 1.066..., which is the fraction 16/15.

Question 7: Geometry of Inscribed Circles

Problem: A circle is inscribed in an equilateral triangle. The perimeter is 90√3. Arc AC is equal to kπ. Find k.

  1. Find Side Length: Perimeter `P = 3s = 90√3`, so side `s = 30√3`.
  2. Use 30-60-90 Triangle: When a circle is inscribed in an equilateral triangle, you can form a 30-60-90 triangle with the radius. The base of this small triangle is half the side length, `15√3`. This is opposite the 60° angle. In a 30-60-90 triangle, the side opposite 60° is `x√3`. So, `x√3 = 15√3`, meaning `x=15`. This `x` is the circle's radius. So, `r = 15`.
  3. Find Arc Length: By symmetry, the central angle for arc AC is 1/3 of the circle, or 120°. The arc length formula is `(Angle/360) * 2πr`. So, Arc AC = `(120/360) * 2π(15) = (1/3) * 30π = 10π`.
  4. Solve for k: We are given Arc AC = kπ. So, `10π = kπ`, meaning k=10.

Conclusion: The Path to Mastery

By reviewing these rigorous problems, several key strategies emerge: deconstruct word problems piece by piece, master foundational formulas, embrace the Desmos calculator for complex computations, watch for hidden traps in the wording, and use logic to work backward when necessary. By integrating these approaches, you can turn even the most intimidating DSAT math problems into a series of logical, conquerable steps.

Ready to apply these strategies to thousands of realistic practice questions? Visit our plans page to get started with The Test Advantage.

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