Ratio, Proportion & Percentages

MYP 1 Maths made clear: proportion, multipliers, and the reasoning traps examiners actually test

A balance scale comparing a shopping bag and coins, with a percentage symbol and a straight-line graph through the origin

Subject

Mathematics

Curriculum

IB MYP

Grade

MYP 1

Topic

Ratio, Proportion & Percentages

Reading

7 min

Difficulty

Foundational

Quick facts

Difficulty

★★☆☆☆

Exam weight

Core Number strand — nearly every unit test

Prerequisites

Fractions, decimals, basic ratio notation

You'll learn

Proportion reasoning, multipliers, reverse %

Revision time

30-40 min

Ratio, proportion and percentages are the maths behind everyday fairness — splitting a bill, resizing a recipe, or checking if a 'sale' is really worth it. In MYP 1 Maths, this topic sits at the heart of the Number strand and turns up in almost every unit test, because it tests reasoning as much as calculation. The 'calculate' parts of a question are usually the easy marks; it's the 'explain why' or 'justify' parts — where Criterion C marks are won or lost — that catch students who can compute but can't say why a method works. This teaser walks through the five ideas that matter most: direct and inverse proportion, the unit rate that powers both, simple versus compound percentage change, the multiplier method, and reverse percentage problems. Master these five, and the rest of the topic falls into place.

What you’ll be able to do

✓Distinguish direct proportion from inverse proportion using ratio and product

✓Find and use a unit rate to scale quantities up or down

✓Apply the multiplier method to increase or decrease an amount by a percentage

✓Explain the difference between simple and compound percentage change

✓Chain multipliers correctly for repeated percentage changes

✓Solve reverse percentage problems by dividing, not subtracting

✓Justify whether a relationship is truly proportional, not just 'bigger when bigger'

✓Avoid the most common mark-losing mistakes in Criterion C reasoning

Direct vs Inverse Proportion

Two quantities are in direct proportion if their ratio $y \div x$ stays the same — double one, the other doubles too. They're in inverse proportion if their product $xy$ stays constant — double one, the other halves. Most 'cost of x kg' or 'pay for x hours' questions at MYP 1 are direct proportion, because a fixed unit price or hourly rate never changes.

Two graphs side by side: a straight line through the origin labelled direct proportion, and a curve labelled inverse proportion

Type	What stays constant	Graph shape
Direct proportion	$y \div x = k$	Straight line through the origin
Inverse proportion	$xy = k$	Curve, never touches either axis

Exam tip

When asked to justify direct proportion, don't say 'the total goes up' — that's true for lots of relationships. Say the RATE ( $y \div x$ ) stays the same, and show two divisions to prove it.

Common mistake

Assuming 'total cost rises when you buy more' proves direct proportion — it doesn't, since the rate could still be changing.

Mini summary

Direct = constant ratio, straight line through origin; inverse = constant product, curve away from both axes.

Unit Rate & the Constant k

The constant of proportionality $k$ IS the unit rate — cost per kg, pay per hour, distance per litre. Find it by dividing one matching pair of values ( $k = y \div x$ ), then multiply to scale up or down to any new amount. Before extrapolating to a value outside your table, always check $k$ against a second pair — a single point can hide a typo.

A table of rice weight vs cost with an arrow showing division to find cost per kg, then multiplication to scale to 8kg

Exam tip

On 'calculate' questions, write the division $k = y \div x$ as its own line, even if you can do it mentally. The method mark is specifically for showing this step — a bare final answer with no working can score 0/2 if it's wrong.

Common mistake

Picking a data point that isn't actually on the proportional line and using it to find $k$ , which throws off every later calculation.

Mini summary

Find $k$ from one verified pair, sanity-check it with a second pair, then scale confidently.

Simple vs Compound Percentage Change

A simple percentage change is applied once to the original amount. A compound change is applied repeatedly, and each new application acts on the most recent amount — not the original. This is the single biggest trip-up in the whole topic: students add the percentages together (10% + 10% = 20%) instead of chaining the multipliers.

A price of $80 going through two 10% increases shown as a chain of two multiplications, compared to one incorrect 20% jump

Exam tip

Whenever you see 'increases by p%, then increases again by p%', chain the multipliers: $(1+p/100)^2$ . Never add the percentage figures.

Common mistake

Adding repeated percentages together (treating 10% then 10% as one 20% change) instead of multiplying $1.10 \times 1.10$ .

Mini summary

Simple = one multiplier once; compound = the same multiplier applied $n$ times, using $(1\pm p/100)^n$ .

Percentage Increase/Decrease: The Multiplier Method

The safest way to change an amount by a percentage is the multiplier method: multiply by $(1+p/100)$ to increase, or $(1-p/100)$ to decrease. This avoids the classic two-step trap where students forget to add/subtract the calculated percentage, or worse, subtract the percentage number itself as if it were a dollar amount.

A $120 jacket price with a correct path showing multiplication by 0.85 giving $102, next to an incorrect path subtracting 15 directly giving $105

Change	Multiplier
Increase by $p\%$	$1 + p/100$
Decrease by $p\%$	$1 - p/100$

Exam tip

For 'find the percentage change' questions (given before and after), divide the change by the ORIGINAL value, then ×100 — not by the new value.

Common mistake

Subtracting the percentage figure directly from the price (e.g. $120 - 15 = 105$ ) instead of finding 15% of $120 first and subtracting that ($ 120 - 18 = 102$).

Mini summary

Convert the percentage to a multiplier first, then apply it to the whole original amount in one step.

Reverse Percentage Problems

Reverse percentage questions give you the value AFTER a percentage change and ask for the original. The number you're given is not the original — it's already the original multiplied by the multiplier — so you must divide by the multiplier, never add or subtract the percentage from the given value.

An arrow showing a new value being divided by a multiplier to recover the original amount, reversing the forward multiplication arrow

Exam tip

Set up $\text{Original} = \text{New} \div (1\pm p/100)$ before you touch a calculator — writing this line down protects your method marks even if the arithmetic slips.

Common mistake

Trying to reverse a percentage by adding or subtracting the percentage from the given (already-changed) value instead of dividing by the multiplier.

Mini summary

New value ÷ multiplier = original — division undoes the multiplier, it doesn't reverse-add the percentage.

Quick formula sheet

y = kx

Direct proportion:

y

k

times

x

— Same direction, straight line through the origin

k = \dfrac{y}{x}

Find the constant of proportionality (unit rate) from any matching pair — k = the 'per one unit' value

y = \dfrac{k}{x} \quad \Leftrightarrow \quad xy = k

Inverse proportion: the product of the two variables never changes — Opposite directions — one up, one down

\text{Amount} = \dfrac{\text{Percentage}}{100} \times \text{Whole}

Finding a percentage of a quantity — Percentage ÷ 100 first, then × the whole

\text{New} = \text{Original} \times \left(1 \pm \dfrac{p}{100}\right)

Multiplier method for a single percentage increase or decrease — + for growth, − for shrink, always multiply the WHOLE amount

\text{Final} = \text{Original} \times \left(1 \pm \dfrac{p}{100}\right)^{n}

Compound percentage change applied

n

times — Same multiplier, chained by the power n — never add the percentages

\%\ \text{change} = \dfrac{|\text{New} - \text{Original}|}{\text{Original}} \times 100

Find the percentage change itself from before-and-after values — Change over ORIGINAL, not over the new value

\text{Original} = \text{New} \div \left(1 \pm \dfrac{p}{100}\right)

Reverse percentage: undo the multiplier by dividing — Forward = multiply, backward = divide

Practice questions

Easy

6 kg of apples costs $12. Find the cost of 1 kg, then the cost of 9 kg.
Find 20% of $150.
Increase $60 by 5% using the multiplier method.

Medium

A $50 jacket is reduced by 12%. Find the new price, showing the multiplier used.
A quantity increases by 10%, then increases again by 10%. Find the overall multiplier and explain why it isn't simply 20%.
Given pairs (3h, $45) and (7h,$ 105), find the pay rate per hour and predict the pay for 10 hours.

Challenge

After a 15% increase, an item costs $92. Find the original price, showing your equation before calculating.
A population grows by 8% each year for 3 years. Write the compound formula and explain what each part represents.
A classmate claims a data point fits a direct proportion table without checking it. Explain, using two verified ratios, how you would prove or disprove this claim.

Frequently asked questions

What's the difference between direct and inverse proportion?+

In direct proportion, the ratio $y \div x$ stays constant — both quantities increase or decrease together. In inverse proportion, the product $xy$ stays constant — as one increases, the other decreases.

Why doesn't a rising total cost prove direct proportion?+

Totals can rise for many reasons. Direct proportion specifically requires the RATE (unit price, unit rate) to stay the same — you must check this by dividing at least two pairs of values.

Why can't I just add percentages together for repeated changes?+

Because each repeated change acts on the newest amount, not the original. Adding (like 10%+10%=20%) under-counts the real effect — you need to chain the multipliers instead, e.g. $1.10 \times 1.10$ .

How do I solve a reverse percentage question?+

Divide the given (already-changed) value by the multiplier $(1\pm p/100)$ — never add or subtract the percentage from the given number, since it isn't the original amount.

What's the quickest way to avoid percentage mistakes?+

Always convert the percentage into a multiplier first and apply it to the whole original amount in one step. This avoids the common error of subtracting the percentage number as if it were a dollar value.

Where do most marks get lost in this topic?+

In the explain/justify parts of questions (Criterion C), not the calculations. Students often calculate correctly but fail to explain why a relationship is proportional or why chaining multipliers is necessary.

Get the full MYP 1 Ratio, Proportion & Percentages notes

Complete worked examples with every trap and fix explained step by step All definitions, formulas, and key takeaways in one organised revision note Extra mock papers and exam-style questions to practise Criterion A and C skills Clear breakdowns of every common mistake examiners see at MYP 1

Get the Ratio, Proportion & Percentages notes on RevisionPrep →

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