Linear & Exponential Growth - PSAT Math

A fish tank has 30 liters of water and leaks 4 liters per hour. Which equation models the amount of water $W$ after $h$ hours?
Answer: $W=30-4h$ The question asks for the equation modeling water W in a leaking tank starting at 30 liters and losing 4 liters per hour. The model is linear decay because it subtracts a constant amount each hour additively, unlike exponential decay with multiplicative factors. The equation is W = 30 - 4h, where the negative slope shows constant loss. This fits as W(0)=30 and each hour reduces by 4. A key error is choosing exponential like 30 $(0.87)^h$ , misinterpreting percentage loss. For decay problems, check if change is fixed amount (linear additive) or fixed percentage (exponential multiplicative).
A coordinate plane shows a line representing the cost $y$ of renting a scooter for $x$ minutes. The line passes through $(0, 1.50)$ and $(20, 6.50)$ . What is the cost for 12 minutes?
Answer: $4.50 The question asks for the cost of renting a scooter for 12 minutes based on a line through (0, 1.50) and (20, 6.50). This is a linear model, with constant additive change in cost per minute, unlike exponential multiplicative changes. Find the slope: (6.50 - 1.50)/(20 - 0) = 5/20 = 0.25 per minute; equation y = 0.25x + 1.50; at x=12, 0.25 × 12 + 1.50 = 3 + 1.50 = $4.50. A common error is ignoring the y-intercept, leading to just 0.25 × 12 = $3.00. Another mistake might be incorrect slope calculation. For linear graphs, calculate slope accurately and use point-slope form to predict values.
A coordinate plane shows an exponential curve passing through $(0, 5)$ and $(1, 15)$ . If the curve follows $y=a(b)^x$ , what is the value of $b$ ?
Answer: $b=3$ The question asks for b in the exponential model y=a $b^x$ passing through (0,5) and (1,15). The model is exponential because it uses a growth factor b for multiplicative change, unlike linear's additive slope. At x=0, y=5=a. At x=1, 15=a b=5b, so b=3. This means y triples each unit x multiplicatively. A key error is setting b=1/3 for decay. For points on exponentials, solve for a at x=0, then b using the ratio of y-values.
A new app had 800 downloads on day 0. • Plan A: The total number of downloads increases by 60 each day. • Plan B: The total number of downloads increases by 9\% each day. After 20 days, which statement about the total downloads is true?
Answer: Plan B will have about 2{,}500 more downloads than Plan A. 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.
A company's number of customers is modeled by $C(t)=1500(1.15)^t$ , where $t$ is in years. Which is the best interpretation of 1.15?
Answer: 15% growth each year The question requires interpreting the 1.15 in the exponential model C(t) = 1500 $(1.15)^t$ for customer growth. This is an exponential model, showing multiplicative growth by 1.15 each year, contrasting with linear additive increases. The 1.15 represents a 15% annual growth rate, as 1 + 0.15 = 1.15. Key errors include mistaking it for additive, like adding 1.15 or 15 customers yearly. Another mistake is confusing the initial value with the growth factor. When analyzing exponential models, identify the base as the multiplicative factor and convert to percentage for interpretation.
A phone battery has 100% charge at noon and loses 12 percentage points per hour while streaming video. Let $t$ be hours after noon and $B$ be battery percent. Which equation models this situation, and what is the predicted battery percent at $t=5$ ?
Answer: $B=100-12t$ ; 40% 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.
A phone battery has 100% charge at $t=0$ hours and then retains 80% of its charge each hour. Which statement best describes the model for battery charge $B$ over time?
Answer: Exponential; multiplies by 0.8 each hour This question asks about identifying and describing an exponential decay model. A battery that retains 80% of its charge each hour loses 20% but keeps 80%, meaning it multiplies by 0.8 each hour. This is exponential decay because we multiply by a constant factor (0.8) rather than subtract a constant amount. The model would be B = 100 $(0.8)^t$ . Linear decay would subtract the same amount each hour, but this problem describes multiplicative change. The key phrase "retains 80% of its charge" signals multiplication, making this exponential.
A plant is 12 cm tall when measured on day 0. It grows at a constant rate of 1.5 cm per day. Which equation gives the plant's height $h$ (cm) after $d$ days?
Answer: $h=1.5d+12$ 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.
The graph shown displays the value of a car over time. The owner models depreciation with $V(t) = V_0(1-r)^t$ , where $t$ is in years. Based on the graph, what is the approximate value of $r$ ?
Answer: 0.15 From the graph, V(0)=24000 and V(4)=12528. So $(1-r)^4$ = 12528/24000 = 0.522. Thus 1-r = $0.522^{1/4}$ ≈ 0.85, giving r ≈ 0.15. (A) gives $(0.9)^4$ ≈ 0.656 → V(4)≈15744, too high. (C) gives $(0.8)^4$ ≈ 0.41 → V(4)≈9830, too low. (D) gives V(4)≈7594, far too low.
A bacteria culture is measured at noon, 1 p.m., and 2 p.m.: 600, 780, and 1,014 cells, respectively. Assuming the pattern continues, which prediction most accurately describes the count at 4 p.m.?
Answer: About 1,700 cells, following exponential growth When you encounter bacteria growth problems, you need to determine whether the population follows linear growth (constant addition) or exponential growth (constant multiplication ratio). Look at the pattern in the data to identify which model applies. Let's examine the given measurements: 600 cells at noon, 780 cells at 1 p.m., and 1,014 cells at 2 p.m. For linear growth, we'd expect constant differences between consecutive measurements. Here, 780 - 600 = 180, but 1,014 - 780 = 234. Since these differences aren't equal, this isn't linear growth. For exponential growth, check if there's a constant ratio between consecutive terms. The ratio from noon to 1 p.m. is 780/600 = 1.3, and from 1 p.m. to 2 p.m. is 1,014/780 = 1.3. This consistent multiplication factor of 1.3 confirms exponential growth. To find the 4 p.m. count, multiply by 1.3 twice more: 1,014 × 1.3 = 1,318 cells at 3 p.m., then 1,318 × 1.3 ≈ 1,713 cells at 4 p.m. Choice A incorrectly assumes linear growth and gives too low a prediction. Choice B also assumes linear growth but arrives at the approximately correct number through flawed reasoning. Choice C correctly identifies exponential growth but severely underestimates the result, possibly from calculation errors. Choice D correctly identifies both the exponential growth pattern and predicts about 1,700 cells. Strategy tip: Always check ratios between consecutive terms first in growth problems. If ratios are constant, you have exponential growth; if differences are constant, you have linear growth.
A charted curve shows exponential decay starting at 300 when $x=0$ and dropping to 75 when $x=2$ . Which equation fits the curve?
Answer: $y=300(0.5)^x$ The question requires finding the exponential equation that fits a curve starting at 300 when x=0 and decaying to 75 when x=2. This is an exponential decay model, identified by constant multiplicative ratios, unlike linear models with additive differences. The general form is y = 300 × $r^x$ ; using x=2, 300 × $r^2$ = 75 gives $r^2$ = 0.25, so r = 0.5, yielding y = 300 $(0.5)^x$ . A key error is confusing it with linear, like choice D, or miscalculating the base as 0.25 instead of 0.5. Another mistake might be using x/2 in the exponent, which doesn't fit the points. Emphasize checking ratios for exponential fits, and verify by plugging in given points.
A taxi charges a flat fee plus a constant amount per mile. The relationship between miles traveled $m$ and total cost $C$ is modeled by $C=2.50+1.75m$ . What does the $2.50$ represent in this context?
Answer: Flat starting fee The question interprets the parameter 2.50 in the linear taxi cost model C=2.50 + 1.75m, where m is miles and C is total cost. This is a linear model with constant additive change per mile, not multiplicative. The 2.50 is the y-intercept, representing the flat starting fee before any miles are traveled (when m=0, C=2.50). The 1.75 is the slope, or cost per mile added on. Errors occur by misidentifying it as cost after 1 mile (D, which is 4.25) or per mile (A). In linear models, distinguish intercepts as initial values from slopes as rates of additive change.
The scatterplot shown displays the number of subscribers (in thousands) to a streaming service over 6 months. A researcher claims the growth is exponential with a monthly growth rate of 25%. Based on the plot, which month's data point is most inconsistent with the researcher's claim?
Answer: Month 4 Starting from 8 (thousand) at Month 1 with 25% monthly growth: predictions are 8, 10, 12.5, 15.625, 19.53, 24.41. Observed: 8, 10, 12.5, 19, 19.5, 24.4. Month 4 shows 19 vs. predicted 15.6 — a large deviation. Months 2, 3, and 5 match the prediction closely. (A), (B), (D) match the model.
A biologist records the mass (in mg) of a growing culture in the table shown. She wants to determine whether the growth is linear or exponential. Which statement best describes the data?
Answer: The growth is exponential because successive ratios of mass are approximately equal. Check ratios: 18/15=1.20, 21.6/18=1.20, 25.92/21.6=1.20, 31.10/25.92≈1.20. The common ratio identifies exponential growth. Differences are 3, 3.6, 4.32, 5.18 — not constant, so not linear. (A) is too vague. (B) is factually wrong. (D) cites doubling, which isn't the defining criterion and isn't even true here (15 to 31 is only about 2.07× — close but the correct reasoning is the constant ratio test).
A bacteria culture starts with 200 cells and doubles every hour. Let $t$ be the number of hours since the start, and let $N$ be the number of cells. Which equation models this situation?
Answer: $N=200(2)^t$ The question requires finding the equation that models a bacteria culture starting with 200 cells and doubling every hour, where t is hours and N is cells. This is an exponential model because the population multiplies by a constant ratio of 2 each hour, emphasizing multiplicative change over additive. Start with the initial value N=200 at t=0, then apply the growth factor: after 1 hour, 200*2; after 2 hours, 200* $2^2$ , leading to N=200 $(2)^t$ . Verify by plugging in t=1 (400 cells) and t=2 (800 cells), which matches doubling. A key error is confusing exponential with linear, like adding 200 each time instead of multiplying, or swapping base and exponent as in choice C. For growth problems, distinguish additive (linear) from multiplicative (exponential) by checking if quantities double or increase by fixed amounts.
Based on the table shown, the value of function $h$ at selected $x$ -values is given. If $h$ is an exponential function of the form $h(x) = a \cdot b^{x}$ , what is the value of $h(5)$ ?
Answer: 486 From the table, h(1)=6 and h(3)=54. So $b^2$ = 54/6 = 9, giving b=3. Then a·3 = 6, so a=2. h(5) = 2· $3^5$ = 2·243 = 486. (C) forgets to multiply by a=2. (B) uses b=3 but computes 2· $3^4$ . (A) uses linear extrapolation.
Monthly sales for a product are recorded as 5{,}000, 5{,}600, 6{,}272, and 7{,}024.64 units. Which statement best describes the pattern?
Answer: The data follow exponential growth with a constant 12\% increase per month. When you see a sequence of numbers representing growth over time, you need to determine whether the pattern is linear (constant addition) or exponential (constant multiplication). The key is checking what's happening between consecutive terms. Let's examine the differences first: From 5,000 to 5,600 is +600. From 5,600 to 6,272 is +672. From 6,272 to 7,024.64 is +752.64. Since these differences aren't constant, this isn't linear growth. Now let's check for exponential growth by calculating the ratios: $\frac{5,600}{5,000} = 1.12$ , $\frac{6,272}{5,600} = 1.12$ , and $\frac{7,024.64}{6,272} = 1.12$ . Each month, the sales are multiplied by 1.12, which represents a 12% increase (since 1.12 = 1 + 0.12). Choice A is wrong because linear growth requires constant differences, but we found 600, 672, and 752.64 - all different values. Choice C incorrectly identifies the growth rate as 20%; a 20% increase would mean multiplying by 1.20, not 1.12. Choice D suggests linear growth with increases of 1,272 units, but this doesn't match any of our calculated differences and ignores the non-constant pattern. The data follow exponential growth with a 12% monthly increase, making B correct. Strategy tip: When analyzing growth patterns, always check ratios first if the differences between terms aren't obviously constant. Exponential growth is more common in real-world scenarios than perfect linear growth.
A linear model for the number of tickets sold is $T(h)=95h+240$ , where $h$ is hours since sales opened. How many tickets are predicted to be sold after 6 hours?
Answer: 810 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.
The table shown gives the value $V$ (in dollars) of an investment account after $t$ years. Which of the following best describes the growth shown?
Answer: Exponential growth with a yearly growth factor of about 1.08 Compute ratios: 3240/3000 = 1.08; 3499.20/3240 = 1.08; 3779.14/3499.20 ≈ 1.08. The differences are NOT constant (240, 259.20, 279.94), ruling out linear. So the account grows exponentially with factor 1.08. (A) and (D) are linear misreads. (B) uses the first difference divided by a later value incorrectly.
A delivery service charges a base fee plus a constant amount per mile. The relationship between miles $m$ and cost $C$ is shown by the line on the graph passing through $(0, 6)$ and $(10, 26)$ . What is the base fee?
Answer: $6 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.

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