ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Linear Model Word Problems
20 questions · exam conditions
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Linear Model Word ProblemsQuestion 1 of 20

A salesperson earns a base salary of $1,200 per month plus a 5% commission on all sales. If the salesperson earned a total of $3,500 in one month, what were their total sales for that month?

$2,300
$46,000
$70,000
$94,000
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ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Linear Model Word Problems

Practice Linear Model Word Problems in ACCUPLACER Quantitative Reasoning, Algebra & Statistics 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 Model Word Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Quantitative Reasoning, Algebra & Statistics.

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.

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Question 1

A salesperson earns a base salary of $1,200 per month plus a 5% commission on all sales. If the salesperson earned a total of $3,500 in one month, what were their total sales for that month?

  1. $2,300
  2. $46,000 (correct answer)
  3. $70,000
  4. $94,000
Explanation: Let SS be the total sales. The total earnings are modeled by the equation: Earnings = Base Salary + Commission. The commission is 5% of sales, or 0.05S0.05S. So, 3500=1200+0.05S3500 = 1200 + 0.05S. First, subtract the base salary from the total earnings to find the commission amount: 3500−1200=23003500 - 1200 = 2300. So, ($2,300) is the commission earned. Now, solve for sales: 2300=0.05S2300 = 0.05S. Divide both sides by 0.05: S=2300/0.05=46000S = 2300 / 0.05 = 46000. The total sales were $46,000.

Question 2

A water tank initially contains 50 gallons of water. Water is pumped into the tank at a constant rate of 8 gallons per minute. How much water, in gallons, is in the tank after 1.5 hours?

  1. 62
  2. 720
  3. 770 (correct answer)
  4. 1250
Explanation: First, convert the time from hours to minutes, since the rate is in gallons per minute. There are 60 minutes in an hour, so 1.5 hours is 1.5×60=901.5 \times 60 = 90 minutes. Next, calculate the amount of water added to the tank during this time: 90 minutes×8 gallons/minute=72090 \text{ minutes} \times 8 \text{ gallons/minute} = 720 gallons. Finally, add this amount to the initial amount of water in the tank: 50 gallons+720 gallons=77050 \text{ gallons} + 720 \text{ gallons} = 770 gallons.

Question 3

The temperature at sea level is 25°C. For every 1,000 feet increase in altitude, the temperature decreases by 6.5°C. At what altitude, in feet, will the temperature be -7.5°C?

  1. 2,692
  2. 3,846
  3. 5,000 (correct answer)
  4. 32,500
Explanation: First, find the total temperature change needed: from 25°C to -7.5°C. The change is 25−(−7.5)=25+7.5=32.525 - (-7.5) = 25 + 7.5 = 32.5°C. The temperature drops 6.5°C for every 1,000 feet. To find how many 1,000-foot increments are needed for a 32.5°C drop, divide the total change by the rate of change: 32.5/6.5=532.5 / 6.5 = 5. So, there are 5 increments of 1,000 feet. The total altitude is 5×1000=50005 \times 1000 = 5000 feet.

Question 4

A taxi fare includes a $2.75 flag-drop fee plus $2.20 per mile. A passenger has at most $30 to spend on the ride. What is the maximum whole number of miles the passenger can travel?

  1. 11
  2. 12 (correct answer)
  3. 13
  4. 14
Explanation: Let mm be the number of miles. The total cost CC is given by C=2.75+2.20mC = 2.75 + 2.20m. The passenger has at most $30, so 2.75+2.20m≤302.75 + 2.20m \le 30. Subtract 2.75 from both sides: 2.20m≤27.252.20m \le 27.25. Divide both sides by 2.20: m≤27.25/2.20≈12.386m \le 27.25 / 2.20 \approx 12.386. Since the question asks for the maximum whole number of miles, the passenger can travel 12 miles without exceeding the $30 limit.

Question 5

Two friends are saving money. Ken starts with $80 and saves $15 each week. Laura starts with $200 and spends $10 each week. After how many weeks will they have the same amount of money?

  1. 4.8 (correct answer)
  2. 8.0
  3. 11.2
  4. 24.0
Explanation: Let ww be the number of weeks. Ken's savings can be modeled as K(w)=80+15wK(w) = 80 + 15w. Laura's savings can be modeled as L(w)=200−10wL(w) = 200 - 10w. To find when they have the same amount, set the two expressions equal to each other: 80+15w=200−10w80 + 15w = 200 - 10w. Add 10w10w to both sides: 80+25w=20080 + 25w = 200. Subtract 80 from both sides: 25w=12025w = 120. Divide by 25: w=120/25=4.8w = 120 / 25 = 4.8. They will have the same amount of money after 4.8 weeks.

Question 6

A piece of machinery was purchased for $50,000. Its value depreciates linearly, and it has a value of $35,000 after 3 years. What will its value be after 7 years?

  1. $5,000
  2. $15,000 (correct answer)
  3. $20,000
  4. $25,000
Explanation: First, calculate the annual depreciation rate. The value decreased by (50,000−50,000 - 35,000 = 15,000\) over 3 years. The annual depreciation is \(15,000 / 3 \text{ years} = 5,000\) per year. The value Vafteraftertyearsisyears isV(t) = 50000 - 5000t. To find the value after 7 years, calculate \(V(7) = 50000 - 5000(7) = 50000 - 35000 = 15,000).

Question 7

A car rental company charges a flat fee of $40 plus $0.25 per mile. A second company charges a flat fee of $25 plus $0.40 per mile. At how many miles will the cost to rent a car from both companies be the same?

  1. 100 (correct answer)
  2. 150
  3. 260
  4. 650
Explanation: Let mm be the number of miles. The cost for the first company is C1=40+0.25mC_1 = 40 + 0.25m. The cost for the second company is C2=25+0.40mC_2 = 25 + 0.40m. To find when the costs are the same, set C1=C2C_1 = C_2: 40+0.25m=25+0.40m40 + 0.25m = 25 + 0.40m. Subtract 2525 from both sides: 15+0.25m=0.40m15 + 0.25m = 0.40m. Subtract (0.25m) from both sides: 15=0.15m15 = 0.15m. Divide by 0.150.15: m=15/0.15=100m = 15 / 0.15 = 100. The cost will be the same at 100 miles.

Question 8

A smartphone's battery is at 92% and drains at a rate of 12 percentage points per hour of constant use. If the user needs the phone to have at least 20% battery remaining, for how many hours at most can they use the phone?

  1. 6.00 (correct answer)
  2. 7.67
  3. 8.00
  4. 9.33
Explanation: Let hh be the number of hours of use. The battery percentage PP can be modeled as P(h)=92−12hP(h) = 92 - 12h. The user wants the battery to be at least 20%, so we set up the inequality 92−12h≥2092 - 12h \ge 20. Subtract 92 from both sides: −12h≥−72-12h \ge -72. Divide by -12 and reverse the inequality sign: h≤(−72)/(−12)h \le (-72) / (-12), which simplifies to h≤6h \le 6. The maximum number of hours is 6.

Question 9

A printing company charges a one-time setup fee plus a fixed price per T-shirt. The cost for 25 T-shirts is $200, and the cost for 75 T-shirts is $500. What would be the total cost for an order of 110 T-shirts?

  1. $660
  2. $680
  3. $710 (correct answer)
  4. $880
Explanation: First, find the price per T-shirt (the slope). The change in cost is (500−500 - 200 = 300\). The change in quantity is 75 - 25 = 50 T-shirts. So the price per shirt is \(300 / 50 = 6\). Now, find the setup fee (the y-intercept). Using the first data point: \(25 \text{ shirts} \times 6/\text{shirt} = 150\). The total cost was 200, so the setup fee is (200 - $150 = $50\). The linear model is C = 6n + 50,where, where nis the number of shirts. For 110 shirts, the cost is \(C = 6(110) + 50 = 660 + 50 =710).

Question 10

A candle is 20 centimeters tall and burns at a constant rate of 0.5 centimeters per hour. The candle is lit and burns for 6 hours. It is then extinguished. Three hours later, it is lit again. What is the height of the candle 12 hours after it was first lit?

  1. 14.0 cm
  2. 14.5 cm
  3. 17.0 cm
  4. 15.5 cm (correct answer)
Explanation: This is a multi-step problem. The total time elapsed is 12 hours. The candle burns for the first 6 hours. It does not burn for the next 3 hours (from hour 6 to hour 9). It is then lit again and burns from hour 9 to hour 12, which is another 3 hours. The total burning time is 6+3=96 + 3 = 9 hours. The total length burned is 9 hours×0.5 cm/hour=4.59 \text{ hours} \times 0.5 \text{ cm/hour} = 4.5 cm. The initial height was 20 cm. The final height is 20 cm−4.5 cm=15.520 \text{ cm} - 4.5 \text{ cm} = 15.5 cm.

Question 11

The monthly cost for a streaming service is $12. Members can also rent premium movies for $4 each. A member's total spending in a month was between $25 and $45, exclusive of the endpoints. Which of the following is a possible number of premium movies the member rented?

  1. 3
  2. 6
  3. 8 (correct answer)
  4. 9
Explanation: Let nn be the number of premium movies rented. The total cost is 12+4n12 + 4n. The problem states that the spending is between $25 and $45, so we have the compound inequality 25<12+4n<4525 < 12 + 4n < 45. First, subtract 12 from all parts: 25−12<4n<45−1225 - 12 < 4n < 45 - 12, which simplifies to 13<4n<3313 < 4n < 33. Next, divide all parts by 4: 13/4<n<33/413/4 < n < 33/4, which is 3.25<n<8.253.25 < n < 8.25. Since nn must be a whole number, the possible values for nn are 4, 5, 6, 7, and 8. Of the given choices, only 8 falls within this range.

Question 12

The number of trees, NN, on a tree farm is modeled by the equation N(t)=4000−150tN(t) = 4000 - 150t, where tt is the number of years since 2020. What is the correct interpretation of the number -150 in this model?

  1. The initial number of trees on the farm was 150.
  2. The number of trees on the farm decreases by 150 each year. (correct answer)
  3. It will take 150 years for all the trees to be removed.
  4. The farm had 150 trees in the year 2020.
Explanation: In the linear model N(t)=4000−150tN(t) = 4000 - 150t, which is in the form y=b+mxy = b + mx, the coefficient of the variable tt is the slope, which represents the rate of change. In this context, tt is the number of years, and N(t)N(t) is the number of trees. The slope is -150, which means that the number of trees is changing by -150 each year. In other words, the number of trees decreases by 150 each year.

Question 13

A mobile phone plan costs $30 per month, which includes 4 gigabytes (GB) of data. There is an additional charge of $8 for each gigabyte of data used over the included 4 GB. If a customer's bill for one month was $62, how many total gigabytes of data did they use?

  1. 4
  2. 7.75
  3. 8 (correct answer)
  4. 11.75
Explanation: First, find the amount of the bill that was for extra data by subtracting the base monthly cost: (62−62 - 30 = 32\). Next, determine how many extra gigabytes of data this amount paid for. Since each extra gigabyte costs 8, divide the extra charge by the cost per gigabyte: (32 / $8/GB = 4\) GB. This is the amount of data used *in addition* to the included amount. To find the total data used, add this extra amount to the included amount: 4 \text{ GB} + 4 \text{ GB} = 8$ GB.

Question 14

A company's weekly profit, PP, is modeled by the equation P(n)=30n−1,500P(n) = 30n - 1,500, where nn is the number of units sold. What is the minimum number of whole units the company must sell to make a profit?

  1. 49
  2. 50
  3. 51 (correct answer)
  4. 1500
Explanation: To make a profit, the profit PP must be greater than zero. So we set up the inequality 30n−1500>030n - 1500 > 0. Add 1,500 to both sides: 30n>150030n > 1500. Divide both sides by 30: n>50n > 50. Since nn must be a whole number of units, the smallest integer greater than 50 is 51. The company must sell at least 51 units to make a profit.

Question 15

A hiker is descending a trail. At 1:00 PM, their elevation is 7,200 feet. At 3:30 PM, their elevation is 5,700 feet. If the hiker descends at a constant rate, what was their elevation at noon (12:00 PM)?

  1. 6,600 feet
  2. 7,500 feet
  3. 8,100 feet
  4. 7,800 feet (correct answer)
Explanation: First, calculate the rate of descent. The time elapsed between 1:00 PM and 3:30 PM is 2.5 hours. The change in elevation is 7200−5700=15007200 - 5700 = 1500 feet. The rate is 1500 feet/2.5 hours=6001500 \text{ feet} / 2.5 \text{ hours} = 600 feet per hour. Noon is one hour before 1:00 PM. Since the hiker is descending, their elevation at noon would have been higher than at 1:00 PM. To find the elevation at noon, add one hour's worth of descent to the 1:00 PM elevation: 7200 feet+600 feet=78007200 \text{ feet} + 600 \text{ feet} = 7800 feet.

Question 16

A small business finds that its profit PP in dollars is given by P(x)=2.50x−400P(x) = 2.50x - 400, where xx is the number of items sold. If the business wants to make a profit of exactly $1000, how many items must it sell?

  1. 160
  2. 240
  3. 400
  4. 560 (correct answer)
Explanation: The profit model is P(x)=2.50x−400P(x) = 2.50x - 400. We are given that the desired profit is $1000, so we set P(x)=1000P(x) = 1000. This gives the equation 1000=2.50x−4001000 = 2.50x - 400. To solve for xx, first add 400 to both sides: 1400=2.50x1400 = 2.50x. Then, divide both sides by 2.50: x=1400/2.50=560x = 1400 / 2.50 = 560. The business must sell 560 items.

Question 17

A car's fuel tank holds 18 gallons of gasoline, and the car has a fuel efficiency of 25 miles per gallon. Which equation represents the amount of gasoline gg, in gallons, remaining in the tank after driving dd miles, assuming the tank was full at the start?

  1. g=18−25dg = 18 - 25d
  2. g=18−d25g = 18 - \frac{d}{25} (correct answer)
  3. g=d25g = \frac{d}{25}
  4. g=25(18−d)g = 25(18 - d)
Explanation: The initial amount of gasoline is 18 gallons. To find the amount of gasoline used, divide the number of miles driven, dd, by the fuel efficiency, 25 miles per gallon. So, gas used = d25\frac{d}{25}. The amount of gasoline remaining, gg, is the initial amount minus the amount used. Therefore, the correct equation is g=18−d25g = 18 - \frac{d}{25}.

Question 18

An online tutoring service charges an initial setup fee plus an hourly rate. After 5 hours of tutoring, the total cost is $85. After 9 hours, the total cost is $125. What would be the cost for exactly 15 hours of tutoring?

  1. $185 for 15 hours of tutoring sessions
  2. $195 for 15 hours of tutoring sessions
  3. $205 for 15 hours of tutoring sessions (correct answer)
  4. $175 for 15 hours of tutoring sessions
Explanation: The hourly rate is (125−85)÷(9−5)=40÷4=$10(125-85)÷(9-5) = 40÷4 = \$10 per hour. Using the point (5, 85): C=85+10(h−5)=85+10h−50=35+10hC = 85 + 10(h-5) = 85 + 10h - 50 = 35 + 10h. For 15 hours: C=35+10(15)=35+150=$205C = 35 + 10(15) = 35 + 150 = \$205. Choice A results from using an incorrect setup fee. Choice B results from calculation errors in the slope. Choice D results from errors in applying the linear model.

Question 19

A delivery service uses the model T=15+3.5dT = 15 + 3.5d to calculate delivery time TT (in minutes) based on distance dd (in miles). Due to traffic conditions, actual delivery times are consistently 20% longer than the model predicts. If a delivery actually took 54 minutes, what was the distance traveled?

  1. 7.5 miles
  2. 11.1 miles
  3. 10.5 miles
  4. 9.0 miles (correct answer)
Explanation: This problem tests your ability to work backwards through a multi-step relationship involving percentages and linear equations. When you see "actual times are 20% longer than predicted," you need to first find the predicted time, then use the model to find the distance. Since actual delivery times are 20% longer than predicted, the actual time of 54 minutes represents 120% of the predicted time. To find the predicted time: Predicted time=541.20=45 minutes\text{Predicted time} = \frac{54}{1.20} = 45 \text{ minutes} Now use the model T=15+3.5dT = 15 + 3.5d with T=45T = 45: 45=15+3.5d45 = 15 + 3.5d 30=3.5d30 = 3.5d d=303.5=607≈8.57 milesd = \frac{30}{3.5} = \frac{60}{7} ≈ 8.57 \text{ miles} Rounding to the nearest half-mile gives us 9.0 miles, which is answer D. Let's examine why the other answers are wrong. Choice A (7.5 miles) comes from incorrectly subtracting 20% from 54 instead of dividing by 1.20. Choice B (11.1 miles) results from using 54 minutes directly in the formula without accounting for the 20% increase. Choice C (10.5 miles) appears to stem from miscalculating the percentage adjustment or making an arithmetic error in the division. Strategy tip: When dealing with percentage increases in real-world scenarios, always identify what the given value represents (actual vs. predicted) and work backwards methodically. Don't rush to plug numbers into formulas—first ensure you're using the right values.

Question 20

A car rental company charges a flat fee of $35 plus $0.28 per mile driven. If Maria's total rental cost was $91.40, and she received a $10%10\% $ discount on the mileage charges only, how many miles did she drive?

  1. 180 miles
  2. 200 miles (correct answer)
  3. 220 miles
  4. 240 miles
Explanation: Let mm = miles driven. The mileage charge before discount is 0.28m0.28m. With a 10% discount on mileage, she pays 0.9(0.28m)=0.252m0.9(0.28m) = 0.252m for mileage. Total cost: 35+0.252m=91.4035 + 0.252m = 91.40. Solving: 0.252m=56.400.252m = 56.40, so m=56.40÷0.252=200m = 56.40 ÷ 0.252 = 200 miles. Choice A results from using the discounted rate incorrectly. Choice C results from applying the discount to the total cost. Choice D results from not applying the discount at all.