Equivalent Ratios
Equivalent ratios make the same comparison with different numbers, so \(3:5\), \(6:10\) and \(30:50\) all describe the same split. In NSW Year 8 maths this subtopic revisits ratios from Year 7 with harder numbers (NESA outcome MA4‑RAT‑C‑01).
You will find the multiplier that links two ratios and use it to fill in a missing term, including decimal multipliers, three-term ratios and ratios with fractions or pronumerals. You will also complete ratio tables, scale up recipes, mixtures and screen sizes, decide which of two drinks is stronger, and join two ratios that share a term into one ratio such as \(A:B:C\). It is part of Year 8 Rates and Ratios.
Theory
Equivalent ratios make the same comparison: multiply or divide every term by the same number. \(5:9 = 30:54\) (each term \(\times 6\)), and in \(4:3 = x:9\) the multiplier is \(3\), so \(x = 12\). This Year 8 maths page shows how to find missing terms.
Equivalent ratios make the same comparison. Every term of one is the same multiple of the matching term of the other: \(5:9 = 30:54\) \((\times 6)\). Dividing every term works the same way.
A missing term. Use a pair of matching terms you know in both ratios. Divide to find the multiplier, then multiply the other terms by it. Any matching pair will do, including the middle term of a three-term ratio.
Multiplier not a whole number? Simplify the ratio first, or use the decimal multiplier: for \(14:22 = 49:\,?\), \(49 \div 14 = 3.5\), so the missing term is \(22 \times 3.5 = 77\).
Decimals and fractions. Multiply every term by \(10\) or \(100\) to clear decimals, and by the lowest common denominator to clear fractions. Change measurements to the same unit before writing the ratio.
Testing equivalence. Equivalent ratios have the same simplest form, so simplify each one and compare.
Joining two ratios. To join \(A:B\) and \(B:C\), make the \(B\) terms the same using their lowest common multiple. With \(A:B = 5:2\) and \(B:C = 6:7\): \(5:2 = 15:6\), so \(A:B:C = 15:6:7\).
Comparing mixtures. Give both ratios the same second term; the one with the larger first term is stronger.
A ratio table: every column is the ratio \(3:8\), because both rows are multiplied by the number at the top. The multiplier need not be whole.
| \(\times 1\) | \(\times 4\) | \(\times 2.5\) | |
|---|---|---|---|
| First term | 3 | 12 | 7.5 |
| Second term | 8 | 32 | 20 |
Finding a missing term
- Match the terms: find a position where both ratios have a number.
- Find the multiplier: divide the new term by the old one (simplify first if it is awkward).
- Multiply the other terms by the multiplier.
- Check by simplifying both ratios to the same simplest form.
| \(9 \div 3\) | \(=\) | \(3\) |
| \(x\) | \(=\) | \(4 \times 3 = 12\) |
\(x = 12\), so \(4:3 = 12:9\). The multiplier \(3\) turns \(3\) into \(9\), so it also turns \(4\) into \(12\).
The first terms give the multiplier: \(4 \to 2\) is \(\div 2\).
| \(x\) | \(=\) | \(3 \div 2 = 1.5\) |
| \(y\) | \(=\) | \(7 \div 2 = 3.5\) |
\(x = 1.5\) and \(y = 3.5\), so \(4:3:7 = 2:1.5:3.5\).
\(66\) does not go into \(39\), so simplify first. The HCF of \(44\) and \(66\) is \(22\): \(44:66 = 2:3\).
| \(39 \div 3\) | \(=\) | \(13\) |
| \(2 \times 13\) | \(=\) | \(26\) |
The missing term is \(26\). Check: \(26:39 = 2:3\) \((\div 13)\).
(ii) Write \(\dfrac{2}{3}:\dfrac{1}{6}:\dfrac{1}{2}\) with whole-number terms.
(i) \(0.9:0.6 = 9:6 = 3:2\). Now \(2 \times 19 = 38\).
(ii) The lowest common denominator is \(6\), so multiply every term by \(6\).
| \(3 \times 19\) | \(=\) | \(57\) |
| \(\dfrac{2}{3} \times 6\) | \(=\) | \(4\) |
| \(\dfrac{1}{6} \times 6\) | \(=\) | \(1\) |
| \(\dfrac{1}{2} \times 6\) | \(=\) | \(3\) |
(i) \(57\)
(ii) \(4:1:3\)
Answers are in grams, so \(1.4\) kg \(= 1400\) g, and sunflower is the first term.
| \(1400 \div 7\) | \(=\) | \(200\) |
| \(5 \times 200\) | \(=\) | \(1000\) |
| \(4 \times 200\) | \(=\) | \(800\) |
\(1000\) g of millet and \(800\) g of canary seed. Check: \(1400:1000:800 = 7:5:4\).
(ii) \((x + 4):2x\) is equivalent to \(3:2\). Find \(x\).
(i) \(7 \times 2 = 14\), so \(3m = 6 \times 2 = 12\).
(ii) Write both ratios as fractions: \(\dfrac{x + 4}{2x} = \dfrac{3}{2}\), so \(2(x + 4) = 6x\).
| \(3m\) | \(=\) | \(12\) |
| \(m\) | \(=\) | \(4\) |
| \(2x + 8\) | \(=\) | \(6x\) |
| \(8\) | \(=\) | \(4x\) |
| \(x\) | \(=\) | \(2\) |
(i) \(m = 4\) (check \(14:12 = 7:6\))
(ii) \(x = 2\) (check \(6:4 = 3:2\))
Same units: \(0.18\) L \(= 180\) mL. Then \(120:500 = 6:25\) and \(180:800 = 9:40\). The LCM of \(25\) and \(40\) is \(200\).
| \(6:25\) | \(=\) | \(48:200\) |
| \(9:40\) | \(=\) | \(45:200\) |
Jug P is stronger: \(48\) parts of cordial to \(200\) of water, against \(45\) for jug Q.
Simplify each one: \(6:9 = 2:3\) \((\div 3)\), \(10:15 = 2:3\) \((\div 5)\), \(8:10 = 4:5\) \((\div 2)\).
\(8:10\) is not equivalent to \(2:3\).
Common pitfalls
Frequently asked questions
What are equivalent ratios?
Ratios that make the same comparison, such as 2:3 and 8:12. Every term is multiplied (or divided) by the same number.
How do I find a missing term in equivalent ratios?
Find the multiplier from a pair of matching terms, then multiply the other term by it.
What if the multiplier is not a whole number?
Simplify the ratio first, or use a decimal multiplier such as 2.5.
How do I check whether two ratios are equivalent?
Simplify both. If they have the same simplest form, they are equivalent.
How do I join A:B and B:C into one ratio?
Make the two B terms equal using their lowest common multiple, then write A:B:C.