Linear growth and decay

Correct answer: C. \(t=8\)

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 \(t\).

Correct answer: B. \(t=8\)

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.

Correct answer: B. 217

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 \(x\).

Correct answer: A. 25

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 \(x\).

Correct answer: B. \(y=18-3x\)

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.

Correct answer: A. \(P(t)=240-12t\)

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.

Correct answer: A. \(-3x+11\)

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.

Correct answer: A. \(12-2x\)

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.

Correct answer: C. 15

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.

Correct answer: C. \(16\)

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.

Correct answer: C. 81

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.

Correct answer: A. \(B(x)=500-12x\)

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.

Correct answer: C. 10

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.

Correct answer: B. 10

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.

Correct answer: A. \(y=29+12x\)

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.

Correct answer: B. \(y=158-16x\)

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.

Correct answer: C. \(t=11\)

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 \(M-N=24\).

Correct answer: B. \(t=11\)

“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.

Correct answer: A. \(b<0\)

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.

Correct answer: C. 16

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.

Correct answer: C. 500

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 \(n\).

Correct answer: A. 23800

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 \(y\).

Correct answer: C. \(x=8\)

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.

Correct answer: B. \(x=15\)

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.

Correct answer: C. \(28t\)

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.