Lesson 22 of 46 · Arithmetic

Ratios and proportions

Turn a ratio into parts with one multiplier, link ratios through a shared term, and solve proportions and changing-ratio problems cleanly.

About 20 minutes

In this lesson

  • Write the quantities in a ratio as multiples of one unknown and find them from a total or a difference.
  • Tell part-to-part from part-to-whole ratios and convert between them.
  • Combine two ratios that share a term into one three-way ratio.
  • Solve proportions, including problems where amounts are added and the ratio changes.

The idea

A ratio compares, it does not count

"The ratio of red to blue marbles is 4:54 : 5" says that for every 4 red marbles there are 5 blue ones. It does not say there are 4 red marbles. There could be 8 and 10, or 40 and 50. A ratio fixes the shape of a group, not its size.

So give the size a name. If red : blue =4:5= 4 : 5, write

red=4k,blue=5k\text{red} = 4k, \qquad \text{blue} = 5k

for some positive number kk, the multiplier. Every ratio problem becomes a one-unknown equation once you do this. Any one actual figure, whether a total, a difference or one of the parts, pins down kk.

Part to part, and part to whole

With red : blue =4:5= 4 : 5, the whole is 4k+5k=9k4k + 5k = 9k. So red marbles are 49\frac{4}{9} of the total, not 45\frac{4}{5}. The ratio 4:54 : 5 is part to part; the fraction 49\frac{4}{9} is part to whole. Read which one a question gives and which one it asks for. "The ratio of red marbles to all marbles" is 4:94 : 9.

Ratios behave like fractions

A ratio a:ba : b can be written ab\frac{a}{b} and simplified the same way: 18:24=3:418 : 24 = 3 : 4. Multiplying or dividing both terms by the same number keeps the ratio. Adding the same number to both terms does not: 3:43 : 4 becomes 4:54 : 5 if you add 1 to each, and 45≠34\frac{4}{5} \ne \frac{3}{4}. Adding the same positive amount to both terms always pulls the ratio towards 1:11 : 1.

Three-way ratios through a shared term

If a:b=2:3a : b = 2 : 3 and b:c=4:5b : c = 4 : 5, the two ratios mention bb with different numbers, 3 and 4. Rescale both so that bb gets the same number, their least common multiple, 12:

a:b=8:12,b:c=12:15,so a:b:c=8:12:15a : b = 8 : 12, \qquad b : c = 12 : 15, \qquad \text{so } a : b : c = 8 : 12 : 15

Now a:c=8:15a : c = 8 : 15. Writing a:c=2:5a : c = 2 : 5, by simply dropping the middle numbers, is the classic error; it ignores that the two bb values were measured on different scales.

Proportions

A proportion says two ratios are equal: ab=cd\frac{a}{b} = \frac{c}{d}. Cross-multiply to solve: ad=bcad = bc. If 3 kg of flour make 40 loaves, then xx kg make 100 loaves where 340=x100\frac{3}{40} = \frac{x}{100}, so x=7.5x = 7.5.

Quantities are directly proportional when their ratio is constant: double one and the other doubles (y=kxy = kx). They are inversely proportional when their product is constant: double one and the other halves (xy=kxy = k). Workers and time to finish a fixed job are the standard inverse pair. Ask before you set up the equation: should the answer go up or down?

When the ratio changes

Problems where items are added or removed are ratio problems in two stages. Write the "before" quantities with a multiplier, apply the change to the actual amounts, and set the new ratio equal to the "after" ratio. The multiplier belongs to the original group; never apply the change to the ratio numbers themselves.

Ratios in comparison questions

A ratio on its own never tells you an actual amount. If the only information is "the ratio of xx to yy is 3:43 : 4", then the size of xx or of x+yx + y cannot be determined; even which of xx and yy is larger can flip if negatives are allowed. Ask whether the stem gives a real figure that fixes kk. If not, look for what the ratio does fix, such as xy\frac{x}{y} or the fraction of the total.

A quick check

When you have the parts, divide them back out: they should simplify to the given ratio, and they should add up to any given total. The check takes seconds and catches a misread part-to-whole.

Worked examples

Example 1: Splitting a total

A bag holds 63 marbles, red and blue only, in the ratio red : blue =4:5= 4 : 5. How many of each colour are there?

  1. Name the multiplier: red =4k= 4k, blue =5k= 5k.

  2. The total is 9k=639k = 63, so k=7k = 7.

  3. Red =28= 28 and blue =35= 35.

  4. Check: 28+35=6328 + 35 = 63, and 2835=45\frac{28}{35} = \frac{4}{5}. Second route: red is 49\frac{4}{9} of 63, which is 28.

Answer

28 red and 35 blue. Taking red as 45\frac{4}{5} of 63 (50.4) confuses the part-to-part ratio with a fraction of the whole.

Example 2: Linking two ratios

If a:b=2:3a : b = 2 : 3 and b:c=4:5b : c = 4 : 5, what is a:ca : c?

  1. bb appears as 3 in one ratio and 4 in the other. Rescale both to give bb the value lcm⁡(3,4)=12\operatorname{lcm}(3, 4) = 12.

  2. a:b=2:3=8:12a : b = 2 : 3 = 8 : 12 (multiply by 4), and b:c=4:5=12:15b : c = 4 : 5 = 12 : 15 (multiply by 3).

  3. So a:b:c=8:12:15a : b : c = 8 : 12 : 15 and a:c=8:15a : c = 8 : 15.

  4. Check by multiplying fractions: ac=ab⋅bc=23⋅45=815\frac{a}{c} = \frac{a}{b} \cdot \frac{b}{c} = \frac{2}{3} \cdot \frac{4}{5} = \frac{8}{15}.

Answer

a:c=8:15a : c = 8 : 15. Dropping the middle term, 2:52 : 5, is the tempting wrong answer.

Example 3: New students join

In a class the ratio of boys to girls is 3:53 : 5. Six more boys join, and the ratio becomes 3:43 : 4. How many girls are in the class?

  1. Before: boys =3k= 3k, girls =5k= 5k.

  2. After: boys =3k+6= 3k + 6, girls still 5k5k. Set the new ratio: 3k+65k=34\frac{3k + 6}{5k} = \frac{3}{4}.

  3. Cross-multiply: 4(3k+6)=15k4(3k + 6) = 15k, so 12k+24=15k12k + 24 = 15k and k=8k = 8.

  4. Girls =5⋅8=40= 5 \cdot 8 = 40. Check: boys go from 24 to 30, and 30:40=3:430 : 40 = 3 : 4.

Answer

There are 40 girls. Adding the 6 to the ratio numbers instead of to the actual number of boys, as in (3+6):5(3 + 6) : 5, leads nowhere: the change applies to people, not to ratio terms.

Common traps

  • Part to part read as part to whole

    With boys : girls =2:3= 2 : 3, boys are 25\frac{2}{5} of the class, not 23\frac{2}{3}. Add the ratio terms to get the whole before you form any fraction of the total.

  • Linking ratios without a common term

    From a:b=2:3a : b = 2 : 3 and b:c=4:5b : c = 4 : 5 you cannot read a:c=2:5a : c = 2 : 5. Rescale until bb has one value in both, or multiply ab⋅bc\frac{a}{b} \cdot \frac{b}{c}.

  • Expecting a ratio to survive addition

    Adding the same amount to both terms changes a ratio (unless it is 1:11 : 1). Multiplying both by the same amount does not. Ages are the usual test: two people whose ages are in the ratio 1:21 : 2 today will not be in that ratio in ten years.

  • Reading an amount into a ratio

    "x:y=3:4x : y = 3 : 4" does not mean x=3x = 3. In a comparison question with no other figure, an actual size such as xx or y−xy - x cannot be determined.

Check yourself

Answer each question before you open its answer.

Question 1: Splitting a fund three ways

Ratio

A fund is divided among three people in the ratio 2:3:72 : 3 : 7. The largest share is $1,500 more than the smallest share. What is the total amount of the fund?

  1. A.$600
  2. B.$2,100
  3. C.$3,600
  4. D.$7,500
  5. E.$18,000
Show the answer

Answer: C. $3,600

Working

Write the shares as 2k2k, 3k3k and 7k7k. The difference between the largest and smallest is 7k−2k=5k=15007k - 2k = 5k = 1500, so k=300k = 300. The total is 12k=$3,60012k = \$3,600. Check: the shares are $600, $900 and $2,100; they add to $3,600, and 2100−600=15002100 - 600 = 1500.

Why the other options are wrong

  • A. This is the smallest share, 2×$3002 \times \$300, not the whole fund.
  • B. This is the largest share, 7×$3007 \times \$300, not the total.
  • D. This multiplies $1,500 by the 5-part difference instead of dividing by it to find one part.
  • E. This treats $1,500 as the value of one part and multiplies by all 12 parts; $1,500 is the value of 5 parts.

What carries over

Express every share as a multiple of one unknown, and use the given figure to find that unknown first.

Question 2: Adding 5 to both terms

Quantitative Comparison

xx and yy are positive numbers, and the ratio of xx to yy is 3:43 : 4.

Quantity A Quantity B
x+5y+5\dfrac{x + 5}{y + 5} 34\dfrac{3}{4}
  1. A.Quantity A is greater.
  2. B.Quantity B is greater.
  3. C.The two quantities are equal.
  4. D.The relationship cannot be determined from the information given.
Show the answer

Answer: A. Quantity A is greater.

Working

Write x=3kx = 3k and y=4ky = 4k with k>0k > 0. Compare 3k+54k+5\frac{3k + 5}{4k + 5} with 34\frac{3}{4} by cross-multiplying (both denominators are positive): 4(3k+5)=12k+204(3k + 5) = 12k + 20 and 3(4k+5)=12k+153(4k + 5) = 12k + 15. Since 12k+20>12k+1512k + 20 > 12k + 15, Quantity A is greater for every kk. Check: x=3x = 3, y=4y = 4 gives 89≈0.89\frac{8}{9} \approx 0.89; x=30x = 30, y=40y = 40 gives 3545≈0.78\frac{35}{45} \approx 0.78; both exceed 0.75.

Why the other options are wrong

  • B. This assumes that adding to both terms shrinks the ratio. For a ratio below 1, adding the same positive amount moves it up, towards 1.
  • C. This assumes adding the same number to both terms keeps the ratio; only multiplying both terms does that.
  • D. Quantity A does change with xx, but for every positive xx it stays above 34\frac{3}{4}.

What carries over

Adding the same positive amount to both terms pulls a ratio towards 1.

Question 3: Students to aides

Ratio

At a school, the ratio of teachers to classroom aides is 5:25 : 2, and the ratio of students to teachers is 12:112 : 1. What is the ratio of students to classroom aides?

  1. A.6:16 : 1
  2. B.12:112 : 1
  3. C.24:124 : 1
  4. D.30:130 : 1
  5. E.60:160 : 1
Show the answer

Answer: D. 30:130 : 1

Working

Make the teacher term match: students : teachers =12:1=60:5= 12 : 1 = 60 : 5, and teachers : aides =5:2= 5 : 2. So students : teachers : aides =60:5:2= 60 : 5 : 2, and students : aides =60:2=30:1= 60 : 2 = 30 : 1. Check by multiplying fractions: studentsaides=121⋅52=30\frac{\text{students}}{\text{aides}} = \frac{12}{1} \cdot \frac{5}{2} = 30.

Why the other options are wrong

  • A. This sets 12 students against 2 aides, as if there were one teacher; the teacher term must first be scaled to 5.
  • B. This copies the students-to-teachers ratio and ignores the aides.
  • C. This multiplies 12 by the aides' term, 2, instead of combining the ratios through the shared teacher term.
  • E. This finds 60 students for 5 teachers but compares them with one aide instead of the two that go with 5 teachers.

What carries over

To link two ratios, scale them until the shared quantity has the same number in both.

Key points

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

  • In a:ba : b, the first part is aa+b\frac{a}{a + b} of the whole.

  • Link two ratios by scaling the shared term to one value.

  • Multiplying both terms keeps a ratio; adding to both changes it.

  • A ratio alone never fixes an actual amount.

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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