Linear growth and decay

Correct answer: B. \(x=10\)

Given \(y=80-5x\) and the condition \(y=3x\), set the two expressions equal: \(80-5x=3x\). Adding \(5x\) to both sides gives \(80=8x\), so \(x=10\). If \(x=8\), then \(y=40\) while \(3x=24\), so it does not satisfy the condition. Exam tip: When two expressions represent the same variable, equate them and solve the resulting linear equation.

Correct answer: B. \(4t\)

Given \(L(t)=2t+11\), we get \(L(3t)=2(3t)+11=6t+11\). Therefore, \(L(3t)-L(t)=(6t+11)-(2t+11)=4t\). \(6t\) is part of the value of \(L(3t)\), not the difference. Exam tip: substitute \(3t\) into the function first, then subtract the two expressions using brackets.

Correct answer: A. \(4t\)

Given \(K(t)=90-4t\). Substituting \(2t\) for \(t\) gives \(K(2t)=90-4(2t)=90-8t\). Therefore, \(K(t)-K(2t)=(90-4t)-(90-8t)=90-4t-90+8t=4t\). Hence, \(4t\) is correct. The distractor \(-4t\) may result from not changing every sign while subtracting the second bracket. Exam tip: when finding a function at \(2t\), replace every \(t\) by \(2t\), then distribute the minus sign carefully.

Correct answer: B. 23

Given \(Y(x)=a+5x\), we get \(Y(2)=a+10\) and \(Y(6)=a+30\). Therefore, \(Y(2)+Y(6)=2a+40=86\). Hence \(2a=46\), so \(a=23\). For example, choosing 28 would give a sum of 96, not 86. Exam tip: Substitute each given \(x\)-value into the function separately before adding.

Correct answer: C. 60

Given \(Z(x)=b-6x\), we get \(Z(1)=b-6\) and \(Z(4)=b-24\). Hence, \(Z(1)+Z(4)=(b-6)+(b-24)=2b-30\). Using \(2b-30=90\), we obtain \(2b=120\), so \(b=60\). If 63 were used, the sum would be 96, not 90. Exam tip: Substitute each given value of \(x\) separately before adding the function values.

Correct answer: C. \(x=12\)

For equality, set \(p(x)=q(x)\): \(10+3x=4x-2\). Subtracting \(3x\) from both sides gives \(10=x-2\), so \(x=12\). On checking, \(p(12)=46\) and \(q(12)=46\). For example, at \(x=10\), their values are 40 and 38, so it is not correct. Exam tip: To find when two polynomials are equal, equate their expressions first and solve for \(x\).

Correct answer: B. \(x=5\)

Since \(p(x)=q(x)\), set \(120-7x=60+5x\). Subtracting 60 from both sides gives \(60-7x=5x\), so \(60=12x\). Hence, \(x=5\). If \(x=6\) is substituted, the two expressions do not have equal values. Exam tip: equate the two linear expressions and collect all \(x\)-terms on one side.

Correct answer: C. \(k>2\)

In \(y=18+(k-2)x\), the coefficient of \(x\), or slope, is \(k-2\). For linear growth, the slope must be positive, so \(k-2>0\). Hence, \(k>2\). When \(k=2\), \(y=18\) is constant, not increasing. Exam tip: in \(y=mx+c\), check \(m>0\) for linear growth.

Correct answer: A. \(k>-3\)

In \(y=100-(k+3)x\), the coefficient of \(x\), or slope, is \(-(k+3)\). For linear decay, the slope must be negative, so \(-(k+3)<0\). This gives \(k+3>0\), hence \(k>-3\). When \(k=-3\), the slope is zero and \(y=100\) is constant, not decay. Exam tip: check the sign of the coefficient of \(x\) to identify linear growth or decay.

Correct answer: A. The condition is true for every real \(s\)

\(S(2)=s-4(2)=s-8\) and \(S(9)=s-4(9)=s-36\). Therefore, \(S(2)-S(9)=(s-8)-(s-36)=28\). The terms containing \(s\) cancel, so the condition is true for every real \(s\). Neither \(s=28\) nor \(s=4\) is a specially required value. Exam tip: In differences of function values, write both values first and check whether the parameter terms cancel.

Correct answer: A. It increases by 3 each step

Find the difference: (B(t)-A(t))=(30+5t)-(18+2t)=12+3t. The coefficient of t is 3, so the difference increases by 3 for every increase of 1 in t. Therefore, option A is correct. The value 12 is the initial difference, not the rate of increase, so option D is incorrect. Exam tip: when subtracting linear expressions, subtract the constant terms and the coefficients of t separately.

Correct answer: A. It increases by 5 each step

Subtracting gives \(C(t)-D(t)=(150-3t)-(120-8t)=30+5t\). Thus, for every increase of 1 in \(t\), the difference increases by 5. Option B is incorrect because \(D(t)\) decreases 5 units faster than \(C(t)\), so the gap between them increases. Exam tip: When subtracting linear expressions, change the signs of every term in the second bracket carefully.

Correct answer: A. \(y=80-5x\)

In \(y=80-5x\), every increase of 1 in x decreases y by a constant 5, so it shows linear decay. Option B has a positive slope and represents growth. Exam tip: linear decay always has a negative coefficient of x.

Correct answer: A. \(P(x)=240-8x\)

In \(P(x)=240-8x\), the constant term 240 is the initial value and the coefficient \(-8\) shows a decrease as x increases. \(240+8x\) represents growth. Exam tip: a negative slope indicates decay.

Correct answer: C. \(x=6\)

Given \(f(x)=5x+20\), we get \(f(2x)=5(2x)+20=10x+20\). Therefore, \(f(2x)-f(x)=(10x+20)-(5x+20)=5x\). Setting \(5x=30\) gives \(x=6\). Hence, option C is correct. For instance, when \(x=5\), the difference is only \(25\), not \(30\). Exam tip: substitute \(2x\) for every \(x\) in \(f(x)\) before simplifying the subtraction.

Correct answer: C. \(x=6\)

Given \(g(x)=96-6x\), we get \(g(2x)=96-6(2x)=96-12x\). Therefore, \(g(x)-g(2x)=(96-6x)-(96-12x)=6x\). Setting \(6x=36\) gives \(x=6\), so option C is correct. For example, at \(x=5\), the difference would be only \(30\). Exam tip: while finding \(g(2x)\), replace every \(x\) in the function by \(2x\).

Correct answer: B. Q(5)

This is a linear function. The midpoint of 1 and 9 is \(\frac{1+9}{2}=5\), so \(Q(1)+Q(9)=2Q(5)\). Checking directly, \(Q(1)=96\) and \(Q(9)=64\), giving a sum of \(160\). Also, \(Q(5)=80\), hence \(2Q(5)=160\). \(Q(4)\) is not at the midpoint, so it is not correct. Exam tip: for a linear function, the sum of the values at two inputs equals twice the value at their midpoint.

Correct answer: C. 7

The change in x is 9−4=5. In the linear expression y=70+ax, the change in y is a×5. Hence, 5a=35, so a=7. Option 5 represents the change in x, not the value of a. Exam tip: In y=mx+c, identify m as the rate of change.

Correct answer: C. \(9\)

The change in \(x\) is \(8-3=5\). In \(y=160-bx\), \(y\) decreases by \(b\) units for every 1-unit increase in \(x\). Hence, the total decrease is \(5b\). Given \(5b=45\), we get \(b=45/5=9\). Therefore, \(9\) is correct. If \(b=15\), the decrease would be \(5\times15=75\), not 45. Exam tip: For a linear relation, total change equals rate per unit times the change in the variable.

Correct answer: C. \(t=10\)

“\(A(t)\) is 4 more than \(B(t)\)” means \(A(t)=B(t)+4\). Thus, \(12+3t=18+2t+4\), which gives \(t=10\). Checking: \(A(10)=42\) and \(B(10)=38\), so the difference is 4. At \(t=9\), the difference is only 3. Exam tip: For “more than” questions, first write the difference as \(A(t)-B(t)\).

Correct answer: B. \(t=9\)

“\(C(t)\) is 12 less than \(D(t)\)” means \(C(t)=D(t)-12\). Thus, \(130-5t=70+3t-12\). Simplifying gives \(72=8t\), so \(t=9\). Option \(t=8\) is close, but the difference between the two values is not 12 there. Exam tip: For “less than” comparisons, subtract the stated amount from the larger expression before forming the equation.

Correct answer: A. 15

Given f(t)=a+4t, we get f(1)=a+4, f(3)=a+12, and f(5)=a+20. Their sum is 3a+36. Hence, 3a+36=81, so 3a=45 and a=15. If a were 17, the sum would be 87, so it is not correct. Exam tip: substitute each given value of t into the function before adding the results.

Correct answer: C. 60

Given \(g(t)=b-3t\), we get \(g(2)=b-6\), \(g(4)=b-12\), and \(g(6)=b-18\). Their sum is \((b-6)+(b-12)+(b-18)=3b-36\). Hence \(3b-36=144\), so \(3b=180\) and \(b=60\). If 54 is used instead, the sum becomes \(126\), not 144. Exam tip: In such questions, first evaluate the function at every given input and then form the sum equation.

Correct answer: A. \(P\) decreases by 7 units in every equal time interval

In linear decay, the change over equal time intervals is constant. Here the change is always \(-7\), so the slope is fixed and the model has the form \(P=a-7t\). A 10% decrease depends on the current value, so it is not linear. Exam tip: check for equal first differences.

Correct answer: A. \(Q(t)=Q_0-rt\)

In \(Q(t)=Q_0-rt\), the quantity decreases by the same amount \(r\) in every unit of time, so it is a first-degree linear decay model. \(Q_0+rt\) shows growth, while \(e^{-rt}\) shows exponential decay. Exam tip: a constant rate of change indicates a linear relation.