Direct and Inverse Proportion: Additional Practice (Set 1)

Find the constant product $k = xy$.

$k = 12 \times 10 = 120$

Then $y = \dfrac{120}{x} = \dfrac{120}{8} = 15$.

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

$84 = k \times 6 \times 2 = 12k \implies k = 7$

Substitute into $y = \dfrac{kx}{z}$.

$9 = \frac{k \times 6}{8} = \frac{3k}{4} \implies k = 12$

The product $xy$ is $20$ in every row ($4 \times 5$, $5 \times 4$, and so on), so the variation is inverse with $k = 20$.

$y = \frac{20}{8} = 2.5$

Find the constant product $k = xy$.

$k = 6 \times 14 = 84$

Then $y = \dfrac{84}{x}$, so $21 = \dfrac{84}{x}$ gives $x = 4$.

Find the constant product $k = mh$, the total mowing work.

$k = 5 \times 6 = 30$

Then $3 = \dfrac{30}{m}$ gives $m = 10$ mowers.

Find $k$ from the known pair.

$k = \frac{48}{6} = 8$

Then $y = 8 \times 11 = 88$.

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

$k = 4 \times 9 = 36$

Then $12 = \dfrac{36}{P}$ gives $P = 3$ atm.

Combined variation means $y = \dfrac{kx}{z}$. Find $k$ first.

$15 = \frac{k \times 10}{4} = 2.5k \implies k = 6$

Then $y = \dfrac{6 \times 8}{6} = 8$.

Inverse variation keeps the product $xy$ constant, and $4 \times 9 = 36$.

$6 \times 6 = 36$

Only $(6, 6)$ has the same product; the others give $144$, $9$, and $45$.

Inverse variation is $y = \dfrac{k}{x}$. Replacing $x$ with $3x$ gives $\dfrac{k}{3x}$.

$y_{\text{new}} = \frac{k}{3x} = \frac{1}{3} \cdot \frac{k}{x} = \frac{y}{3}$

So $y$ becomes one third as large.

Substitute the known values into $P = \dfrac{kT}{V}$.

$20 = \frac{k \times 300}{6} = 50k \implies k = 0.4$