Constructing direct/inverse proportional equations

(a) $y = 2x^2$

1. Write the proportion with a constant.
   $$y = kx^2$$
2. Substitute.
   $$18 = k \times 9 \Rightarrow k = 2$$

• M1 Writing $y = kx^2$ and substituting.
• A1 The correct answer, $y = 2x^2$.

(b) $50$

1. $y = 2 \times 25 = 50$.

• B1 The correct answer, $50$.

(a) $192$

1. $$75/5^{2}$$
2. $$3\times 8^{2}$$
3. Write y = $kx^{2}.$ Substitute 75 = 25k to get k = 3.
4. For x = 8, y = 3 $\times$ $8^{2}$ = 192.

• P1 Establishing $75/5^{2}$ or an equivalent valid method.
• P1 Establishing $3\times 8^{2}$ or an equivalent valid method.
• A1 Correct answer: $192$

(a) $27$

1. $$15/\sqrt{25}$$
2. $$3\sqrt{81}$$
3. Write y = $k\sqrt{}x.$ Then 15 = 5k, so k = 3.
4. At x = 81, y = 3 $\times$ 9 = 27.

• P1 Establishing $15/\sqrt{25}$ or an equivalent valid method.
• P1 Establishing $3\sqrt{81}$ or an equivalent valid method.
• A1 Correct answer: $27$

(a) $15$

1. $$12\times 5/8$$
2. $$7.5\times 14/7$$
3. The combined relationship is z = kx/y.
4. 12 = 8k/5 gives k = 7.5.
5. z = 7.5 $\times$ 14/7 = 15.

• P1 Establishing $12\times 5/8$ or an equivalent valid method.
• P1 Establishing $7.5\times 14/7$ or an equivalent valid method.
• A1 Correct answer: $15$

(a) $4$

1. Write the proportion with a constant.
   $$y = \frac{k}{\sqrt{x}}$$
2. Find $k$.
   $$6 = \frac{k}{4} \Rightarrow k = 24$$
3. $$y = \frac{24}{\sqrt{36}} = \frac{24}{6} = 4$$

• M1 Writing $y = \frac{k}{\sqrt{x}}$.
• M1 Finding $k = 24$.
• A1 The correct answer, $4$.

(b) $4$

1. $12 = \frac{24}{\sqrt{x}} \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4$.

• M1 Rearranging to $\sqrt{x} = 2$.
• A1 The correct answer, $4$.

(a) $8$

1. $y = kx^3$. Doubling $x$ multiplies $x^3$ by $2^3 = 8$.

• B1 The correct answer, $8$.

(b) $20\%$

1. $p = \frac{k}{q}$. The new $q$ is $1.25q$, so $p$ is multiplied by $\frac{1}{1.25} = 0.8$.
2. That is a $20\%$ decrease.

• M1 Finding the multiplier $\frac{1}{1.25} = 0.8$.
• A1 The correct answer, $20\%$.

(a) $3$

1. $$(x+3)^{2}=4x^{2}$$
2. $$x+3=2x$$
3. The ratio of the y values is the square of the ratio of the x values: (x + 3)$^{2}$/$x^{2}$ = 4.
4. Because x is positive, (x + 3)/x = 2.
5. Thus x + 3 = 2x and x = 3.

• P1 Establishing $(x+3)^{2}=4x^{2}$ or an equivalent valid method.
• P1 Establishing $x+3=2x$ or an equivalent valid method.
• A1 Correct answer: $3$

(a) $20$

1. $$I = \frac{k}{d^2}, \qquad 45 = \frac{k}{4} \Rightarrow k = 180$$
2. $$I = \frac{180}{9} = 20$$

• M1 Writing $I = \frac{k}{d^2}$ and finding $k = 180$.
• A1 The correct answer, $20$.

(b) $55.6\%$

1. The new distance is $1.5d$, so the brightness is multiplied by
   $$\frac{1}{1.5^2} = \frac{1}{2.25} = 0.444\ldots$$
2. The decrease is $1 - 0.444\ldots = 0.5555\ldots$, which is $55.6\%$.

• P1 Recognising the multiplier for the distance is 1.5.
• P1 Finding the brightness multiplier $\frac{1}{1.5^2} = 0.444\ldots$ (or using values: $d = 2$ to $d = 3$ gives $45 \to 20$).
• A1 The correct answer, $55.6\%$.