Lesson 24 of 36 · Quantitative reasoning

Fractions, ratios and proportions

Handle fractions by hand, split totals in a ratio, link two ratios through a shared term, and tell direct from inverse proportion.

About 20 minutes

In this lesson

  • Add, multiply, divide and compare fractions quickly without a calculator.
  • Split a total in a given ratio and solve ratio problems with a single multiplier.
  • Combine two ratios that share a term, and recognise direct and inverse proportion.

The idea

Fractions by hand

Without a calculator, fractions are usually faster than decimals, because they let you cancel before you multiply.

  • Add or subtract over a common denominator: 23+34=812+912=1712\frac{2}{3} + \frac{3}{4} = \frac{8}{12} + \frac{9}{12} = \frac{17}{12}.
  • Multiply tops and bottoms, cancelling first: 1415×2528=13×52=56\frac{14}{15} \times \frac{25}{28} = \frac{1}{3} \times \frac{5}{2} = \frac{5}{6}, after cancelling 14 with 28 and 25 with 15.
  • Divide by multiplying by the reciprocal: 34÷910=34×109=3036=56\frac{3}{4} \div \frac{9}{10} = \frac{3}{4} \times \frac{10}{9} = \frac{30}{36} = \frac{5}{6}.

To compare two fractions, cross-multiply: ab>cd\frac{a}{b} > \frac{c}{d} (with positive denominators) exactly when ad>bcad > bc. Is 712\frac{7}{12} or 1119\frac{11}{19} larger? 7×19=1337 \times 19 = 133 and 11×12=13211 \times 12 = 132, so 712\frac{7}{12} is larger, by a hair. Benchmarks help too: a fraction whose top is more than half its bottom is above 12\frac{1}{2}.

A few decimal equivalents are worth knowing by heart: 18=0.125\frac{1}{8} = 0.125, 16≈0.167\frac{1}{6} \approx 0.167, 15=0.2\frac{1}{5} = 0.2, 14=0.25\frac{1}{4} = 0.25, 13≈0.333\frac{1}{3} \approx 0.333, 38=0.375\frac{3}{8} = 0.375.

A ratio is a recipe

A ratio 3:53 : 5 of orange to apple juice says nothing about amounts; it says that for every 3 parts of orange there are 5 parts of apple. The whole has 3+5=83 + 5 = 8 parts, so orange is 38\frac{3}{8} of the blend and apple 58\frac{5}{8}. Keep the two readings apart:

Reading From 3:53 : 5
part to part orange is 35\frac{3}{5} of the apple
part to whole orange is 38\frac{3}{8} of the blend

The multiplier method

Most ratio problems become one equation if you call the size of one part kk. With orange to apple 3:53 : 5, the amounts are 3k3k and 5k5k. Whatever the question tells you (the total, the difference, one of the amounts, or what happens after a change) becomes an equation in kk. For example, if the total is 24 litres, 8k=248k = 24, so k=3k = 3 and there are 9 litres of orange.

The method shines when a ratio changes. If boys and girls are in the ratio 5:35 : 3 and 6 girls join to make it 5:45 : 4, write boys =5k= 5k, girls =3k= 3k, and translate the new ratio: 5k3k+6=54\frac{5k}{3k + 6} = \frac{5}{4}. Cross-multiplying gives 20k=15k+3020k = 15k + 30, so k=6k = 6: 30 boys, and 18 girls who become 24. Check: 30:24=5:430 : 24 = 5 : 4.

Linking two ratios

Given a:b=3:4a : b = 3 : 4 and b:c=6:5b : c = 6 : 5, you cannot read off a:ca : c directly, because the two bb values are on different scales. Rescale both ratios so that bb has the same number: the LCM of 4 and 6 is 12, so a:b=9:12a : b = 9 : 12 and b:c=12:10b : c = 12 : 10. Now a:b:c=9:12:10a : b : c = 9 : 12 : 10, and a:c=9:10a : c = 9 : 10. Equivalently, ac=ab×bc=34×65=1820=910\frac{a}{c} = \frac{a}{b} \times \frac{b}{c} = \frac{3}{4} \times \frac{6}{5} = \frac{18}{20} = \frac{9}{10}.

Direct and inverse proportion

Two quantities are directly proportional when their ratio is constant, y=kxy = kx: double one and the other doubles. Price and quantity at a fixed unit price behave this way. They are inversely proportional when their product is constant, xy=kxy = k: double one and the other halves. Speed and time over a fixed distance behave this way, as do workers and days for a fixed job.

Before you set up a proportion, ask which kind it is: "more of this means more of that" or "more of this means less of that". Then use a ratio for the first and a product for the second. If 4 identical printers finish a job in 9 hours, then the job is 4×9=364 \times 9 = 36 printer-hours, so 6 printers take 366=6\frac{36}{6} = 6 hours, not 64×9=13.5\frac{6}{4} \times 9 = 13.5.

Additive changes break ratios

Adding the same amount to both terms of a ratio changes it: 1:21 : 2 becomes 3:43 : 4 after adding 2 to each. Only multiplying both terms by the same number leaves a ratio unchanged. This is why "the ratio after a change" questions need the multiplier method rather than guesswork.

Worked examples

Example 1: Splitting a profit three ways

A profit of €7,200 is shared among three partners in the ratio 2:3:42 : 3 : 4. How much does each receive?

  1. Count the parts: 2+3+4=92 + 3 + 4 = 9.

  2. One part is worth 72009=800\frac{7200}{9} = 800 euros.

  3. Multiply: 2×800=16002 \times 800 = 1600, 3×800=24003 \times 800 = 2400 and 4×800=32004 \times 800 = 3200.

  4. Check the shares add back to the total: 1600+2400+3200=72001600 + 2400 + 3200 = 7200.

Answer

The partners receive €1,600, €2,400 and €3,200. Dividing the total by 3, the number of partners, ignores the ratio.

In a club, the ratio of juniors to adults is 3:43 : 4 and the ratio of adults to seniors is 6:56 : 5. What is the ratio of juniors to seniors?

  1. Adults appear in both ratios, as 4 parts and as 6 parts. Rescale so adults have the same number of parts: the LCM of 4 and 6 is 12.

  2. Juniors : adults =3:4=9:12= 3 : 4 = 9 : 12, and adults : seniors =6:5=12:10= 6 : 5 = 12 : 10.

  3. So juniors : adults : seniors =9:12:10= 9 : 12 : 10, and juniors : seniors =9:10= 9 : 10.

  4. Second route: juniorsseniors=34×65=1820=910\frac{\text{juniors}}{\text{seniors}} = \frac{3}{4} \times \frac{6}{5} = \frac{18}{20} = \frac{9}{10}.

Answer

Juniors to seniors is 9:109 : 10. Reading the outer numbers of the two ratios, 3:53 : 5, is the tempting error; it treats the 4 and the 6 as if they were the same number of adults.

Example 3: A ratio that changes

A class has boys and girls in the ratio 5:35 : 3. After 6 more girls join, the ratio becomes 5:45 : 4. How many boys are there?

  1. Call one part kk: boys =5k= 5k, girls =3k= 3k. The number of boys does not change.

  2. After the change: 5k3k+6=54\frac{5k}{3k + 6} = \frac{5}{4}.

  3. Cross-multiply: 20k=15k+3020k = 15k + 30, so 5k=305k = 30 and k=6k = 6.

  4. Boys =30= 30, girls =18= 18, becoming 24. Check: 30:24=5:430 : 24 = 5 : 4.

Answer

There are 30 boys. A tempting shortcut is "the girls' share rose by 1 part, and that part is 6 people", which happens to work here only because the boys' term stayed at 5 in both ratios; when the unchanged term differs between the two ratios, the shortcut fails.

Common traps

  • Part-to-part read as part-to-whole

    In a 3:53 : 5 mixture, the first ingredient is 38\frac{3}{8} of the whole, not 35\frac{3}{5}. Add the terms to get the whole before you take a fraction of a total.

  • Linking ratios without a common term

    From a:b=3:4a : b = 3 : 4 and b:c=6:5b : c = 6 : 5, the answer is not 3:53 : 5. Rescale until the shared quantity has the same number in both ratios, or multiply the fractions ab×bc\frac{a}{b} \times \frac{b}{c}.

  • Using direct proportion where it is inverse

    More workers means fewer days, so workers and days multiply to a constant. Scaling the days up with the workers gives an answer that moves in the wrong direction; a quick "should this be bigger or smaller?" catches it.

  • Adding to both terms and expecting the same ratio

    Adding 10 to both terms of 2:32 : 3 gives 12:1312 : 13, a different ratio. Only multiplying both terms keeps a ratio; any added amount must go through the multiplier method.

Check yourself

Answer each question before you open its answer.

Question 1: Orange in the blend

Quantitative reasoning: ratios

A juice blend mixes orange juice and apple juice in the ratio 3:53 : 5. How many litres of orange juice are in 24 litres of the blend?

  1. A.8
  2. B.9
  3. C.14.4
  4. D.15
Show the answer

Answer: B. 9

Working

The blend has 3+5=83 + 5 = 8 parts, so one part is 248=3\frac{24}{8} = 3 litres. Orange is 3×3=93 \times 3 = 9 litres and apple 5×3=155 \times 3 = 15 litres, which add to 24.

Why the other options are wrong

  • A. Divides 24 by 3, the orange term, as if the orange term were the number of parts in the whole.
  • C. Takes 35\frac{3}{5} of 24, reading the part-to-part ratio as a fraction of the whole blend.
  • D. This is the apple juice, 58\frac{5}{8} of 24, not the orange juice.

What carries over

Add the terms of a ratio to find the whole before taking a share of a total.

Quantitative reasoning: ratios

If x:y=2:3x : y = 2 : 3 and y:z=4:5y : z = 4 : 5, what is x:zx : z?

  1. A.2:52 : 5
  2. B.3:43 : 4
  3. C.5:65 : 6
  4. D.8:158 : 15
Show the answer

Answer: D. 8:158 : 15

Working

Make yy the same in both: x:y=8:12x : y = 8 : 12 and y:z=12:15y : z = 12 : 15, so x:y:z=8:12:15x : y : z = 8 : 12 : 15 and x:z=8:15x : z = 8 : 15. Check: xz=23×45=815\frac{x}{z} = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}.

Why the other options are wrong

  • A. Takes the outer numbers of the two ratios, but yy is 3 parts in one and 4 parts in the other, so they are on different scales.
  • B. Adds the terms, (2+4):(3+5)(2 + 4) : (3 + 5); ratios combine by multiplying, not adding.
  • C. Divides 23\frac{2}{3} by 45\frac{4}{5} instead of multiplying; xy×yz\frac{x}{y} \times \frac{y}{z} is what cancels yy.

What carries over

Rescale ratios until the shared quantity matches, or multiply the fractions so it cancels.

Question 3: Adding red counters

Quantitative reasoning: ratios

A bag holds red and blue counters in the ratio 4:74 : 7. After 10 red counters are added, the ratio becomes 6:76 : 7. How many counters were in the bag at first?

  1. A.20
  2. B.35
  3. C.55
  4. D.65
Show the answer

Answer: C. 55

Working

Let red =4k= 4k and blue =7k= 7k. Then 4k+107k=67\frac{4k + 10}{7k} = \frac{6}{7}, so 28k+70=42k28k + 70 = 42k, giving 14k=7014k = 70 and k=5k = 5. At first there are 2020 red and 3535 blue, 5555 in all. Check: 30:35=6:730 : 35 = 6 : 7.

Why the other options are wrong

  • A. This is the number of red counters at first, not the total.
  • B. This is the number of blue counters, which does not change; the question asks for the total.
  • D. This is the total after the 10 red counters were added; the question asks for the bag at first.

What carries over

Name one part k, write every amount in terms of k, and turn the changed ratio into one equation.

Key points

  • Compare ab\frac{a}{b} and cd\frac{c}{d} by comparing adad with bcbc.

  • In a ratio, add the terms to find the whole; each share is its term over that sum.

  • Call one part k and every ratio problem becomes one equation.

  • Link two ratios by rescaling the shared term to the same number.

  • Direct proportion keeps a ratio constant; inverse proportion keeps a product constant.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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