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Using ratios

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visibility 90update 6 months agobookmarkshare

🎯 In this topic you will

  • Use ratios in a range of contexts
 

You already know the method to share an amount in a given ratio. For example:

Question
Share $120$ between Ali, Bea and Cas in the ratio $4:5:6$.

Answer

$4+5+6=15$ and $120 \div 15 = 8$ (value of one part).

Ali gets $4\times 8 = 32$

Bea gets $5\times 8 = 40$

Cas gets $6\times 8 = 48$

Check:$32+40+48=120$

You also need to be able to reverse this method to solve similar problems when you are given different pieces of information. For example:

Question
Ali, Bea and Cas share some money in the ratio $4:5:6$.
Ali gets $32$.
How much money do they share?

 
Worked example

A fruit drink contains orange juice and mango juice in the ratio $2:3$.

There are $500 \text{ mL}$ of orange juice in the drink.

a. How much mango juice is in the drink?
b. How much fruit juice is in the drink altogether?

Answer:

a. $1$ part = $500 \div 2 = 250 \text{ mL}$

Mango juice = $3 \times 250 = 750 \text{ mL}$

b. Total = $500 + 750 = 1250 \text{ mL}$

$= 1.25 \text{ litres}$

For a. Since $2$ parts of the drink is orange juice ($500 \text{ mL}$), divide to find $1$ part: $500 \div 2 = 250 \text{ mL}$. Multiply by $3$ to get the mango juice: $750 \text{ mL}$.

For b. Add the amounts of orange and mango juice: $500 + 750 = 1250 \text{ mL}$. Convert into litres: $1.25 \text{ L}$.

 

🧠 PROBLEM-SOLVING Strategy

Use Ratios in a Range of Contexts

Pick the right pathway: share a total, work backwards from one share, or use a known component in a mixture. Always keep the order of the ratio.

  1. Forward sharing (total known).
    Total parts: $P=a+b(+c+\cdots)$
    Value of one part: $u=\dfrac{\text{total}}{P}$
    Each share: e.g., first share $=a\,u$, second share $=b\,u$, etc.
    Check: add all shares to get the original total.
  2. Reverse sharing (one person’s amount known).
    If the $k$-th person has $S_k$ and their ratio part is $r_k$:
    One part: $u=\dfrac{S_k}{r_k}$, Total: $P\,u$.
  3. Reverse sharing (difference known).
    If the difference between two people is $\Delta$ and their parts differ by $|r_i-r_j|$:
    One part: $u=\dfrac{\Delta}{\,|r_i-r_j|\,}$, then find each share with $a\,u,b\,u,\dots$ and sum for the total.
  4. Mixtures/parts known.
    With ratio $a:b$ and a known component amount, find one part:
    If the amount of the first component is $A$, then $u=\dfrac{A}{a}$, second component $=b\,u$, total $=(a+b)u$.
  5. Keep ratios tidy.
    Use whole-number parts; convert units before using them; keep the given order throughout.

Mini worked examples

  1. Forward share (total known): Share $120$ in ratio $4:5:6$.
    $P=4+5+6=15,\quad u=120\div15=8$
    Shares: $4u=32,\ 5u=40,\ 6u=48$ (check: $32+40+48=120$).
  2. Reverse share (one part known): Ratio $4:5:6$, Ali gets $32$ (that’s $4$ parts).
    $u=32\div4=8,\quad \text{total}=15\times8=120$.
  3. Mixture from one component: Orange : Mango $=2:3$, orange is $500\,\text{mL}$.
    $u=500\div2=250$ → Mango $=3u=750\,\text{mL}$, Total $=1250\,\text{mL}=1.25\,\text{L}$.
Quick reminders:
  • Find $u$ (one part) first — it unlocks everything.
  • For “difference” problems, divide the money difference by the difference in parts.
  • Always keep the ratio order and check with a quick sum at the end.
Algebra connection: For ratio parts $r_1,r_2,\dots,r_n$ and unit part $u$, shares are $r_i u$ and total is $u\sum r_i$. Given any one share or a difference, solve for $u$ and build the rest.
 

📘 Exercise

1. Marco uses sultanas and cherries in the ratio 5:2. He uses 80 g of cherries. Answer these:

1a. What mass of sultanas does Marco use?
1b. What is the total mass of sultanas and cherries?

👀 Show answer

2. A fruit dessert contains raspberries and strawberries in the ratio 1:2. There are 400 g of strawberries.

2a. How many grams of raspberries are there?
2b. What is the total mass of fruit in the dessert?

👀 Show answer

2a. 2 parts = 400 g → 1 part = 200 g. Raspberries = 200 g.

2b. Total = 400 + 200 = 600 g.

3. Xavier and Alicia share some money in the ratio 3:5. Xavier gets $75.

3a. How much money does Alicia get?
3b. What is the total amount of money they share?

👀 Show answer

4. Kaya and Akiko share their electricity bill in the ratio 3:4. Akiko pays $24.

4a. How much does Kaya pay?
4b. What is their total bill?

👀 Show answer

5. Jerry makes concrete using cement, sand, and gravel in the ratio 1:2:4. He uses 15 kg of sand.

5a. How much cement and gravel does he use?
5b. What is the total mass of the concrete?

👀 Show answer

6. Three children share sweets in the ratio 4:7:9. The child with the most sweets gets 54 sweets.

6a. How many sweets do the other two children get?
6b. What is the total number of sweets?

👀 Show answer

9 parts = 54 → 1 part = 6. Other two: 4 parts = 24, 7 parts = 42. ✅ Total = 24 + 42 + 54 = 120 sweets.

 

🧠 Think like a Mathematician

Task: Compare different methods for solving a ratio-sharing problem. Reflect on which method is more efficient and why.

Scenario: Jan, Kai, and Li share a water bill in the ratio $2:3:5$. Li pays $36.25. How much is the total bill?

Methods:

Nia’s method:
Total parts = 2 + 3 + 5 = 10
Li = 5 parts = $36.25
1 part = $36.25 ÷ 5 = $7.25
Total = $7.25 × 10 = $72.50
Rhys’s method:
Li = 5 parts = $36.25 ⇒ 1 part = $7.25
Jan = 2 parts = $14.50
Kai = 3 parts = $21.75
Total = $36.25 + $14.50 + $21.75 = $72.50

Questions:

a) Critique their methods. Which do you find clearer or more efficient?
b) Can you think of a better method to use? If so, describe it.
c) Compare your answers and reflect on which strategy is best in general.
👀 show answer
  • a) Both methods are correct. - Nia’s method goes straight to the total, which is quicker. - Rhys’s method breaks it into shares, which gives more detail but takes longer.
  • b) Another method: use proportions. If 5 parts = $36.25, then 10 parts (the total) = $36.25 \times \tfrac{10}{5} = 72.50$. This is essentially Nia’s method in a single step.
  • c) The most efficient strategy is to scale directly from Li’s 5 parts to 10 parts. However, Rhys’s breakdown is useful if you also need to know each person’s share.
 

📘 Exercise

8. Xavier makes some green paint by mixing yellow, blue, and white paint in the ratio 4:5:1. He uses 600 mL of yellow paint.

8a. How much blue paint does he use?
8b. How much green paint does he make in total? (Give your answer in litres.)

👀 Show answer

8a. Yellow = 4 parts = 600 mL → 1 part = 150 mL. Blue = 5 parts = 5 × 150 = 750 mL.

8b. White = 1 part = 150 mL. Total = 600 + 750 + 150 = 1500 mL = 1.5 litres.

9. Purple gold is made from gold and aluminium in the ratio 4:1. A purple gold bracelet has 39 g more gold than aluminium. What is the mass of the bracelet?

The cards are not in the correct order. Write the cards in the correct order and check all the steps.

👀 Show answer

Step 1. Difference in number of parts = 4 − 1 = 3 parts.

Step 2. 3 parts = 39 g.

Step 3. 1 part = 39 ÷ 3 = 13 g.

Step 4. Gold = 4 parts = 13 × 4 = 52 g.

Step 5. Aluminium = 1 part = 13 g.

Step 6. Total mass = 52 + 13 = 65 g.

10. Moira and Non share money in the ratio 3:7. Non gets $28 more than Moira.

10a. What is the total amount of money that they share?
10b. How much money do they each get?

👀 Show answer

Step 1 – Find the difference: Ratio difference = 7 − 3 = 4 parts. This difference = $28.

Step 2 – Value of 1 part: 1 part = 28 ÷ 4 = $7.

Step 3 – Work out shares: Moira = 3 × 7 = $21. Non = 7 × 7 = $49.

Step 4 – Total: Total = 21 + 49 = $70.

 

🧠 Think like a Mathematician

Task: Investigate the ratio $2:3$ when one number is given as 6. Explore all possibilities and justify your reasoning.

Questions:

a) Two numbers are in the ratio $2:3$. One of the numbers is 6. What is the other number?
b) Look back at your solution to part a. i. Were you able to answer the question or did you need more information? ii. How many possible answers are there to the question? Explain why. iii. Show how to check that your answers are correct.
c) Reflect on your answers. Which mathematical skills and reasoning strategies helped you solve this problem?
👀 show answer
  • a) If the numbers are in the ratio $2:3$, then one number = $2k$ and the other = $3k$ for some integer $k$. - If $2k = 6$, then $k = 3$, and the other number = $3k = 9$. - If $3k = 6$, then $k = 2$, and the other number = $2k = 4$. So the other number could be 9 or 4.
  • b i) The question does not specify which part of the ratio corresponds to 6, so there are two possible answers.
  • b ii) There are 2 possible answers: 9 or 4, depending on whether 6 corresponds to the “2 parts” or the “3 parts.”
  • b iii) Check: - 6:9 simplifies to $2:3$. - 4:6 simplifies to $2:3$. Both satisfy the condition.
  • c) This problem required reasoning about ratios, using algebra ($2k, 3k$), and checking solutions by simplifying back to the original ratio.
 

📘 Exercise

12. Two numbers are in the ratio 8:3. One of the numbers is 0.48. Work out the two possible values for the other number. Show how to check that your answers are correct.

👀 Show answer

13. Sofia makes oat biscuits using syrup, butter, and oats in the ratio 1:2:4. She has 250 g of butter and 440 g of oats, with plenty of syrup available. She makes as many biscuits as she can with these ingredients. How much of each ingredient does she use?

👀 Show answer

Ratio = 1 (syrup) : 2 (butter) : 4 (oats).

Butter = 2 parts = 250 g → 1 part = 125 g. Oats required = 4 parts = 4 × 125 = 500 g. But only 440 g oats are available, so oats are the limiting ingredient.

Oats = 440 g → 4 parts = 440 → 1 part = 110 g. Butter used = 2 × 110 = 220 g. Syrup used = 1 × 110 = 110 g.

✅ She uses: 110 g syrup, 220 g butter, 440 g oats.

14. White gold is made from gold, palladium, nickel and zinc in the ratio 15:2:2:1. A ring contains 9 g of gold. What is the mass of the ring?

👀 Show answer

15. Largest angle in a triangle is 75°. The difference between the two other angles is 15°. Write the ratio of the angles from smallest to biggest in simplest form.

👀 Show answer

16. The table shows child:staff ratios and the numbers of children in each age group. One room per age group.

Age group Child : staff Children
up to 18 months 3 : 1 10
18 months up to 3 years 4 : 1 18
3 years up to 5 years 8 : 1 15
5 years up to 7 years 14 : 1 24

Erin thinks 12 staff are needed. What do you think? Show your working.

🔎 Reasoning Tip

The child : staff ratio shows the maximum number of children allowed in the room for each member of staff.

For example, a ratio of 3 : 1 means that there can be no more than three children for one member of staff.

👀 Show answer

Staff needed per room = ceiling(children ÷ children-per-staff).

  • Up to 18 months: 10 ÷ 3 = 3.33 → 4 staff.
  • 18 months–3 yrs: 18 ÷ 4 = 4.5 → 5 staff.
  • 3–5 yrs: 15 ÷ 8 = 1.875 → 2 staff.
  • 5–7 yrs: 24 ÷ 14 = 1.71… → 2 staff.

Total staff required = 4 + 5 + 2 + 2 = 13. Erin’s estimate of 12 is too low.

 

⚠️ Be careful! Using Ratios in Context

  • Keep the order exactly as stated. “orange : mango = 2:3” means orange is the 2-part, mango is the 3-part.
  • Find the unit part first. If one share is known, \(u=\frac{\text{known amount}}{\text{its ratio part}}\); if the total is known, \(u=\frac{\text{total}}{\text{sum of parts}}\).
  • Given a difference? Use the parts’ difference: \(u=\frac{\Delta}{|r_i-r_j|}\); then build each share with \(r_k u\).
  • Mixtures/components. If one component amount is given, \(u=\frac{\text{component}}{\text{its part}}\). Total \(=(\text{sum of parts})\times u\).
  • Unify units before you start. e.g., \(1.2\text{ L}=1200\text{ mL}\). Ratios simplify only after units match.
  • Ratios in simplest form use whole numbers only. Clear decimals/fractions, then divide through by the HCF.
  • Check with a quick add-back. Your shares should sum to the original total; re-substitute to verify.
  • “One number is 6” in \(2:3\) has two cases. Either \(2k=6\Rightarrow k=3\) (other is \(3k=9\)) or \(3k=6\Rightarrow k=2\) (other is \(2k=4\)).
  • Staffing/“at most” ratios. For child:staff = \(c:1\), staff needed is \(\lceil \tfrac{\text{children}}{c}\rceil\) (always round up).
  • Don’t mix up ratio parts with amounts. You may add shares (amounts), but never add parts to amounts.
Mini checks:
• Share $120 in 4:5:6 → parts sum 15, \(u=8\), shares 32, 40, 48 ✓
• Orange:mango = 2:3, orange 500 mL → \(u=250\) mL, mango 750 mL, total 1.25 L ✓
• Non gets $28 more in 3:7 → parts diff 4 ⇒ \(u=7\); total \((3+7)\times7=70\) ✓
• Child:staff 8:1 with 15 children ⇒ \(\lceil 15/8\rceil=2\) staff ✓
 

📘 What we've learned — Ratios in Contexts

  • Read the order carefully: a:b means “first : second”. Keep units consistent before using the ratio.
  • Unit part method: Total parts P=sum of ratio numbers; one part u=total÷P; each share = (ratio part)×u.
  • Work backwards from a share: if someone with part r has amount S, then one part u=S÷r; rebuild all amounts and totals.
  • Work from a difference: if two shares differ by Δ and their parts differ by |r₁−r₂|, then one part u=Δ÷|r₁−r₂|.
  • Mixtures/components known: with ratio a:b and known amount of one component A, one part u=A÷a; the other = bu; total =(a+b)u.
  • Simplify before using: reduce ratios to whole-number simplest form (no fractions/decimals) to make arithmetic cleaner.
  • Check: sums of shares must equal the total; amounts should be proportional (divide any share by its ratio part to get the same u).
  • Algebra link: with parts r₁,…,rₙ and unit u, shares are rᵢu and total is u(r₁+…+rₙ); scaling all parts by the same factor leaves the ratio unchanged.

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