Linear growth and decay

In \(P(t)=120-5t\), the power of \(t\) is 1 and its coefficient is negative, so the value decreases by 5 per unit time. \(120-5t^2\) is not linear. Exam tip: a linear model always has degree 1.

In linear decay, the decrease per equal time interval is constant, so the graph has a fixed negative slope. Check differences between consecutive values; equal percentage decrease indicates exponential decay.

For the models to be equal, set \(36+9t=126-6t\). Moving \(6t\) to the left gives \(15t=90\), so \(t=6\). If \(t=5\) is substituted, the two model values are not equal. Exam tip: to find when two linear models meet, equate their expressions and solve for \(t\).

For equality, set \(52+4t=148-8t\). Moving \(8t\) to the left gives \(12t=96\), so \(t=8\). On checking, \(P(8)=52+32=84\) and \(Q(8)=148-64=84\). At \(t=10\), the two values are not equal. Exam tip: while equating linear expressions, collect all terms containing \(t\) on one side and constants on the other.

Given f(x)=a+12x and f(4)=95, we get a+12(4)=95. Thus, a+48=95, so a=47. Now, f(10)=47+12(10)=47+120=167. Therefore, option C is correct. The value 155 can result from not determining the constant a correctly. Exam tip: first use the given function value to find the unknown constant, then substitute the required value of x.

Given \(g(x)=b-11x\). Substituting \(x=5\), \(g(5)=b-55=103\), so \(b=158\). Now, for \(x=12\), \(g(12)=158-11(12)=158-132=26\). Hence, 26 is correct. An option such as 37 can result from an incorrect subtraction. Exam tip: first find the unknown constant \(b\) from the given function value, then substitute the required value of \(x\).

In \(y=18-2.5x\), the constant term 18 is the initial value and the coefficient \(-2.5\) shows an equal decrease per unit of x. Option C is not linear because it contains \(x^2\). Exam tip: for linear decay, x has power 1 and a negative coefficient.

In \(y=250-7x\), the coefficient of x is \(-7\), so y decreases by 7 for every 1-unit increase in x. \(y=250+7x\) shows growth, while an \(x^2\) relation is not linear. Exam tip: linear decay has a negative coefficient of x.

In linear growth, equal increases in x produce the same change in y each time. For example, when x rises by 1, y-differences may be 5, 5, 5. A constant ratio indicates exponential growth. Exam tip: check first differences first.

\(7-3x\) is a linear polynomial with coefficient \(-3\) of x. Thus, for every increase of 1 in x, its value decreases by 3. In \(x^2-3x+7\), the rate of change is not constant. Exam tip: constant growth or decay requires degree 1.

Given \(C(t)=28+kt\), we have \(C(10)=28+10k\) and \(C(4)=28+4k\). Hence, \(C(10)-C(4)=(28+10k)-(28+4k)=6k\). As the difference is 72, \(6k=72\), so \(k=12\). For example, choosing 14 would give a difference of \(6\times14=84\), not 72. Exam tip: when subtracting two values of a linear expression, the constant term cancels out.

Given \(D(t)=240-kt\), we have \(D(5)=240-5k\) and \(D(13)=240-13k\). Hence, \(D(5)-D(13)=(240-5k)-(240-13k)=8k\). Therefore, \(8k=96\), so \(k=12\). If \(k=16\), the difference would be \(128\), not \(96\). Exam tip: when subtracting two values of a linear expression, the constant terms cancel.

The initial value is the constant term, so it is 50. A decrease of 4 per time unit gives the coefficient of \(t\) as \(-4\), hence \(Q(t)=50-4t\). \(50-4t^2\) is not linear. Exam tip: constant change means degree 1.

Putting \(t=0\) in \(15-7t\) gives 15, and the coefficient of \(t\) is \(-7\), so the quantity falls by 7 per unit time. \(15+7t\) indicates growth. Exam tip: in a linear model, the variable has power 1.

Given P(t)=p+10t, substitute P(4)=92 first: p+10(4)=92, so p+40=92 and p=52. Now use P(s)=152: 52+10s=152. Thus 10s=100, giving s=10. If s were 9, then P(9)=52+90=142, not 152. Exam tip: In a linear model, first find the constant term from one known value, then substitute the second value to solve for the variable.

Given \(R(t)=r-12t\), substitute \(t=3\): \(r-12(3)=134\). Thus, \(r-36=134\), so \(r=170\). Now \(R(s)=50\) gives \(170-12s=50\). Hence \(12s=120\), and \(s=10\). If \(s=9\), then \(R(9)=62\), not 50. Exam tip: first find the constant \(r\) from the given value, then use the second condition to solve for the variable.

An increase of 9 per step requires a linear expression with coefficient 9, so \(y=a+9x\). Using \(y(4)=61\), we get \(a+9(4)=61\), hence \(a=25\). Therefore, \(y=25+9x\) is correct. Although \(y=61+9x\) also has rate 9, it gives \(y(4)=97\), not 61. Exam tip: substitute the given \(x\)-value into an option to check the condition quickly.

A decrease of 7 per step means that the coefficient of \(x\) must be \(-7\). So the equation has the form \(y=a-7x\). Using \(y(8)=94\), we get \(a-7(8)=94\), so \(a=150\). Therefore, \(y=150-7x\) is correct. Option A has the correct rate of decrease, but it gives \(y(8)=38\). Exam tip: first identify the sign of the rate, then substitute the given condition to find the constant term.

“20 more” means \(M(t)=N(t)+20\). Thus, \(62+5t=142-5t+20\), or \(62+5t=162-5t\). This gives \(10t=100\), so \(t=10\). At \(t=8\), the difference is 0, so it is a close but incorrect option. Exam tip: for “more than” questions, equate the larger quantity to the smaller quantity plus the stated difference.

“12 more” means \(A(t)=B(t)+12\). Thus, \(190-8t=46+4t+12\), which gives \(132=12t\) and hence \(t=11\). Therefore, \(t=11\) is correct. At \(t=10\), the difference between the two quantities is \(24\), not \(12\). Exam tip: In “more than” questions, set \(A(t)-B(t)\) equal to the stated difference first.

Given \(F(3)=F(11)-80\), we get \(F(11)-F(3)=80\). For the linear function \(F(t)=50+rt\), the change in \(t\) from 3 to 11 is \(8\), so \(F(11)-F(3)=8r\). Thus \(8r=80\), giving \(r=10\). Option 8 is the time interval, not the rate \(r\). Exam tip: subtract two function values in a linear expression to eliminate the constant term 50.

Given \(G(t)=220-rt\), we have \(G(4)=220-4r\) and \(G(13)=220-13r\). Hence, \(G(4)-G(13)=9r\). The condition \(G(4)=G(13)+99\) gives \(G(4)-G(13)=99\), so \(9r=99\) and \(r=11\). Option \(10\) is close, but it would give a difference of \(9\times10=90\), not 99. Exam tip: when subtracting values of a linear function, the constant term cancels out.

The price increases by 20 for every increase of 1 in \(n\). From 6 to 13, there are 7 steps, so the increase is \(7\times20=140\). Therefore, \(P(13)=230+140=370\). Option 350 can result from incorrectly using 6 steps instead of 7. Exam tip: for a linear function, use \(P(13)-P(6)=20(13-6)\) to find the change quickly.

The machine’s value decreases by 2500 each year. From year 4 to year 10, there are 6 years, so the total decrease is \(6\times2500=15000\). Therefore, \(V(10)=38000-15000=23000\). Option 28000 results from subtracting the depreciation for only 4 years, so it is incorrect. Exam tip: In linear decay, first find the difference between the two time values and multiply it by the yearly decrease.

Given \(y=30+6x\) and the condition \(y=11x\), we get \(30+6x=11x\). Subtracting \(6x\) from both sides gives \(30=5x\), so \(x=6\). For example, at \(x=5\), the first expression gives \(60\), whereas \(11x=55\), so it is not correct. Exam tip: first equate the two expressions for \(y\), then collect like terms on one side.