Linear growth and decay

For the models to be equal, set \(64+9t=184-6t\). Moving \(6t\) to the left gives \(15t=120\), so \(t=8\). Although \(t=7\) is a close distractor, the two model values are not equal at that time. Exam tip: when one linear model increases and the other decreases, equate their expressions and solve for \(t\).

For equality, set \(75+8t=219-10t\). Moving \(10t\) to the left gives \(18t=144\), so \(t=8\). Substituting \(t=6\) does not give equal values, so it is incorrect despite being a close distractor. Exam tip: When equating linear expressions, collect all terms containing \(t\) on one side first.

Given \(f(x)=a+15x\) and \(f(4)=112\), we get \(a+15(4)=112\). Thus, \(a+60=112\), so \(a=52\). Now \(f(11)=52+15(11)=52+165=217\). Hence, option B is correct. The value 232 can result from incorrectly adding the growth term. Exam tip: first find the unknown constant using the given function value, then substitute the required value of \(x\).

Given \(g(x)=b-13x\), use \(g(5)=116\): \(b-13(5)=116\), so \(b-65=116\) and hence \(b=181\). Now \(g(12)=181-13(12)=181-156=25\). Therefore, 25 is correct. A value such as 38 results from an incorrect subtraction. Exam tip: first find the unknown constant \(b\) using the given function value, then substitute the required value of \(x\).

In \(y=18-3x\), the coefficient of \(x\) is \(-3\), so \(y\) decreases by 3 for every increase of 1 in \(x\). This is linear decay at a constant rate. \(18-3x^2\) is not linear because it contains \(x^2\). Exam tip: in \(y=a+bx\), \(b<0\) indicates decay.

For \(P(0)=240\), the constant term must be 240. Uniform decay requires a negative slope, so \(-12t\) is correct. \(240-12t^2\) is not linear. Exam tip: in a linear model, the variable has power 1.

In a linear polynomial \(p(x)=ax+b\), increasing x by 1 changes p(x) by \(a\). Here \(a=-3\), so the value falls by 3 each step. \(-3x^2+11\) is not linear. Exam tip: a negative coefficient of x indicates linear decay.

In \(12-2x\), the coefficient of x is \(-2\), so the quantity decreases by 2 for every increase of 1 in x. In \(12-2x^2\), the decrease is not constant. Exam tip: linear decay has degree 1 and a negative coefficient of x.

Given \(C(t)=55+kt\), we have \(C(12)=55+12k\) and \(C(5)=55+5k\). Thus, \(C(12)-C(5)=(55+12k)-(55+5k)=7k\). Since this difference is 105, \(7k=105\), so \(k=15\). If \(k=21\), the difference would be \(7\times21=147\), not 105. Exam tip: when subtracting two values of a linear expression, the constant term 55 cancels out.

Given \(D(t)=340-kt\), we have \(D(6)=340-6k\) and \(D(14)=340-14k\). Hence, \(D(6)-D(14)=(340-6k)-(340-14k)=8k\). Since \(8k=128\), \(k=16\). The value \(14\) is a time input, not the rate \(k\). Exam tip: while subtracting linear expressions, use brackets to handle negative signs correctly.

The difference between the inputs is
(x+7)-(x-2)=9
. In the linear function A(x)=52+9x, the value increases by 9 for every increase of 1 in x. Therefore, for an input difference of 9, the difference in function values is 9×9=81. Choosing 72 would incorrectly treat the input difference as 8. Exam tip: in differences of a linear function, the constant term 52 cancels; use the coefficient of x and the difference between the inputs.

At \(x=0\), the constant term gives the initial value, so it must be 500. The coefficient \(-12\) shows a constant decrease. A term containing \(x^2\) is not linear. Exam tip: linear decay always has a negative slope.

Given \(P(t)=p+16t\), use \(P(4)=125\): \(p+16(4)=125\), so \(p=61\). Now \(P(s)=221\) gives \(61+16s=221\). Hence \(16s=160\) and \(s=10\). For example, \(s=12\) would give \(61+16(12)=253\), not 221. Exam tip: In a linear model, find the unknown constant \(p\) first and then use the second condition to find the variable.

Given R(t)=r-18t. Substituting R(3)=166 gives r-18(3)=166, so r-54=166 and r=220. Now, using R(s)=40, we get 220-18s=40. Thus, 18s=180 and s=10. If s=9, then R(9)=58, not 40. Exam tip: first find the constant r from one condition, then substitute it into the second condition to solve for the variable.

An increase of 12 per step means that the coefficient of \(x\) must be \(12\). Let \(y=a+12x\). Using \(y(6)=101\), we get \(a+12\times6=101\), so \(a=29\). Hence, \(y=29+12x\) is correct. Option C does give \(y(6)=101\), but its coefficient is \(-12\), so it represents decay rather than growth. Exam tip: first check the sign of the rate of change, then substitute the given value to find the constant term.

A decrease of 16 per step means that the coefficient of \(x\) must be \(-16\). So write the equation as \(y=a-16x\). Using \(y(5)=78\), we get \(a-16(5)=78\), hence \(a=158\). Therefore, \(y=158-16x\) is correct. In option A, the constant is 78, so \(y(5)=-2\), not 78. Exam tip: substitute the given \(x\)-value into each likely equation to check the condition quickly.

The required condition is \(M(t)=N(t)+24\). Thus, \(95+6t=203-6t+24\), so \(12t=132\) and \(t=11\). Checking: \(M(11)=161\) and \(N(11)=137\), giving a difference of \(24\). At \(t=10\), the difference is only \(12\). Exam tip: translate “24 more than” as \(M-N=24\).

“17 more” means \(A(t)=B(t)+17\). Therefore, \(280-13t=65+5t+17\), which gives \(198=18t\) and hence \(t=11\). Substituting \(t=10\) gives a difference of 35, not 17. Exam tip: In “more than” questions, first write the difference between the two quantities.

When \(x\) rises by 1, \(P(x+1)-P(x)=b\). In decay, the quantity must decrease, so this change is negative and \(b<0\). A value of \(b=0\) represents no change. Exam tip: read \(b\) as the rate of change.

Given \(G(4)=G(13)+144\), we get \(G(4)-G(13)=144\). Now \(G(4)=320-4r\) and \(G(13)=320-13r\). Hence \((320-4r)-(320-13r)=9r=144\), so \(r=16\). Option 18 can result from using an incorrect time difference; here \(13-4=9\). Exam tip: taking the difference of function values cancels the constant term in a linear function.

Given \(P(n)=a+30n\). Using \(P(5)=260\), we get \(a+30(5)=260\), so \(a+150=260\) and hence \(a=110\). Therefore, \(P(13)=110+30(13)=110+390=500\). Thus, 500 is correct. A value such as 470 results from not applying the growth rate or the initial value correctly. Exam tip: first find the constant \(a\) from the given condition, then substitute the required value of \(n\).

Given \(V(y)=c-4200y\). Using \(V(4)=53200\), we get \(c-4200\times4=53200\), so \(c=70000\). Now \(V(11)=70000-4200\times11=23800\). Therefore, 23800 is correct. A nearby option such as 28000 can result from using an incorrect number of years or making an arithmetic error in the depreciation term. Exam tip: first find the constant \(c\) from the given value, then substitute the required value of \(y\).

Given \(y=56+8x\) and the condition \(y=15x\), we equate the two expressions: \(56+8x=15x\). Subtracting \(8x\) from both sides gives \(56=7x\), so \(x=8\). Hence, option C is correct. For instance, if \(x=7\), then \(y=112\) but \(15x=105\), so it does not satisfy the condition. Exam tip: equate the two expressions for the same variable and solve the resulting linear equation.

Given \(y=255-14x\) and the condition \(y=3x\), set the two expressions equal: \(255-14x=3x\). Thus, \(255=17x\), so \(x=15\). For the nearby option \(x=17\), \(y=17\) but \(3x=51\), so it does not satisfy the condition. Exam tip: equate the two expressions and solve the resulting linear equation.

Given \(L(t)=7t+41\), substitute \(6t\) for \(t\): \(L(6t)=7(6t)+41=42t+41\). Similarly, \(L(2t)=7(2t)+41=14t+41\). Therefore, \(L(6t)-L(2t)=(42t+41)-(14t+41)=28t\). \(42t\) is part of \(L(6t)\), not the difference. Exam tip: Use brackets while subtracting function values so that constant terms and signs are handled correctly.