MYP 3 · Maths

PERCENTAGE

Finding the original amount

QUESTION 1 [3 marks] — Criterion A Medium
After a 15% increase, a salary becomes \$46,000.
a. Write an equation connecting the original salary $x$ to the new salary, using a multiplier.
[1]
Show Solution
$$1.15x = 46000$$
b. Solve for $x$ to find the original salary.
[2]
Show Solution
$$x = 46000 \div 1.15 = \$40000$$
QUESTION 2 [3 marks] — Criterion A Medium
After a 20% discount, a jacket costs \$68.
a. Write an equation connecting the original price $x$ to the sale price, using a multiplier.
[1]
Show Solution
$$0.80x = 68$$
b. Solve for $x$ to find the original price.
[2]
Show Solution
$$x=68\div0.80=\$85$$
QUESTION 3 [6 marks] — Criterion B Medium
Investigate the relationship between a percentage increase/decrease and the multiplier needed to REVERSE it.
a. A value increases by 25% (multiplier 1.25). If the new value is 125, find the original value by dividing by 1.25.
[2]
Show Solution
$$125 \div 1.25 = 100$$
b. Now suppose, incorrectly, someone tried to reverse a 25% increase by simply decreasing the new value by 25%. Calculate $125\times0.75$ and compare to the correct original value (100).
[2]
Show Solution
$125\times0.75=93.75$, which does NOT match the correct original value of 100 — 'decreasing by the same percentage' is not the correct way to reverse an increase.
c. Explain why dividing by the ORIGINAL multiplier (1.25) is the correct way to reverse an increase, rather than applying the 'opposite' percentage as a decrease.
[2]
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Since original $\times 1.25 = $ new value, to reverse the operation you must do the inverse operation: divide by 1.25 (not multiply by a different number like 0.75). Multiplying by 0.75 undoes a DIFFERENT operation (a 25% decrease from a different starting point), not the original 25% increase.
QUESTION 4 [5 marks] — Criterion C Medium
A student wants to find the original price before a 30% discount, given a sale price of \$63. They calculate: 'discount amount = 30% of 63 = 18.90, so original = 63+18.90 = 81.90'.
a. Verify whether \$81.90 is actually correct, by checking whether a 30% discount on \$81.90 gives \$63.
[2]
Show Solution
$30\%$ of $81.90 = 24.57$. $81.90-24.57=57.33 \ne 63$. So \$81.90 is NOT correct.
b. Explain the student's error, and find the CORRECT original price.
[3]
Show Solution
The error: the student calculated 30% of the SALE price (63), but the discount was actually 30% of the ORIGINAL price, which is a different (larger) amount — you can't find 30% of a number you don't know yet using the wrong base. Correct method: $0.70x=63 \Rightarrow x=63\div0.70=\$90$.
QUESTION 5 [5 marks] — Criterion D Medium
A real estate agent reports that after a 12% price increase over the past year, a house is now valued at \$392,000.
a. Find the house's value one year ago (before the increase).
[2]
Show Solution
$$392000 \div 1.12 = \$350000$$
b. The owner is considering selling now versus waiting another year, IF prices continue rising at the same 12% rate. Estimate the value in one more year, and find the total increase in value over the full 2 years (from \$350,000).
[3]
Show Solution
Value in 1 more year: $392000\times1.12=\$439040$. Total 2-year increase: $439040-350000=\$89040$.
QUESTION 6 [3 marks] — Criterion A Medium
After a 12% increase, an engineer's annual salary became \$156.80 thousand.
a. Write an equation connecting the original salary $x$ (in thousands) to the new salary.
[1]
Show Solution
$$1.12x=156.80$$
b. Solve for $x$.
[2]
Show Solution
$$x=156.80\div1.12=140 \text{ (i.e. \$140{,}000)}$$
QUESTION 7 [3 marks] — Criterion A Medium
After a 15% discount, a laptop's price became \$680.20.
a. Write an equation connecting the original price $x$ to the sale price, and solve for $x$, correct to 2 decimal places.
[3]
Show Solution
$0.85x=680.20 \Rightarrow x=680.20\div0.85\approx\$800.24$.
QUESTION 8 [6 marks] — Criterion B Hard
Investigate the RELATIVE ERROR introduced when someone mistakenly reverses a percentage DECREASE using the wrong operation — e.g. adding the percentage back on, instead of dividing by the correct multiplier.
a. A price of \$85 resulted from a 15% decrease. Find the TRUE original price using the correct method (dividing by 0.85).
[2]
Show Solution
$$85\div0.85=\$100$$
b. A student instead tries to 'reverse' the decrease by simply ADDING 15% to \$85. Find their (incorrect) answer, and calculate the PERCENTAGE ERROR of their method relative to the true original price of \$100.
[4]
Show Solution
Student's method: $85\times1.15=\$97.75$. This is INCORRECT (true value is \$100). Percentage error: $\frac{100-97.75}{100}\times100=2.25\%$ — a fairly small-looking error in this case, but the method itself is fundamentally flawed and the error size would grow for larger percentage decreases.
QUESTION 9 [6 marks] — Criterion B Hard
Investigate how the SAME reversal error (adding back the percentage instead of using the correct divisor) behaves differently for a LARGE percentage decrease, using a 60% decrease resulting in a price of \$40.
a. Find the TRUE original price (dividing by the correct multiplier, $1-0.60=0.40$).
[2]
Show Solution
$$40\div0.40=\$100$$
b. Find the INCORRECT answer using the 'just add back 60%' method, and calculate the percentage error THIS time. Compare it to the smaller error found in the 15%-decrease case, and explain why the error gets WORSE for bigger percentage decreases.
[4]
Show Solution
Incorrect method: $40\times1.60=\$64$. True value is \$100, so percentage error: $\frac{100-64}{100}\times100=36\%$ — MUCH larger than the 2.25% error found for the 15% case. This happens because for larger decreases, the gap between the correct divisor ($1-p$) and the incorrect 'add-back' multiplier ($1+p$) grows disproportionately larger as $p$ increases, making the flawed method increasingly inaccurate for bigger percentage changes.
QUESTION 10 [5 marks] — Criterion C Medium
A student solves 'after a 20% increase, a value became 480; find the original' by writing: '480 is 20% more, so 20% of 480 is 96, and 480-96=384 is the original.'
a. Verify whether \$384 is actually correct, by checking whether a 20% increase on 384 gives 480.
[2]
Show Solution
$384\times1.20=460.80\ne480$. So \$384 is NOT correct.
b. Explain the student's error precisely, and find the CORRECT original value.
[3]
Show Solution
The student calculated 20% of the NEW value (480) and subtracted it — but the 20% increase was based on the ORIGINAL (unknown) value, not the new one, so subtracting 20% of 480 doesn't correctly reverse the operation. Correct method: $1.20x=480 \Rightarrow x=480\div1.20=400$.
QUESTION 11 [4 marks] — Criterion C Medium
A student solves a 'find the original price' reverse-percentage problem correctly, but writes their entire solution as: '650/0.82=792.68'.
a. Rewrite this as a complete, clearly communicated solution: define what the original problem must have been (a price reduced by what percentage, resulting in what sale price), state your equation clearly with defined variables, and present the final answer in a full sentence with appropriate rounding.
[4]
Show Solution
Suppose a price was decreased by 18% (since $1-0.82=0.18$), resulting in a sale price of \$650. Let $x$ represent the original price. Then: $$0.82x=650$$ Solving: $$x=650\div0.82\approx\$792.68$$ Therefore, the original price was approximately \$792.68, before the 18% discount was applied.
QUESTION 12 [4 marks] — Criterion C Medium
Two students both correctly find that an original value was \$250, but present their solutions differently: Student 1 writes '$1.35x=337.50, x=250$'. Student 2 writes '$337.50\div1.35=250$' with no other explanation.
a. Evaluate which student's solution demonstrates BETTER mathematical communication, referencing what information is present in one solution but missing from the other (e.g. does either show what $x$ represents, or state the final answer in context?).
[4]
Show Solution
Student 1's solution is SLIGHTLY better organized, showing an equation before solving — but BOTH solutions share a key weakness: neither explicitly defines what $x$ represents, what the 337.50 and 1.35 refer to in context, or states the final answer as a complete sentence (e.g. 'the original price was \$250'). A genuinely well-communicated solution would include: a clear statement of what is being found, defined variables, the equation, the working, AND a concluding sentence connecting the numerical answer back to the original question's context.