Linear modelling - Answer key

Fluency

$F = 2.2d + 4.5$ .
$F = 2.2(12) + 4.5 = 26.4 + 4.5 = 30.90$ dollars.
$I = 3000 \times 0.05 \times 2 = 300$ dollars.
$I = 8000 \times 0.035 \times 4 = 1120$ dollars. Total: $A = 8000 + 1120 = 9120$ dollars.
$B = -8t + 100$ .
$20 = -8t + 100$ , so $8t = 80$ , $t = 10$ hours.
$V = -500t + 12000$ .
$E = 18.5h + 25$ .

Reasoning

$30 + 0.1n = 15 + 0.4n$ , so $15 = 0.3n$ , $n = 50$ texts. At $60$ texts: Plan X $= 36$ , Plan Y $= 39$ . Plan X is cheaper.
$I = 4800 - 4000 = 800$ . $800 = 4000 \times r \times 5$ . $r = 0.04 = 4\%$ p.a.
$V = -3000t + 25000$ . Set $V = 7000$ : $7000 = -3000t + 25000$ , $3000t = 18000$ , $t = 6$ years.
Runner A: $d_A = 6t + 20$ . Runner B: $d_B = 8t$ . Set equal: $8t = 6t + 20$ , $2t = 20$ , $t = 10$ s. Position: $80$ m from the start.
The $150 is the fixed hire cost (charged regardless of time). The $45 is the hourly rate (cost per additional hour).
Gradient: $m = \dfrac{5100 - 2700}{7 - 3} = \dfrac{2400}{4} = 600$ L/h. Using $(3, 2700)$ : $2700 = 600(3) + c$ , $c = 900$ . Model: $W = 600t + 900$ . The pool initially had $900$ litres.

Reasoning

$A = P(1 + rT) = PrT + P$ . This has the form $y = mx + c$ with $m = Pr$ (constant interest per year) and $c = P$ (initial deposit). The gradient is $Pr$ .
Bank A after $8$ years: $A = 10000(1 + 0.04 \times 8) = 10000 \times 1.32 = 13200$ dollars. Bank B after $8$ years: $A = 10000(1 + 0.03 \times 8) + 200 = 10000 \times 1.24 + 200 = 12600$ dollars. Bank A gives $600 more.
For $d \geq 1$ : first day costs $120, each additional day costs $80. Total: $C = 80(d - 1) + 120 = 80d + 40$ . Yes, this is linear in $d$ : gradient $= 80$ (daily rate), $y$ -intercept $= 40$ (but note the model only applies for $d \geq 1$ , so the actual starting cost is $C(1) = 120$ ).
Car A from Town X: $d_A = 60t$ . Car B from Town X: $d_B = 120 - 40t$ . Set equal: $60t = 120 - 40t$ , $100t = 120$ , $t = 1.2$ hours ( $1$ h $12$ min). Position: $60 \times 1.2 = 72$ km from Town X.

Reasoning

Interest per year: $15000 \times 0.06 = 900$ dollars. Net reduction per year: $500 - 900 = -400$ dollars (the debt grows by $400 each year because repayments do not cover interest). Actually, since the question states interest is on the original principal: amount owed after $n$ years $= 15000 + 900n - 500n = 15000 + 400n$ . The debt increases, so the loan is never repaid under these terms. Alternatively, if we interpret “repays $500 after interest” as reducing the balance: total interest over $n$ years is $900n$ . Total repaid: $500n$ . Amount owed: $15000 + 900n - 500n = 15000 + 400n$ . The repayments ($500/year) are less than the annual interest ($900), so the loan balance grows. The repayment amount would need to exceed $900 per year to reduce the debt.
Plan A vs B: $0.8m = 20 + 0.3m$ , $0.5m = 20$ , $m = 40$ min. Plan B vs C: $20 + 0.3m = 50$ , $0.3m = 30$ , $m = 100$ min. Plan A cheapest: $0$ to $40$ min. Plan B cheapest: $40$ to $100$ min. Plan C cheapest: over $100$ min.
Candle: $h_c = -2t + 20$ , burns out at $t = 10$ min. Sparkler: $h_s = -4t + 12$ , burns out at $t = 3$ min. Same height: $-2t + 20 = -4t + 12$ , $2t = -8$ , $t = -4$ . Since $t = -4$ is negative, they are never the same height during the time both are burning (the sparkler starts shorter and burns faster, so the candle is always taller while both exist).

Year 9 core - answers