Direct and Inverse Proportion: Additional Practice (Set 2)

Joint variation means $y = kxz$. Substitute the known values.

$60 = k \times 3 \times 4 = 12k \implies k = 5$

Find $k$ from the known pair.

$k = \frac{63}{9} = 7$

Then $y = 7x = 7 \times 4 = 28$.

Find the constant product $k = xy$.

$k = 9 \times 8 = 72$

Then $y = \dfrac{72}{x}$, so $6 = \dfrac{72}{x}$ gives $x = 12$.

Substitute the known values into $R = \dfrac{kL}{A}$.

$4 = \frac{k \times 100}{5} = 20k \implies k = 0.2$

Substitute the known values into $y = \dfrac{kxz}{w}$.

$10 = \frac{k \times 4 \times 3}{6} = 2k \implies k = 5$

Substitute the values directly.

$R = \frac{0.2 \times 150}{3} = \frac{30}{3} = 10$

Find $k$ from the known pair.

$k = \frac{18}{24} = 0.75$

Then $y = 0.75 \times 40 = 30$.

Find $k$ from $y = kxz$.

$90 = k \times 5 \times 6 = 30k \implies k = 3$

Then $y = 3 \times 4 \times 9 = 108$.

Find the constant product $k = (\text{pressure}) \times (\text{volume})$.

$k = 2 \times 30 = 60$

Then the volume is $\dfrac{60}{5} = 12$ L. More pressure means less volume.

Direct variation keeps $y/x$ constant, and $\dfrac{15}{6} = 2.5$.

$\frac{5}{2} = 2.5$

Only $(2, 5)$ has the same ratio; the others give $2$, $0.4$, and $0.4$.

Inverse variation is $y = \dfrac{k}{x}$. Replacing $x$ with $\dfrac{x}{2}$ gives $\dfrac{k}{x/2}$.

$y_{\text{new}} = \frac{k}{x/2} = \frac{2k}{x} = 2y$

So $y$ doubles.

The load is $L = \dfrac{kw}{\ell}$. Replace $w$ with $2w$ and $\ell$ with $2\ell$.

$L_{\text{new}} = \frac{k(2w)}{2\ell} = \frac{2}{2} \cdot \frac{kw}{\ell} = L$

The factor is $\dfrac{2}{2} = 1$, so the load is unchanged.