Proportions

Keep minutes on top and liters on the bottom on both sides. $$\frac{2}{18}=\frac{8}{x}$$ Cross-multiply, so $2x=18\cdot 8=144$, and dividing by $2$ gives $\boxed{{\displaystyle 72}}$ liters. Quick sanity check, the flow is $9$ liters per minute and $9\times 8=72$.

Cups on top, campers on the bottom on both sides. $$\frac{2}{5}=\frac{x}{15}$$ The campers went from $5$ to $15$, a jump of $\times 3$, so the cups take the same $\times 3$, giving $2\times 3=\boxed{{\displaystyle 6}}$ cups. Cross-multiplying gets there too, since $5x=2\times 15=30$. Scaling is just faster when the multiplier is a clean whole number.

Jars on top, liters on the bottom in both ratios. $$\frac{8}{14}=\frac{20}{x}$$ Multiply both sides by $14$ and by $x$. The $14$ cancels on the left and the $x$ cancels on the right, leaving $8x=14\cdot 20=280$, so $x=\frac{280}{8}=\boxed{{\displaystyle 35}}$ liters. That step, each numerator times the other side's denominator, is what cross-multiplication is. It is the fraction-clearing move from 6.2 done to both denominators at once.

Grams on top, layers on the bottom on both sides. $$\frac{5}{8}=\frac{x}{6}$$ Cross-multiply to get $8x=5\times 6=30$, then divide by $8$, so $x=\frac{30}{8}=\boxed{{\displaystyle 15\mathrm{/}4}}$ grams, which is $3.75$. No whole-number multiplier carries $8$ to $6$, so scaling has nothing to grab, but cross-multiplication never needs one.

Real kilometers on top, map centimeters on the bottom in both ratios. $$\frac{10}{6}=\frac{9}{x}$$ Cross-multiply to get $10x=6\times 9=54$, so $x=\frac{54}{10}=\boxed{{\displaystyle 27\mathrm{/}5}}$ centimeters, which is $5.4$. The unknown sat in a denominator this time, and cross-multiplication treated it exactly like a numerator.

Dollars on top, shells on the bottom in both ratios. $$\frac{7}{91}=\frac{11}{x}$$ Cross-multiply to get $7x=91\cdot 11=1001$, so $x=\frac{1001}{7}=\boxed{{\displaystyle 143}}$ shells. The rate confirms it, since $\frac{7}{91}=\frac{1}{13}$ and $\frac{11}{143}=\frac{1}{13}$, thirteen shells per dollar both times.

Put real inches on top and model inches on the bottom on both sides. $$\frac{90}{3}=\frac{x}{5}$$ Cross-multiply to get $3x=90\times 5=450$, so $x=\frac{450}{3}=\boxed{{\displaystyle 150}}$ inches. The student's setup had model over real on one side and real over model on the other, which sets two different rates equal and gives a wrong answer.

Oats on top, cashews on the bottom on both sides. $$\frac{5}{2}=\frac{12}{x}$$ Cross-multiply to get $5x=2\cdot 12=24$, so $x=\boxed{{\displaystyle 24\mathrm{/}5}}$ scoops, which is $4.8$. Going from $5$ oats to $12$ is not a whole-number jump, so scaling is no help and cross-multiplication does the work.

Height over shadow is the same for both, since the sun hits the rod and the antenna at the same slant. $$\frac{6}{8}=\frac{h}{28}$$ Cross-multiply to get $8h=6\cdot 28=168$, so $h=\frac{168}{8}=\boxed{{\displaystyle 21}}$ feet. A rod you can reach gives you the ratio, and that ratio measures anything else standing in the same sunlight.

Bundles on top, dollars on the bottom in both ratios. $$\frac{6}{15}=\frac{10}{x}$$ Cross-multiply to get $6x=15\cdot 10=150$, so $x=\frac{150}{6}=\boxed{{\displaystyle 25}}$ dollars. The unit rate from 7.3 lands in the same place, since one bundle is $15÷6=2.5$ dollars and $2.5\times 10=25$.

Small height over small width equals big height over big width. $$\frac{5}{12}=\frac{20}{x}$$ Cross-multiply to get $5x=12\cdot 20=240$, so $x=\frac{240}{5}=\boxed{{\displaystyle 48}}$ feet wide. Both $\frac{5}{12}$ and $\frac{20}{48}$ reduce to $\frac{5}{12}$, so the copy really is the same shape.

Width over height on both sides. $$\frac{8}{5}=\frac{20}{x}$$ Cross-multiply to get $8x=5\cdot 20=100$, so $x=\frac{100}{8}=\boxed{{\displaystyle 25\mathrm{/}2}}$ inches, which is $12.5$. Keeping the same shape just means keeping width over height fixed, which is what the proportion says.

Juice on top, water on the bottom in both fractions. $$\frac{4}{10}=\frac{6}{x}$$ Cross-multiply to get $4x=10\cdot 6=60$, so $x=\frac{60}{4}=\boxed{{\displaystyle 15}}$ cups of water. The cross-products agree, $4\cdot 15=60$ and $10\cdot 6=60$.

Cross-multiply $\frac{3}{7}=\frac{x}{21}$ to get $7x=3\cdot 21=63$, so $x=\frac{63}{7}=\boxed{{\displaystyle 9}}$ ounces of teal pigment. Scaling checks it. The base tripled from $7$ to $21$, so the pigment triples too, $3\times 3=9$.

Keep the ratio of scoops of clay powder to cups of water the same. Let $x$ be the scoops needed for 40 cups, so $$\frac{6}{5}=\frac{x}{40}\mathrm{.}$$ Cross-multiply to clear the denominators, giving $5x=6\times 40$, so $5x=240$. Divide both sides by 5 to get $x=48$. The batch needs $\boxed{{\displaystyle 48}}$ scoops of clay powder.

Grams on top, rolls on the bottom in both fractions, so $\frac{7}{10}=\frac{x}{4}$. Cross-multiply to get $10x=7\times 4=28$, then divide by $10$ and reduce, $x=\frac{28}{10}=\boxed{{\displaystyle 14\mathrm{/}5}}$ grams, the same as $2.8$.

Both mixes keep the same ratio of rose oil to sandalwood oil, so write a proportion with rose on top and sandalwood on the bottom in each fraction. $$\frac{12}{21}=\frac{8}{x}$$ Cross-multiply to clear the denominators. $$12\cdot x=21\cdot 8$$ $$12x=168$$ Divide both sides by $12$. $$x=\frac{168}{12}=14$$ The tester vial holds $\boxed{{\displaystyle 14}}$ drops of sandalwood oil.

Put tulips on top and daffodils on the bottom in both fractions so the units match in matching positions. Let $x$ be the number of daffodils in the second bed. $$\frac{9}{12}=\frac{15}{x}$$ Cross-multiply to clear the denominators. $$9\cdot x=12\cdot 15$$ $$9x=180$$ Divide both sides by $9$. $$x=\frac{180}{9}=20$$ The second bed has $\boxed{{\displaystyle 20}}$ daffodils.

Because the rate is fixed, the ratio of dollars to tokens is the same for both swaps. Keep dollars on top and tokens on the bottom in both ratios. $$\frac{5}{65}=\frac{8}{x}$$ Cross-multiply to clear the denominators. $$5\times x=65\times 8$$ $$5x=520$$ Divide both sides by $5$. $$x=\frac{520}{5}=104$$ So $8$ dollars gives you $\boxed{{\displaystyle 104}}$ tokens.

The flour to sugar ratio stays the same, so set the recipe ratio equal to the big-batch ratio with flour on top and sugar on the bottom in both places. $$\frac{5}{2}=\frac{x}{12}$$ Cross-multiply to clear the denominators. $$2x=5\times 12$$ $$2x=60$$ Divide both sides by $2$. $$x=\frac{60}{2}=30$$ The giant batch needs $\boxed{{\displaystyle 30}}$ cups of flour.

Keep map centimeters on top and real kilometers on the bottom in both ratios, with $x$ standing for the real distance you are after. $$\frac{2}{45}=\frac{5}{x}$$ Cross-multiply to clear the denominators. $$2\cdot x=45\cdot 5$$ $$2x=225$$ Divide both sides by $2$. $$x=\frac{225}{2}=112.5$$ So the two stations are $\boxed{{\displaystyle 112.5}}$ kilometers apart in real life.

Both objects share the same height-to-shadow ratio, so put height on top and shadow on the bottom on each side. $$\frac{4}{5}=\frac{h}{50}$$ Cross-multiply to get $5h=4\times 50=200$, so $h=\frac{200}{5}=\boxed{{\displaystyle 40}}$ feet tall.

The speed is steady, so miles over hours is the same both times. $$\frac{60}{4}=\frac{x}{7}$$ Cross-multiply to get $4x=60\times 7=420$, so $x=\frac{420}{4}=\boxed{{\displaystyle 105}}$ miles.

Width on top and height on the bottom for both panels. $$\frac{6}{9}=\frac{22}{x}$$ Cross-multiply to get $6x=9\times 22=198$, so $x=\frac{198}{6}=\boxed{{\displaystyle 33}}$ inches tall. The check is that $22:33$ reduces to $2:3$, the same shape as $6:9$.

Starter on top, flour on the bottom in both ratios. $$\frac{3}{8}=\frac{x}{5}$$ Cross-multiply to get $8x=3\times 5=15$, so $x=\boxed{{\displaystyle 15\mathrm{/}8}}$ grams of starter, which is $1.875$.

Because the sails are similar, the bottom edge and the slanted edge keep a constant ratio. Put bottom edges on top and slanted edges on the bottom in both ratios. $$\frac{12}{8}=\frac{27}{x}$$ Cross-multiply to clear the denominators. $$12x=8\times 27$$ $$12x=216$$ Divide both sides by $12$. $$x=\frac{216}{12}=18$$ The slanted edge of the enlarged sail is $\boxed{{\displaystyle 18}}$ centimeters.