What this quiz covers
This quiz focuses on Equations With One Variable, giving you a quick way to practice the rules, question types, and explanations that matter most for PSAT Math.
A phone plan charges $18 plus x dollars per gigabyte. If 7 gigabytes cost $46, the situation is modeled by 18+7x=46. What is the value of x?
PSAT Math Quiz
Practice Equations With One Variable in PSAT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
This quiz focuses on Equations With One Variable, giving you a quick way to practice the rules, question types, and explanations that matter most for PSAT Math.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
A phone plan charges $18 plus x dollars per gigabyte. If 7 gigabytes cost $46, the situation is modeled by 18+7x=46. What is the value of x?
Explanation: The problem asks to find x from the phone plan equation 18+7x=46, where 18 is the base charge and 7x represents 7 gigabytes at x dollars each. Subtract 18 from both sides: 7x=46−18=28. Divide both sides by 7: x=28÷7=4. A common mistake would be dividing the wrong terms or making arithmetic errors in subtraction. Always isolate the variable term first, then divide by its coefficient.
A gym charges a one-time sign-up fee plus a monthly fee. The total cost after 6 months is modeled by 18+6x=3x+54, where x is the monthly fee in dollars. What is the value of x?
Explanation: This problem asks us to find the monthly fee x when the total cost after 6 months equals 18+6x=3x+54. To solve, we first subtract 3x from both sides: 18+6x−3x=54, which simplifies to 18+3x=54. Next, we subtract 18 from both sides: 3x=54−18=36. Finally, dividing both sides by 3 gives us x=12. A common error is incorrectly combining the x terms or making arithmetic mistakes when subtracting. When solving equations with variables on both sides, always collect like terms on one side first.
A company's total cost C, in dollars, to produce x units of a product is given by the function C(x)=15x+500. The company sells each unit for $25.Whatisthenumberofunitsthecompanymustselltoachieveaprofitofexactly$4,500?
Explanation: This problem tests your ability to set up and solve profit equations. When you see questions about company costs, revenues, and profits, remember that profit equals revenue minus costs. First, let's establish the key relationships. The cost function is C(x)=15x+500, where x is the number of units produced. Since each unit sells for $25, the revenue function is $R(x)=25x .Profitiscalculatedas P(x)=R(x)−C(x)=25x−(15x+500)=10x−500 $. To find when profit equals $4,500, set up the equation: 10x−500=4500. Adding 500 to both sides gives 10x=5000, so x=500. This confirms answer choice D is correct. Let's examine why the other answers are wrong. Choice A (300 units) would yield a profit of 10(300)−500=2500, which is 2,000shortofthetarget.ChoiceB(400units)gives 10(400)−500=3500 ,still1,000 below the goal. Choice C (450 units) results in 10(450)−500=4000, which is $500 less than needed. Each incorrect answer represents a common calculation error: either mistakes in setting up the profit equation or arithmetic errors when solving. Study tip: Always write out the profit equation explicitly as Revenue - Costs before substituting numbers. This prevents confusion about which values represent income versus expenses, and double-check your arithmetic by substituting your answer back into the original profit equation.
The perimeter of a rectangle is equal to the perimeter of a regular pentagon. The length of the rectangle is 4 cm more than its width, w. The side length of the regular pentagon is equal to the width of the rectangle. What is the length of the rectangle?
Explanation: This problem tests your ability to set up equations using perimeter formulas and solve for unknown variables. When you see problems involving equal perimeters of different shapes, focus on writing expressions for each perimeter and setting them equal. Let's define our variables. The rectangle has width w and length w+4. The pentagon has side length w (equal to the rectangle's width). The perimeter of a rectangle is 2(length+width), so our rectangle's perimeter is 2(w+4+w)=2(2w+4)=4w+8. The perimeter of a regular pentagon is 5×side length, so our pentagon's perimeter is 5w. Since the perimeters are equal: 4w+8=5w Solving for w: 8=w Therefore, the rectangle's length is w+4=8+4=12 cm. Looking at the wrong answers: Choice A (8 cm) gives you the width, not the length—a common mistake when solving multi-step problems. Choice C (20 cm) results from incorrectly setting up the pentagon's perimeter as 4w instead of 5w, leading to w=12 and length = 16, but this doesn't match any calculation. Choice D (40 cm) comes from computational errors in the equation setup. The correct answer is B. Strategy tip: In perimeter problems with multiple shapes, always write out the perimeter formula for each shape explicitly before setting up your equation. Double-check that you're solving for the right quantity—here, length, not width.
The formula for the surface area, A, of a right circular cylinder is A=2πr2+2πrh, where r is the radius of the base and h is the height of the cylinder. Which of the following equations correctly expresses the height, h, in terms of A and r?
Explanation: When you encounter a formula that needs to be solved for a different variable, you're working with algebraic manipulation—a core skill tested throughout the PSAT Math section. To solve for h in the surface area formula A=2πr2+2πrh, you need to isolate h on one side. Start by subtracting 2πr2 from both sides: A−2πr2=2πrh. Then divide both sides by 2πr to get: h=2πrA−2πr2. You can simplify this by splitting the fraction: h=2πrA−2πr2πr2=2πrA−r. This confirms that choice A is correct. Choice B contains a critical error: it shows h=2πrA−r2, which incorrectly simplifies 2πr2πr2. When you divide 2πr2 by 2πr, you get r, not r2. Choice C gives h=2πA−2πr2, which represents the intermediate step before dividing by r. This student forgot to complete the division by the full denominator 2πr. Choice D shows h=A−r, which completely ignores the coefficients and demonstrates a fundamental misunderstanding of algebraic manipulation. Remember: when solving literal equations, perform the same operations you would with numbers, and always check your algebra by substituting back into the original equation.
A water tank starts with 120 liters and drains at x liters per minute. After 9 minutes, 66 liters remain, modeled by 120−9x=66. What is x?
Explanation: The problem describes a water tank draining equation 120 - 9x = 66, where 120 is initial volume, 9x is amount drained in 9 minutes at x liters per minute, and 66 is final volume. Subtract 120 from 66: -9x = 66 - 120 = -54. Divide by -9: x = 6. A common mistake would be incorrect subtraction or sign handling. When solving real-world problems, ensure the equation makes physical sense before solving.
A class collects $180 total from a fixed fee of $30 plus x dollars per student. If there are 10 students, then 30+10x=180. What is x?
Explanation: The problem describes a class collection equation 30 + 10x = 180, where 30 is a fixed fee and 10x represents 10 students at x dollars each. Subtract 30 from both sides: 10x = 150, then divide by 10: x = 15. A common mistake would be dividing by the wrong coefficient or making subtraction errors. When solving real-world problems, identify the fixed and variable components clearly before isolating the variable.
A taxi fare is modeled by 3.50+2.25x=21.50, where x is the number of miles. What is the value of x?
Explanation: The problem asks to solve the taxi fare equation 3.50 + 2.25x = 21.50 for x, where x represents miles traveled. Subtract 3.50 from both sides: 2.25x = 18. Divide both sides by 2.25: x = 18 ÷ 2.25 = 8. A common error would be incorrect decimal division or forgetting to subtract the base fare first. When solving real-world linear equations, identify the fixed cost and variable cost components before isolating the variable.
A contractor charges $50 for a visit plus x dollars per hour. If a 3-hour job costs $170, then 50+3x=170. What is x?
Explanation: The problem describes a contractor's billing equation 50+3x=170, where 50 is the visit fee and 3x represents 3 hours at x dollars per hour. Subtract 50 from both sides: 3x=120, then divide by 3: x=40. A common mistake would be dividing by the wrong coefficient or making subtraction errors. When solving real-world linear equations, clearly identify the fixed and variable components before isolating the variable.
If 4x−31=65, what is the value of x? Combine the fractions on the right side before dividing.
Explanation: The problem asks to solve 4x - 1/3 = 5/6 for x. Add 1/3 to both sides: 4x = 5/6 + 1/3. Convert to common denominator: 4x = 5/6 + 2/6 = 7/6. Divide by 4: x = 7/6 ÷ 4 = 7/24. A common error is incorrect fraction arithmetic when adding or dividing fractions. When combining fractions, always find a common denominator first, then divide carefully.
If 2x+5=23(x−1), what is the value of x? Multiply to clear the fraction before isolating x.
Explanation: The problem asks to solve 2x+5=23(x−1) for x. First, multiply both sides by 2 to clear the fraction: 4x+10=3(x−1)=3x−3. Subtract 3x from both sides: x+10=−3, then subtract 10: x=−13. A common mistake is not properly clearing fractions or making sign errors during distribution. When fractions appear in equations, multiply by the denominator to eliminate them before proceeding.
A store owner buys an item for c dollars. To set a selling price, the owner marks up the cost by 50%. After two weeks, the item has not sold, so the owner offers a 20% discount off the current selling price. If the final price after the discount is $36,whatwastheoriginalcost,c$$?
Explanation: When you encounter markup and discount problems, you're working with percentage changes applied in sequence. The key is to track how the price changes step by step, then work backwards from the final price. Let's trace through the price changes. The store owner starts with cost c, then marks it up 50%. This gives a selling price of c+0.50c=1.50c. Next comes a 20% discount off this selling price: 1.50c−0.20(1.50c)=1.50c(1−0.20)=1.50c×0.80=1.20c. Since the final price is $36, we have $1.20c=36 ,so c=36÷1.20=30 $. The original cost was $30. Looking at the wrong answers: Choice A (22.50)representsacommonerrorwherestudentsmightsubtractbothpercentagesfrom10037.50) could result from confusing which percentage to apply when, perhaps treating the discount as applying to the original cost. Choice D ($45.00) might come from incorrectly adding the percentages or applying them in the wrong direction. The correct answer is B. Study tip: For sequential percentage problems, convert each percentage to a multiplier (50% markup means multiply by 1.50, 20% discount means multiply by 0.80), then combine them before solving. This prevents calculation errors and makes the algebra cleaner.
A streaming service bills $9 plus x dollars per movie. If the bill for 5 movies is $24, then 9+5x=24. What is x?
Explanation: The problem describes a streaming service billing equation 9 + 5x = 24, where 9 is the base fee and 5x represents 5 movies at x dollars each. Subtract 9 from both sides: 5x = 15, then divide by 5: x = 3. A common mistake would be dividing by the wrong coefficient or making arithmetic errors in subtraction. When solving real-world problems, identify the fixed and variable costs before isolating the variable.
A phone plan's cost in dollars is modeled by 12+0.75x=0.5x+18, where x is the number of gigabytes used. What is the value of x that makes the two sides equal?
Explanation: We need to solve 12+0.75x=0.5x+18 for the number of gigabytes x. Subtracting 0.5x from both sides: 12+0.75x−0.5x=18, which simplifies to 12+0.25x=18. Subtracting 12 from both sides: 0.25x=6. Dividing by 0.25: x=6÷0.25=24. A common error is making arithmetic mistakes with decimals or incorrectly combining the x terms. When working with decimal coefficients, consider converting to fractions to avoid calculation errors.
Solve the equation 32x−1=2x+5. What is the value of x?
Explanation: The problem asks to solve (2x - 1)/3 = (x + 5)/2 for x. Cross-multiply to get 2(2x - 1) = 3(x + 5), which gives 4x - 2 = 3x + 15. Subtract 3x from both sides: x - 2 = 15, then add 2: x = 17. A common error is making mistakes during cross-multiplication or not properly distributing. When solving rational equations, cross-multiplication is often the most efficient method to eliminate fractions.
A lab mixture has x milliliters of solution A. After adding 12 mL, the total becomes 45 mL, modeled by x+12=45. What is the value of x?
Explanation: The problem describes a lab mixture equation x + 12 = 45, where x is the initial amount and 12 mL is added to get 45 mL total. Subtract 12 from both sides: x = 45 - 12 = 33. A common mistake would be adding 12 instead of subtracting or making arithmetic errors. This is a simple one-step equation requiring only subtraction to isolate the variable.
Solve for x: 7x−3=2x+217. What is the value of x?
Explanation: The problem asks to solve 7x - 3 = 2x + 17/2 for x. Subtract 2x from both sides: 5x - 3 = 17/2, then add 3: 5x = 17/2 + 6/2 = 23/2. Finally, divide by 5: x = 23/10. A common error is incorrect fraction arithmetic when combining terms. When adding fractions to whole numbers, convert the whole number to a fraction with the same denominator first.
If 2x+1+3x+2=316, what is the value of x?
Explanation: When you encounter an equation with fractions, your goal is to clear the denominators and solve for the variable systematically. To solve 2x+1+3x+2=316, find a common denominator for the left side. The LCD of 2 and 3 is 6, so multiply each term appropriately: 63(x+1)+62(x+2)=316 This simplifies to: 63x+3+2x+4=316 65x+7=316 Cross-multiply to eliminate fractions: 3(5x+7)=6⋅16 15x+21=96 15x=75 x=5 You can verify: 25+1+35+2=26+37=3+37=316 ✓ Choice A (−31) likely results from sign errors during fraction manipulation. Choice B (1) might come from incorrectly combining like terms or making arithmetic mistakes when finding the common denominator. Choice D (25) could result from errors in cross-multiplication, such as multiplying 15x=75 incorrectly or confusing the final division step. Strategy tip: When solving equations with multiple fractions, always find a common denominator first, then cross-multiply to eliminate fractions entirely. This reduces the chance of arithmetic errors and makes the algebra more straightforward. Always substitute your answer back into the original equation to verify it works.
What is the solution to the equation 3(x−2(x+1))=4(x−5)+2?
Explanation: This equation requires careful algebraic manipulation, particularly when dealing with nested parentheses. The key is to work systematically from the inside out, then combine like terms to isolate the variable. Start with the left side: 3(x−2(x+1)). First, distribute the −2 inside the inner parentheses: x−2(x+1)=x−2x−2=−x−2. Now distribute the 3: 3(−x−2)=−3x−6. For the right side: 4(x−5)+2=4x−20+2=4x−18. The equation becomes: −3x−6=4x−18 Collect like terms by adding 3x to both sides: −6=7x−18 Add 18 to both sides: 12=7x Therefore: x=712 Choice A (3) likely comes from making errors in the distribution process and getting a simpler equation. Choice B (716) probably results from a sign error when combining the constant terms—getting 16=7x instead of 12=7x. Choice C (316) suggests confusion in both the coefficient of x and the constant term, possibly mixing up the 7 and 3 from the distribution steps. When solving multi-step equations with nested parentheses, always distribute from the innermost parentheses outward, then carefully track positive and negative signs as you combine like terms. Double-check by substituting your answer back into the original equation.
For what value of a does the equation 5(x−2)−x=2(ax−3) have no solution?
Explanation: When you encounter an equation with a parameter and are asked when it has "no solution," you're dealing with a situation where the equation becomes contradictory—like 3=5. Let's simplify both sides of 5(x−2)−x=2(ax−3) and see what happens. On the left side: 5(x−2)−x=5x−10−x=4x−10. On the right side: 2(ax−3)=2ax−6. So our equation becomes 4x−10=2ax−6. Rearranging to get all terms on one side: 4x−2ax=−6+10, which gives us (4−2a)x=4. For this equation to have a unique solution, we'd divide both sides by (4−2a). But if 4−2a=0, we can't divide, and we get 0⋅x=4, or 0=4—which is impossible. This happens when 4−2a=0, so a=2. Looking at the wrong answers: Choice A (-2) would give us 8x=4, which has solution x=21. Choice B (0) would give us 4x=4, so x=1. Choice D (4) would give us −4x=4, so x=−1. All of these produce valid solutions. Only choice C creates the contradiction that results in no solution. Strategy tip: When asked about "no solution" scenarios, look for cases where you end up with a false statement like 0=nonzero number after simplifying.