Linear & Exponential Growth - SAT Math

A candle's height is modeled by $h(t)=12-0.8t$ , where $t$ is hours since it was lit. What is the meaning of the $12$ in this model?
Answer: The initial height is $12$ inches The candle height model h(t) = 12 - 0.8t is linear, where t is hours since lighting. In the form y = mx + b, the constant term (12) represents the y-intercept, which is the value when t = 0. Therefore, 12 represents the initial height of the candle in inches before any burning occurs. We can verify: at t = 0, h(0) = 12 - 0.8(0) = 12 inches. The coefficient -0.8 tells us the candle burns 0.8 inches per hour. A common error is confusing the constant term with the rate of change or the time duration. In linear models, always identify what happens at time zero to understand the constant term.
A laboratory sample cools according to $T(t)=20+60(0.75)^t$ , where $T$ is temperature in °C and $t$ is time in minutes. What does the 20 represent in context, and what is $T(2)$ to the nearest tenth?
Answer: Ambient temp; 53.8°C The question asks about the cooling model T(t) = 20 + 60 $(0.75)^t$ and what the constant 20 represents. In Newton's Law of Cooling models, the equation typically has the form T = T_ambient + (T_initial - T_ambient)(decay factor)^t. Here, 20 represents the ambient (room) temperature that the sample approaches as t increases. To find T(2): T(2) = 20 + 60 $(0.75)^2$ = 20 + 60(0.5625) = 20 + 33.75 = 53.75°C, which rounds to 53.8°C. The key insight is recognizing that exponential decay models often include a horizontal asymptote representing the limiting value.
Solve for $x$ : $3^x=27$ .
Answer: 3 Since $27=3^3$ , it follows that $x=3$ . The other options are powers or reciprocals that do not satisfy the equation.
A bacteria culture starts with 500 cells and doubles every 3 hours. Which equation models the number of cells, $N$ , after $h$ hours?
Answer: $N=500(2)^{h/3}$ We need to model bacteria that start at 500 cells and double every 3 hours. Since the population doubles (multiplies by 2), this is exponential growth with base 2. However, doubling happens every 3 hours, not every hour, so we need to adjust the exponent. After h hours, the number of 3-hour periods is h/3, giving us N = 500 $(2)^{h/3}$ . We can verify: at h=0, N=500 $(2)^0$ =500; at h=3, N=500 $(2)^1$ =1000; at h=6, N=500 $(2)^2$ =2000. A common error is using N = 500 $(2)^h$ , which would mean doubling every hour instead of every 3 hours. When the growth period doesn't match the time unit, divide the time variable by the growth period in the exponent.
A town's water usage, $W$ , is modeled by $W(t)=8000(1.03)^t$ , where $t$ is years. Approximately what is the growth rate per year?
Answer: $3\%$ We need to find the growth rate from the model W(t) = 8000 $(1.03)^t$ . In exponential growth models of the form y = $a(1+r)^t$ , the growth rate is r. Here, 1.03 = 1 + 0.03, so r = 0.03 = 3%. This means water usage grows by 3% each year. To verify: if usage grows by 3%, then after one year it becomes 8000 × 1.03, which matches our model at t=1. Don't confuse the growth factor (1.03) with the growth rate (0.03 or 3%). The growth factor is always 1 plus the growth rate for exponential growth.
A water tank contains 300 liters and is being drained at a constant rate of 12 liters per minute. Which equation gives the amount of water $W(t)$ after $t$ minutes, and how much water remains after 15 minutes?
Answer: $W=300-12t$ ; 120 L The question involves a water tank draining at a constant rate, which represents linear decrease. Starting with 300 liters and draining 12 liters per minute gives the equation W(t) = 300 - 12t. This follows the linear model y = mx + b where the negative slope (-12) represents the drainage rate. After 15 minutes: W(15) = 300 - 12(15) = 300 - 180 = 120 liters. The key concept is that constant rate problems result in linear models, not exponential ones. A common error would be using an exponential decay model when the problem states a constant rate.
Solve $(3^x)^2=27$ for $x$ .
Answer: $\frac{3}{2}$ $3^{2x}=27=3^3$ , so $2x=3$ and $x=\frac{3}{2}$ . A negative root is extraneous because $3^x>0$ for all real $x$ .
For $g(t)=120(1.2)^t$ , by what factor does $g$ increase when $t$ increases by 3?
Answer: $1.728$ The factor over 3 units is $(1.2)^3=1.728$ . Choices B, C, and D incorrectly add percentages or mix factor with rate.
A taxi charges a base fare plus a per-mile fee. The relationship between miles $x$ and total cost $y$ is $y=2.75x+4.50$ . What does the slope represent, and what is the total cost for a 6-mile ride?
Answer: $2.75$ per mile; $21.00$ The question provides the taxi fare equation y = 2.75x + 4.50 and asks for the meaning of the slope and the cost of a 6-mile ride. In this linear equation, the slope 2.75 represents the rate of change - the cost per mile ($2.75/mile). The y-intercept 4.50 is the base fare. For a 6-mile ride: y = 2.75(6) + 4.50 = 16.50 + 4.50 = 21.00 dollars. The key concept is interpreting slope as rate of change in context - here it's the variable cost per unit distance. Common errors include confusing the roles of slope and y-intercept in real-world applications.
A car's value is modeled by $V(t)=24000(0.85)^t$ , where $t$ is years after purchase. What does the number $0.85$ represent in this context?
Answer: The car keeps $85%$ each year This question asks about the meaning of 0.85 in the exponential decay model V(t) = 24000 $(0.85)^t$ . In exponential models of the form y = $a(b)^t$ , the base b represents the decay factor when 0 < b < 1. A base of 0.85 means the car retains 85% of its value each year (or loses 15%). To verify: if the car keeps 85% of its value, then after 1 year it's worth 24000(0.85) = 0.85 × 24000, which is 85% of the original. The common error is thinking 0.85 means losing 85%, but remember that the base represents what remains, not what is lost. For percent problems, convert the remaining percentage to decimal form for the base.
A streaming service's subscribers follow the model $S=12000+800t$ , where $t$ is months since launch. What does the number 800 represent in context?
Answer: Increase in subscribers per month This question tests interpretation of parameters in a linear model for subscriber growth. In S = 12000 + 800t, the coefficient 800 represents the rate of change - specifically, the increase in subscribers per month. The constant term 12000 is the initial number of subscribers at launch (when t = 0). Students often confuse these roles or misinterpret 800 as a percentage rather than an absolute increase. In linear models of the form y = mx + b, m always represents the rate of change per unit of the independent variable. When solving context problems, always identify what each parameter means in real-world terms.
In the equation $y = 7x + 2$ , what does the number 7 represent?
Answer: Slope This question asks us to identify what the coefficient of x represents in a linear equation. In the linear equation y = 7x + 2, the number 7 is the coefficient of x, which represents the slope of the line. The slope tells us how much y changes for each unit increase in x - specifically, that y increases by 7 units when x increases by 1 unit. The slope represents the rate of change, which is the defining characteristic of linear relationships (constant rate of change). Students might confuse slope with other terms, but remember that in y = mx + b form, m is always the slope and b is the y-intercept. For linear functions, the coefficient of the variable always represents the constant rate of change or slope.
A population of bacteria doubles every 3 hours. If the initial population is 100, what will the population be after 9 hours?
Answer: 800 This problem asks for the population after 9 hours when bacteria doubles every 3 hours. Since the population doubles at regular intervals, this follows an exponential growth model. Starting with 100 bacteria, after 3 hours we have 100 × 2 = 200, after 6 hours we have 200 × 2 = 400, and after 9 hours we have 400 × 2 = 800.
Solve for $x$ : $3^{2x-1}=27$ .
Answer: 2 Since $27=3^3$ , set exponents equal: $2x-1=3$ so $x=2$ . The other choices result from incorrect exponent algebra or arithmetic.
A population is modeled by $P(t)=1200(1.08)^t$ , where $t$ is measured in years. What is the percent increase per year?
Answer: 8% The growth factor is 1.08, which corresponds to an 8% increase. 108% confuses the factor with the rate, and 0.08% or 0.8% are decimal-place errors.
A bacteria culture starts with 500 cells and doubles every 3 hours. Let $t$ be time in hours and $N(t)$ be the number of cells. Which equation represents this situation?
Answer: $N(t)=500(2)^{t/3}$ We need to find an equation for bacteria that start at 500 cells and double every 3 hours. Since the population multiplies by a constant factor (doubles), this is exponential growth. The general form is N(t) = N₀ · (growth factor)^(t/doubling time), where N₀ = 500, growth factor = 2, and doubling time = 3 hours. This gives us N(t) = 500 $(2)^{t/3}$ . We can verify: at t = 0, N(0) = 500 $(2)^0$ = 500; at t = 3, N(3) = 500 $(2)^1$ = 1000 (doubled). A common mistake is using N(t) = 500 $(2)^t$ , which would mean doubling every hour instead of every 3 hours.
A savings account starts with $500 and grows by 6% each month. Which equation gives the balance $B$ after $m$ months (no additional deposits or withdrawals)?
Answer: $B=500(1.06)^m$ This problem asks for an exponential growth equation modeling compound interest. Since the account grows by 6% each month, the growth factor is 1 + 0.06 = 1.06, and we multiply the initial amount by this factor raised to the power of months: B = 500 $(1.06)^m$ . Linear models like B = 500 + 0.06m would only add 6% of the original $500 each month, not 6% of the current balance. The key distinction is that exponential growth compounds - each month's 6% is calculated on the new, larger balance. Always use exponential models for percentage-based growth or decay.
A taxi fare is shown on a coordinate plane with points $(1,6)$ and $(5,14)$ , where $x$ is miles and $y$ is total cost in dollars. Assuming a linear model, what is the cost for a 9-mile ride?
Answer: $22 This problem requires finding a linear equation from two points and then extrapolating. Given points (1,6) and (5,14), first find the slope: m = (14-6)/(5-1) = 8/4 = 2. Using point-slope form with (1,6): y - 6 = 2(x - 1), which simplifies to y = 2x + 4. For a 9-mile ride, y = 2(9) + 4 = 18 + 4 = 22 dollars. The key steps are calculating slope from two points, finding the equation, then evaluating at the desired x-value. A common error is using the slope incorrectly or forgetting the y-intercept when writing the final equation.
A quantity is modeled by $A(t)=200(1.21)^{t/3}$ , where $t$ is measured in years. What is the annual percent increase?
Answer: 6.60% The yearly growth factor is $(1.21)^{1/3}\approx 1.0656$ , which is about a 6.6% increase per year. 21% confuses the 3-year factor with a yearly rate, and 7% or 3% are rough or incorrect heuristics.
A savings account has an initial balance of $200 and increases by $5 each week. Let $t$ be the number of weeks and $B(t)$ the balance in dollars. What is $B(12)$ ?
Answer: $260$ We need to find the balance after 12 weeks given an initial balance of $200 that increases by $5 each week. Since the balance increases by a constant amount ($5), this is linear growth with the model B(t) = 200 + 5t. To find B(12), we substitute: B(12) = 200 + 5(12) = 200 + 60 = 260. The key is recognizing that "increases by $5 each week" indicates additive change (linear), not multiplicative change (exponential). Always check your arithmetic carefully when substituting values.

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