Proportions: Additional Practice (Set 2)

Keep model inches over real feet on both sides, with $m$ for the model length: $\frac{2}{5} = \frac{m}{85}$. Cross-multiply the known diagonal $2 \times 85$ and divide by $5$.

$m = \frac{2 \times 85}{5} = \frac{170}{5} = 34$

So the model is $34$ inches long.

Set up cups over muffins on both sides, with $f$ for the unknown cups: $\frac{5}{12} = \frac{f}{42}$. Cross-multiply the known diagonal $5 \times 42$ and divide by $12$.

$f = \frac{5 \times 42}{12} = \frac{210}{12} = 17.5$

So $17.5$ cups of flour are needed.

Keep pounds over dollars on both sides, in the same order. The first side is $\frac{4 \text{ lb}}{9 \text{ dollars}}$, so the second must be pounds over dollars too, with $c$ as the unknown cost.

$\frac{4}{9} = \frac{10}{c}$

The other setups flip the order or misplace the unknown.

The bottom went from $5$ to $20$, a scale factor of $20 \div 5 = 4$. Multiply the top by $4$.

$n = 3.5 \times 4 = 14$

So the missing number is $14$.

Set up defective over total on both sides, with $x$ for the unknown defective count: $\frac{7}{200} = \frac{x}{5000}$. Cross-multiply the known diagonal $7 \times 5000$ and divide by $200$.

$x = \frac{7 \times 5000}{200} = \frac{35000}{200} = 175$

So about $175$ parts are defective.

Heights and shadows grow together, so set up height over shadow on both sides, with $h$ for the tree height: $\frac{7}{3} = \frac{h}{21}$. Cross-multiply the known diagonal $7 \times 21$ and divide by $3$.

$h = \frac{7 \times 21}{3} = \frac{147}{3} = 49$

So the tree is $49$ feet tall.

Set up miles over gallons on both sides, with $d$ for the unknown distance: $\frac{165}{5} = \frac{d}{13}$. Cross-multiply the known diagonal $165 \times 13$ and divide by $5$.

$d = \frac{165 \times 13}{5} = \frac{2145}{5} = 429$

So the car can travel $429$ miles.

Set up gallons over minutes on both sides, with $g$ for the unknown gallons: $\frac{250}{4} = \frac{g}{30}$. Cross-multiply the known diagonal $250 \times 30$ and divide by $4$.

$g = \frac{250 \times 30}{4} = \frac{7500}{4} = 1875$

So it moves $1875$ gallons.

Keep real feet over photo centimeters on both sides, with $H$ for the real house height: $\frac{24}{8} = \frac{H}{11}$. Cross-multiply the known diagonal $24 \times 11$ and divide by $8$.

$H = \frac{24 \times 11}{8} = \frac{264}{8} = 33$

So the house is $33$ feet tall.

Check $\frac{8}{20} = \frac{10}{24}$ with cross-products.

$8 \times 24 = 192 \qquad 20 \times 10 = 200$

Since $192 \neq 200$, this proportion is false. The other three have equal cross-products.

The known diagonal is $9$ and $22$, with $n$ over the $9$, opposite the $6$. Multiply the known diagonal and divide by $6$.

$n = \frac{9 \times 22}{6} = \frac{198}{6} = 33$

So the missing number is $33$.

Set up notebooks over dollars on both sides, with $c$ for the unknown cost: $\frac{3}{8} = \frac{27}{c}$. Cross-multiply the known diagonal $8 \times 27$ and divide by $3$.

$c = \frac{8 \times 27}{3} = \frac{216}{3} = 72$

So $27$ notebooks cost $72$ dollars.