Math // Ratio & Proportion

Percentages (age 12+)

Solve problems involving percentage increase, percentage decrease, finding the original value after a percentage change, and calculating simple interest

Ages
12 to 14
Type
Method
Builds on
3 topics
Leads to
1 topic
Standards
2 codes

Quick check

If a bike originally costs £250 and its price goes up by 8%, can the learner work out the new price — and if a savings account earns 3% simple interest per year, how much interest would £500 earn in 4 years?

Ask this to see whether the topic is secure.

Topic // Mastery

What mastery looks like

What a learner can do once this topic is secure.
  • Calculate a 15% increase on £240 using a decimal multiplier (× 1.15)
  • Find the original price before a 20% discount resulted in a sale price of £64
  • Calculate simple interest on £500 at 3% per annum for 4 years

Topic // Map

Where this fits

What to learn first and what this topic opens up, with the reason for each link. Required links come first.
  • RequiredMust come first
  • HelpfulHelps, but not essential

Before this

3 topics
  • Decimals and fractions (age 10+)Required

    Fractions // Ages 10 to 11

    Using decimal multipliers for percentage change requires fraction-decimal fluency

  • Calculating PercentagesRequired

    Ratio & Proportion // Ages 10 to 11

    Percentage increase/decrease extends KS2 percentage calculations

  • Proportional Reasoning VocabularyHelpful

    Ratio & Proportion // Ages 11 to 14

    Percentage increase/decrease problems use 'rate', 'proportion', and 'multiplicative relationship' vocabulary

You are here

Percentages (age 12+)

Ages 12 to 14 // Method

Leads to

1 topic

Topic // Research-informed

Teaching notes

From the Marble Skill Taxonomy. These notes have not been validated in classrooms, so treat them as a starting point for your own judgement.

The approach

Percentage change taught as a single multiplication by a scale factor: increase by r% is ×(1 + r/100), decrease by r% is ×(1 − r/100). This reframing makes reverse percentages tractable (divide by the multiplier), exposes why an increase then an equal-percentage decrease does not return to the start (1.2 × 0.8 = 0.96, because the second change acts on a changed base), and distinguishes simple interest (a fixed percentage of the original principal each period) from compound. Learners build the decimal multiplier explicitly, solve forward and reverse problems, confront the 'it cancels out' intuition with computation, and compute simple interest on a fixed base.

Common misconceptions

  • Treats a percentage change as addition/subtraction rather than a single multiplication (e.g. 240 + 0.15 or 240 + 1.15 rather than ×1.15)
  • Reverse-percentage error: to find the original before a 20% discount, adds 20% of the SALE price back on (64 + 20% of 64) instead of dividing by 0.8
  • Assumes an increase then a decrease by the same percentage returns to the original (subtracts percents: 20% up then 20% down = no change), not seeing the base has changed
  • Builds the wrong multiplier: 5% increase as ×1.5, or 6% as ×0.6 (decimal misplacement)
  • Computes simple interest on a growing balance (compound) or forgets to multiply by the number of years

Useful representations

  • Decimal-multiplier form of change: ×(1 + r/100) and ×(1 − r/100)
  • Paired multiplier cards showing reverse = reciprocal (×1.25 ↔ ×0.8, product 1)
  • Bar model where the original is 100% and the change adds/removes a shaded strip
  • Reverse-percentage arrow: original →(×0.8)→ new, so new →(÷0.8)→ original
  • Simple-interest table: same interest (r% of P) added each year for t years

Teaching practices

  • Teach every percentage change as a single multiplier: increase by r% is ×(1 + r/100), decrease by r% is ×(1 − r/100) — e.g. +15% is ×1.15, −20% is ×0.8.
  • Explicitly correct the multiplier: a 5% increase is ×1.05, NOT ×1.5; a 6% amount is ×0.06, not ×0.6 — decimal placement is the classic slip.
  • Teach reverse percentages as the inverse operation: if the new value is the original ×0.8, then original = new ÷ 0.8 — never 'add the percentage back onto the new value'.
  • Confront 'increase then decrease by the same % returns to the start' directly: show 1.2 × 0.8 = 0.96, and the classic case that a price up 50% must then fall by 33⅓% (not 50%) to return — because the base has changed.
  • Distinguish simple interest (a fixed percentage of the ORIGINAL principal each period, so total = P × r × t) from compound (interest on a growing balance); keep the base fixed for simple interest.
  • Use a bar/100-grid and pair decimal multipliers with their reverse (×1.25 pairs with ×0.8) so the reciprocal relationship of reverse percentages is visible.

Activities to try

  1. 1

    The price-changer: every change is a multiplier

    Why it works: A percentage increase or decrease is one multiplication by (1 ± r/100), not an add-on or subtract-off, and the decimal multiplier must be built carefully: a 5% increase is ×1.05, not ×1.5; and finding 6% OF a value is ×0.06, not ×0.6 (distinct from a 6% decrease, which is ×0.94).

    1. Start with a value shown as a bar (drawn on paper or on screen), e.g. £240, apply a 15% increase, write the change as a single multiplication ×1.15, and read off the new value (£276).
    2. Apply a 20% decrease to a value and write it as ×0.8; write a 6% decrease as ×0.94 (100% − 6% = 94%).
    3. Deliberately contrast ×1.15 with ×1.5 and ×0.15 for a 15% increase, stating which is correct and why (100% + 15% = 115% = 1.15).
    4. Match pairs of multiplier cards and check which pairs multiply to 1 (e.g. ×1.25 ↔ ×0.8), noticing the multiplier that UNDOES a change is its reciprocal.
    5. OPTIONAL interaction if a live tool is available: drag or set the percentage and watch the bar grow/shrink while the displayed multiplier and new value update together.
  2. 2

    Reverse percentages: undo by dividing

    Why it works: To find an original value before a percentage change, divide by the multiplier that produced the change; the percentage is taken of the original, not of the new/sale value, so adding the percentage back onto the new value is wrong.

    1. Given a sale price of £64 that is already 20% off, find the original price by recognising sale = original ×0.8, so original = £64 ÷ 0.8 = £80.
    2. Check the answer: reduce the found original by 20% and confirm it returns to the sale price (20% of £80 is £16, so £80 → £64).
    3. Explicitly compare the correct method (÷0.8) with the common wrong method (£64 + 20% of £64 = £76.80) and check which one survives the reversal test (£76.80 reduced by 20% gives £61.44, not £64).
    4. Draw a reverse arrow: original →(×0.8)→ new, so new →(÷0.8)→ original.
  3. 3

    Does it cancel out? Chaining changes on a changing base

    Why it works: Equal up-then-down percentages do not cancel, because each percentage acts on whatever base is current; the two multipliers are not reciprocals (1.2 × 0.8 = 0.96, not 1), so the net result is a fall. (Note this differs from the reciprocal pair that UNDOES a single change, e.g. ×1.25 then ×0.8 = 1.)

    1. Predict first: if a price goes up 20% and then down 20%, is it back where it started?
    2. Compute the chain as multipliers: ×1.2 then ×0.8 = ×0.96, a 4% net fall; test on a concrete value, e.g. £100 → £120 → £96.
    3. Identify why: after the increase the base is larger (£120), so the 20% decrease removes more (£24) than the 20% increase added (£20).
    4. Explore the classic case: a £20 item up 50% becomes £30; find what percentage fall returns it to £20 (£10 off £30 is 33⅓%, not 50%).
    5. OPTIONAL interaction if a chain tool is available: apply +20% then −20% and watch the bar land short of the start, with the post-increase base highlighted.

Topic // Standards

Curriculum standards

Codes for the curriculum statements this topic maps to. Each publisher owns its standard text.
Common Core State Standards for Mathematics
7.RP.3
The national curriculum in England: Key stages 1 and 2 framework document
KS3.Maths.Ratio.8

Math // Ratio & Proportion

More in Ratio & Proportion

All 18 topics