Linear growth and decay

Given \(H(t)=18+4t\). Substituting \(t=5\), \(H(5)=18+4(5)=18+20=38\) cm. Hence, 38 cm is correct. Getting 32 cm would mean calculating the growth term \(4t\) incorrectly. Exam tip: To find a function value, substitute the given value for the variable and then simplify.

Given \(W(t)=120-6t\). Substituting \(t=8\), \(W(8)=120-6(8)=120-48=72\) L. Therefore, 72 L is correct. 48 L is only the amount of water removed, not the water remaining. Exam tip: substitute the value of \(t\) first, then multiply and subtract.

In \(P(n)=40-3n\), the power of \(n\) is 1 and its coefficient is \(-3\), so the quantity decreases by 3 for every one-unit increase. \(40-3n^2\) is not linear. Exam tip: linear decay has degree 1 with a negative variable coefficient.

In \(12-3t\), the power of \(t\) is 1, so the change is linear. The coefficient \(-3\) means the quantity decreases by 3 per unit time. \(12+3t\) shows growth. Exam tip: check the sign of the slope.

To make the amount ₹500, put S(m)=500: 200+50m=500. Thus, 50m=300, so m=6. Therefore, the amount becomes ₹500 after 6 months. After 5 months, the amount is only ₹450, so it is a close but incorrect option. Exam tip: Set the given linear expression equal to the required amount and solve for m.

For the battery to be at 60%, put \(B(h)=60\) in the model: \(100-8h=60\). Thus, \(8h=40\), so \(h=5\) hours. At 4 hours, the battery would be \(100-8(4)=68\%\), so that option is not correct. Exam tip: In linear decay questions, equate the function to the required target value first, 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, the initial value is 35 and the growth is 6, so the rule is \(y=35+6x\). In option B, 6 is subtracted, so it represents decay. Exam tip: put \(x=0\) to identify the initial value.

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 140 and the value decreases by 10 each step, so the change is \(-10\). Therefore, the rule is \(y=140-10x\). Option A represents an increase of 10 per step. Exam tip: words such as “decrease” or “decay” indicate a negative slope.

In \(y=12-2x\), the coefficient of x is \(-2\), so y falls by 2 units for every 1-unit rise in x. Option B represents growth. Exam tip: linear decay has x to power 1 with a negative coefficient.

In option B, the coefficient of \(x\) is \(-7\). So, when \(x\) increases by 1, \(y\) decreases by 7 each time; this is constant-rate linear decay. Relations involving \(x^2\) or \(1/x\) are not linear. Exam tip: a negative slope indicates decay.

Given A(t)=a+4t and A(3)=32, substitute t=3: a+4×3=32. Thus, a+12=32, so a=20. If a were 24, then A(3)=24+12=36, not the given value 32. Exam tip: when a function value is given, substitute the corresponding value of t directly into the function.

Given \(R(t)=b-6t\). Substituting \(t=5\) gives \(R(5)=b-6(5)=b-30\). Since \(R(5)=50\), \(b-30=50\), so \(b=80\). Option 50 is the value of \(R(5)\), not the value of \(b\). Exam tip: When a function value is given, first substitute the stated input into the formula and form an equation.

In \(120-8t\), the highest power of \(t\) is 1 and its coefficient is \(-8\), so the quantity decreases by 8 per time unit. \(120+8t\) shows growth. Exam tip: linear decay has a negative coefficient of the variable.

In \(P(t)=12+3t\), the coefficient of \(t\) is \(+3\), so the quantity increases by 3 for every unit of time. Option B is linear but shows decay because its coefficient is negative. Exam tip: for linear growth, the power of \(t\) must be 1 and its coefficient must be positive.

For linear growth the coefficient must be positive and the initial value must be (12). (y=12+5x) satisfies both.

A linear decay rule has the form y=a-bx with b>0: it is linear, its slope is negative, and its initial value at x=0 is a. In option A, y=60-4x, the initial value is 60 and the slope is -4, so the quantity decreases steadily. Option B grows, option C starts at 4, and option D is quadratic with initial value -60.

Given \(M(t)=m+5t\). Substituting \(t=2\), we get \(M(2)=m+10=25\), so \(m=15\). Now, for \(t=6\), \(M(6)=15+5\times6=45\). Hence, 45 is correct. The nearby option 40 can result from incorrectly counting the growth; from \(t=2\) to \(t=6\), the increase is \(5\times4=20\). Exam tip: first use the given condition to find the unknown constant \(m\), then substitute the required value.

Given \(N(t)=n-3t\). Substituting \(t=4\), \(N(4)=n-12=28\), so \(n=40\). Now, for \(t=9\), \(N(9)=40-3\times9=40-27=13\). Hence, 13 is correct. A value such as 16 can result from using the rate of decrease or the time value incorrectly. Exam tip: first use the given condition to find the unknown constant \(n\), then substitute the required value of \(t\).

At \(x=3\), the value of \(y\) is 35. Substituting these values in \(y=20+kx\) gives \(35=20+3k\). Hence, \(3k=15\), so \(k=5\). Option 15 is the value of \(3k\), not of \(k\). Exam tip: In a linear expression, substitute the given values of \(x\) and \(y\) first, then isolate the unknown coefficient.

Given \(x=4\) and \(y=76\), substitute these values in \(y=100-kx\): \(76=100-4k\). Hence, \(4k=24\), so \(k=6\). Option 24 is the value of \(4k\), not of \(k\). Exam tip: In a linear expression, substitute the given values of \(x\) and \(y\) first, then isolate the unknown term.