QuanticaPrealgebraOpen in the courseWriting \(\frac{a}{b} = \frac{c}{d}\) is a proportion. Two quantities are proportional when their ratio stays constant. The coming shortcut, cross-multiplication, is just clearing denominators from Chapter 6.
Problem
A fountain pours at 9 liters per minute. You timed it for 2 minutes and caught 18 liters, and a friend measured 45 liters in 5 minutes. Keeping that same flow, how many liters pour out in 8 minutes?
Show a hint
- Because the flow is proportional, the ratio of minutes to liters stays the same at every moment. Write two equal ratios with minutes on top and liters on the bottom in both, using \(x\) for the unknown liters. One side is the \(2\) minute reading, the other is the \(8\) minute pour. That gives you \(\frac{2}{18} = \frac{8}{x}\).
- To solve a proportion, cross-multiply. For \(\frac{a}{b} = \frac{c}{d}\) the cross products are equal, so \(a\cdot d = b\cdot c\). Apply that to \(\frac{2}{18} = \frac{8}{x}\) to clear the denominators, then solve the simple equation for \(x\).
Show the full solution
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 \(\boxed{72}\) liters. Quick sanity check, the flow is \(9\) liters per minute and \(9 \times 8 = 72\).
Problem
A camp kitchen uses 2 cups of rice for every 5 campers. Tonight 15 campers arrive. How many cups of dry rice are needed?
Show a hint
- Set two equal ratios side by side, keeping cups on top and campers on the bottom in both. That gives \(\frac{2}{5} = \frac{x}{15}\), where \(x\) is the cups you want. Now look at the camper side. What did \(5\) get multiplied by to become \(15\)?
- Notice that \(5 \times 3 = 15\), so the campers grew by a factor of \(3\). To keep the ratio equal, the cups have to grow by that same \(3\). If scaling feels shaky, cross-multiply instead, which turns \(\frac{2}{5} = \frac{x}{15}\) into \(5x = 2 \times 15\).
Show the full solution
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 \(2 \times 3 = \boxed{6}\) cups. Cross-multiplying gets there too, since \(5x = 2 \times 15 = 30\). Scaling is just faster when the multiplier is a clean whole number.
Problem
A paint mix calls for 8 jars of pigment in 14 liters of base. A big order needs 20 jars. Set up \(\frac{8}{14} = \frac{20}{x}\) and solve. How many liters of base are needed?
Show a hint
- Write a proportion with the same units in the same spots on both sides. Keep jars on top and liters on the bottom, so \(\frac{8 \text{ jars}}{14 \text{ liters}} = \frac{20 \text{ jars}}{x \text{ liters}}\).
- You can clear both denominators at once. Multiply each numerator by the other side's denominator, which gives \(8 \cdot x = 14 \cdot 20\), then solve the equation \(8x = 280\).
Show the full solution
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 \(x = \frac{280}{8} = \boxed{35}\) 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.
Problem
A 3D printer uses 5 grams of filament for 8 layers. How many grams does a 6-layer piece need? (There is no clean whole-number scale from 8 to 6, so use cross-multiplication.)
Show a hint
- Both ratios compare the same two things, grams and layers, so keep grams on top and layers on the bottom in both fractions. The test gives \(\frac{5}{8}\), and the unknown grams over the \(6\) layers gives \(\frac{x}{6}\). Set them equal as \(\frac{5}{8} = \frac{x}{6}\).
- You cannot scale \(8\) to \(6\) cleanly, so cross-multiply instead. Multiply each numerator by the other denominator to clear the fractions, which turns \(\frac{5}{8} = \frac{x}{6}\) into \(8x = 5 \times 6\). Then divide both sides by \(8\).
Show the full solution
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 \(x = \frac{30}{8} = \boxed{15/4}\) 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.
Problem
A map uses the scale 6 cm = 10 km. Two overlooks are 9 km apart on the trail. How many centimeters apart are they on the map?
Show a hint
- Write two ratios that must be equal, with real kilometers on top and map centimeters on the bottom in both. The first pair gives \(\frac{10}{6}\), and the second pair gives \(\frac{9}{x}\), so your proportion is \(\frac{10}{6} = \frac{9}{x}\) with the unknown sitting in a denominator.
- Cross-multiply just like always, top times opposite bottom equals top times opposite bottom, which gives \(10x = 6 \times 9\). A denominator unknown is no harder, since cross-multiplication treats both slots the same way. Work out the right side, then divide both sides by \(10\) to free \(x\).
Show the full solution
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 \(x = \frac{54}{10} = \boxed{27/5}\) centimeters, which is \(5.4\). The unknown sat in a denominator this time, and cross-multiplication treated it exactly like a numerator.
Problem
A harbor kiosk trades at one fixed rate: 7 dollars gets 91 shells. How many shells does the traveler get for 11 dollars?
Show a hint
- The rate is the same both times, so the ratio of dollars to shells is constant. Write one proportion with dollars on top and shells on the bottom in both fractions, \(\frac{7}{91} = \frac{11}{x}\), so the matching units line up.
- Now cross-multiply. Multiply \(7\) by \(x\) and set it equal to \(91 \times 11\), then divide to get \(x\) by itself.
Show the full solution
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 \(x = \frac{1001}{7} = \boxed{143}\) 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.
Problem
A 3-inch mast represents 90 real inches. The boom is 5 inches. A student writes \(\frac{3}{90} = \frac{x}{5}\) with mixed units. Set it up correctly. What is the real boom length in inches?
Show a hint
- Keep matching units in matching positions so each side reads the same rate. Put real inches on top and model inches on the bottom for both ratios. The mast gives \(\frac{90}{3}\), and the boom gives \(\frac{x}{5}\), so write \(\frac{90}{3} = \frac{x}{5}\).
- Clear the fractions by cross-multiplying. From \(\frac{90}{3} = \frac{x}{5}\) you get \(3x = 90 \times 5\). Work out the right side, then divide both sides by \(3\) to find \(x\).
Show the full solution
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 \(x = \frac{450}{3} = \boxed{150}\) 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.
Problem
A trail-snack recipe uses 5 scoops oats for every 2 scoops cashews. A big batch needs 12 scoops of oats. How many scoops of cashews keep the ratio right?
Show a hint
- Write the proportion with matching units in matching spots. Put oats on top and cashews on the bottom on both sides, \(\frac{5}{2} = \frac{12}{x}\). The recipe ratio sits on the left, and the big batch sits on the right with the unknown cashews as \(x\).
- Jumping from \(5\) oats to \(12\) oats is not a clean whole-number scale, so cross-multiply instead of guessing. Multiply each top by the opposite bottom to get \(5 \cdot x = 2 \cdot 12\), then solve for \(x\).
Show the full solution
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 \(x = \boxed{24/5}\) 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.
Problem
A park ranger measures a 6-foot rod casting an 8-foot shadow. Across the field, a fire-watch antenna casts a 28-foot shadow at the same instant. How tall is the antenna in feet?
Show a hint
- The rod and the antenna make the same shape, so height over shadow is the same for both. Write one ratio for the rod and set it equal to the same kind of ratio for the antenna, keeping height on top and shadow on the bottom in both. That gives \(\frac{6}{8} = \frac{h}{28}\).
- Now clear the fractions by cross-multiplying. Multiply the \(6\) by the \(28\) and set it equal to \(8\) times \(h\), then divide both sides to get \(h\) by itself.
Show the full solution
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 \(h = \frac{168}{8} = \boxed{21}\) feet. A rod you can reach gives you the ratio, and that ratio measures anything else standing in the same sunlight.
Problem
At a craft fair, 6 bundles cost 15 dollars at a fixed price. You want 10 bundles. How many dollars do 10 bundles cost?
Show a hint
- Two ratios that match describe the same fixed price, so set them equal. Keep the units lined up, bundles over dollars on each side, with the unknown cost as \(x\), which gives \(\frac{6}{15} = \frac{10}{x}\).
- Clear the fractions by cross-multiplying. Multiply each numerator by the other side's denominator, so \(6 \cdot x = 15 \cdot 10\), then divide to get \(x\) by itself.
Show the full solution
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 \(x = \frac{150}{6} = \boxed{25}\) dollars. The unit rate from 7.3 lands in the same place, since one bundle is \(15 \div 6 = 2.5\) dollars and \(2.5 \times 10 = 25\).
Problem
A mural is 5 ft tall and 12 ft wide. A scaled-up copy is 20 ft tall. How many feet wide is the big copy?
Show a hint
- Keep matching positions in matching positions. Put the two heights together and the two widths together, so the small mural's height over its width equals the big copy's height over its width. With \(x\) as the unknown width, that is \(\frac{5}{12} = \frac{20}{x}\).
- Now clear the fractions by cross-multiplying. The product across one diagonal equals the product across the other, so \(5 \cdot x = 12 \cdot 20\). Work out the right side, then divide both sides by \(5\) to get \(x\).
Show the full solution
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 \(x = \frac{240}{5} = \boxed{48}\) feet wide. Both \(\frac{5}{12}\) and \(\frac{20}{48}\) reduce to \(\frac{5}{12}\), so the copy really is the same shape.
Problem
A design is 8 inches wide and 5 inches tall. The printed banner will be 20 inches wide. What height keeps the same shape without stretching?
Show a hint
- Keep the same kind of measurement in the same spot. Put width over height for both pictures and set them equal, so \(\frac{8}{5} = \frac{20}{x}\), where \(x\) is the height you want.
- Cross-multiply to clear the fractions. From \(\frac{8}{5} = \frac{20}{x}\) you get \(8x = 5 \cdot 20\), then divide both sides by \(8\) to find \(x\).
Show the full solution
Width over height on both sides. $$\frac{8}{5} = \frac{20}{x}$$ Cross-multiply to get \(8x = 5 \cdot 20 = 100\), so \(x = \frac{100}{8} = \boxed{25/2}\) inches, which is \(12.5\). Keeping the same shape just means keeping width over height fixed, which is what the proportion says.
Practice these ideas
Practice
A fruit punch keeps juice to water in the ratio \(4:10\). A bigger batch uses 6 cups of juice. Set up \(\frac{4}{10} = \frac{6}{x}\) and solve. How many cups of water keep the same ratio?
Show the solution
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 \(x = \frac{60}{4} = \boxed{15}\) cups of water. The cross-products agree, \(4 \cdot 15 = 60\) and \(10 \cdot 6 = 60\).
Practice
A stage paint uses 3 oz pigment for every 7 oz base. A bigger batch uses 21 oz base. Set up \(\frac{3}{7} = \frac{x}{21}\). How many ounces of teal pigment are needed?
Show the solution
Cross-multiply \(\frac{3}{7} = \frac{x}{21}\) to get \(7x = 3 \cdot 21 = 63\), so \(x = \frac{63}{7} = \boxed{9}\) ounces of teal pigment. Scaling checks it. The base tripled from \(7\) to \(21\), so the pigment triples too, \(3 \times 3 = 9\).
Practice
A pottery glaze uses 6 scoops of clay powder for every 5 cups of water. A big batch uses 40 cups of water. How many scoops of clay powder are needed?
Show the solution
Keep the ratio of scoops of clay powder to cups of water the same. Let \(x\) be the scoops needed for 40 cups, so $$\frac{6}{5} = \frac{x}{40}.$$ 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 \(\boxed{48}\) scoops of clay powder.
Practice
A recipe uses 7 g of yeast for every 10 dinner rolls. The baker wants just 4 rolls. How many grams of yeast are needed?
Show the solution
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, \(x = \frac{28}{10} = \boxed{14/5}\) grams, the same as \(2.8\).
Practice
A perfume sample uses 12 drops rose oil for every 21 drops sandalwood. A tester vial uses only 8 drops of rose oil. How many drops of sandalwood are in the tester?
Show the solution
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 \(\boxed{14}\) drops of sandalwood oil.
Practice
Two flower beds keep the same ratio of 9 tulips to 12 daffodils. The second bed has 15 tulips. How many daffodils are in the second bed?
Show the solution
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 \(\boxed{20}\) daffodils.
Practice
A board game café trades 5 dollars for 65 tokens at a fixed rate. You hand over 8 dollars. How many tokens do you get?
Show the solution
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 \(\boxed{104}\) tokens.
Practice
A pancake batter uses 5 cups flour for every 2 cups sugar. A big batch calls for 12 cups of sugar. How many cups of flour does it need?
Show the solution
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 \(\boxed{30}\) cups of flour.
Practice
A hiking map: 2 cm stands for 45 km. Two ranger stations sit 5 cm apart on the map. How many kilometers apart are they in real life?
Show the solution
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 \(\boxed{112.5}\) kilometers apart in real life.
Practice
A garden gnome (4 ft tall) casts a 5-foot shadow. A nearby water tower casts a 50-foot shadow at the same moment. How tall is the water tower in feet?
Show the solution
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 \(h = \frac{200}{5} = \boxed{40}\) feet tall.
Practice
A van holds a steady speed, covering 60 miles in 4 hours. At that same speed, how many miles does it cover in 7 hours?
Show the solution
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 \(x = \frac{420}{4} = \boxed{105}\) miles.
Practice
A glassworker enlarges a 6-inch-wide by 9-inch-tall panel so the new width is 22 inches. Keeping the same shape, how tall is the enlarged panel in inches?
Show the solution
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 \(x = \frac{198}{6} = \boxed{33}\) inches tall. The check is that \(22:33\) reduces to \(2:3\), the same shape as \(6:9\).
Practice
A sourdough starter uses 3 g starter for every 8 g flour. Tomorrow the baker uses 5 g flour. How many grams of starter should go in?
Show the solution
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 \(x = \boxed{15/8}\) grams of starter, which is \(1.875\).
Practice
Two similar triangular sails. The small sail has a 12 cm bottom edge and 8 cm slanted edge. The enlarged sail has a 27 cm bottom edge. How long is its slanted edge in centimeters?
Show the solution
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 \(\boxed{18}\) centimeters.