QuanticaPrealgebraOpen in the courseA smoothie owner keeps the flavor steady by using the same comparison every time, three mango chunks for every two yogurt scoops. That balance between two quantities is a ratio, and it holds whether she is making one cup or filling a pitcher.
Problem
Each scoop drops 5 almonds and 3 dried cherries into a bag. A customer orders 4 scoops. How many dried cherries land in the bag?
Show a hint
- Every scoop is identical, and the cherries from one scoop do not depend on the almonds at all. Just count the cherries one scoop brings, then ask how many scoops there are.
Show the full solution
One scoop drops \(3\) cherries, and there are \(4\) scoops, so \(4 \times 3 = 12\). The bag holds \(\boxed{12}\) dried cherries. The almond count never enters the work, since each side of a pairing carries its own steady count.
Problem
At a kite festival the rule is 3 yellow kites for every 5 red ones. One cluster has 9 yellow kites. How many red kites are in the cluster?
Show a hint
- The phrase for every 3 yellow there are 5 red is a ratio of \(3:5\). The yellow part grew from \(3\) to \(9\). How many times bigger is that? That same multiplier has to act on the red part too, or the comparison would change.
- To go from \(3\) yellow to \(9\) yellow you multiply by \(3\), since \(3\times 3 = 9\). Apply that same \(\times 3\) to the \(5\) reds.
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To go from \(3\) yellow to \(9\) yellow you multiply by \(3\), so the reds triple too, giving \(5 \times 3 = 15\). The cluster has \(\boxed{15}\) red kites. Scaling both parts by the same number leaves the comparison alone, which is why \(9:15\) reduces right back to \(3:5\).
Problem
A garden plants basil and mint in the ratio \(4:5\). One full repeat of the pair uses \(4+5=9\) plants, so every valid total is a multiple of \(9\). What is the only possible total between 30 and 40?
Show a hint
- One full repeat of the comparison is \(4\) basil and \(5\) mint, so each bundle is \(4+5=9\) plants. The real total has to be made of whole bundles, so it must be a multiple of \(9\).
- List the multiples of \(9\) and find the one that lands between \(30\) and \(40\). You have \(9, 18, 27, 36, 45\). Which single one fits?
Show the full solution
Each full repeat of \(4:5\) is \(4+5=9\) plants, and the multiples of \(9\) are \(9, 18, 27, 36, 45\). Only \(36\) sits between \(30\) and \(40\), so the total is \(\boxed{36}\). That is \(4\) bundles, \(16\) basil and \(20\) mint, which still compares as \(4:5\).
Problem
A board game ships tokens in the ratio \(18:24\). Reduce it to lowest terms. Write your answer as a:b.
Show a hint
- List the factors that \(18\) and \(24\) share. Both are even, so \(2\) works, and both are divisible by \(3\), so \(6\) divides each. Is there anything bigger than \(6\) that divides both?
- The greatest common factor is \(6\). Divide each part by \(6\): \(18\div 6\) gives the new first number and \(24\div 6\) gives the new second number. Then check the two results share no common factor larger than \(1\).
Show the full solution
The greatest common factor of \(18\) and \(24\) is \(6\). Divide each part by it, \(18\div 6 = 3\) and \(24\div 6 = 4\), so the simplest form is \(\boxed{3:4}\). Since \(3\) and \(4\) share no factor larger than \(1\), nothing can shrink it further. That is exactly what simplest form means.
Problem
A school library returned books on time to late in the ratio \(45:75\). Reduce to simplest form. Write as a:b.
Show a hint
- Simplifying a ratio works the same way as simplifying a fraction. You want the greatest common factor of \(45\) and \(75\), the biggest number that divides both cleanly. Both end in a \(5\) or a \(0\), so \(5\) divides each, but check whether something larger does too.
- Both numbers are multiples of \(15\), since \(45 = 3 \times 15\) and \(75 = 5 \times 15\). Divide each part of the ratio by \(15\), then check that the two results have no common factor left bigger than \(1\).
Show the full solution
Break each part apart, \(45 = 3 \times 3 \times 5\) and \(75 = 3 \times 5 \times 5\). They share one \(3\) and one \(5\), so the greatest common factor is \(15\). Dividing gives \(45 \div 15 = 3\) and \(75 \div 15 = 5\), so the ratio is \(\boxed{3:5}\). Stopping at the easy \(5\) would leave \(9:15\), which still reduces, so always take the greatest common factor.
Problem
A cooling fluid should compare water to coolant as \(2:3\). Six ratios were recorded: \(4:6\), \(8:10\), \(6:9\), \(9:12\), \(10:15\), \(14:21\). How many are equivalent to \(2:3\)?
Show a hint
- Test each ratio by dividing both numbers by their greatest common factor, the same move you used to reduce a fraction to lowest terms. Then see whether what is left is \(2:3\).
- Try \(4:6\). Both numbers divide by \(2\), giving \(2:3\), so that one matches. Now do the same simplifying step to \(8:10\) and notice it becomes \(4:5\), which is not \(2:3\). Work through all six and keep a running count of the ones that land on \(2:3\).
Show the full solution
Reduce each one and compare with \(2:3\). \(4:6\) divides by \(2\) to give \(2:3\), a match. \(8:10\) divides by \(2\) to give \(4:5\), no. \(6:9\) divides by \(3\) to give \(2:3\), a match. \(9:12\) divides by \(3\) to give \(3:4\), no. \(10:15\) divides by \(5\) to give \(2:3\), a match. \(14:21\) divides by \(7\) to give \(2:3\), a match. That is \(\boxed{4}\) matching ratios. Two ratios are equivalent exactly when they reduce to the same simplest form.
Problem
A trail map records two legs as \(0.75\,\text{km}\) and \(1.25\,\text{km}\). Write \(0.75:1.25\) as a whole-number ratio in lowest terms, in the form a:b.
Show a hint
- The decimals reach the hundredths place, so multiply both parts by \(100\). That turns \(0.75\) into \(75\) and \(1.25\) into \(125\), giving the whole-number ratio \(75:125\). Stretching both legs by the same factor of \(100\) keeps the comparison identical, the same way \(\tfrac{75}{125}\) and \(\tfrac{0.75}{1.25}\) are equal fractions.
- Now reduce \(75:125\) the way you reduced fractions in Chapter 4. The largest number that divides both \(75\) and \(125\) is \(25\). Divide each part by \(25\) and see what whole-number ratio is left.
Show the full solution
Multiply both parts by \(100\) to clear the decimals. $$0.75:1.25 = 75:125$$ The greatest common factor of \(75\) and \(125\) is \(25\), so divide each part by it. $$75:125 = 3:5$$ The two legs compare as \(\boxed{3:5}\). Scaling both parts by the same number never changes the comparison, so clearing decimals first is always safe.
Problem
A bakery uses \(\tfrac{2}{3}\) cup almond flour and \(\tfrac{4}{9}\) cup tapioca starch. Write the ratio in lowest terms as a:b.
Show a hint
- Both fractions sit over a denominator that divides into \(9\). Multiply each part of the ratio by \(9\). Multiplying both parts by the same number gives an equivalent ratio, the same way \(\tfrac{6}{4}\) and \(\tfrac{3}{2}\) name the same fraction.
- After scaling you get \(6:4\). Now finish like you would simplify a fraction. Divide both parts by the largest number that goes into each.
Show the full solution
Multiply both parts by \(9\) to clear the fractions. $$\tfrac{2}{3}\times 9 = 6 \qquad \tfrac{4}{9}\times 9 = 4$$ That makes the ratio \(6:4\), and dividing both parts by \(2\) leaves \(\boxed{3:2}\). Clear the fractions first, then reduce. Trying to reduce \(\tfrac{2}{3}:\tfrac{4}{9}\) as it stands is much harder to see.
So far a ratio has fixed only the relative sizes, never the totals. To get real amounts, read a ratio as a recipe built from equal parts, \(a\) parts of one thing and \(b\) parts of the other, with every part the same size. Once you know what a single part is worth, every quantity in the problem follows by multiplication.
Problem
A bag holds 56 pieces of dried fruit mixed as raisins to pumpkin seeds in the ratio \(5:3\). How many pumpkin seeds are in the bag?
Show a hint
- The ratio \(5:3\) means the bag is built from \(5\) equal parts of raisins and \(3\) equal parts of seeds. Add those up to see how many equal parts make the whole bag.
- There are \(5+3=8\) equal parts in all, and they must total \(56\) pieces. So one part is \(56\div 8\). The seeds are the \(3\) parts, so multiply that one part value by \(3\).
Show the full solution
The ratio \(5:3\) makes \(5+3=8\) equal parts, so one part is \(56 \div 8 = 7\) pieces. The seeds are \(3\) parts, which is \(3 \times 7 = 21\). The bag holds \(\boxed{21}\) pumpkin seeds. Check it with the raisins, \(5 \times 7 = 35\), and \(35+21=56\). Find what one part is worth and every amount follows.
Problem
A box of 84 stickers is split between two friends in the ratio \(5:7\), bigger share to the friend who paid more. How many stickers are in the smaller share?
Show a hint
- Add the two numbers in the ratio to find how many equal parts the box is divided into. Then figure out how many stickers sit in a single part.
- One part is \(84 \div 12 = 7\) stickers. The smaller share is the \(5\)-part side, so it holds five of those parts.
Show the full solution
The ratio \(5:7\) splits the box into \(5+7=12\) equal parts, so one part is \(84 \div 12 = 7\) stickers. The smaller share is the \(5\)-part side, which is \(5 \times 7 = \boxed{35}\) stickers. The larger share is \(7 \times 7 = 49\), and \(35+49=84\), the whole box.
Problem
A lemonade recipe uses lemon juice to water in the ratio \(2:9\). One batch uses 6 ounces of lemon juice. How many ounces does the whole batch hold?
Show a hint
- The lemon side of the ratio is \(2\) parts, and those \(2\) parts landed on six ounces. What number do you multiply \(2\) by to get \(6\)? That single number is the multiplier for every part of the ratio.
- One full repeat of the recipe is \(2 + 9 = 11\) parts. Multiply those \(11\) parts by the multiplier you found, and you have the whole batch in ounces.
Show the full solution
The lemon juice is \(2\) parts and comes to six ounces, so one part is \(6 \div 2 = 3\) ounces. The whole batch is \(2+9=11\) parts, which is \(11 \times 3 = 33\) ounces. The batch holds \(\boxed{33}\) ounces. Knowing just one quantity is enough, because it pins down the value of a part and every other amount follows.
Problem
An aquarium opens with 40 fish, blue to green in \(3:2\). The keeper adds only green fish until the ratio reads \(6:5\). How many green fish must the keeper add?
Show a hint
- The blue total never moves, it stays at \(24\). So your job is to find how much green pairs with \(24\) blue when the comparison is \(6:5\), then compare that to the \(16\) green already swimming there.
- In the new comparison \(6:5\), the blue is \(6\) parts and equals \(24\) fish, so one part is \(24\div 6=4\). The green is \(5\) parts. Find the green count, then subtract the \(16\) you started with.
Show the full solution
The opening ratio \(3:2\) makes \(5\) parts, so \(40 \div 5 = 8\) fish per part, giving \(24\) blue and \(16\) green. Only green is added, so blue stays at \(24\). In the target \(6:5\), blue is the \(6\) side, so one part is \(24 \div 6 = 4\) and green must reach \(5 \times 4 = 20\). That means adding \(20-16 = \boxed{4}\) green fish. The quantity that never changes is what sets the new part size, so anchor everything to the blue.
Practice these ideas
Practice
A vendor packs gift bags with 4 glitter stickers and 6 mini erasers each. She fills 5 bags. How many mini erasers does she use in total?
Show the solution
Each bag holds six mini erasers and she fills five bags, so \(5 \times 6 = 30\). She uses \(\boxed{30}\) mini erasers. The sticker count does not matter here, since the question asks about only one side of the pairing.
Practice
A swim club keeps 2 coaches for every 9 swimmers. This morning 36 swimmers arrived. How many coaches are on deck?
Show the solution
The swimmers went from \(9\) to \(36\), and \(36 \div 9 = 4\), so the coaches scale by \(4\) as well, giving \(2 \times 4 = 8\). There are \(\boxed{8}\) coaches on deck. Both parts of a ratio move by the same multiplier, which is why \(8:36\) reduces right back to \(2:9\).
Practice
A flower stand bundles tulips and daffodils in the ratio \(5:6\). One full repeat is \(5+6=11\) stems, so every valid total is a multiple of 11. What is the only total between 70 and 80?
Show the solution
Every valid total is a multiple of \(11\). Near the range, \(11 \times 6 = 66\) is too small and \(11 \times 8 = 88\) is too big, so the total is \(11 \times 7 = \boxed{77}\). That is \(7\) repeats, meaning \(35\) tulips and \(42\) daffodils.
Practice
A fabric border repeats wide stripes to narrow stripes in the ratio \(9:12\). Reduce first, then scale up so the wide count reaches 24. How many narrow stripes go with 24 wide ones?
Show the solution
Reduce \(9:12\) by dividing both parts by \(3\), which gives \(3:4\). To get from \(3\) wide stripes to \(24\) you multiply by \(8\), so the narrow part becomes \(4 \times 8 = 32\). That is \(\boxed{32}\) narrow stripes. You could also scale the original \(9:12\) by \(\tfrac{24}{9} = \tfrac{8}{3}\), since \(12 \times \tfrac{8}{3} = 32\) too. Reducing first just keeps the numbers friendlier.
Practice
A juice bar restocks mango and guava pods in the ratio \(24:36\). Simplify to lowest terms. Write your answer as a:b.
Show the solution
The greatest common factor of \(24\) and \(36\) is \(12\). Dividing gives \(24 \div 12 = 2\) and \(36 \div 12 = 3\), so the simplified ratio is \(\boxed{2:3}\). The only factor \(2\) and \(3\) share is \(1\), so it is fully reduced.
Practice
A bakery sells sourdough to rye in the ratio \(42:54\). Simplify to lowest terms. Write your answer as a:b.
Show the solution
Both \(42\) and \(54\) are divisible by \(2\) and by \(3\), so their greatest common factor is \(6\). Dividing gives \(42 \div 6 = 7\) and \(54 \div 6 = 9\), so the ratio is \(\boxed{7:9}\). Now \(7\) and \(9\) share no factor larger than \(1\), so this is lowest terms.
Practice
A smoothie uses \(\tfrac{5}{6}\) cup berries and \(\tfrac{5}{9}\) cup banana. Clear the fractions and reduce to lowest terms. Write the ratio of berries to banana as a:b.
Show the solution
Both \(6\) and \(9\) divide evenly into \(18\), so multiply each part by \(18\). That gives \(\tfrac{5}{6}\times 18 = 15\) and \(\tfrac{5}{9}\times 18 = 10\), so the ratio is \(15:10\). Dividing both by \(5\) leaves \(\boxed{3:2}\). Any common multiple of the denominators will clear the fractions, and then you reduce what is left.
Practice
A window glaze uses \(\tfrac{1}{2}\) cup fast-drying liquid and \(\tfrac{3}{8}\) cup slow-drying liquid. Multiply both by 8, then reduce. Write the ratio in the form a:b.
Show the solution
Multiply both parts by \(8\). That gives \(\tfrac{1}{2} \times 8 = 4\) and \(\tfrac{3}{8} \times 8 = 3\), so the ratio is \(\boxed{4:3}\). The factors of \(4\) are \(1, 2, 4\) and the factors of \(3\) are \(1, 3\), so nothing larger than \(1\) divides both and it is already in lowest terms.
Practice
Two garden rows measure \(0.6\) m and \(1.5\) m. Rewrite \(0.6:1.5\) as a whole-number ratio in simplest form. Write your answer as a:b.
Show the solution
Multiply both parts of \(0.6:1.5\) by \(10\) to clear the decimals, giving \(6:15\). Both divide by \(3\), so \(6 \div 3 = 2\) and \(15 \div 3 = 5\), which is \(\boxed{2:5}\). Clearing decimals this way is safe because scaling both parts by the same number leaves the comparison alone.
Practice
A soup base uses \(2.4\) cups broth and \(0.9\) cups cream. Multiply by 10 to clear decimals, then reduce. Write the ratio as a:b.
Show the solution
Multiply both parts of \(2.4:0.9\) by \(10\) to clear the decimals, giving \(24:9\). The greatest common factor is \(3\), so \(24 \div 3 = 8\) and \(9 \div 3 = 3\), leaving \(\boxed{8:3}\). That is eight parts broth for every three parts cream.
Practice
A bracelet has 72 beads, copper to jade in the ratio \(4:5\). How many jade beads are on the bracelet?
Show the solution
The ratio \(4:5\) cuts the bracelet into \(4+5=9\) equal parts, so one part is \(72 \div 9 = 8\) beads. Jade is the five-part color, which is \(5 \times 8 = \boxed{40}\) beads. Copper gets \(4 \times 8 = 32\), and \(32+40=72\), the whole bracelet.
Practice
A 90-minute playlist splits jazz to folk in the ratio \(7:3\). The jazz piece is the larger share. How many minutes long is the jazz piece?
Show the solution
The ratio \(7:3\) splits the playlist into \(7+3=10\) equal parts, so one part is \(90 \div 10 = 9\) minutes. Jazz is seven parts, which is \(7 \times 9 = \boxed{63}\) minutes. Folk gets \(3 \times 9 = 27\) minutes, and \(63+27=90\), the full playlist.
Practice
A potting soil blend mixes compost to sand in the ratio \(3:8\). One batch uses 15 kg of compost, where one part equals one kilogram. What is the total mass of the batch?
Show the solution
The compost is the \(3\) part and it weighs \(15\) kilograms, so the multiplier is \(15 \div 3 = 5\). One full repeat of the mix is \(3+8=11\) kilograms, so the batch is \(11 \times 5 = \boxed{55}\) kilograms. Check by parts, \(3 \times 5 = 15\) kilograms of compost and \(8 \times 5 = 40\) of sand, and \(15+40=55\).
Practice
A reef tank holds 35 fish in the ratio \(3:4\) blue to gold. The keeper adds only blue fish until the count is \(1:1\). How many blue fish must she add?
Show the solution
The ratio \(3:4\) makes \(3+4=7\) equal parts, so one part is \(35 \div 7 = 5\) fish, giving \(15\) blue and \(20\) gold. Only blue is added, so gold stays locked at \(20\), and \(1:1\) means blue has to climb to \(20\) as well. It started at \(15\), so she adds \(20-15 = \boxed{5}\) blue fish. Anchor to the count that does not move.