Prealgebra Quiz: Linear Equation Word Problems
4 questions · exam conditions
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Linear Equation Word ProblemsQuestion 1 of 4

Marcus has been saving money for a new bike. He started with some initial amount and saves the same amount each week. After 6 weeks, he had $127. After 10 weeks, he had $167. How much money did he start with?

$67
$70
$73
$77
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Prealgebra Quiz

Prealgebra Quiz: Linear Equation Word Problems

Practice Linear Equation Word Problems in Prealgebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Linear Equation Word Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for Prealgebra.

How to use this quiz

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.

All questions

Question 1

Marcus has been saving money for a new bike. He started with some initial amount and saves the same amount each week. After 6 weeks, he had $127. After 10 weeks, he had $167. How much money did he start with?

  1. $67 (correct answer)
  2. $70
  3. $73
  4. $77
Explanation: Let ss = initial amount and ww = weekly savings. We have: s+6w=127s + 6w = 127 and s+10w=167s + 10w = 167. Subtracting the first from the second: 4w=404w = 40, so w=10w = 10. Substituting back: s+6(10)=127s + 6(10) = 127, so s=127−60=67s = 127 - 60 = 67. Choice B (70)mightresultfromacalculationerror.ChoiceC(70) might result from a calculation error. Choice C (73) might come from using the wrong equation setup. Choice D ($77) might result from adding instead of subtracting when solving for the initial amount.

Question 2

A taxi company charges an initial fee when you get in the cab, plus a rate per mile traveled. The total cost for a 8-mile trip is $19.20, and the total cost for a 15-mile trip is $30.30. What is the cost per mile?

  1. $1.45
  2. $1.50
  3. $1.58 (correct answer)
  4. $1.62
Explanation: Let ff = initial fee and rr = rate per mile. We have: f+8r=19.20f + 8r = 19.20 and f+15r=30.30f + 15r = 30.30. Subtracting the first equation from the second: 7r=30.30−19.20=11.107r = 30.30 - 19.20 = 11.10. Therefore r=11.10÷7=1.585...r = 11.10 ÷ 7 = 1.585... which rounds to 1.58.ChoiceA(1.58. Choice A (1.45) might result from incorrectly calculating 11.10÷811.10 ÷ 8. Choice B ($1.50) might come from rounding $11.10÷711.10 ÷ 7 $ incorrectly. Choice D ($1.62) might result from a calculation error in the subtraction step.

Question 3

A gym membership has a one-time enrollment fee plus a monthly charge. After 4 months, the total amount paid is $290. After 9 months, the total amount paid is $490. What will be the total amount paid after exactly 1 year of membership?

  1. $610 (correct answer)
  2. $630
  3. $650
  4. $670
Explanation: Let ee = enrollment fee and mm = monthly charge. We have: e+4m=290e + 4m = 290 and e+9m=490e + 9m = 490. Subtracting: 5m=2005m = 200, so m=40m = 40. Then e=290−160=130e = 290 - 160 = 130. After 12 months: 130+12(40)=130+480=610130 + 12(40) = 130 + 480 = 610. Choice B (630)mightresultfrommiscountingmonthsorcalculationerrors.ChoiceC(630) might result from miscounting months or calculation errors. Choice C (650) might come from using an incorrect monthly rate. Choice D ($670) might result from adding the enrollment fee twice.

Question 4

A tutoring service charges a registration fee plus a rate per hour of tutoring. The total cost for 8 hours of tutoring is $156, and the total cost for 12 hours is $204. A student wants to prepay for exactly enough hours so that their average cost per hour (including the registration fee) is $15.00. How many hours should they purchase?

  1. 16 hours
  2. 18 hours
  3. 20 hours (correct answer)
  4. 22 hours
Explanation: Let rr = registration fee and hh = hourly rate. We have: r+8h=156r + 8h = 156 and r+12h=204r + 12h = 204. Subtracting: 4h=484h = 48, so h=12h = 12. Then r=156−96=60r = 156 - 96 = 60. For average cost of $15 per hour over $nn hours:hours: 60+12nn=15\frac{60 + 12n}{n} = 15 .Solving:. Solving: 60+12n=15n60 + 12n = 15n ,so, so 60=3n60 = 3n ,giving, giving n=20n = 20 .ChoiceA(16hours)givesaveragecost. Choice A (16 hours) gives average cost 60+19216=15.75\frac{60 + 192}{16} = 15.75 .ChoiceB(18hours)gives. Choice B (18 hours) gives 60+21618=15.33\frac{60 + 216}{18} = 15.33 .ChoiceD(22hours)gives. Choice D (22 hours) gives 60+26422=14.73\frac{60 + 264}{22} = 14.73 $.