Lesson 4 of 35 · Quantitative Reasoning: arithmetic

Ratios and mixtures

Write ratios as parts, link ratios through a common term, follow a ratio as quantities change, and solve mixtures with the balance method.

About 20 minutes

In this lesson

  • Turn a ratio into actual quantities with one multiplier, and tell part-to-part from part-to-whole.
  • Combine two ratios that share a term, and solve problems where a ratio changes.
  • Solve mixture and weighted-average questions by tracking the pure ingredient or with the balance method.

The idea

A ratio is a recipe of parts

A ratio of 3:53 : 5 says nothing about size; it says the quantities are 3k3k and 5k5k for some positive number kk. Introducing that one multiplier turns every ratio question into ordinary algebra. The total is 8k8k, so the first quantity is 38\frac{3}{8} of the whole and the second 58\frac{5}{8}. A ratio alone never fixes the actual numbers; you need one real quantity (a total, a difference, one of the parts) to find kk.

Part-to-part and part-to-whole

"Boys to girls is 3:53 : 5" is part-to-part. "Three eighths of the class are boys" is part-to-whole. They carry the same information, but mixing them up is a classic slip: with a ratio of 3:53 : 5, boys are 38\frac{3}{8} of the class, not 35\frac{3}{5}. When a question gives a total, convert to part-to-whole; when it gives a difference, use the parts: the difference between the groups is 5k−3k=2k5k - 3k = 2k.

Linking ratios through a common term

If a:b=3:4a : b = 3 : 4 and b:c=6:5b : c = 6 : 5, the two ratios describe bb with different numbers of parts. Scale each so that bb gets the same number, the LCM of 4 and 6, which is 12: 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. Joining the ratios without this step, as 3:4:53 : 4 : 5, is wrong because the "4" and the "6" are not the same amount of bb.

When the quantities change

Questions often add or remove items and give the new ratio. Write the original quantities with kk, apply the change, set the new ratio and cross-multiply. Whatever does not change is your anchor. If men and women are 5:35 : 3 and only women join, the number of men is fixed, so the new ratio tells you how many women there must be per man.

Mixtures: track the pure ingredient

A concentration is a part-to-whole ratio: amount of ingredienttotal amount\frac{\text{amount of ingredient}}{\text{total amount}}. Before writing any equation, ask which amount stays fixed.

  • Adding water (or any other diluting liquid) leaves the amount of the ingredient unchanged. 40 litres of a 20% salt solution hold 8 litres of salt; to make it 16%, the total must become 80.16=50\frac{8}{0.16} = 50 litres, so add 10 litres of water.
  • Adding pure ingredient raises both the ingredient and the total, so both appear in the equation.
  • Removing water (evaporation) keeps the ingredient and lowers the total.

Weighted averages and the balance method

When two groups with different values are combined, the result is a weighted average: it lies between the two values, closer to the value of the larger group. Picture a seesaw. With low value LL, high value HH and mixed value MM,

amount at Lamount at H=H−MM−L\frac{\text{amount at } L}{\text{amount at } H} = \frac{H - M}{M - L}

The amounts are in the inverse ratio of the distances. Mixing 30% and 75% solutions to get 45%: the distances are 45−30=1545 - 30 = 15 and 75−45=3075 - 45 = 30, so the amounts of 30% and 75% solution are in the ratio 30:15=2:130 : 15 = 2 : 1. The 30% solution is the one the mixture sits closer to, so there must be more of it. The same method handles prices of blended goods, average scores of two classes and average speeds over equal times.

Two checks before you choose

Is the mixed value between the two ingredient values, and is it closer to the one there is more of? Does your answer give the stated total? Both catch reversed ratios, the most common mistake in this topic.

Worked examples

Example 1: Linking two ratios

In a box of pens, the ratio of black to blue is 3:43 : 4, and the ratio of blue to red is 6:56 : 5. There are 155 pens in all. How many are red?

  1. Blue appears in both ratios as 4 parts and 6 parts. Scale both so blue is lcm(4,6)=12\text{lcm}(4, 6) = 12 parts: black : blue =9:12= 9 : 12, blue : red =12:10= 12 : 10.

  2. Combined: black : blue : red =9:12:10= 9 : 12 : 10, a total of 31 parts.

  3. Each part is 155÷31=5155 \div 31 = 5 pens, so red is 10×5=5010 \times 5 = 50.

  4. Check: 45 black, 60 blue, 50 red add to 155; 45:60=3:445 : 60 = 3 : 4 and 60:50=6:560 : 50 = 6 : 5.

Answer

There are 50 red pens. Chaining the ratios without scaling, as 3:4:53 : 4 : 5, gives 12 parts and a non-whole number of pens per part.

Example 2: A ratio that changes

A club's members are men and women in the ratio 5:35 : 3. After 12 more women join and nobody leaves, the ratio becomes 5:45 : 4. How many men are in the club?

  1. Let the men be 5k5k and the women 3k3k. After the change: 5k:(3k+12)=5:45k : (3k + 12) = 5 : 4.

  2. Cross-multiply: 4×5k=5(3k+12)4 \times 5k = 5(3k + 12), so 20k=15k+6020k = 15k + 60 and k=12k = 12.

  3. Men: 5×12=605 \times 12 = 60. Women: 36 before, 48 after, and 60:48=5:460 : 48 = 5 : 4.

  4. Second route: the men are fixed. In the new ratio the women are 45\frac{4}{5} of the men, up from 35\frac{3}{5}; the gain of 15\frac{1}{5} of the men equals 12, so there are 60 men.

Answer

The club has 60 men. Reading the ratios as actual numbers, 5 men and 3 women, is the trap: then a single woman joining would already give 5:45 : 4.

Example 3: Mixing two solutions

How many litres of a 30% acid solution must be added to 12 litres of a 75% acid solution to obtain a 45% acid solution?

  1. Distances from the target: 45−30=1545 - 30 = 15 and 75−45=3075 - 45 = 30.

  2. Amounts are in the inverse ratio of the distances: 30% solution : 75% solution =30:15=2:1= 30 : 15 = 2 : 1.

  3. There are 12 litres of the 75% solution, so 2×12=242 \times 12 = 24 litres of the 30% solution.

  4. Check by tracking the acid: 0.3×24+0.75×12=7.2+9=16.20.3 \times 24 + 0.75 \times 12 = 7.2 + 9 = 16.2 litres of acid in 36 litres, and 16.236=0.45\frac{16.2}{36} = 0.45.

Answer

24 litres. Using the distances directly instead of inversely gives 6 litres, which would leave the mixture closer to 75% than to 30%.

Common traps

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

    With boys and girls in the ratio 3:53 : 5, the boys are 38\frac{3}{8} of the class. Using 35\frac{3}{5} of a stated total is the slip. Write the total as 8k8k and the confusion disappears.

  • Joining ratios without a common term

    a:b=3:4a : b = 3 : 4 and b:c=6:5b : c = 6 : 5 do not give 3:4:53 : 4 : 5 or 3:6:53 : 6 : 5. The shared quantity must have the same number of parts in both before the ratios can be chained.

  • Averaging concentrations without weights

    Mixing equal volumes of 20% and 50% solutions gives 35%, but unequal volumes do not. A mixture's concentration is total ingredient over total volume, so the larger volume pulls the result towards its own value.

  • Reversing the balance ratio

    In the balance method the amounts are in the inverse ratio of the distances. A quick test: the result must be closer to the ingredient there is more of. If your answer puts it closer to the smaller amount, swap the ratio.

Check yourself

Answer each question before you open its answer.

Question 1: Marbles added to a bag

Problem Solving

The ratio of red to blue marbles in a bag is 4:74 : 7. After 6 red marbles are added, the ratio of red to blue is 2:32 : 3. How many blue marbles are in the bag?

  1. A.36
  2. B.42
  3. C.63
  4. D.99
  5. E.105
Show the answer

Answer: C. 63

Working

Let red be 4k4k and blue 7k7k. Then 4k+67k=23\frac{4k + 6}{7k} = \frac{2}{3}, so 12k+18=14k12k + 18 = 14k and k=9k = 9. Blue: 7×9=637 \times 9 = 63. Check: red goes from 36 to 42, and 42:63=2:342 : 63 = 2 : 3.

Why the other options are wrong

  • A. 36 is the original number of red marbles, not the blue ones.
  • B. 42 is the number of red marbles after the 6 are added.
  • D. 99 is the total number of marbles before the red ones were added.
  • E. 105 is the total number of marbles after the red ones were added.

What carries over

Name the quantities with one multiplier, apply the change, then answer exactly the quantity asked for.

Question 2: A coffee blend

Problem Solving

A grocer blends coffee costing $12\$12 per kilogram with coffee costing $20\$20 per kilogram to make 40 kilograms of a blend costing $15\$15 per kilogram. How many kilograms of the $20\$20 coffee are in the blend?

  1. A.3
  2. B.5
  3. C.8
  4. D.15
  5. E.25
Show the answer

Answer: D. 15

Working

Distances from the blend price: 15−12=315 - 12 = 3 and 20−15=520 - 15 = 5. The amounts are in the inverse ratio, so cheap : dear =5:3= 5 : 3. The 40 kilograms make 8 parts of 5 kilograms, so the $20\$20 coffee is 3×5=153 \times 5 = 15 kilograms. Check: 25×12+15×20=300+300=600=40×1525 \times 12 + 15 \times 20 = 300 + 300 = 600 = 40 \times 15.

Why the other options are wrong

  • A. 3 is the price gap 15−1215 - 12, a distance on the balance, not an amount of coffee.
  • B. 5 is the price gap 20−1520 - 15, a distance on the balance, not an amount of coffee.
  • C. 8 is the full price range 20−1220 - 12; it is the number of parts, not kilograms.
  • E. 25 kilograms is the amount of the $12\$12 coffee; the ratio was applied the wrong way round.

What carries over

On the balance, amounts are in the inverse ratio of the distances from the mixed value.

Question 3: Raising a juice concentration

Problem Solving

A 60-litre tank holds a drink that is 25% juice. How many litres of pure juice must be added to make the drink 40% juice?

  1. A.9
  2. B.15
  3. C.22.5
  4. D.24
  5. E.37.5
Show the answer

Answer: B. 15

Working

The drink holds 0.25×60=150.25 \times 60 = 15 litres of juice. Adding xx litres of pure juice: 15+x60+x=25\frac{15 + x}{60 + x} = \frac{2}{5}, so 75+5x=120+2x75 + 5x = 120 + 2x and x=15x = 15. Check: 3075=0.4\frac{30}{75} = 0.4. Balance route: the drink is 25% and pure juice is 100%; the target 40% is 15 from one and 60 from the other, so drink : juice =60:15= 60 : 15, and with 60 litres of drink that is 15 litres of juice.

Why the other options are wrong

  • A. This applies the 15-point gap to the original 60 litres, ignoring that the added juice also increases the total volume.
  • C. This is the water that would have to be removed to reach 40% juice, a different operation from adding juice.
  • D. 24 litres is the juice a 60-litre drink at 40% would contain; adding juice changes the volume, so this is not the amount to add.
  • E. This is the total volume at which the original 15 litres of juice would make up 40%, not an amount to add.

What carries over

Decide which amounts change before writing the concentration equation.

Key points

  • Write a ratio a:ba : b as akak and bkbk; one real quantity fixes kk.

  • Give a shared term the same number of parts before chaining two ratios.

  • In mixtures, track the pure ingredient and ask which amounts change.

  • Amounts in a blend are in the inverse ratio of their distances from the mixed value.

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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