PSAT Math Quiz: Linear And Exponential Growth

What this quiz covers

This quiz focuses on Linear And Exponential Growth, giving you a quick way to practice the rules, question types, and explanations that matter most for PSAT Math.

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.

Question 1

1. 5 years
2. 7 years
3. 10 years (correct answer)
4. 12 years

Explanation: P(t) = 2000 + 300t (linear through (0,2000) and (10,5000)). Q(t) = 1000·(1.18)^t. Q(10) ≈ 1000·5.23 ≈ 5230, while P(10) = 5000. Checking Q(9) ≈ 4435 < P(9)=4700. So Q first exceeds P near t=10. (A), (B) occur before crossover; (D) is too late.

Question 2

1. $2
2. $6 (correct answer)
3. $20
4. $26

Explanation: The question asks for the base fee from the linear graph of cost C versus miles m passing through (0,6) and (10,26). The straight line indicates a linear model with constant additive cost per mile. The y-intercept at (0,6) is the base fee of $6. This emphasizes additive changes, unlike exponential multiplication. A common error is mistaking the total at m=10 for the base. Use the intercept directly for fixed costs in linear models.

Question 3

1. 0.03%
2. 3% (correct answer)
3. 30%
4. 103%

Explanation: The question asks for the annual percent increase in water usage from W(t)=12000(1.03)^t. This is exponential growth with multiplicative factor 1.03 per year. The 1.03 means 3% increase annually. Emphasize multiplicative over additive. A common error is misreading as 0.03% or adding 103%. Identify the base minus 1 for percent in exponential models.

Question 4

1. $V=4t+15$ (correct answer)
2. $V=6t+15$
3. $V=15(4)^t$
4. $V=15(1.4)^t$

Explanation: The question asks for the equation modeling the linear relationship between volume V and time t for a tank filled at a constant rate. The straight-line graph indicates a linear model with constant additive changes in volume per minute. The slope is calculated as (39 - 15) / (6 - 0) = 4 gallons per minute, and the y-intercept is 15, yielding V = 4t + 15. This emphasizes additive growth, unlike exponential which would show multiplicative changes. A key error is confusing the points and selecting an exponential form like V = 15(1.4)^t, which doesn't fit the straight line. For graph-based questions, calculate the slope using two points to verify linear models and avoid assuming exponential without checking ratios.

Question 5

1. 570
2. 810 (correct answer)
3. 900
4. 1,110

Explanation: The question asks for the predicted number of tickets sold after 6 hours using the linear model T(h) = 95h + 240. This is a linear model, characterized by constant additive increases of 95 tickets per hour, in contrast to exponential models where changes are multiplicative. Calculate by substituting h = 6: 95 × 6 = 570, plus 240 equals 810 tickets. Common errors include omitting the +240 to get 570 or adding incorrectly to reach 900 or 1110. Remember, linear growth adds a fixed amount each time, not a percentage. For test-taking, double-check substitutions in linear equations to ensure all terms are accounted for.

Question 6

1. about 25 years
2. about 28 years
3. about 31 years (correct answer)
4. They will never be equal.

Explanation: When you encounter problems comparing linear and exponential functions, you're looking for intersection points where the two functions have equal values. This requires setting the equations equal and solving for the variable. To find when the balances are equal, set $F(t) = G(t)$: $2{,}000 + 150t = 2{,}000(1.04)^t$ This equation can't be solved algebraically, so you need to test values or use a graphing approach. Since Plan F grows by a constant $150 per year while Plan G grows by 4% annually, the exponential plan will eventually overtake the linear one. Testing strategic values:

• At $$t = 25$ $:$ $F(25) = 5{,}750$ $and$ $G(25) ≈ 5{,}332$ $
• At $$t = 30$$: $$F(30) = 6{,}500$$ and $$G(30) ≈ 6{,}493$$
• At $$t = 31$$: $$F(31) = 6{,}650$$ and $$G(31) ≈ 6{,}752$$

Question 7

1. $B=100(0.88)^t$; 52%
2. $B=100-12t$; 40% (correct answer)
3. $B=12t-100$; 40%
4. $B=100-5t$; 75%

Explanation: The question models phone battery percent B losing 12 points per hour after 100% at noon, and predicts at t=5. This is linear decay with constant subtraction of 12 per hour, emphasizing additive change over multiplicative. The equation is B=100-12t; at t=5, B=100-60=40%. Verify: at t=1, 88%; t=2, 76%, linear. Errors include exponential decay (A) or positive slope (C). Distinguish: constant differences mean linear, ratios mean exponential.

Question 8

1. $1.02 added each day
2. 2% increase per day (correct answer)
3. $800 increase per day
4. 2-day doubling time

Explanation: The question asks what 1.02 represents in the exponential revenue model R(d)=800(1.02)^d. The model is exponential growth because 1.02 is a constant multiplicative factor greater than 1, unlike linear's additive increases. The 1.02 means revenue multiplies by 1.02 daily, or grows 2% per day. This is calculated as growth rate = factor - 1 = 0.02. A key error is interpreting it as additive $1.02 daily. In exponential models, subtract 1 from the base to find the percentage multiplicative change.

Question 9

1. 3 minutes
2. 4 minutes (correct answer)
3. 5 minutes
4. 6 minutes

Explanation: When you encounter problems comparing linear and exponential functions, you need to find where one function overtakes another by testing values systematically. Here you have a linear function $L(t) = 5 + 2t$ versus an exponential function $E(t) = 5(1.3)^t$. The linear function grows at a constant rate of 2 liters per minute, while the exponential function starts slower but accelerates over time. Let's calculate both functions at each answer choice: At $t = 3$: $L(3) = 5 + 2(3) = 11$ liters, $E(3) = 5(1.3)^3 = 5(2.197) ≈ 11.0$ liters At $t = 4$: $L(4) = 5 + 2(4) = 13$ liters, $E(4) = 5(1.3)^4 = 5(2.856) ≈ 14.3$ liters At 4 minutes, hose E finally produces more water than hose L for the first time. Choice A (3 minutes) is incorrect because at $t = 3$, both functions produce approximately the same amount—hose E hasn't exceeded hose L yet. Choice C (5 minutes) and D (6 minutes) are wrong because while hose E does exceed hose L at these times, the question asks for the first time this happens, which occurs earlier at 4 minutes. The key strategy here is recognizing that exponential functions eventually outpace linear functions, but you need to test values methodically to find the exact crossover point. Always check the time period just before your suspected answer to confirm it's truly the "first" occurrence.

Question 10

1. Plan A will have about 1{,}500 more downloads than Plan B.
2. Plan A will have about 400 more downloads than Plan B.
3. Both plans will result in about the same number of downloads.
4. Plan B will have about 2{,}500 more downloads than Plan A. (correct answer)

Explanation: When you encounter problems comparing different growth patterns, you need to distinguish between linear growth (constant amount added) and exponential growth (constant percentage increase). This distinction becomes crucial over longer time periods. For Plan A (linear growth), the app gains exactly 60 downloads each day. After 20 days: $800 + (60 \times 20) = 800 + 1{,}200 = 2{,}000$ total downloads. For Plan B (exponential growth), the total downloads multiply by 1.09 each day. After 20 days: $800 \times (1.09)^{20}$. Since $(1.09)^{20} \approx 5.604$, this gives us $800 \times 5.604 = 4{,}483$ total downloads. Plan B exceeds Plan A by $4{,}483 - 2{,}000 = 2{,}483$ downloads, which is about 2,500 more downloads. Choice A incorrectly suggests Plan A has more downloads and uses an unrealistic difference. Choice B also wrongly favors Plan A with a much smaller gap that doesn't reflect exponential growth over 20 days. Choice C assumes the plans yield similar results, but exponential growth significantly outpaces linear growth over this timeframe. Choice D correctly identifies that Plan B generates about 2,500 more downloads than Plan A. Strategy tip: When comparing linear versus exponential growth over multiple time periods, exponential growth typically dominates after sufficient time, even when it starts slower. Always calculate both scenarios completely rather than estimating—the exponential function $(1.09)^{20}$ creates a much larger multiplier than intuition suggests.

Question 11

1. $P = 12{,}500 + 4.5t$
2. $P = 12{,}500(1 + 4.5t)$
3. $P = 12{,}500(1.045)^t$
4. $P = 12{,}500 + 450t$ (correct answer)

Explanation: When you encounter word problems about population growth, the key is identifying whether the growth is linear (constant amount added each period) or exponential (constant percentage increase each period). The phrase "gaining about 450 residents each year" signals linear growth. For linear growth, you need a base amount plus a constant rate times time. Here, the town starts with 12,500 residents and adds 450 residents per year. After $t$ years, the population will be the initial population plus 450 times the number of years that have passed. This gives us $P = 12{,}500 + 450t$, which matches choice D. Choice A incorrectly uses 4.5 instead of 450 for the yearly increase. This represents a common decimal placement error—perhaps confusing 450 residents with 4.5 hundred residents, but the problem clearly states 450. Choice B uses the form for exponential growth but with the wrong structure. The expression $12{,}500(1 + 4.5t)$ would mean the population is multiplied by $(1 + 4.5t)$ each year, which isn't what's happening here. Choice C shows exponential growth with the correct initial value but assumes a 4.5% annual growth rate. Exponential models use the form $P = P_0(1 + r)^t$ where $r$ is the growth rate as a decimal. However, the problem describes adding a fixed number of residents, not a percentage increase. Remember: "adds/gains a fixed amount" means linear growth (addition), while "grows by a percentage" means exponential growth (multiplication with exponents).

Question 12

1. $y=10x+100$
2. $y=100(1.1)^x$ (correct answer)
3. $y=100(0.9)^x$
4. $y=100+1.1x$

Explanation: The question asks for the equation best modeling points (0,100), (1,110), (2,121), (3,133.1). The model is exponential because ratios are constant at 1.1, showing multiplicative growth of 10% per unit, unlike linear constant differences. The equation y=100(1.1)^x fits: at x=1, 100×1.1=110; x=2, 110×1.1=121; x=3, 121×1.1=133.1. Differences increase (10,11,12.1), confirming not linear. A key error is choosing linear like 100+10x, ignoring accelerating change. Plot or check ratios for points to identify multiplicative exponential patterns over additive linear.

Question 13

1. Hourly wage
2. Starting amount earned at 0 hours (correct answer)
3. Percent increase per hour
4. Hours needed to earn $12

Explanation: The question asks what the 12 represents in the linear model y=1.8x +12 for earnings. The model is linear because earnings increase additively by $1.80 per hour, unlike exponential percentage growth. The 12 is the y-intercept, meaning $12 earned at 0 hours, perhaps a base pay. Substituting x=0 gives y=12 directly. A common error is thinking it's the hourly wage, which is the slope. In linear fits, interpret intercept as the additive starting value when input is zero.

Question 14

1. $h=12(1.5)^d$
2. $h=1.5d+12$ (correct answer)
3. $h=12d+1.5$
4. $h=12-1.5d$

Explanation: The question asks which equation gives the plant's height h after d days, growing at a constant rate. This is a linear model with constant additive growth of 1.5 cm per day, not exponential. Starting at 12 cm, h=12 + 1.5d fits. Build the model by adding the daily increase to the initial height. A key error is using exponential like (1.5)^d for additive scenarios. Check for constant differences to confirm linear growth.

Question 15

1. Bacteria added each hour
2. Initial bacteria at $t=0$ (correct answer)
3. Doubling time in hours
4. Growth factor per hour

Explanation: The question asks what the number 200 represents in the exponential model $B(t) = 200(2)^{t/3}$ for bacteria growth. The model is exponential, with multiplicative doubling every 3 hours, not linear additive changes. Substituting $t = 0$ gives $B(0) = 200 * 1 = 200$, showing it as the initial number of bacteria. The exponent t/3 adjusts for the tripling period, emphasizing the growth factor. A common error is misinterpreting 200 as the growth factor or added amount, confusing exponential multiplication with linear addition. When interpreting models, evaluate at t=0 to identify initial values and distinguish additive from multiplicative parameters.

Question 16

1. Correct; 6 means 6% growth
2. Incorrect; should be $y=200+6t$ (correct answer)
3. Incorrect; should be $y=200(6)^t$
4. Correct; exponential always fits

Explanation: The question requires critiquing a student's exponential model y=200(1.06)^t for a quantity starting at 200 and increasing by 6 each day. The situation describes linear growth with constant additive increases of 6, not multiplicative exponential change. The correct model is y=200 + 6t, so the student's use of 1.06 (implying 6% growth) is incorrect. A key error is confusing fixed amount increases with percentage increases. Another mistake might be thinking exponential always fits growth. Distinguish by checking if changes are additive (linear) or proportional (exponential), and choose models accordingly.

Question 17

1. 300/month; constant ratio
2. 12%; exponential growth (correct answer)
3. 300/month; exponential growth
4. 12%; constant difference

Explanation: The question asks which service will have more users after 6 months and why. The first is linear with additive 300/month, the second exponential with multiplicative 12% growth. Calculate linear: 2000 + 3006=3800; exponential: 2000(1.12)^6 ≈3946, so exponential has more due to compounding. This highlights exponential multiplicative advantage over linear additive. A key error is swapping model types. Compare final values after identifying growth types.

Question 18

1. Linear decay
2. Exponential decay (correct answer)
3. Linear growth
4. Exponential growth

Explanation: The question asks which model best describes the battery percentage over time from the given values. The percentages show constant ratios of 0.85 each hour, indicating exponential decay with multiplicative changes, not linear additive subtraction. Calculate ratios: 85/100=0.85, 72.25/85=0.85, confirming exponential. This emphasizes multiplicative reduction per hour over constant differences. A key error is seeing decreasing values and assuming linear without checking ratios. Always compute successive ratios for potential exponential patterns in decay scenarios.

Question 19

1. $10$
2. $20$ (correct answer)
3. $30$
4. $40$

Explanation: This question asks for the rate of change (slope) of a linear relationship given two points. Using points (2, 50) and (6, 130), we calculate slope = (change in p)/(change in t) = (130 - 50)/(6 - 2) = 80/4 = 20 pages per hour. This represents how many pages the student reads each hour. Common errors include reversing the subtraction order or mixing up which variable goes in the numerator versus denominator. Remember that rate of change in a word problem context tells you "how much y changes per unit of x."

Question 20

1. The cost per hour
2. The total cost for 6 hours
3. The fixed fee (starting cost) (correct answer)
4. The number of hours included free

Explanation: This question asks about the meaning of the constant term in a linear equation representing bike rental costs. The equation C = 6 + 4h follows the form y = mx + b, where b is the y-intercept. In this context, when h = 0 (no hours rented), the cost C = 6 + 4(0) = 6 dollars, which represents the fixed fee charged regardless of rental time. The coefficient 4 represents the hourly rate ($4 per hour), not the constant 6. A common error is confusing the roles of the slope and y-intercept in real-world contexts. When interpreting linear models, always check what happens when the independent variable equals zero.