Linear growth and decay

For equality, set \(p(x)=q(x)\): \(35+7x=10x-4\). Subtracting \(7x\) from both sides gives \(35=3x-4\). Adding 4 gives \(39=3x\), so \(x=13\). Hence, option C is correct. At \(x=12\), the values of the two polynomials are not equal. Exam tip: after equating two linear expressions, collect the \(x\)-terms on one side and constants on the other.

Equating the two expressions gives \(220-12x=76+4x\). Therefore, \(220-76=4x+12x\), so \(144=16x\). Hence, \(x=9\). Substituting \(x=8\) does not make the two expressions equal, so it is a close but incorrect distractor. Exam tip: Collect the variable terms on one side and constants on the other carefully.

In a linear equation, the coefficient of \(x\) represents the rate of increase or decrease. Here the slope is \(k-7\). For linear growth, it must be positive: \(k-7>0\), so \(k>7\). If \(k=7\), the slope is zero and \(y=55\) is constant, not growth. Exam tip: check whether the slope is positive for growth and negative for decay.

In the linear form \(y=mx+c\), \(m\) is the slope. Here, the coefficient of \(x\) is \(-(k-4)=4-k\). For decay, the slope must be negative, so \(4-k<0\), which gives \(k>4\). At \(k=4\), the slope is zero, so the value remains constant rather than decaying. Exam tip: check the sign of the coefficient of \(x\) to identify linear growth or decay.

Here (R(0)=90) is directly the initial value. The difference gives (m=13), but the initial value remains the same.

\(S(4)=s-9(4)=s-36\) and \(S(12)=s-9(12)=s-108\). Hence, \(S(4)-S(12)=(s-36)-(s-108)=72\). The terms containing \(s\) cancel, so the given condition is true for every real \(s\). Neither \(s=72\) nor \(s=9\) is a specially required value. Exam tip: while subtracting function values, use brackets carefully to avoid sign errors.

B(t)-A(t)=(70+11t)-(44+6t)=26+5t. The coefficient of t is 5, so the difference increases by 5 for every increase of 1 in t. Here, 26 is only the initial difference, not the rate of increase, so option D is incorrect. Exam tip: when subtracting linear expressions, subtract constant terms and t-terms separately.

Find the difference: \(C(t)-D(t)=(260-8t)-(180-15t)=80+7t\). The coefficient of \(t\) is \(+7\), so the difference increases by 7 for every 1-unit increase in \(t\). Option B results from a sign error: \(-8t-(-15t)=-8t+15t=+7t\), not \(-7t\). Exam tip: when subtracting an expression, change the signs of every term in the second expression.

In \(y=200-5t\), the coefficient of \(t\) is \(-5\), so \(y\) decreases by exactly 5 per time unit. \(200(0.95)^t\) shows exponential decay because its decrease is percentage-based. Exam tip: in linear decay, the variable has power 1.

In \(y=p-rx\), increasing x by 1 changes y by \(-r\), so the decrease is constant; putting \(x=0\) gives y=p. \(p+rx\) represents growth instead. Exam tip: identify decay by a negative slope.

Given \(f(x)=9x+28\), we get \(f(3x)=9(3x)+28=27x+28\). Therefore, \(f(3x)-f(x)=(27x+28)-(9x+28)=18x\). Setting \(18x=108\) gives \(x=6\). For \(x=5\), the value would be \(90\), so the nearest distractor is not correct. Exam tip: while finding \(f(3x)\), replace every \(x\) in the function by \(3x\).

Given \(g(x)=240-12x\), we get \(g(4x)=240-12(4x)=240-48x\). Hence, \(g(x)-g(4x)=(240-12x)-(240-48x)=36x\). Setting \(36x=144\) gives \(x=4\). Therefore, option B is correct. For instance, at \(x=3\), the difference is only \(108\), not \(144\). Exam tip: while finding \(g(4x)\), substitute the entire expression \(4x\) for \(x\).

Here, \(P(t)=60+5t\) is a linear polynomial. The midpoint of \(5\) and \(15\) is \(10\), so for a linear function, \(P(5)+P(15)=2P(10)\). Checking directly, \(P(5)=85\) and \(P(15)=135\), giving a sum of \(220\); also, \(2P(10)=2\times110=220\). \(P(9)\) is close, but \(9\) is not the midpoint of \(5\) and \(15\). Exam tip: for a linear function, the function value at the average of two inputs equals the average of their function values.

Given Q(t)=250-10t, Q(4)=210 and Q(16)=90. Hence, Q(4)+Q(16)=300. Now, Q(10)=250-100=150, so 2Q(10)=2×150=300. Therefore, Q(10) is correct. For example, Q(12)=130, and twice this value is 260, not 300. Exam tip: For a linear function, the function value at the average of two inputs equals the average of the two function values.

The change in x is 15−7=8. In the linear expression y=150+ax, the change in y equals a times the change in x. Thus, 8a=104, so a=104/8=13. If 12 were chosen, the increase in y would be 8×12=96, not 104. Exam tip: the constant term 150 does not affect the change; only the ax term determines it.

The change in \(x\) is \(14-6=8\). In \(y=300-bx\), every increase of 1 unit in \(x\) decreases \(y\) by \(b\) units. Therefore, the total decrease is \(8b\). Given \(8b=120\), we get \(b=120/8=15\). If \(b=14\), the decrease would be \(8\times14=112\), not 120. Exam tip: For a linear relation, total change equals rate of change times the change in input.

The condition is \(A(t)=B(t)+9\). So, \(40+6t=76+3t+9\), which gives \(3t=45\) and hence \(t=15\). Checking: \(A(15)=130\) and \(B(15)=121\), so the difference is \(9\). At \(t=14\), the difference is only \(6\), so it is not correct. Exam tip: translate “9 more than” as \(A=B+9\).

“\(C(t)\) is 12 less than \(D(t)\)” means \(C(t)=D(t)-12\). Thus, \(260-9t=80+5t-12\), or \(260-9t=68+5t\). This gives \(192=14t\), so \(t=\frac{96}{7}\). At \(t=12\), the difference between the two functions is 24, not 12. Exam tip: For “less than by” questions, write the smaller quantity as the larger quantity minus the stated difference.

The sum is ((a+14)+(a+42)+(a+70)=3a+126). From (3a+126=225), (a=33), which is not in the options.

The sum is ((b-18)+(b-42)+(b-66)=3b-126). From (3b-126=234), (b=120), so none of the options is correct.