Sec 2 Maths: Direct & Inverse Proportion, practice questions & worked solutions

(a)

Divide each cost by its number of pages:

$$\frac{2.40}{20}=0.12,\qquad \frac{6.00}{50}=0.12,\qquad \frac{9.60}{80}=0.12,\qquad \frac{12}{100}=0.12$$

Since $\dfrac{C}{n}=0.12$ is the same constant for every pair, $C$ and $n$ are directly proportional.

(b)

$$C=0.12n$$

(c)

$$\begin{aligned}27&=0.12n\\ n&=\frac{27}{0.12}\\ &=225\end{aligned}$$

$225$ pages were printed.

(a)

$x$ is in metres, so convert first: $5$ cm $=0.05$ m.

$$\begin{aligned}F&=kx\\ 2.5&=k(0.05)\\ k&=50\\ \therefore F&=50x\end{aligned}$$

(b)

$8$ cm $=0.08$ m.

$$\begin{aligned}F&=50(0.08)\\ &=4\text{ N}\end{aligned}$$

(a)

$$\begin{aligned}y&=k\sqrt{x}\\ 10&=k\sqrt{25}\\ 10&=5k\\ k&=2\\ \therefore y&=2\sqrt{x}\end{aligned}$$

(b)

$$\begin{aligned}12&=2\sqrt{x}\\ \sqrt{x}&=6\\ x&=36\end{aligned}$$

(c)

$$\begin{aligned}y&=2\sqrt{64}\\ &=2(8)\\ &=16\end{aligned}$$

(a)

$$\begin{aligned}y&=kx^2\\ 64&=k(4^2)\\ k&=4\\ \therefore y&=4x^2\end{aligned}$$

When $x=6.5$:

$$\begin{aligned}y&=4(6.5)^2\\ &=4(42.25)\\ &=169\end{aligned}$$

(b)

Starting from the given pair $y=64$ when $x=4$, the new value of $y$ is $64\times 9=576$.

$$\begin{aligned}4x^2&=576\\ x^2&=144\\ x&=12\quad(x=-12\text{ is rejected since }x\gt 0)\end{aligned}$$

Check: multiplying $y$ by $9=3^2$ multiplies $x$ by $3$, and $4\times 3=12$.

(a)

$$\begin{aligned}y&=kt^2\\ 4&=k(6^2)\\ k&=\frac{4}{36}=\frac19\\ \therefore y&=\frac19 t^2\end{aligned}$$

(b)(i)

$$\begin{aligned}y&=\frac19(12^2)\\ &=16\end{aligned}$$

(b)(ii)

Increasing $t=12$ by $50\%$ gives $t=18$.

$$\begin{aligned}\text{New }y&=\frac19(18^2)=36\\ \text{Percentage increase}&=\frac{36-16}{16}\times 100\%\\ &=125\%\end{aligned}$$

Check: $t$ is multiplied by $1.5$, so $y$ is multiplied by $1.5^2=2.25$, an increase of $125\%$.

Let the original values be $x$ and $y=kx^2$. Increasing $x$ by $300\%$ gives $x+3x=4x$ (not $3x$).

$$\begin{aligned}\text{New }y&=k(4x)^2\\ &=16kx^2\\ &=16y\end{aligned}$$
$$\begin{aligned}\text{Percentage increase}&=\frac{16y-y}{y}\times 100\%\\ &=1500\%\end{aligned}$$

The number of workers is inversely proportional to the number of days, so workers $\times$ days is constant.

$$\begin{aligned}\text{Total work}&=12\times 10=120\text{ worker-days}\\ \text{Workers needed for 8 days}&=\frac{120}{8}=15\\ \text{Additional workers}&=15-12=3\end{aligned}$$

(a)

$$\begin{aligned}T&=\frac{k}{N}\\ 25&=\frac{k}{6}\\ k&=150\\ \therefore T&=\frac{150}{N}\end{aligned}$$

(b)

With two pipes faulty, $N=6-2=4$.

$$\begin{aligned}T&=\frac{150}{4}=37.5\text{ minutes}\\ \text{Extra time}&=37.5-25\\ &=12.5\text{ minutes}\end{aligned}$$

(a)(i)

Let $w$ be the number of workers and $D$ the number of days. Then $wD=k$ is constant.

$$\begin{aligned}k&=12\times 100=1200\\ D&=\frac{1200}{16}\\ &=75\text{ days}\end{aligned}$$

(a)(ii)

All the workers work at the same rate (for example, every worker builds the same amount each day, and conditions such as the weather do not change).

(b)

$y=\dfrac{k}{x^2}$. Reducing $x$ by $50\%$ gives $0.5x$.

$$\begin{aligned}\text{New }y&=\frac{k}{(0.5x)^2}\\ &=\frac{k}{0.25x^2}\\ &=4\left(\frac{k}{x^2}\right)=4y\end{aligned}$$
$$\begin{aligned}\text{Percentage change}&=\frac{4y-y}{y}\times 100\%\\ &=300\%\end{aligned}$$

$y$ increases by $300\%$.

(a)

Let the particular value of $x$ be $a$, so $y=\dfrac{k}{x^2}$ with $y=12$ when $x=a$:

$$\begin{aligned}12&=\frac{k}{a^2}\\ k&=12a^2\end{aligned}$$

Increasing $x$ by $100\%$ doubles it, so the new value is $x=2a$:

$$\begin{aligned}y&=\frac{12a^2}{(2a)^2}\\ &=\frac{12a^2}{4a^2}\\ &=3\end{aligned}$$

(b)

$$\begin{aligned}\text{Percentage decrease}&=\frac{12-3}{12}\times 100\%\\ &=75\%\end{aligned}$$

(c)

No, I disagree. If $q$ were directly proportional to $p$, then $q=kp$ and the graph would be a straight line passing through the origin. This line cuts the $q$-axis above $O$, so it does not pass through the origin, and $q$ is not directly proportional to $p$.