Mathematics • Year 8 • Unit 1 • Lesson 11
Successive % Changes — Mixed Challenge
Pull together everything from Lesson 11: combined multipliers, equivalent single percentage changes, and the “different bases” trap. Six mixed problems, one “find the mistake”, and one open-ended challenge.
1. Mixed problems — choose the right move
Each question uses a different combination of ideas from Lesson 11. Decide which move applies before you start writing. Show your working. 3 marks each
1.1 A $\$250$ bike has its price increased by 10%, then a further 20%. Find the final price and the equivalent single percentage change.
1.2 Find the combined multiplier and the equivalent single percentage change for the sequence: +25%, then −20%.
1.3 A $\$1000$ investment falls 25% in year 1, then rises 25% in year 2. Find the value at the end of year 2 and the overall percentage change from $\$1000$.
1.4 Two discounts of 15% and 15% are applied in succession. Show that the equivalent single discount is not 30%, and find what it actually is.
1.5 A $\$60$ shirt has three changes applied: +20%, −10%, −5%. Find the final price and the combined multiplier.
1.6 A combined multiplier is 0.84. (a) Does it represent an overall rise or fall? (b) What is the equivalent single percentage change? (c) Apply it to a starting value of $\$2500$ to find the final value.
2. Find the mistake
Another student is trying to find the final price after a $\$200$ item gets two successive 10% discounts. Their working is below. Exactly one line contains the key mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — “10% off, then 10% off” on $\$200$:
Line 1: Two 10% discounts = 20% off altogether.
Line 2: 20% of $\$200 = \$40$.
Line 3: Final price = $\$200 - \$40 = \$160$.
Line 4: So the customer saves $\$40$ — same as one 20% discount.
(a) Which line contains the key mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, using multipliers, including the corrected final price and saving.
Stuck? The second 10% is taken off the price AFTER the first discount — a smaller base — so it removes fewer dollars than the first 10%. Use 0.90 × 0.90 = 0.81.3. Open-ended challenge — design a discount that beats 30%
This question has more than one valid answer. 4 marks
3.1 A shop wants to advertise “X% off, then Y% off” (two whole-number percentages, each between 5% and 25%) that is actually cheaper than a flat “30% off”.
For your two-discount combination:
(i) Write down your chosen X% and Y%.
(ii) Find the combined multiplier.
(iii) Find the equivalent single percentage discount.
(iv) Confirm it beats 30% off (i.e., your equivalent single % is greater than 30%).
(v) On a $\$200$ item, find both final prices and the dollar saving over the flat 30%.
Bonus: Try to make your combination one where the X% + Y% sum is LESS than 30% but the combined effect still beats it. (Hint: that's impossible — explain in one line why.)
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $\$250$, +10%, +20%
$250 \times 1.10 \times 1.20 = 250 \times 1.32 = \textbf{\$330}$. Equivalent single = +32% (not +30%).
1.2 — +25%, then −20%
Combined = $1.25 \times 0.80 = \textbf{1.00}$. Equivalent single change = 0% (these two happen to exactly cancel — rare!).
1.3 — $\$1000$, −25%, +25%
$1000 \times 0.75 \times 1.25 = 1000 \times 0.9375 = \textbf{\$937.50}$. Overall change = −6.25% (a loss, even though the percentages look symmetric).
1.4 — Two 15% discounts
Combined multiplier = $0.85 \times 0.85 = 0.7225$. Equivalent single discount = $1 - 0.7225 = 0.2775 = \textbf{27.75%}$ off — not 30% off. The second 15% acts on a smaller base, so less is removed than naive addition predicts.
1.5 — $\$60$, +20%, −10%, −5%
Combined = $1.20 \times 0.90 \times 0.95 = \textbf{1.026}$. Final price = $60 \times 1.026 = \textbf{\$61.56}$. Overall +2.6%.
1.6 — Combined multiplier 0.84
(a) Fall, because 0.84 < 1.
(b) Equivalent single change = $0.84 - 1 = -0.16 = \textbf{16% decrease}$.
(c) $\$2500 \times 0.84 = \textbf{\$2100}$.
2 — Find the mistake
(a) The key mistake is on Line 1 (which then makes Lines 2 and 4 wrong too).
(b) The student added the percentages. Successive percentage changes don't add — the second 10% is on the smaller, already-discounted price, so the actual saving is less than 20%.
(c) Corrected working with multipliers:
10% off → multiplier 0.90. Two in a row → combined multiplier $0.90 \times 0.90 = 0.81$.
Final price = $\$200 \times 0.81 = \textbf{\$162}$ (not $\$160$).
Saving = $\$200 - \$162 = \textbf{\$38}$ (not $\$40$).
Equivalent single discount = $1 - 0.81 = \textbf{19% off}$ (not 20%).
3 — Open-ended challenge (sample solution)
We need two percentages X% and Y% (each 5–25%) whose combined multiplier is less than 0.70.
Choice 1: X = 20%, Y = 15%.
Combined multiplier = $0.80 \times 0.85 = 0.68$.
Equivalent single = $1 - 0.68 = 32%$ off.
On $\$200$: stacked = $\$200 \times 0.68 = \$136$; flat 30% = $\$200 \times 0.70 = \$140$. Saving over flat 30% = $\$4$. ✓
Choice 2: X = 25%, Y = 10%.
Combined = $0.75 \times 0.90 = 0.675 \to 32.5%$ equivalent — even better.
On $\$200$: stacked = $\$135$; flat 30% = $\$140$. Saving = $\$5$. ✓
Bonus: If X + Y < 30, the combined discount must also be less than 30%. The combined multiplier $(1-X/100)(1-Y/100) = 1 - X/100 - Y/100 + XY/10\,000$, which is GREATER than $1 - (X+Y)/100$ by the small positive term $XY/10\,000$. So the combined discount is always LESS than X + Y. If X + Y < 30, the combined discount is even smaller — it cannot beat 30%.
Marking: 1 mark for a valid (X%, Y%) pair within range; 1 mark for the correct combined multiplier and equivalent single %; 1 mark for the dollar comparison on $\$200$; 1 bonus mark for the impossibility explanation.