Linear Function Word Problems - Accuplacer Aaf

1. A cell phone plan charges a monthly fee of $25 plus $0.15 for each text message sent. If Maria's bill for last month was $43, and she wants to reduce her next month's bill to $35 by sending fewer text messages, by how many text messages must she reduce her usage?
   Answer: 53 text messages fewer First, find how many texts Maria sent last month: $43 - $25 = $18 in text charges, so $18 ÷ $0.15 = 120 texts. For a $35 bill: $35 - $25 = $10 in text charges, so $10 ÷ $0.15 = 66.67, which rounds to 67 texts maximum. She must reduce by 120 - 67 = 53 texts. Choice B (67) is the target number of texts for next month, not the reduction needed. Choice C (120) is her current usage. Choice D (233) results from incorrectly calculating the total texts across both months.
2. A gym membership has an enrollment fee plus a monthly charge. The total cost after 4 months is $290, and the total cost after 9 months is $440. A new member wants to budget for exactly $500 total. For how many complete months can this member afford the membership?
   Answer: 11 complete months maximum affordable This is a linear equation problem where you need to find the enrollment fee and monthly charge, then determine how many months $500 will cover. Start by setting up the cost structure: Total Cost = Enrollment Fee + (Monthly Charge × Number of Months), or $C = E + M \times n$. You have two data points: after 4 months the cost is $290, and after 9 months it's $440. This gives you two equations: $290 = E + 4M$ and $440 = E + 9M$. Subtracting the first from the second: $440 - 290 = (E + 9M) - (E + 4M)$, which simplifies to $150 = 5M$. Therefore, $M = 30$ (monthly charge is $30). Substituting back: $290 = E + 4(30)$, so $E = 290 - 120 = 170$ (enrollment fee is $170). Now you can find how many months $500 covers: $500 = 170 + 30n$. Solving: $330 = 30n$, so $n = 11$ months exactly. Answer D is correct because the member can afford exactly 11 complete months with $500. Answer A (10 months) would only cost $470, leaving money unused. Answer B (13 months) would cost $560, exceeding the budget. Answer C (12 months) would cost $530, also over budget. When solving linear cost problems, always identify the fixed cost (enrollment fee) versus the variable cost (monthly charges), then use the given data points to solve the system of equations systematically.
3. A delivery service charges based on package weight. The cost for a 3-pound package is $8.50, and the cost for a 7-pound package is $14.50. If the pricing follows a linear model, what would be the cost to ship a 12-pound package?
   Answer: $22.00 for the 12-pound package The rate is ($14.50 - $8.50) ÷ (7 - 3) = $6.00 ÷ 4 = $1.50 per pound. Using point (3, 8.50): Cost = 1.50w + b, so 8.50 = 1.50(3) + b, giving b = 4. Therefore Cost = 1.50w + 4. For 12 pounds: Cost = 1.50(12) + 4 = 18 + 4 = $22. Choice A ($20) forgets the base fee. Choice C ($24) uses $2 per pound rate. Choice D ($27) results from using the wrong base calculation.
4. A student's meal plan starts with a balance of $1,500. Each lunch the student buys costs $8.50, and each dinner costs $12.50. If the student bought 40 lunches, how many dinners can the student buy to have a remaining balance of exactly $260?
   Answer: 72 First, calculate the total amount spent. The starting balance was $1,500 and the desired remaining balance is $260, so the total amount spent is $1500 - 260 = 1240$. Next, calculate the amount spent on lunches: $40 \times 8.50 = 340$. The remaining amount available for dinners is $1240 - 340 = 900$. Finally, divide this amount by the cost of one dinner to find the number of dinners: $900 / 12.50 = 72$. The student can buy 72 dinners.
5. The temperature of a cooling object decreases linearly over time. At 2:00 PM, the temperature is 180°F. At 2:45 PM, the temperature is 135°F. At what time will the temperature reach 90°F?
   Answer: 3:30 PM on the same day The cooling rate is (135 - 180) ÷ 45 = -45 ÷ 45 = -1°F per minute. Using T = 180 - 1(t), where t is minutes after 2:00 PM: 90 = 180 - t, so t = 90 minutes = 1.5 hours after 2:00 PM = 3:30 PM. Choice A (3:15 PM) results from calculating 75 minutes instead of 90. Choice C (4:00 PM) uses 2 hours. Choice D (4:15 PM) results from adding instead of subtracting in the equation setup.
6. The pressure in a certain liquid increases by 10.2 kilopascals (kPa) for every 1 meter increase in depth. At a depth of 2 meters, the total pressure is measured to be 121.9 kPa. What is the pressure at the surface (a depth of 0 meters)?
   Answer: 101.5 kPa Let $P(d)$ be the pressure at depth $d$. The rate of increase is the slope, so $m=10.2$. The function is $P(d) = 10.2d + b$, where $b$ is the pressure at the surface (the y-intercept). We are given the point (2, 121.9). Substitute this into the equation: $121.9 = 10.2(2) + b \implies 121.9 = 20.4 + b$. Solving for $b$ gives $b = 121.9 - 20.4 = 101.5$. The pressure at the surface is 101.5 kPa.
7. A factory produces widgets at a constant rate. After 4 hours of production, there are 320 widgets in the warehouse. After 7 hours of production, there are 470 widgets in the warehouse. If the warehouse initially contained some widgets before production began, how many widgets were in the warehouse at the start?
   Answer: 120 widgets initially present The production rate is (470 - 320) ÷ (7 - 4) = 150 ÷ 3 = 50 widgets per hour. Using the point (4, 320): if 50 widgets are produced per hour, then after 4 hours, 200 widgets were produced. So the initial amount was 320 - 200 = 120 widgets. Choice A (50) is the production rate, not the initial amount. Choice B (80) results from using the wrong rate calculation. Choice D (150) is the total production over 3 hours, not the initial amount.
8. A candle burns at a constant rate. When lit, it is 9 inches tall. After burning for 2 hours, it is 7.5 inches tall. If the candle continues burning at this rate, what will be its height after burning for a total of 8 hours?
   Answer: 3.0 inches tall after 8 hours The burning rate is (9 - 7.5) ÷ 2 = 1.5 ÷ 2 = 0.75 inches per hour. The height function is h(t) = 9 - 0.75t. After 8 hours: h(8) = 9 - 0.75(8) = 9 - 6 = 3 inches. Choice A (4.5) uses 0.5625 inches per hour rate. Choice B (5.0) subtracts only 4 inches total. Choice D (6.0) uses 0.375 inches per hour rate.
9. Two cyclists are in a race. Cyclist A starts at the starting line and travels at a constant speed of 18 miles per hour. Cyclist B starts with a 5-mile head start and travels at a constant speed of 15 miles per hour along the same route. How long after the start of the race will it take for Cyclist A to catch up to Cyclist B?
   Answer: 1 hour and 40 minutes Let $t$ be the time in hours. The distance traveled by Cyclist A is $D_A = 18t$. The distance traveled by Cyclist B is $D_B = 15t + 5$ (initial distance plus distance traveled). Cyclist A catches up when their distances are equal: $18t = 15t + 5$. Subtracting $15t$ from both sides gives $3t = 5$, so $t = \frac{5}{3}$ hours. To convert this to hours and minutes: $\frac{5}{3} \text{ hours} = 1 \frac{2}{3} \text{ hours}$. Since $\frac{2}{3}$ of an hour is $\frac{2}{3} \times 60 = 40$ minutes, the time is 1 hour and 40 minutes.
10. A company has fixed monthly costs of $22,000. It costs the company $6.50 to produce each unit of its product, which it sells for $17.50 per unit. How many units must the company sell in a month to break even (i.e., for total revenue to equal total cost)?
    Answer: 2,000 Let $x$ be the number of units. The total cost function is $C(x) = 22000 + 6.50x$. The total revenue function is $R(x) = 17.50x$. To break even, $C(x) = R(x)$. So, $22000 + 6.50x = 17.50x$. Subtract $6.50x$ from both sides: $22000 = 11x$. Divide by 11: $x = 2000$. The company must sell 2,000 units to break even.
11. A submarine is at a depth of 450 meters below sea level and ascends toward the surface at a constant rate of 75 meters per minute. The function $D(t) = 450 - 75t$ gives the submarine's depth in meters after $t$ minutes. What does the t-intercept of the graph of this function represent?
    Answer: The time it takes for the submarine to reach the sea surface. The t-intercept is the value of $t$ for which the function's value is zero. In this context, it's the time $t$ when the depth $D(t)$ is 0. A depth of 0 meters represents the sea surface. Therefore, the t-intercept represents the time it takes for the submarine to reach the surface.
12. To host a banquet, a hotel charges a one-time reservation fee plus a cost per person. A banquet for 50 people costs a total of $2,900. A banquet for 80 people costs a total of $4,400. What is the hotel's one-time reservation fee?
    Answer: $400 This problem describes a linear relationship where the reservation fee is the y-intercept. Let the points be (number of people, total cost). So we have (50, 2900) and (80, 4400). First, find the slope (cost per person): $m = \frac{4400 - 2900}{80 - 50} = \frac{1500}{30} = 50$. Now use the point-slope form $y - y_1 = m(x - x_1)$ to find the equation of the line: $y - 2900 = 50(x - 50)$. The reservation fee is the cost when $x=0$. $y - 2900 = 50(0 - 50) \implies y - 2900 = -2500 \implies y = 400$. The fee is $400.
13. The value, $V$, in dollars, of a piece of industrial equipment after $t$ years is modeled by the function $V(t) = 45,000 - 2,500t$. Which of the following is the best interpretation of the number 2,500 in this context?
    Answer: The value of the equipment decreases by $2,500 each year. In the linear model $y = b + mx$, $m$ represents the rate of change. In $V(t) = 45,000 - 2,500t$, the slope is -2,500. This means that for each one-year increase in $t$, the value $V$ changes by -2,500. Therefore, the value of the equipment decreases by $2,500 each year.
14. A candle is initially 25 centimeters tall. After burning for 4 hours, its height is 19 centimeters. The candle burns at a constant rate. What will be the height of the candle after it has been burning for a total of 6 hours?
    Answer: 16 cm The initial height (at $t=0$) is 25 cm. After 4 hours, the height is 19 cm. The amount burned is $25 - 19 = 6$ cm in 4 hours. The rate of burning is $6 \text{ cm} / 4 \text{ hours} = 1.5$ cm per hour. The linear model for the height $H$ after $t$ hours is $H(t) = 25 - 1.5t$. To find the height after 6 hours, calculate $H(6)$: $H(6) = 25 - 1.5(6) = 25 - 9 = 16$. The height will be 16 cm.
15. An employee earns $18 per hour for the first 40 hours worked in a week and $27 per hour for any hours worked over 40. Which function $E(h)$ represents the employee's total weekly earnings for working $h$ hours, where $h > 40$?
    Answer: $E(h) = 18(40) + 27(h-40)$ For $h > 40$, the employee's earnings consist of two parts. First, the pay for the initial 40 hours, which is $18 \times 40 = 720$. Second, the pay for overtime hours. The number of overtime hours is $h - 40$, and the rate for these hours is $27. So, the overtime pay is $27(h-40)$. The total earnings are the sum of these two parts: $E(h) = 18(40) + 27(h-40)$.
16. A salesperson earns a weekly base salary of $400 plus an 8% commission on all sales revenue over $1,000. This week, the salesperson's total earnings were $820. What was the total sales revenue for the week?
    Answer: $6,250 Let $S$ be the total sales revenue. The commission is earned on the amount $S - 1000$. The earnings function is $E(S) = 400 + 0.08(S - 1000)$. We are given $E(S) = 820$. So, $820 = 400 + 0.08(S - 1000)$. Subtract 400: $420 = 0.08(S - 1000)$. Divide by 0.08: $5250 = S - 1000$. Add 1000: $S = 6250$. The total sales revenue was $6,250.
17. A rental car company charges $35 per day plus $0.25 per mile driven. A customer's total bill was $127.50 for a 3-day rental. What linear equation represents the relationship between the total cost $C$ and the number of miles $m$ driven for this specific rental period?
    Answer: $C = 105 + 0.25m$ For a 3-day rental, the fixed cost is $35 × 3 = $105. The variable cost is $0.25 per mile. Therefore, the equation is C = 105 + 0.25m. Choice A uses the daily rate instead of the 3-day total. Choice C incorrectly uses the customer's total bill as the fixed cost. Choice D shows the calculation but isn't simplified to standard form.
18. A water tank contains 800 gallons and is being drained at a constant rate. After 25 minutes, 650 gallons remain. After how many total minutes from the start will the tank contain exactly 200 gallons?
    Answer: 100 minutes from the start When you encounter a problem about constant rates of change, you're working with linear relationships. The key is to find the rate and use it to predict future values. First, let's determine the drainage rate. The tank started with 800 gallons and had 650 gallons after 25 minutes, so it lost $800 - 650 = 150$ gallons in 25 minutes. This gives us a rate of $\frac{150 \text{ gallons}}{25 \text{ minutes}} = 6$ gallons per minute. Now we can find when the tank will contain 200 gallons. Starting from 800 gallons, the tank needs to lose $800 - 200 = 600$ gallons total. At 6 gallons per minute, this will take $\frac{600 \text{ gallons}}{6 \text{ gallons/minute}} = 100$ minutes. Looking at the wrong answers: Choice A (75 minutes) would leave $800 - (75 \times 6) = 350$ gallons, not 200. Choice B (150 minutes) would result in $800 - (150 \times 6) = -100$ gallons, meaning the tank would be empty. Choice C (125 minutes) would give $800 - (125 \times 6) = 50$ gallons remaining. The correct answer is D) 100 minutes from the start. Strategy tip: For constant rate problems, always identify what changed over what time period to find your rate, then use that rate to answer the question. Double-check by working backwards from your answer to see if the math holds up.
19. An online streaming service offers two monthly plans. Plan X costs $25 per month plus $1.50 per movie rental. Plan Y costs $10 per month plus $2.25 per movie rental. At what number of movie rentals per month does Plan X become the less expensive option?
    Answer: 21 rentals or more Let $m$ be the number of movie rentals. The cost of Plan X is $C_X = 25 + 1.50m$. The cost of Plan Y is $C_Y = 10 + 2.25m$. We want to find when Plan X is less expensive, so $C_X < C_Y$. $25 + 1.50m < 10 + 2.25m$. Subtract (1.50m) from both sides: $25 < 10 + 0.75m$. Subtract 10 from both sides: $15 < 0.75m$. Divide by 0.75: $m > \frac{15}{0.75} \implies m > 20$. Plan X becomes less expensive when the number of rentals is greater than 20. The smallest integer number of rentals for which this is true is 21.
20. The total cost to manufacture a product is a linear function of the number of units produced. The cost to produce 100 units is $5,000, and the cost to produce 400 units is $14,000. What is the cost per unit to produce this product?
    Answer: $30 The cost per unit is the rate of change of the linear function, which is its slope. Using the two given points, (100, 5000) and (400, 14000), we can calculate the slope: $m = \frac{\text{change in cost}}{\text{change in units}} = \frac{14000 - 5000}{400 - 100} = \frac{9000}{300} = 30$. The cost per unit is $30.

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