Linear growth and decay

In \(P=15+4t\), the power of \(t\) is 1 and 4 is the constant growth rate, so \(P\) rises by 4 per time unit. \(15t^2\) has a changing rate. Exam tip: identify linear forms as \(a+bt\).

Given H(t)=36+4t, substitute t=10: H(10)=36+4(10)=36+40=76. Therefore, option B is correct. The value 72 would result from incorrectly adding 36 and 36; here, 4t becomes 4×10=40. In exams, substitute the input value first and then simplify carefully.

Substitute 8 for t: \(K(8)=144-12\times 8=144-96=48\). Therefore, the correct answer is 48. The value 96 is only \(12\times 8\); it must still be subtracted from 144. Exam tip: To evaluate a function, substitute the given value for the variable and follow the order of operations.

The coefficient of \(x\) is \(-3\), so \(y\) decreases as \(x\) increases; this is linear decay. A positive slope would indicate growth. Exam tip: the coefficient of \(x\) is the slope.

In \(y=50-3x\), the coefficient of \(x\) is \(-3\), so \(y\) falls by 3 for each one-unit rise in \(x\). Relations involving \(x^2\) or \(1/x\) are not linear. Exam tip: identify decay by a constant negative slope.

The initial number is 5, so it is the constant term. Since the number rises by 6 at each step, the total increase after \(x\) steps is \(6x\). Hence, the rule is \(y=5+6x\). In \(y=6+5x\), the initial value and the rate of increase are interchanged. Exam tip: in a linear rule, the constant term is the starting value and the coefficient of \(x\) is the change per step.

Given \(y=42+3x\), substitute \(x=6\): \(y=42+3(6)=42+18=60\). Hence, 60 is correct. The value 48 results from doing \(42+6\) and incorrectly omitting multiplication by 3. Exam tip: after substitution in a linear expression, perform multiplication before addition or subtraction.

Given \(y=85-5x\), substitute \(x=9\): \(y=85-5(9)=85-45=40\). Therefore, the correct value is 40. The value 45 is only \(5\times9\); it must be subtracted from 85. Exam tip: When evaluating a linear expression, substitute the given value first, multiply, and then perform addition or subtraction.

In \(y=12-2x\), the coefficient of x is \(-2\). When x increases by 1, y decreases by 2, so it shows decay. \(y=x^2+1\) is not linear. Exam tip: a negative slope indicates decrease.

In \(20-3x\), the coefficient of x is \(-3\), so y falls by 3 units for every increase of 1 in x. \(7x+4\) represents growth. Exam tip: linear decay has a negative coefficient of x.

Given \(H(t)=30+6t\). Substituting \(t=5\), \(H(5)=30+6(5)=30+30=60\) cm. Therefore, 60 cm is correct. 66 cm would result from using \(t=6\), but the given value is \(t=5\). Exam tip: after substitution, perform multiplication before addition.

Given \(W(t)=180-12t\). Substituting \(t=9\), \(W(9)=180-12(9)=180-108=72\) litres. Therefore, 72 litres is correct. The nearby distractor 84 litres does not result from substituting 9 correctly in the given expression. Exam tip: in a linear decay expression, substitute the time value first, multiply, and then subtract the decrease from the initial amount.

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.