Linear growth and decay

In \(P(n)=72-5n\), the coefficient of n is −5, so increasing n by 1 changes P by −5. \(72-5n^2\) is not linear because it contains n². Exam tip: a linear expression has the variable to power 1.

The coefficient of \(t\) is \(-5\), so when time increases by 1 unit, \(R\) falls by 5 units. Here 90 is the initial amount, not the rate of growth. Exam tip: a negative coefficient indicates decay.

The savings model is \(S(m)=400+60m\). Put \(S(m)=760\): \(400+60m=760\). Thus, \(60m=360\), so \(m=6\). Therefore, the amount will be ₹760 after 6 months. After 5 months, the amount would be ₹700, so that option is incorrect. Exam tip: To find the time for a given value in a linear model, equate the model to the target value first.

Set \(B(h)=50\) because the battery level is required to be 50%. Thus, \(120-10h=50\), so \(10h=70\) and \(h=7\). At \(h=6\), the battery level would be 60%, so that option is not correct. Exam tip: In linear decay questions, equate the model to the required value first and then solve for the variable.

A linear rule has the form \(y=a+bx\), where \(a\) is the initial value and \(b\) is the change per step. Here, \(a=42\) and \(b=+5\), so the rule is \(y=42+5x\). In option A, \(-5\) represents decay rather than growth. Exam tip: identify the constant term for the initial value and the coefficient of \(x\) for the rate of change.

A linear rule has the form \(y=a+bx\), where \(a\) is the initial value and \(b\) is the change per step. Here, the initial value is \(210\), and the value decreases by \(14\) each step, so \(b=-14\). Therefore, the rule is \(y=210-14x\). Option A represents an increase, not a decrease. Exam tip: Words such as “decrease” or “decay” indicate a negative coefficient of \(x\).

In \(y=40-3x\), the coefficient of x is \(-3\), so y falls by 3 for every increase of 1 in x. \(40+3x\) represents growth, not decay. Exam tip: linear decay has a constant negative coefficient of x.

In \(y=42-5x\), the coefficient of x is the constant \(-5\), so y falls by an equal amount each time x increases. A relation containing \(x^2\) is not linear. Exam tip: for linear decay, look for x to power 1 with a negative coefficient.

Given \(A(t)=a+9t\), substitute \(t=3\): \(A(3)=a+9\times3=a+27\). Since \(A(3)=57\), we get \(a+27=57\), so \(a=30\). Option 27 is the amount added over 3 time units, not the initial value \(a\). Exam tip: substitute the specified value of \(t\) first, then solve the resulting equation.

Given R(t)=b-11t and R(4)=76. Substituting t=4 gives b-11(4)=76, so b-44=76. Adding 44 to both sides gives b=120. Hence, 120 is the correct option. If b were 110, then R(4)=110-44=66, not 76. Exam tip: In a linear expression, first substitute the given value of the variable and then isolate the unknown term.

The coefficient \(b\) is the rate of change, or slope, per unit increase in \(x\). If \(b<0\), \(Q(x)\) decreases as \(x\) rises, so it represents decay. In contrast, \(b=0\) represents a constant quantity. Exam tip: use the sign of the slope to identify growth or decay.

In linear decay, a quantity decreases by a fixed amount in equal time intervals. Losing 5 litres every hour follows this pattern. Becoming half each hour is exponential decay. Exam tip: look for a fixed difference, not a fixed ratio.

The direct answer is option C, \(y=24+7x\). A linear rule has the form \(y=mx+c\), where \(m\) is the constant rate of change and \(c\) is the value when \(x=0\). Growth means \(m>0\), and the initial value 24 means \(y=24\) at \(x=0\). In C, the intercept is 24 and the slope is +7, so both conditions hold. Option A has initial value 24 but slope −7, so it describes linear decay. Option B has positive growth but starts at 7, not 24. Option D starts at 24 but contains \(x^2\), so its rate is not constant and it is not linear. Memory cue: positive coefficient of \(x\) means growth; the constant term is the initial value.

A linear rule has the form \(y=a+bx\), where the initial value is \(a\), found by putting \(x=0\). In \(y=150-12x\), the initial value is 150 and the coefficient of \(x\) is \(-12\), so the value decreases linearly. \(y=12-150x\) also represents decay, but its initial value is 12. Exam tip: put \(x=0\) to check the initial value, and look for a negative slope to identify decay.

Given \(M(t)=m+8t\). Substituting \(t=2\), \(M(2)=m+16=44\), so \(m=28\). Now, at \(t=7\), \(M(7)=28+8\times7=28+56=84\). Therefore, 84 is the correct answer. The value 76 can result from incorrectly counting the increase. Exam tip: first use the given condition to find the unknown constant \(m\), then substitute the required value of \(t\).

Given \(N(t)=n-6t\). At \(t=5\), \(N(5)=n-30=70\), so \(n=100\). Therefore, \(N(12)=100-6\times12=100-72=28\). Hence, 28 is correct. A value such as 34 can result from an error in subtraction. Exam tip: first find the unknown constant \(n\) from the given value, then substitute the required value of \(t\).

Substituting the given values into y=45+kx gives 81=45+6k. Hence, 6k=36 and k=6. Option 36 is only the difference 81-45; it must be divided by x=6 to find k. Exam tip: In a linear equation, substitute the given x and y values before solving for the unknown coefficient.

At \(x=8\), the value of \(y\) is 136. Substituting in the equation gives \(136=200-8k\). Hence, \(8k=64\), so \(k=8\). Note that 64 is the value of \(8k\), not of \(k\). Exam tip: Substitute the given values first, then isolate the unknown term carefully.

Substituting t=7 gives A(7)=20+5(7)=55 and B(7)=50+2(7)=64. Since 64 is greater than 55, the value of B(t) is greater. They are not equal because the two values are different. Exam tip: To compare linear expressions at a given value, substitute that value into both expressions first.

Substituting \(t=6\), \(C(6)=140-8\times6=92\) and \(D(6)=110-3\times6=92\). Hence, the two model values are equal, so \(C(6)=D(6)\). Options A and B are incorrect because neither value is greater than the other, and both values are not zero. Exam tip: evaluate each expression separately at the given value of \(t\) before comparing them.

Putting \(T(t)=102\) gives \(54+6t=102\). Hence, \(6t=48\), so \(t=8\). Therefore, option C is correct. For example, at \(t=7\), \(T(t)=54+42=96\), not 102. Exam tip: equate the function to the target value first, then isolate and solve for the variable.

Given \(U(t)=150-15t\), put \(U(t)=45\). This gives \(150-15t=45\), so \(-15t=-105\) and hence \(t=7\). If \(t=9\), the value is \(15\), not 45. Exam tip: In linear decay questions, substitute the target value into the function and solve the resulting equation for \(t\).

First, f(x+4)=25+6(x+4)=25+6x+24. Therefore, f(x+4)-f(x)=(25+6x+24)-(25+6x)=24. Hence, 24 is the correct option. 18 may seem tempting, but an increase of 4 units at a rate of 6 gives 6×4=24. Exam tip: For a linear function ax+b, increasing x by h always changes the function by ah.

Linear decay has the form \(a-bx\), where \(b>0\). In \(P(x)=80-6x\), the value decreases by 6 for every increase of 1 in \(x\). \(80+6x\) shows linear growth. In exams, check the sign of the \(x\)-coefficient.

In a linear expression \(y=a+bx\), \(a=y(0)\), while \(b\) gives the change in \(y\) for every increase of 1 in \(x\). Here, \(y(0)=18\) and the increase is 7, so \(y=18+7x\). In \(y=7+18x\), the initial value is 7 and the rate of increase is 18, so it is not correct. Exam tip: Substitute \(x=0\) to check the constant term quickly.