QuanticaPrealgebraOpen in the courseA kombucha brewer keeps the same blend of tea, juice, and ginger every batch, writing it \(6:4:1\). Three quantities compared at once work exactly like a two-way ratio, the same idea with one more part.
Problem
A garden potting mix follows the ratio \(5:3:2\), compost to peat to sand. Using exactly one unscaled batch, how many scoops of potting mix is that in total?
Show a hint
- A multi-way ratio like \(5:3:2\) names the actual amount of each part in one batch. So this batch really is 5 scoops of compost, 3 scoops of peat, and 2 scoops of sand.
- Every scoop ends up in the same pile, so the total is just the parts of the ratio added together. Add \(5+3+2\).
Show the full solution
One unscaled batch is 5 scoops of compost, 3 scoops of peat, and 2 scoops of sand. Add them, \(5+3+2=10\), so the batch is \(\boxed{10}\) scoops. A ratio always carries a hidden total, the sum of its parts.
Problem
A marching band lines up in the ratio \(5:3:1\), brass to woodwinds to percussion. Reading the ratio in that order, how many parts belong to the woodwinds?
Show a hint
- The labels are read in the same order as the numbers. The first number goes with brass, the second number goes with woodwinds, the third number goes with percussion. So which number sits in the woodwinds spot?
- Line them up. Brass matches the \(5\), woodwinds matches the \(3\), and percussion matches the \(1\). The woodwinds part is the middle number.
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Read the labels in the same order as the numbers. Brass gets the \(5\), woodwinds gets the \(3\), and percussion gets the \(1\), so the woodwinds share is \(\boxed{3}\) parts. In a multi-way ratio the position carries the meaning, so a number tells you nothing until you know which label it sits under.
Problem
A florist arranges roses to lilies to ferns in the ratio \(3:2:4\). What is the smallest total number of stems that fits this comparison exactly?
Show a hint
- The three numbers \(3\), \(2\), \(4\) tell you the size of each share inside one full set. Add them up to see how many stems a single set of shares uses, then ask whether you could ever use fewer than that and still keep at least one of each.
- One full set is \(3\) roses, \(2\) lilies, and \(4\) ferns. Bigger windows just stack more identical sets, so every valid total is a multiple of that one set's size. The smallest total is the size of a single set, which is why \(8\) cannot work.
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One full set of the ratio is \(3\) roses, \(2\) lilies, and \(4\) ferns, which is \(3+2+4=9\) stems. Any bigger arrangement just stacks more copies of that set, so every valid total is a multiple of \(9\). The smallest is one set, \(\boxed{9}\) stems. Anything smaller cannot even hold one full set, so it cannot match the comparison.
Problem
A muralist mixes 18 oz blue, 12 oz white, and 30 oz black. Find the GCF of all three parts and write the simplified ratio of blue to white to black in the form a:b:c.
Show a hint
- With a three-way ratio you reduce the same way you would a two-way one, except the factor has to divide every part at once. Ask what is the largest whole number that goes evenly into \(18\), into \(12\), and into \(30\) all three.
- List the factors of each part and hunt for the biggest one they share. \(18\) splits as \(6\times 3\), \(12\) as \(6\times 2\), and \(30\) as \(6\times 5\), so the common factor is \(6\). Divide all three parts by \(6\).
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The largest number dividing \(18\), \(12\), and \(30\) is \(6\), since \(18=6\times 3\), \(12=6\times 2\), and \(30=6\times 5\). Divide every part by \(6\) to get \(\boxed{3:2:5}\). The factor has to divide all three parts, not just two of them, and \(3\), \(2\), \(5\) share nothing above \(1\), so this is fully reduced.
Problem
A board game ships action, coin, and gem tokens in the ratio \(6:10:15\). Write the ratio in lowest terms as a:b:c.
Show a hint
- To simplify a multi-way ratio you cannot just look at two parts at a time. You need a number that divides into every part at once. Check each part: does \(2\) go into \(6\), \(10\), and \(15\)? Does \(3\)? Does \(5\)? A factor only counts if it works for all three.
- List the factors of each part. \(6 = 2\times 3\), \(10 = 2\times 5\), \(15 = 3\times 5\). The pairs each share something, but no single number divides all three. So the greatest common factor of all three is \(1\), and dividing every part by \(1\) changes nothing.
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Factor each part. \(6=2\times 3\), \(10=2\times 5\), and \(15=3\times 5\). Each pair shares something, but no number above \(1\) divides all three, so the greatest common factor is \(1\) and nothing reduces. The ratio stays \(\boxed{6:10:15}\). A factor shared by two parts is not enough, it has to divide every part at once.
Problem
A baker\'s blend uses \(1\tfrac{1}{3}\) cups oat flour, \(2\) cups rice flour, and \(\tfrac{2}{3}\) cup almond meal. Clear the fractions and reduce. Write the ratio of oat to rice to almond as a:b:c.
Show a hint
- Every part is either a whole number or a number of thirds, so think about what single multiplier would turn all three into whole numbers at once. Multiplying every part by \(3\) does it, because \(1\tfrac{1}{3}=\tfrac{4}{3}\) and \(\tfrac{2}{3}\) both clear when tripled.
- After scaling by \(3\) you get \(4:6:2\). Now find the greatest common factor of all three parts and divide every part by it to reach lowest terms.
Show the full solution
Write every part in thirds. Oat flour is \(1\tfrac{1}{3}=\tfrac{4}{3}\), rice flour is \(2=\tfrac{6}{3}\), and almond meal is \(\tfrac{2}{3}\). Multiplying all three by \(3\) clears the denominators and gives \(4:6:2\). The greatest common factor of those is \(2\), so divide through to get \(\boxed{2:3:1}\). Scaling every part by the same number never changes the comparison, so you can clear fractions first and reduce after.
Problem
A terrarium is layered sand to gravel to soil in the ratio \(3:5:7\), filling 60 scoops total. How many scoops of gravel does the terrarium use?
Show a hint
- The ratio \(3:5:7\) means the 60 scoops are shared into \(3+5+7\) equal parts. Add those to find the total number of parts.
- There are \(15\) parts in all, so one part is \(60\div 15\) scoops. Gravel is the \(5\) in the ratio, so it gets \(5\) of those parts.
Show the full solution
The ratio \(3:5:7\) splits the jar into \(3+5+7=15\) equal parts, so one part is \(60\div 15=4\) scoops. Gravel takes \(5\) parts, giving \(5\times 4=20\). The terrarium uses \(\boxed{20}\) scoops. Checking the rest, \(12+20+28=60\), the full jar.
Problem
A choir keeps sopranos to altos to tenors in the ratio \(6:5:4\). Tonight 18 sopranos arrived. How many tenors are at the rehearsal?
Show a hint
- The sopranos hold the soprano part of the comparison, which is 6. So 18 sopranos fill up 6 equal parts. Divide to find what just one part is worth, then remember the tenors hold 4 of those same parts.
- One part is \(18 \div 6 = 3\) singers. Tenors take the \(4\) part, so multiply \(4 \times 3\).
Show the full solution
Sopranos hold the \(6\) in \(6:5:4\), and 18 sopranos fill those \(6\) parts, so one part is \(18 \div 6 = 3\) singers. Tenors hold \(4\) parts, so there are \(4 \times 3 = \boxed{12}\) tenors. You never need the choir total here. Once one part is known, every section scales from it.
Problem
An orchard plants apple to pear to plum in the ratio \(4:3:2\). This season there are 24 plum trees. How many trees are in the whole orchard?
Show a hint
- The plum trees are the \(2\) part of the ratio \(4:3:2\). So those \(24\) plum trees are spread evenly across \(2\) equal parts. Divide to find what just one part is worth.
- One part is \(24 \div 2 = 12\) trees. Now count how many parts the whole orchard has, \(4+3+2\), and multiply that many parts by \(12\).
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Plums are the \(2\) in \(4:3:2\), so the 24 plum trees fill \(2\) parts and one part is \(24 \div 2 = 12\) trees. The whole orchard is \(4+3+2 = 9\) parts, so it holds \(9 \times 12 = \boxed{108}\) trees. Checking, \(48 + 36 + 24 = 108\).
Sometimes you get two ratios that share a quantity, like \(A:B\) and \(B:C\), and you want all three in one ratio \(a:b:c\). The trouble is that the shared quantity usually shows up as a different number in each ratio, so they do not line up. The fix is to scale each ratio until both give the shared quantity the same number, and then the two snap together into one.
Problem
In a gear train, front to middle is \(4:3\) and middle to rear is \(2:5\). Rescale so the middle numbers match, then write front to rear as a:b in lowest terms.
Show a hint
- The middle gear is counted as \(3\) in the first ratio and \(2\) in the second. To stitch the two ratios together you need the middle number to be the same in both. Pick a common multiple of \(3\) and \(2\), and \(6\) is the smallest one.
- Scale \(4:3\) up by \(2\) to get \(8:6\), and scale \(2:5\) up by \(3\) to get \(6:15\). Now both say \(6\) for the middle, so line them up as front to middle to rear and look at just the front and rear numbers.
Show the full solution
The middle gear reads \(3\) in \(4:3\) and \(2\) in \(2:5\), so scale both until the middle is \(6\). Doubling the first gives \(8:6\), and tripling the second gives \(6:15\). The chain is \(8:6:15\), front to middle to rear, so front to rear is \(\boxed{8:15}\), already in lowest terms. The shared gear is the hinge. Once its number matches in both ratios, the two comparisons join into one.
Problem
Yellow to blue is \(3:2\) and blue to white is \(4:7\). Rescale so the blue numbers agree, then write the ratio of yellow to white as a:b in lowest terms.
Show a hint
- Blue is \(2\) in the first ratio and \(4\) in the second. To make them match, double everything in the first ratio so its blue becomes \(4\). What does \(3:2\) become when you double both parts?
- After doubling, yellow to blue is \(6:4\) and blue to white is \(4:7\). The blue \(4\) is now the same in both, so stack them into one chain \(6:4:7\). Read off just the yellow and the white parts, then reduce.
Show the full solution
Blue is \(2\) in \(3:2\) and \(4\) in \(4:7\), so double the first ratio to get \(6:4\). Both now call blue \(4\), and the chain is \(6:4:7\), yellow to blue to white. Yellow to white is \(\boxed{6:7}\), already in lowest terms. Doubling both parts leaves the comparison unchanged, which is what lets the two ratios line up.
Problem
In a sourdough recipe, flour to water is \(5:3\), water to starter is \(6:5\). Water appears as 3 in one and 6 in the other. Rescale and combine. How many grams of flour are in a 630-gram batch?
Show a hint
- The water amount is 3 in the first ratio and 6 in the second. Rescale one ratio so both read the same number for water. Doubling \(5:3\) gives \(10:6\), which now matches the \(6\) in the water to starter ratio, so you can stitch them into one flour to water to starter ratio.
- Once the three-way ratio is \(10:6:5\), add the parts to see how many equal shares the \(630\) grams splits into, find what one share weighs, then take the flour parts.
Show the full solution
Water reads \(3\) in \(5:3\) and \(6\) in \(6:5\), so double the first ratio to \(10:6\). The chain is \(10:6:5\), flour to water to starter, which is \(10+6+5=21\) parts. One part is \(\tfrac{630}{21}=30\) grams, and flour is \(10\) parts, so the batch has \(10\times 30=\boxed{300}\) grams of flour. Checking, \(300+180+150=630\).
Practice these ideas
Practice
A birdseed recipe follows the ratio \(7:4:2\), millet to sunflower to cracked corn. Using exactly one unscaled batch, how many parts of birdseed are there in total?
Show the solution
One unscaled batch is \(7\) parts millet, \(4\) parts sunflower, and \(2\) parts cracked corn. Add them, \(7 + 4 + 2 = \boxed{13}\) parts. Unscaled means you use the ratio numbers exactly as written, with no multiplier.
Practice
A printmaker logs a blend as \(20:50:35\), cyan to magenta to yellow. Divide every part by the GCF of all three. Write the simplified ratio in the form a:b:c.
Show the solution
The factors of \(20\) are \(1, 2, 4, 5, 10, 20\), of \(50\) are \(1, 2, 5, 10, 25, 50\), and of \(35\) are \(1, 5, 7, 35\). The largest on all three lists is \(5\), so divide every part by \(5\) to get \(\boxed{4:10:7}\). Since \(7\) is prime and divides neither \(4\) nor \(10\), there is nothing left to divide out.
Practice
A trail mix uses oats, raisins, and seeds in the ratio \(10:9:25\). Reduce to lowest terms, if nothing divides all three, the original is the answer. Write your answer as a:b:c.
Show the solution
Start with \(9\), the most restrictive part. It is odd, so \(2\) is out, and its only other factors are \(3\) and \(9\), neither of which divides \(10\). So the greatest common factor of \(10\), \(9\), and \(25\) is \(1\) and nothing reduces. The ratio stays \(\boxed{10:9:25}\).
Practice
A flower farmer counts 42 marigolds, 14 basil plants, and 28 thyme plants. Simplify the ratio \(42:14:28\) to lowest terms. Write as a:b:c.
Show the solution
The factors of \(14\) are \(1, 2, 7, 14\), and \(14\) itself divides the others, since \(42=14\times 3\) and \(28=14\times 2\). Divide every part by \(14\), keeping the order marigolds, basil, thyme, and you get \(\boxed{3:1:2}\). Checking the smallest part first is a shortcut, since the greatest common factor can never be bigger than it.
Practice
A smoothie uses \(\tfrac{1}{3}\) cup mango, \(\tfrac{1}{6}\) cup spinach, and \(\tfrac{1}{2}\) cup yogurt. Clear fractions with a common denominator and reduce. Write the ratio as a:b:c.
Show the solution
The denominators are 3, 6, and 2, and 6 is the smallest number all three divide into. Multiply every part by 6, giving \(\tfrac{1}{3}\times 6 = 2\), \(\tfrac{1}{6}\times 6 = 1\), and \(\tfrac{1}{2}\times 6 = 3\). That is \(\boxed{2:1:3}\), and 2, 1, and 3 share no factor above 1. Multiplying every part by the same number keeps the comparison the same, so clearing fractions costs you nothing.
Practice
A candle recipe uses \(2\tfrac{1}{2}\) oz soy wax, \(1\tfrac{1}{4}\) oz beeswax, and \(\tfrac{5}{8}\) oz fragrance oil. Clear fractions, then reduce. Write the ratio as a:b:c in lowest terms.
Show the solution
Every part is a number of eighths, so multiply all three by 8. Soy wax gives \(2\tfrac{1}{2}\times 8 = 20\), beeswax gives \(1\tfrac{1}{4}\times 8 = 10\), and fragrance oil gives \(\tfrac{5}{8}\times 8 = 5\). The greatest common factor of \(20\), \(10\), and \(5\) is \(5\), so divide through to get \(\boxed{4:2:1}\). Clear the fractions first, then reduce, and every step stays in whole numbers.
Practice
A mosaic border of 96 tiles uses gold, copper, and slate in the ratio \(1:2:5\). How many copper tiles are in the border?
Show the solution
The ratio \(1:2:5\) splits the border into \(1+2+5=8\) equal parts, so one part is \(96\div 8=12\) tiles. Copper is the \(2\), so it takes \(2\times 12=\boxed{24}\) tiles. Checking, \(12+24+60=96\), the whole border.
Practice
A 75-bead bracelet uses jade, amber, onyx, and pearl in the ratio \(2:4:3:6\). How many pearl beads are on the bracelet?
Show the solution
Add the parts, \(2+4+3+6 = 15\), so one share is \(75 \div 15 = 5\) beads. Pearl holds \(6\) shares, giving \(6 \times 5 = \boxed{30}\) pearl beads. Four shares work exactly like three. The only change is that you add one more number at the start.
Practice
A robotics club has builders, coders, and designers in the ratio \(8:5:3\). Tonight 35 coders showed up. How many builders does the club have?
Show the solution
Coders hold \(5\) parts, and 35 coders fill them, so one part is \(35 \div 5 = 7\) members. Builders hold \(8\) parts, so the club has \(8 \times 7 = \boxed{56}\) builders. Once one part is known, every group scales from it, and you never need the club total.
Practice
A produce crate is packed plums to apples to pears in the ratio \(2:9:4\). The crate holds 28 pears. How many apples are in the crate?
Show the solution
In the comparison \(2:9:4\) the pears are the third number, so they are worth 4 parts. The crate holds 28 pears, and those 28 pears fill 4 equal parts, so one part is \(28 \div 4 = 7\) pieces of fruit. Now look at the apples. They are the middle number, worth 9 parts. Each part is 7, so the apples come to \(9 \times 7 = 63\). That gives \(\boxed{63}\).
Practice
A relay team splits training as run to bike to swim to row in the ratio \(6:5:4:3\). The bike leg is 20 km. How many kilometers is the whole training distance?
Show the solution
The bike leg is the \(5\) in \(6:5:4:3\), and 20 km fills those \(5\) parts, so one part is \(20 \div 5 = 4\) km. The whole distance covers \(6+5+4+3 = 18\) parts, so it is \(18 \times 4 = \boxed{72}\) kilometers.
Practice
A spice jar blends cumin, paprika, and chili in the ratio \(7:2:1\). The jar has 18 grams of paprika. How many grams of spice are in the jar altogether?
Show the solution
Paprika is the \(2\) in \(7:2:1\), and 18 grams fills those 2 parts, so one part is \(18\div 2 = 9\) grams. The whole jar is \(7+2+1 = 10\) parts, so it holds \(10\times 9 = \boxed{90}\) grams. Checking, \(63+18+9 = 90\).
Practice
A clockwork toy has three gears. Gear 1 to gear 2 turns at \(5:6\). Gear 2 to gear 3 turns at \(4:3\). Rescale so the gear-2 numbers match, then write gear 1 to gear 3 as a:b in lowest terms.
Show the solution
Gear 2 reads \(6\) in \(5:6\) and \(4\) in \(4:3\), and both divide \(12\). Scale \(5:6\) by \(2\) to get \(10:12\), and scale \(4:3\) by \(3\) to get \(12:9\). The chain is \(10:12:9\), so gear 1 to gear 3 is \(\boxed{10:9}\), already in lowest terms.
Practice
Nitrogen to phosphorus is \(2:3\) and phosphorus to potash is \(9:8\). Rescale the first ratio so its phosphorus matches 9, combine, then write nitrogen to potash as a:b in lowest terms.
Show the solution
Phosphorus reads \(3\) in \(2:3\) and \(9\) in \(9:8\), so multiply the first ratio by \(3\) to get \(6:9\). The chain is \(6:9:8\), nitrogen to phosphorus to potash, so nitrogen to potash is \(6:8\), which divides by \(2\) down to \(\boxed{3:4}\). Reduce at the end, after you have pulled out the two parts you actually want.