SAT Prep is an important milestone, but you do not need to approach it with fear! To help students build consistency, confidence, and critical thinking skills, we post two new SAT-style math questions every single Saturday.
These problems are completely original, designed by me using official SAT test materials as a blueprint to guarantee realistic practice without infringing on copyright laws. They serve as excellent weekly training for the PSAT and ACT as well!
Here is how it works: Try your hand at this week’s new challenges (Questions 13 and 14), and then scroll down to review the step-by-step solutions for last week’s challenges (Questions 11 and 12).
Our local mason will build a retention wall on your property. He charges $220 for the first two hours of building, plus an hourly fee for each additional hour. The total cost for 5 hours of building is $520. Which function f gives the total cost, in dollars, for $x$ hours of labor in building your wall, when x ≥ 2?
In the given equation, c is a positive constant. Which of the following is one of the solutions of the given equation?
A. -c
B. 1 – c
C. c + 1
D. -c² – 33²
Think you found the answers? Drop your guesses in the comments below, or check back next Saturday when we release the full step-by-step mathematical explanations!
Question 11: C (3√5)
Question 12: B (4$)
The Problem: In right triangle FGH, sin(x) = 2/3. If the side opposite angle F has a length of 6√3, what is the length of adjacent side FG?
The Explanation: By definition, the sine ratio in a right triangle is:
We know the side opposite the x is 6√3, but we don’t know the hypotenuse. Set up the ratio of the fractions and solve for h.
2/3 = 6√3/h
2h = 18
h = 9√3
Now that we know the hypotenuse is 9√3 and the opposite leg is 6√3, we use the Pythagorean Theorem (a2 + b2 = c2) to find the remaining adjacent leg, FG:
(6√3)2 + b2 = (9√3)2
108 + b2 = 243
b = √135 = √9 · √15 = 3√15 option C
Question 12 Solution: Algebra 2 (Polynomial Graphs)
The Problem: The function f(x) is defined by f(x) = x(x + 1)(x2 – 4). How many different x-intercepts does the graph have?
Other common names for x-intercepts are solutions, zeros, or roots. These specific points happen on a graph whenever y = 0 (or f(x) = 0).
To make a fully multiplied function equal zero, at least one of its individual factors needs to equal zero. Because this function is already nearly completely factored for us, we can evaluate each part individually:
x = 0
x + 1 = 0 ⇒ x = – 1
x2 – 4 = 0 ⇒ (x + 2) (x – 2) = 0 ⇒ x = – 2 or x = 2
Counting them up, we have found exactly 4 distinct, real x-intercept solutions: -2, -1, 0, and 2.This matches Option B..
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