Algebraic Relationships Topic Test

C, the point of intersection. The solution must satisfy both equations at the same time, and the only point that lies on both lines is the point where they cross.

A, $x = 4,\ y = 10$. Set the right-hand sides equal: $3x - 2 = x + 6 \Rightarrow 2x = 8 \Rightarrow x = 4$. Then $y = 4 + 6 = 10$. Check: $3(4) - 2 = 10$ ✓ and $4 + 6 = 10$ ✓.

B, $x = 2,\ y = 3$. The $x$-coefficients match with the same sign, so subtract: $(2x + 5y) - (2x + y) = 19 - 7 \Rightarrow 4y = 12 \Rightarrow y = 3$. Then $2x + 3 = 7 \Rightarrow x = 2$. Check: $2(2) + 5(3) = 4 + 15 = 19$ ✓.

D, none. Both lines have gradient 2 but different $y$-intercepts (3 and $-1$), so they are parallel and never intersect. A system of parallel, distinct lines has no solution.

B, the fixed cost. In a linear cost model $C = F + vn$, the constant term $F$ is the fixed cost (rent, wages, insurance) that must be paid regardless of output. The $3.50n$ term is the variable cost, which grows with each milkshake made.

C, 25 items. Break-even occurs when $R = C$: $\ 12n = 200 + 4n \Rightarrow 8n = 200 \Rightarrow n = 25$. Check: $C = 200 + 4(25) = 300$ and $R = 12(25) = 300$, equal, so profit is zero at 25 items.

A, 34 items. $21n = 500 + 6n \Rightarrow 15n = 500 \Rightarrow n = 33.3\ldots$ At $n = 33$: $R = 693$, $C = 698$, still a $\$5$ loss. At $n = 34$: $R = 714$, $C = 704$, a $\$10$ profit. Always round up to the first whole unit that produces a profit.

D, 120 adult tickets. Let $a$ = adults, $c$ = children: $a + c = 200$ and $15a + 9c = 2520$. Substitute $c = 200 - a$: $15a + 1800 - 9a = 2520 \Rightarrow 6a = 720 \Rightarrow a = 120$. Check: $15(120) + 9(80) = 1800 + 720 = 2520$ ✓.

B, $\$50$. $R = 8 \times 60 = \$480$ and $C = 250 + 3 \times 60 = \$430$. Profit $= R - C = 480 - 430 = \$50$. Choosing $\$230$ is the common error of subtracting only the fixed cost and forgetting the variable cost.

A, $1.04$. For growth at $r\%$ per period, $b = 1 + \dfrac{r}{100} = 1 + \dfrac{4}{100} = 1.04$. The value $0.96$ would model a 4% decay, and $1.4$ would be 40% growth.

C, decays by 12% each year. Since $b = 0.88 = 1 - 0.12$, the machine keeps 88% of its value each year, which means it loses 12% each year. A decay model always has $0 < b < 1$.

B, $(0, 5)$. At $x = 0$: $y = 5 \times 3^0 = 5 \times 1 = 5$. For any exponential $y = Ab^x$, the $y$-intercept is always $A$, the initial value.

D, $y$ approaches 0 but never reaches it. Since $0 < b < 1$, each step multiplies $y$ by 0.8, so $y$ keeps shrinking but stays positive. The line $y = 0$ is a horizontal asymptote of every exponential $y = Ab^x$ with $A > 0$.

C, $2315$. $N = 2000 \times 1.05^3 = 2000 \times 1.157625 = 2315.25 \approx 2315$. Note $2300$ is the linear error ($3 \times 5\% = 15\%$ of 2000 added once), $2205$ is only 2 hours of growth, and $2431$ is 4 hours.

A, opens downward with a maximum vertex. The sign of $a$ controls concavity: $a > 0$ gives an upward-opening parabola with a minimum, while $a < 0$ gives a downward-opening parabola whose vertex is the highest point, a maximum.

C, $7$. Substitute $x = 0$: $y = 2(0)^2 - 3(0) + 7 = 7$. For any quadratic $y = ax^2 + bx + c$, the $y$-intercept is always the constant term $c$.

B, $x = 2$. The axis of symmetry passes through the midpoint of the $x$-intercepts: $x = \dfrac{-2 + 6}{2} = \dfrac{4}{2} = 2$. The answer $x = 4$ comes from the common error of averaging $2$ and $6$ (dropping the negative sign).

D, quadratic. Constant second differences identify a quadratic. Linear relationships have constant first differences, exponentials have a constant ratio between successive $y$ values, and inverse variation has a constant product $xy$.

C, $20$ m. $h = 0$ gives $5t(4 - t) = 0$, so $t = 0$ or $t = 4$. The axis of symmetry is $t = \dfrac{0 + 4}{2} = 2$, and $h(2) = -5(4) + 20(2) = -20 + 40 = 20$ m. The maximum occurs at the vertex.

A, $0 \leq t \leq 4$. The ball leaves the ground at $t = 0$ and lands at $t = 4$ (the $t$-intercepts). Outside this interval the model predicts negative heights, which are not physically meaningful, so the valid domain runs between the intercepts.

A, the break-even points. On a profit graph, the horizontal intercepts are where $P = 0$: $-n^2 + 12n - 20 = -(n - 2)(n - 10) = 0$ at $n = 2$ and $n = 10$. Profit is positive between them and the maximum profit occurs at the vertex, $n = 6$.

D. Setting $h = 0$: $d(-0.02d + 1.6) = 0$ gives $d = 0$ or $d = 80$. Beyond those intercepts the parabola dips below the axis, predicting the track is underground. Real quadratic models are only valid on a restricted domain, and you should state that domain in HSC answers.

B, $y = \dfrac{24}{x}$. Test the products: $2 \times 12 = 24$, $4 \times 6 = 24$, $8 \times 3 = 24$. A constant product $xy = k$ means inverse variation with $k = 24$, so $y = \dfrac{24}{x}$.

C, $160$ km/h. $1.5 = \dfrac{240}{s} \Rightarrow s = \dfrac{240}{1.5} = 160$ km/h. Check with the constant product: $160 \times 1.5 = 240$ ✓.

D, a hyperbola. $y = \dfrac{k}{x}$ has two branches (in quadrants 1 and 3 when $k > 0$) that approach but never touch the axes. The line $x = 0$ is the vertical asymptote and $y = 0$ is the horizontal asymptote; the curve has no intercepts at all.