MYP 3 · Maths
PERCENTAGE
71 questions across 7 sub-topics
Use the Sub-Topic filter above to focus on one.
Percentage
Expressing one quantity as a percentage of another
Finding a percentage of a quantity
The unitary method in percentage
Percentage increase and decrease
Finding the original amount
Simple interest
Percentage 10 questions
QUESTION 1
4 marks
Criterion A
Medium
Convert each of the following:
a.
$35\%$ to a fraction in simplest form.
[1] b.
$\frac{9}{25}$ to a percentage.
[1] c.
$0.625$ to a percentage.
[1] d.
$120\%$ to a decimal.
[1] Show complete worked solution
(a)
$$35\% = \frac{35}{100} = \frac{7}{20}$$
(b)
$$\frac{9}{25} = \frac{36}{100} = 36\%$$
(c)
$$0.625 = 62.5\%$$
(d)
$$120\% = 1.2$$
QUESTION 2
5 marks
Criterion A
Medium
Order the following from smallest to largest: $\frac{3}{8}$, $42\%$, $0.39$, $\frac{2}{5}$.
a.
Convert all four values to percentages.
[3] b.
Write the original four values in order from smallest to largest.
[2] Show complete worked solution
(a)
$\frac{3}{8}=37.5\%$. $42\%$ (already). $0.39=39\%$. $\frac{2}{5}=40\%$.
(b)
$$\frac{3}{8}, \ 0.39, \ \frac{2}{5}, \ 42\%$$
QUESTION 3
5 marks
Criterion B
Medium
Investigate the relationship between a fraction $\frac{n}{100}$ and its percentage form.
a.
Write $\frac{17}{100}$, $\frac{53}{100}$, and $\frac{8}{100}$ as percentages.
[2] b.
State the rule connecting a fraction with denominator 100 and its percentage.
[1] c.
Using this rule, explain why $\frac{n}{100} \times 100\% = n\%$ for ANY fraction $\frac{n}{100}$, algebraically.
[2] Show complete worked solution
(a)
$17\%$, $53\%$, $8\%$.
(b)
The numerator of the fraction (out of 100) IS the percentage number directly.
(c)
$\frac{n}{100}\times100 = \frac{100n}{100}=n$, so $\frac{n}{100}$ expressed as a percentage is always exactly $n\%$ — multiplying any fraction by 100% is really just multiplying by 100 and attaching the % symbol.
QUESTION 4
4 marks
Criterion C
Medium
A student says: '50% means half of 50, so 50% of 80 is 40... wait, that's the same as just finding half of 80. So percent doesn't really matter, I can just think in halves and quarters.'
a.
Explain what 'percent' actually means (breaking down the word itself), and why the student's reasoning happens to work for 50% specifically.
[2] b.
Show why the student's 'just use fractions like half and quarter' approach breaks down for a percentage like 37%.
[2] Show complete worked solution
(a)
'Percent' comes from Latin 'per centum', meaning 'per hundred' — so $50\%$ literally means '50 out of every 100', which happens to simplify to exactly $\frac{1}{2}$ (half). This is why the student's shortcut works for 50%, but only because $50/100$ simplifies neatly to $1/2$.
(b)
$37\%=\frac{37}{100}$, which does NOT simplify to a nice fraction like $\frac{1}{2}$ or $\frac{1}{4}$ — there's no simple 'halves and quarters' shortcut for 37%, so the general percentage method (multiplying by $\frac{37}{100}$ or $0.37$) is still needed.
QUESTION 5
4 marks
Criterion D
Medium
A store's price tag shows: 'Was \$85, Now 30% OFF'. A shopper mentally estimates the discount as 'about \$25' before checking exactly.
a.
Calculate the EXACT discount amount and the new price.
[2] b.
Was the shopper's mental estimate of '\$25' reasonable? Explain your reasoning.
[2] Show complete worked solution
(a)
Discount $=30\% \times 85 = 0.30\times85=\$25.50$. New price $=85-25.50=\$59.50$.
(b)
Yes, it was a reasonable estimate — the exact discount (\$25.50) is very close to \$25, showing the shopper's quick mental approximation (perhaps using 'about 30% of about 85') was accurate enough for a quick real-world check.
QUESTION 6
5 marks
Criterion A
Medium
A renewable energy report states that solar panel efficiency has 'increased by 40% relative to a decade ago', and that current panels convert 22% of sunlight into electricity.
a.
If the CURRENT efficiency (22%) represents a 40% increase over the efficiency a decade ago, find the efficiency a decade ago, correct to 1 decimal place.
[3] b.
Express both efficiencies, 22% now and approximately 15.7% a decade ago, as fractions.
[2] Show complete worked solution
(a)
$1.40x=22 \Rightarrow x=22\div1.40\approx15.7\%$.
(b)
$22\%=\frac{22}{100}=\frac{11}{50}$. $15.7\%\approx\frac{157}{1000}$ (does not simplify further, since $157$ is prime).
QUESTION 7
6 marks
Criterion A
Medium
True or False: 'Increasing a value by 50%, then decreasing the RESULT by 50%, always returns the original value.' Justify your answer fully with a worked example.
a.
Test the statement using a starting value of \$80: increase by 50%, then decrease the new value by 50%.
[3] b.
State whether the claim is TRUE or FALSE based on your test, and explain in general terms (without redoing the arithmetic) why a percentage increase followed by the SAME percentage decrease can never exactly return to the start (except when the percentage is 0%).
[3] Show complete worked solution
(a)
After $+50\%$: $80\times1.5=120$. After $-50\%$ of 120: $120\times0.5=60$.
(b)
FALSE — the result (\$60) is less than the original (\$80). In general, since the decrease is applied to the LARGER (increased) amount, the same percentage removes MORE actual value than the percentage added back — this asymmetry means equal-percentage up-then-down changes always result in a net decrease, for any nonzero percentage.
QUESTION 8
9 marks
Criterion B
Hard
Investigate the pattern in successive percentage discounts, e.g. 'take 20% off, then take a FURTHER 20% off the new price' (a common retail strategy), compared to a single 40% discount.
a.
Starting with a \$200 item, apply 20% off, then a further 20% off the new price. Find the final price.
[3] b.
Compare this to applying a SINGLE 40% discount directly to \$200. Which gives the customer a better deal, and by how much?
[3] c.
Explain algebraically why two successive $p\%$ discounts are always LESS beneficial to the customer than a single $2p\%$ discount (for $p>0$), by comparing the multipliers $(1-\frac{p}{100})^2$ and $(1-\frac{2p}{100})$.
[3] Show complete worked solution
(a)
After first 20%: $200\times0.8=160$. After second 20%: $160\times0.8=128$.
(b)
Single 40% discount: $200\times0.6=120$. The single 40% discount ($120) is BETTER for the customer than the two successive 20% discounts ($128), by $128-120=\$8$.
(c)
Two successive discounts multiply the price by $(1-\frac{p}{100})^2$, while a single $2p\%$ discount multiplies by $(1-\frac{2p}{100})$. Expanding $(1-\frac{p}{100})^2 = 1-\frac{2p}{100}+\frac{p^2}{10000}$ — this has an EXTRA positive term $\frac{p^2}{10000}$ compared to the single-discount multiplier, meaning the two-step discount always leaves a HIGHER final price (less benefit to the customer) than the equivalent single discount, for any $p>0$.
QUESTION 9
5 marks
Criterion C
Medium
A classmate calculates a 15% tip on a \$64 restaurant bill by finding 10% (\$6.40) then simply adding \$1 to 'estimate the extra 5%', getting \$7.40 total tip.
a.
Calculate the EXACT 15% tip, and compare it to the classmate's estimate to determine how accurate their shortcut was.
[3] b.
Explain a MORE reliable mental-maths shortcut for finding 15% (hint: 15% = 10% + 5%, and 5% is exactly half of 10%), and show it gives the exact answer.
[2] Show complete worked solution
(a)
Exact: $15\%\times64=0.15\times64=\$9.60$. The classmate's estimate (\$7.40) is significantly LOWER than the exact value — a notable underestimate of $9.60-7.40=\$2.20$.
(b)
Reliable method: 10% of 64 is \$6.40; half of that (5%) is \$3.20; adding these gives $6.40+3.20=\$9.60$ — matching the exact value, unlike the classmate's rough 'add \$1' guess.
QUESTION 10
6 marks
Criterion D
Hard
An airline reports that a particular flight route's on-time performance was 78% last year, based on 2,400 total flights on that route.
a.
Find the number of flights that were on-time, and the number that were delayed.
[2] b.
The airline sets a target to IMPROVE on-time performance to 85% next year, while flight volume is expected to grow by 10% (to 2,640 flights). Find the number of ADDITIONAL on-time flights needed next year compared to this year's actual on-time count, to hit both targets simultaneously.
[4] Show complete worked solution
(a)
On-time: $0.78\times2400=1872$. Delayed: $2400-1872=528$.
(b)
Next year's target on-time flights: $0.85\times2640=2244$. Additional on-time flights needed vs this year: $2244-1872=372$ more on-time flights.
Expressing one quantity as a percentage of another 7 questions
QUESTION 1
6 marks
Criterion B
Medium
Investigate what happens to the percentage when you SWAP which quantity is treated as the 'part' and which is the 'whole'.
a.
Find what percentage 15 is of 60, then find what percentage 60 is of 15.
[3] b.
These two answers are very different. Explain why swapping the 'part' and 'whole' changes the result so dramatically, and what a percentage over 100% actually means.
[3] Show complete worked solution
(a)
15 as a % of 60: $\frac{15}{60}\times100=25\%$. 60 as a % of 15: $\frac{60}{15}\times100=400\%$.
(b)
Swapping part and whole essentially flips the fraction (from $\frac{15}{60}$ to $\frac{60}{15}$), giving a completely different value — division is not commutative. A percentage over 100% simply means the 'part' is actually LARGER than the 'whole' being compared to (here, 60 is 4 times as large as 15).
QUESTION 2
4 marks
Criterion C
Medium
A student calculates 'what percentage is 40 of 25' by computing $\frac{25}{40}\times100$, getting $62.5\%$.
a.
Identify the student's error, and give the correct calculation and answer.
[2] b.
Explain, using the phrase itself ('what percentage is 40 OF 25'), a reliable way to remember which number goes on top of the fraction.
[2] Show complete worked solution
(a)
The student divided the wrong way around — to find what percentage 40 IS OF 25, the 'part' (40) must go on top: $\frac{40}{25}\times100=160\%$, not $\frac{25}{40}\times100$.
(b)
The number right after 'is' (or being described) goes on TOP (the part), and the number right after 'of' goes on the BOTTOM (the whole) — so 'what percentage IS 40 OF 25' becomes $\frac{40}{25}$.
QUESTION 3
5 marks
Criterion D
Medium
In a basketball season, Player A scored 84 points out of 120 shot attempts. Player B scored 105 points out of 150 shot attempts.
a.
Find each player's shooting success rate as a percentage.
[3] b.
A commentator claims 'Player B is clearly the better shooter since they scored more total points (105 vs 84)'. Evaluate this claim using your percentages.
[2] Show complete worked solution
(a)
Player A: $\frac{84}{120}\times100=70\%$. Player B: $\frac{105}{150}\times100=70\%$.
(b)
The claim is misleading — both players have exactly the same shooting SUCCESS RATE (70%). Player B simply took more shots overall (150 vs 120), which is why they scored more total points, not because they were more accurate.
QUESTION 4
5 marks
Criterion A
Medium
A water conservation project reduced a city's daily water usage from 240 million litres to 156 million litres over 5 years.
a.
Find the NEW usage as a percentage of the ORIGINAL usage.
[2] b.
Express the REDUCTION itself (84 million litres) as a percentage of the original usage, and verify your two percentages (from this part and part a) sum to 100%.
[3] Show complete worked solution
(a)
$$\frac{156}{240}\times100=65\%$$
(b)
$\frac{84}{240}\times100=35\%$. Check: $65\%+35\%=100\%$ (makes sense, since 'new usage %' and 'reduction %' together must account for the whole original amount).
QUESTION 5
7 marks
Criterion B
Medium
Investigate what happens to 'what percentage is $a$ of $b$' as $a$ gets closer and closer to $b$, and as $a$ gets closer to 0.
a.
Calculate what percentage 45 is of 50, then what percentage 49 is of 50, then what percentage 50 is of 50.
[3] b.
Calculate what percentage 5 is of 50, then what percentage 1 is of 50, then what percentage 0 is of 50.
[2] c.
Based on both patterns, state (without further calculation) what percentage would result if $a$ EQUALS $b$ exactly, and explain why this makes intuitive sense.
[2] Show complete worked solution
(a)
$\frac{45}{50}\times100=90\%$. $\frac{49}{50}\times100=98\%$. $\frac{50}{50}\times100=100\%$.
(b)
$\frac{5}{50}\times100=10\%$. $\frac{1}{50}\times100=2\%$. $\frac{0}{50}\times100=0\%$.
(c)
When $a=b$, the percentage is always $100\%$ — this makes sense because 'what percentage is $a$ of $b$' is really asking 'how much of the whole ($b$) does $a$ represent', and when $a$ equals the whole amount exactly, it represents ALL of it, i.e. 100%.
QUESTION 6
6 marks
Criterion B
Hard
Investigate whether 'what percentage is $a$ of $b$' plus 'what percentage is $b$ of $a$' always sums to 100%, using $a=30, b=70$.
a.
Calculate both percentages: what percentage is 30 of 70, and what percentage is 70 of 30.
[3] b.
Do these two percentages sum to 100%? Explain why the '$a$ of $b$' plus '$b$ of $a$' pattern does NOT generally work the same way as the 'part + remainder = 100%' pattern investigated earlier (which specifically involved a part and the REMAINDER of the same whole).
[3] Show complete worked solution
(a)
$\frac{30}{70}\times100\approx42.9\%$. $\frac{70}{30}\times100\approx233.3\%$.
(b)
No — they sum to $42.9+233.3=276.2\%$, nowhere near 100%. This is different from the earlier pattern because here, $a$ and $b$ are each being treated as the 'whole' in turn (two DIFFERENT wholes), rather than $a$ being a part of ONE fixed whole $b$ with the remainder being $b-a$ — these are fundamentally different mathematical relationships, so there's no reason to expect them to sum to 100%.
QUESTION 7
3 marks
Criterion C
Medium
A student calculating 'what percentage is 18 of 45' writes $\frac{45}{18}\times100\approx250\%$, then says 'that seems too big, so I think I need to just flip it', getting the right final answer by trial and error rather than by understanding WHY they needed to flip it.
a.
Calculate the CORRECT answer, and explain — using the actual SIZES of 18 and 45 — why an answer over 100% should have immediately signalled an error, without needing trial and error.
[3] Show complete worked solution
(a)
Correct: $\frac{18}{45}\times100=40\%$. Since 18 is SMALLER than 45, the percentage 18 represents OF 45 must be under 100% (a part can't exceed 100% of a whole that's bigger than it) — recognizing this BEFORE calculating would have immediately flagged the first (250%) attempt as using the numbers the wrong way round, without needing to guess-and-check.
Finding a percentage of a quantity 7 questions
QUESTION 1
5 marks
Criterion B
Medium
Investigate a mental-maths shortcut method for finding percentages, using 10%, 5%, and 1% as 'building blocks'.
a.
Find 10% of \$240 and 1% of \$240 using simple mental division.
[2] b.
Using only your answers from (a) (adding or combining them, without recalculating from scratch), find 23% of \$240.
[3] Show complete worked solution
(a)
10% of 240 = 24. 1% of 240 = 2.4.
(b)
$23\% = 2\times10\% + 3\times1\% = 2(24)+3(2.4)=48+7.2=55.2$. So 23% of \$240 is \$55.20.
QUESTION 2
5 marks
Criterion C
Medium
A student finds 45% of 80 using decimals ($0.45\times80$) and gets 36. A classmate finds it using fractions ($\frac{45}{100}\times80$) and also gets 36.
a.
Show both methods in full, confirming both give 36.
[3] b.
Explain why these two methods always give the same result, using the definition of a decimal.
[2] Show complete worked solution
(a)
Decimal method: $0.45\times80=36$. Fraction method: $\frac{45}{100}\times80=\frac{3600}{100}=36$. Both confirm 36.
(b)
A decimal like 0.45 IS simply the fraction $\frac{45}{100}$ written in a different form — they represent exactly the same value, so multiplying by either form must always give an identical result.
QUESTION 3
4 marks
Criterion D
Medium
A restaurant bill totals \$68.50 before tax and tip. Sales tax is 8%, and a customary tip of 18% is added to the pre-tax amount.
a.
Calculate the sales tax amount.
[1] b.
Calculate the tip amount (based on the pre-tax bill).
[1] c.
Find the total amount the customer pays (bill + tax + tip).
[2] Show complete worked solution
(a)
$$8\% \times 68.50 = \$5.48$$
(b)
$$18\% \times 68.50 = \$12.33$$
(c)
$$68.50+5.48+12.33 = \$86.31$$
QUESTION 4
6 marks
Criterion A
Medium
A property developer pays an annual property tax of 0.85% on a building valued at \$185,000.
a.
Calculate the annual property tax.
[2] b.
The building's value is expected to increase by 12% over the next 2 years. Find the NEW building value, then the NEW annual tax at the SAME 0.85% rate, and the total INCREASE in tax paid per year as a result.
[4] Show complete worked solution
(a)
$$0.0085\times185000=\$1572.50$$
(b)
New value: $185000\times1.12=\$207200$. New tax: $0.0085\times207200=\$1761.20$. Increase: $1761.20-1572.50=\$188.70$ per year.
QUESTION 5
7 marks
Criterion D
Hard
A hospital uses a standard medication dosage of $25\text{ mg}$ per kilogram of a patient's body mass.
a.
For a patient with body mass 72kg, calculate the dosage in mg, showing your unit conversion clearly.
[3] b.
The maximum safe dosage is $2000\text{ mg}$. Find the greatest patient body mass for which the standard dosage remains safe, and explain why a maximum cap is useful.
[4] Show complete worked solution
(a)
$$72\text{ kg}\times25\text{ mg/kg}=\boxed{1800\text{ mg}}.$$
(b)
Let $m$ be the mass in kilograms.
$$25m\le2000$$
$$m\le\frac{2000}{25}=\boxed{80\text{ kg}}.$$
A maximum cap prevents a proportional rule from producing an unsafe absolute dose for a patient with a large body mass.
QUESTION 6
5 marks
Criterion B
Medium
Investigate a mental-maths method for finding 'awkward' percentages like 17.5% (a common tax rate in some countries), by breaking it into 10% + 5% + 2.5%.
a.
For a bill of \$340, find 10%, 5%, and 2.5% separately.
[3] b.
Add these together to find 17.5% of \$340, then verify by calculating $0.175\times340$ directly.
[2] Show complete worked solution
(a)
10%: \$34. 5%: \$17. 2.5%: \$8.50.
(b)
$34+17+8.50=\$59.50$. Direct: $0.175\times340=\$59.50$.
QUESTION 7
7 marks
Criterion D
Hard
A solar farm generates enough electricity to power 3,400 homes at full capacity, but currently operates at only 68% of a target output due to a technical issue.
a.
If the 3,400 homes figure represents the TARGET (100%) capacity, find the number of homes actually being powered at the current 68% output.
[2] b.
Engineers restore output at 4 percentage points per week. Find the number of whole weeks needed to reach at least 95% capacity, then calculate the additional number of homes powered after that many complete weeks.
[5] Show complete worked solution
(a)
$$0.68\times3400=2312 \text{ homes}$$
(b)
$95-68=27$ percentage points are needed. Since $27\div4=6.75$, 7 whole weeks are required.
$$68+7(4)=96\%$$
$$0.96(3400)=3264$$
$$3264-2312=\boxed{952\text{ additional homes}}.$$
The unitary method in percentage 9 questions
QUESTION 1
2 marks
Criterion A
Medium
In a survey, 65% of respondents said they prefer tea over coffee, and this represented 130 people.
a.
Using the unitary method, find how many people represent 1% of respondents.
[1] b.
Hence find the total number of people surveyed (100%).
[1] Show complete worked solution
(a)
$$130 \div 65 = 2 \text{ people per 1\%}$$
(b)
$$2 \times 100 = 200 \text{ people}$$
QUESTION 2
3 marks
Criterion A
Medium
A shop states that 8 items cost \$36, and this represents 40% of a customer's total weekly grocery budget.
a.
Find the price of 1 item.
[1] b.
Using the unitary method, find the customer's total weekly grocery budget (100%).
[2] Show complete worked solution
(a)
$$36 \div 8 = \$4.50 \text{ per item}$$
(b)
1% of budget $=\$36\div40=\$0.90$. Total budget $=\$0.90\times100=\$90$.
QUESTION 3
6 marks
Criterion B
Medium
Investigate the unitary method as a general two-step process.
a.
If 35% of a quantity is 105, use the unitary method (find 1%, then find 100%) to find the whole quantity.
[2] b.
Now find the whole quantity directly using division: $105 \div 0.35$. Compare this to your unitary method answer.
[2] c.
Explain, in general terms using $p\%$ and value $v$, why 'finding 1% then multiplying by 100' always gives the same result as 'dividing by $p$ then multiplying by 100', i.e. dividing by $\frac{p}{100}$.
[2] Show complete worked solution
(a)
1% $=105\div35=3$. Whole quantity (100%) $=3\times100=300$.
(b)
$105\div0.35=300$ — identical to the unitary method answer, confirming both approaches are mathematically equivalent, just organised differently.
(c)
Finding 1% means dividing by $p$ (since $p\%$ corresponds to value $v$, 1% corresponds to $v/p$). Multiplying by 100 to reach 100% gives $\frac{v}{p}\times100 = \frac{100v}{p} = v \div \frac{p}{100}$ — showing both methods are algebraically the same operation.
QUESTION 4
4 marks
Criterion C
Medium
A classmate finds the whole amount when 'given that 20% of a number is 50' by calculating $50\times20=1000$.
a.
Explain the classmate's error, and demonstrate the correct unitary method approach.
[3] b.
Verify the correct answer by checking: does 20% of 250 actually equal 50?
[1] Show complete worked solution
(a)
The classmate multiplied by 20 instead of first finding 1% and then scaling to 100% (or dividing by the decimal 0.20). Correct approach: 1% $=50\div20=2.5$. Whole (100%) $=2.5\times100=250$.
(b)
$20\%\times250=0.20\times250=50$ — confirms 250 is correct.
QUESTION 5
5 marks
Criterion D
Medium
A charity fun-run raised \$3,750, which was reported as being 75% of their fundraising target.
a.
Use the unitary method to find the charity's full fundraising target (100%).
[3] b.
The charity needs to decide whether to extend the campaign by one more week. Based on your answer, how much more money do they need to reach their target?
[2] Show complete worked solution
(a)
1% $=3750\div75=50$. Target (100%) $=50\times100=\$5000$.
(b)
$$5000-3750=\$1250 \text{ more needed}$$
QUESTION 6
3 marks
Criterion A
Medium
A crowdfunding campaign for a community water well raised \$4,200, which its organizers announced represents exactly 70% of the total goal.
a.
Using the unitary method, find 1% of the target, then find the full target amount (100%).
[3] Show complete worked solution
(a)
1% $=4200\div70=60$. Target $=60\times100=\$6000$.
QUESTION 7
5 marks
Criterion B
Medium
Investigate the unitary method as applied to finding a percentage that ISN'T a whole number, using 'if 8.5% of a quantity is 68, find the whole quantity'.
a.
Using the unitary method (find 1% first), solve for the whole quantity.
[3] b.
Verify your answer using DIRECT division ($68\div0.085$) instead, and explain why both methods must give the same result.
[2] Show complete worked solution
(a)
1% $=68\div8.5=8$. Whole (100%) $=8\times100=800$.
(b)
$68\div0.085=800$ — matches. Both methods are mathematically equivalent: 'find 1% then scale to 100%' is really just dividing by $8.5$ then multiplying by $100$, which is the exact same overall operation as dividing directly by $\frac{8.5}{100}=0.085$.
QUESTION 8
5 marks
Criterion C
Medium
A classmate uses the unitary method to solve 'if 45% of a number is 90, find the number' by writing: '1% = 90 ÷ 45 = 2, so the number is... 45 × 2 = 90? That's just the same number I started with, something's wrong.'
a.
Identify the classmate's error, and clearly explain the CORRECT final step of the unitary method.
[3] b.
Verify the correct answer (200) by checking that 45% of 200 does indeed equal 90.
[2] Show complete worked solution
(a)
The classmate correctly found 1% = 2, but then multiplied by 45 (the ORIGINAL percentage) instead of by 100 (to reach the WHOLE, 100%). The unitary method's final step ALWAYS scales up to 100%, regardless of what percentage was originally given: correct answer $=2\times100=200$.
(b)
$45\%\times200=0.45\times200=90$, confirming 200 is correct.
QUESTION 9
6 marks
Criterion C
Hard
A national park's annual visitor revenue was \$84,000 last year, which reportedly represented a 14% increase from the previous year's revenue (i.e. last year's revenue is 114% of the previous year's).
a.
Using the unitary method (treating last year's \$84,000 as 114%), find 1%, and hence find the PREVIOUS year's revenue (100%).
[3] b.
Communicate your full solution as a clear, well-organized written explanation (in full sentences, not just calculations), suitable for inclusion in a park management report, explicitly stating what '114%' represents and why this specific percentage (rather than simply 14%) was used as the basis for the unitary method calculation.
[3] Show complete worked solution
(a)
1% $=84000\div114\approx736.84$. Previous year's revenue $\approx736.84\times100\approx\$73{,}684$.
(b)
Since revenue increased by 14%, this year's revenue represents the original 100% PLUS an additional 14%, totalling $100\%+14\%=114\%$ of the previous year's figure. Using the unitary method with 114% (not 14%) as the reference percentage is essential, because \$84,000 corresponds to the ENTIRE new revenue figure — including the original base amount — not just the increase portion alone. Dividing \$84,000 by 114 gives the value of 1% of the previous year's revenue, which when multiplied by 100 recovers the previous year's full (100%) revenue of approximately \$73,684.
Percentage increase and decrease 9 questions
QUESTION 1
3 marks
Criterion A
Medium
A shirt's price increases from \$40 to \$46.
a.
Find the amount of increase.
[1] b.
Express this increase as a percentage of the ORIGINAL price.
[2] Show complete worked solution
(a)
$$46-40=\$6$$
(b)
$$\frac{6}{40}\times100=15\%$$
QUESTION 2
4 marks
Criterion A
Medium
A laptop's price decreases from \$900 to \$720.
a.
Find the amount of decrease, and express it as a percentage of the original price.
[2] b.
A second laptop, originally \$650, is also reduced by 20%. Find its new price.
[2] Show complete worked solution
(a)
Decrease $=900-720=\$180$. Percentage decrease $=\frac{180}{900}\times100=20\%$.
(b)
$650 \times (1-0.20) = 650\times0.80=\$520$.
QUESTION 3
6 marks
Criterion B
Medium
Investigate what happens when a value is increased by 10%, then the RESULT is decreased by 10%.
a.
Start with \$200. Increase it by 10%, then decrease the new amount by 10%. Find the final value.
[3] b.
Compare the final value (\$198) to the original (\$200). Is a 10% increase followed by a 10% decrease the same as 'no change overall'?
[1] c.
Explain algebraically why this happens, using $x$ for the original value.
[2] Show complete worked solution
(a)
After +10%: $200\times1.10=220$. After $-10\%$ of 220: $220\times0.90=198$.
(b)
No — the final value (\$198) is LESS than the original (\$200), even though the percentages (+10%, then $-10\%$) might seem like they should cancel out.
(c)
$x \times 1.10 \times 0.90 = 0.99x$ — the combined effect is always a $1\%$ DECREASE from the original, because the second percentage (the decrease) is calculated on the ALREADY-INCREASED amount, which is larger than the original, so 10% of it is a bigger 'chunk' being removed than was added back proportionally.
QUESTION 4
5 marks
Criterion C
Medium
A student says a 25% increase followed by a 25% decrease returns a value to its original amount, since '25% up and 25% down cancel out'.
a.
Test the student's claim using an original value of \$80.
[3] b.
Explain why percentage increases and decreases of the SAME percentage never exactly cancel out (except when the percentage is 0%).
[2] Show complete worked solution
(a)
After $+25\%$: $80\times1.25=100$. After $-25\%$ of 100: $100\times0.75=75$. Final value is \$75, NOT \$80 — the claim is false.
(b)
The percentage decrease is applied to the NEW (increased) amount, which is larger than the original — so the same percentage represents a larger actual amount being subtracted than was added. This asymmetry means the two changes can never perfectly cancel, unless the percentage itself is 0%.
QUESTION 5
7 marks
Criterion D
Medium
A city's population was 240,000 at the start of Year 1. It grew by 5% during Year 1, then grew by a further 3% during Year 2.
a.
Find the population at the end of Year 1.
[2] b.
Find the population at the end of Year 2.
[2] c.
A city planner estimates the 2-year growth as simply '5%+3%=8% total growth'. Compare this estimate to the actual overall percentage growth, and explain the small discrepancy.
[3] Show complete worked solution
(a)
$$240000 \times 1.05 = 252000$$
(b)
$$252000 \times 1.03 = 259560$$
(c)
Actual overall growth: $\frac{259560-240000}{240000}\times100 = 8.15\%$. The planner's estimate (8%) is close but not exact — this is because the second year's 3% growth was calculated on the ALREADY-larger population (252,000), not the original 240,000, adding a small extra amount beyond a simple 8% sum.
QUESTION 6
6 marks
Criterion A
Hard
A country's population was 42.8 million, growing at 6.5% per year (compounding annually).
a.
Find the population after 1 year, and after 2 years, to 3 significant figures.
[3] b.
Find the OVERALL percentage growth over the 2 years (comparing the final population to the original), and explain why this is NOT simply $6.5\%\times2=13\%$.
[3] Show complete worked solution
(a)
Retain full precision until the final rounding. After one year, $$42.8(1.065)=45.582\approx\boxed{45.6\text{ million}}.$$ After two years, $$42.8(1.065)^2=48.54483\approx\boxed{48.5\text{ million}}.$$ Both answers are given to three significant figures.
(b)
Using the unrounded values, the two-year growth factor is $(1.065)^2=1.134225$. Hence the overall percentage growth is $$100(1.134225-1)=13.4225\%\approx\boxed{13.4\%}.$$ It is greater than $13\%$ because the second year's increase is calculated from the already increased population.
QUESTION 7
6 marks
Criterion B
Hard
Investigate whether percentage changes are 'additive' when applied to DIFFERENT base quantities — using two separate investment accounts.
a.
Account A (\$5,000) grows by 8%. Account B (\$12,000) grows by 3%. Find the dollar growth in EACH account, and the growth of the COMBINED total.
[3] b.
Find the OVERALL percentage growth of the COMBINED total (\$17,000 originally), and explain why this overall percentage is NOT simply the average of 8% and 3% (i.e. not 5.5%).
[3] Show complete worked solution
(a)
Account A growth: $5000\times0.08=\$400$. Account B growth: $12000\times0.03=\$360$. Combined growth: $400+360=\$760$.
(b)
Overall: $\frac{760}{17000}\times100\approx4.47\%$. This is not the simple average (5.5%) because the overall percentage is a WEIGHTED average, weighted by the relative SIZE of each account — since Account B (\$12,000, growing at only 3%) is much larger than Account A (\$5,000, growing at 8%), the overall rate is pulled closer to Account B's lower rate, not sitting exactly between the two rates.
QUESTION 8
8 marks
Criterion D
Hard
A shipping company's fuel costs form a major expense. A shipment's fuel cost was \$156,000, having DECREASED by 8.5% due to a new fuel-efficient fleet.
a.
Find the ORIGINAL fuel cost before the 8.5% decrease.
[3] b.
The company wants to know: if fuel PRICES themselves rise by 12% next year (independent of the fleet efficiency), while the fleet efficiency saving remains the SAME 8.5% relative reduction, estimate next year's fuel cost, and comment on whether the fleet upgrade will have 'paid for itself' in terms of avoided cost, compared to a scenario with NO fleet upgrade and the same 12% price rise applied to the ORIGINAL cost.
[5] Show complete worked solution
(a)
$$\frac{156000}{1-0.085}=\frac{156000}{0.915}\approx\$170{,}492$$
(b)
With the efficient fleet, next year's cost is $$156000(1.12)=\$174{,}720.$$ Without the upgrade, the corresponding cost is approximately $$170491.80(1.12)=\$190{,}951.$$ The avoided fuel cost is therefore about $$190951-174720=\boxed{\$16{,}231}.$$ Whether the fleet upgrade has paid for itself cannot be decided without knowing its purchase and installation cost.
QUESTION 9
6 marks
Criterion D
Medium
An agricultural cooperative reports crop yield per hectare INCREASED by 22% this season due to new irrigation, reaching 4.88 tonnes/hectare.
a.
Find last season's yield per hectare (before the 22% increase), to 2 decimal places.
[3] b.
The cooperative farms 340 hectares. Find the TOTAL additional tonnes of crop gained this season (compared to what WOULD have been produced at last season's rate across the same 340 hectares).
[3] Show complete worked solution
(a)
$$\frac{4.88}{1.22}\approx4.00 \text{ tonnes/hectare}$$
(b)
Additional yield per hectare: $4.88-4.00=0.88$ tonnes/hectare. Total additional tonnes: $0.88\times340\approx299.2$ tonnes.
Finding the original amount 12 questions
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] b.
Solve for $x$ to find the original salary.
[2] Show complete worked solution
(a)
$$1.15x = 46000$$
(b)
$$x = 46000 \div 1.15 = \$40000$$
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] b.
Solve for $x$ to find the original price.
[2] Show complete worked solution
(a)
$$0.80x = 68$$
(b)
$$x=68\div0.80=\$85$$
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] 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] 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] Show complete worked solution
(a)
$$125 \div 1.25 = 100$$
(b)
$125\times0.75=93.75$, 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)
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] b.
Explain the student's error, and find the CORRECT original price.
[3] Show complete worked solution
(a)
$30\%$ of $81.90 = 24.57$. $81.90-24.57=57.33 \ne 63$. So \$81.90 is NOT correct.
(b)
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] 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 complete worked solution
(a)
$$392000 \div 1.12 = \$350000$$
(b)
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] b.
Solve for $x$.
[2] Show complete worked solution
(a)
$$1.12x=156.80$$
(b)
$$x=156.80\div1.12=140 \text{ (i.e. \$140{,}000)}$$
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 complete worked solution
(a)
$0.85x=680.20 \Rightarrow x=680.20\div0.85\approx\$800.24$.
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] 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 complete worked solution
(a)
$$85\div0.85=\$100$$
(b)
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] 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 complete worked solution
(a)
$$40\div0.40=\$100$$
(b)
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] b.
Explain the student's error precisely, and find the CORRECT original value.
[3] Show complete worked solution
(a)
$384\times1.20=460.80\ne480$. So \$384 is NOT correct.
(b)
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 complete worked solution
(a)
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 complete worked solution
(a)
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.
Simple interest 17 questions
QUESTION 1
3 marks
Criterion A
Medium
\$2,500 is invested at a simple interest rate of 4.5% per year, for 6 years.
a.
Find the interest earned, using $I = \dfrac{PRT}{100}$.
[2] b.
Find the total amount in the account after 6 years.
[1] Show complete worked solution
(a)
$$I = \frac{2500\times4.5\times6}{100} = \$675$$
(b)
$$2500+675=\$3175$$
QUESTION 2
4 marks
Criterion A
Medium
A loan of \$8,000 is taken out at a simple interest rate of 7% per year.
a.
Find the interest owed after 3 years.
[2] b.
Find the total amount owed after 3 years, and the amount that would need to be repaid monthly if paid off in 36 equal instalments.
[2] Show complete worked solution
(a)
$$I=\frac{8000\times7\times3}{100}=\$1680$$
(b)
Total owed $=8000+1680=\$9680$. Monthly instalment $=9680\div36\approx\$268.89$.
QUESTION 3
6 marks
Criterion B
Medium
Investigate the pattern in the total amount in a simple-interest account over successive years.
a.
\$1,000 is invested at 6% simple interest per year. Find the total amount after 1, 2, 3, and 4 years.
[3] b.
Describe the pattern in the totals (1060, 1120, 1180, 1240, ...).
[1] c.
Explain why simple interest always produces this kind of pattern (constant amount added each year), unlike compound interest.
[2] Show complete worked solution
(a)
Year 1: $1000+60=1060$. Year 2: $1000+120=1120$. Year 3: $1000+180=1180$. Year 4: $1000+240=1240$.
(b)
This is an arithmetic sequence — the total increases by exactly \$60 every year (the same fixed amount).
(c)
Simple interest is always calculated on the ORIGINAL principal only, never on previously-earned interest — so the same fixed dollar amount ($6\%$ of the original \$1000, i.e. \$60) is added every single year, creating a straight arithmetic (linear) pattern rather than a growing (compounding) one.
QUESTION 4
4 marks
Criterion C
Medium
A student calculates the interest on \$5,000 at 8% simple interest for 2.5 years by first finding 1 year's interest (\$400), then says 'so for 2.5 years it's 400+400+half of 400 = 1000'.
a.
Verify the student's answer using the formula $I=\dfrac{PRT}{100}$ directly.
[2] b.
Explain why the student's step-by-step reasoning (one year, plus another year, plus half a year) is actually a valid way to think about simple interest, even without using the formula directly.
[2] Show complete worked solution
(a)
$$I=\frac{5000\times8\times2.5}{100}=\$1000$$ — matches the student's answer.
(b)
Because simple interest adds the SAME fixed amount each year (unlike compound interest), it's mathematically valid to just add up whole and partial years directly — \$400 for each full year, plus half of \$400 for the half year, giving the same total as the formula.
QUESTION 5
4 marks
Criterion D
Medium
A person deposits \$12,000 into a simple-interest savings account at 3.5% per year. They plan to withdraw enough at the end of 4 years to buy a car costing \$13,500.
a.
Calculate the total amount in the account after 4 years.
[2] b.
Will they have enough to buy the \$13,500 car? If so, how much will be left over; if not, how much more do they need?
[2] Show complete worked solution
(a)
$$I=\frac{12000\times3.5\times4}{100}=1680. \quad \text{Total}=12000+1680=\$13680$$
(b)
Yes, \$13,680 is enough. They will have $13680-13500=\$180$ left over after the purchase.
QUESTION 6
3 marks
Criterion A
Easy
\$3,400 is invested at 4.5% simple interest per year, for 7 years.
a.
Find the interest earned.
[2] b.
Find the total amount in the account after 7 years.
[1] Show complete worked solution
(a)
$$I=\frac{3400\times4.5\times7}{100}=\$1071$$
(b)
$$3400+1071=\$4471$$
QUESTION 7
3 marks
Criterion A
Easy
\$15,000 is invested at 3.8% simple interest per year, for 5 years.
a.
Find the interest earned, and the total amount after 5 years.
[3] Show complete worked solution
(a)
$I=\frac{15000\times3.8\times5}{100}=\$2850$. Total: $15000+2850=\$17850$.
QUESTION 8
6 marks
Criterion B
Hard
Investigate the relationship between the interest RATE and the TIME needed to double an investment under simple interest.
a.
At a rate of 5% simple interest, find how many years it takes for ANY principal to double (hint: interest earned must equal the original principal, i.e. $I=P$, so $\frac{P\times5\times T}{100}=P$).
[3] b.
Repeat for a rate of 8%, and describe the general pattern connecting rate and doubling time under simple interest.
[3] Show complete worked solution
(a)
$\frac{5T}{100}=1 \Rightarrow T=20$ years — this works for ANY principal $P$, since $P$ cancels out of the equation entirely.
(b)
$\frac{8T}{100}=1 \Rightarrow T=12.5$ years. Pattern: doubling time $T=\frac{100}{\text{rate}}$ — as the rate increases, the doubling time decreases proportionally (specifically, it's inversely proportional to the rate).
QUESTION 9
5 marks
Criterion B
Hard
Investigate whether DOUBLING the interest rate always HALVES the time needed to earn a FIXED target amount of interest, under simple interest.
a.
At 4% simple interest, find the time needed for \$5,000 to earn exactly \$800 interest.
[2] b.
Now at 8% (DOUBLE the rate), find the time needed for the SAME \$5,000 to earn the SAME \$800 interest. Does doubling the rate exactly halve the time, confirming the relationship?
[3] Show complete worked solution
(a)
$\frac{5000\times4\times T}{100}=800 \Rightarrow 200T=800 \Rightarrow T=4$ years.
(b)
$\frac{5000\times8\times T}{100}=800 \Rightarrow 400T=800 \Rightarrow T=2$ years — exactly half of 4 years, confirming that (for a FIXED target interest amount and fixed principal) doubling the rate does exactly halve the required time, since time and rate are inversely proportional in the simple interest formula when $I$ and $P$ are held constant.
QUESTION 10
4 marks
Criterion C
Medium
A student calculates simple interest on \$6,000 at 5% for 3 years as '$6000\times5\times3=90000$', forgetting to divide by 100.
a.
Explain the error, and give the correct interest amount.
[2] b.
Use estimation to show why \$90,000 is an obviously unreasonable answer for 3 years of interest on a \$6,000 investment, without redoing the full calculation.
[2] Show complete worked solution
(a)
The student forgot the $\div100$ in the formula $I=\frac{PRT}{100}$ — the rate 5% must be converted to a decimal-equivalent proportion of the principal, which requires dividing by 100. Correct: $I=\frac{6000\times5\times3}{100}=\$900$.
(b)
\$90,000 interest on a \$6,000 investment would mean the investment grew to 16 TIMES its original value in just 3 years — an absurdly high return rate for a simple 5% interest rate; real-world simple interest at modest rates like 5% should yield interest that's a small FRACTION of the principal over a few years, not many times larger than it.
QUESTION 11
4 marks
Criterion C
Medium
A student presents a complete, correct simple-interest calculation as a single unexplained line: '2400×3.5×6/100=504, 2400+504=2904'.
a.
Rewrite this as a well-communicated solution: state what values $P$, $R$, and $T$ represent (inventing a plausible context), show the formula being used with labels, and conclude with a full sentence.
[4] Show complete worked solution
(a)
Suppose \$2,400 (the principal, $P$) is invested at an annual simple interest rate of 3.5% ($R$) for 6 years ($T$). Using the formula $I=\frac{PRT}{100}$: $$I=\frac{2400\times3.5\times6}{100}=\$504$$ The total amount in the account after 6 years is therefore $2400+504=\$2904$.
QUESTION 12
7 marks
Criterion D
Medium
A microfinance organization offers small business loans of \$8,000 at 6% simple interest per year, to be repaid after 4 years.
a.
Find the total amount the borrower must repay after 4 years.
[3] b.
The borrower's business is projected to generate \$2,600 profit per year. Determine whether the business can fully repay the loan (total, in one lump sum) using EXACTLY 4 years of accumulated profit, and if there's a surplus or shortfall, state the amount.
[4] Show complete worked solution
(a)
$I=\frac{8000\times6\times4}{100}=\$1920$. Total repayment: $8000+1920=\$9920$.
(b)
4 years of profit: $2600\times4=\$10{,}400$. Since $10400>9920$ (the repayment amount), the business CAN repay the loan using 4 years of profit, with a surplus of $10400-9920=\$480$.
QUESTION 13
7 marks
Criterion D
Hard
A community savings group pools \$45,000 and invests it at 4.2% simple interest annually, planning to fund a \$12,000 scholarship each year using ONLY the interest earned (not touching the principal).
a.
Find the annual interest earned on the \$45,000.
[2] b.
Determine whether the annual interest alone is sufficient to fund the \$12,000 scholarship each year. If not, calculate the MINIMUM additional principal the group would need to invest (at the same 4.2% rate) to generate enough interest to fully cover the scholarship from interest alone.
[5] Show complete worked solution
(a)
$$I=\frac{45000\times4.2\times1}{100}=\$1890 \text{ per year}$$
(b)
\$1,890 is far short of the \$12,000 needed — INSUFFICIENT. Required total principal for \$12,000 annual interest: $\frac{P\times4.2\times1}{100}=12000 \Rightarrow P=\frac{1200000}{4.2}\approx\$285{,}714$. Additional principal needed: $285714-45000\approx\$240{,}714$.
QUESTION 14
7 marks
Criterion D
Medium
A vehicle financing company offers a car loan of \$18,000 at 7.5% simple interest, with repayment over 5 years via EQUAL monthly instalments.
a.
Find the total interest, and the total amount to be repaid over the 5 years.
[3] b.
Find the required EQUAL monthly instalment (over 60 months), correct to the nearest dollar, and comment on whether this monthly amount seems reasonable relative to a typical monthly household budget (using your own general knowledge of living costs, no specific figure given).
[4] Show complete worked solution
(a)
$I=\frac{18000\times7.5\times5}{100}=\$6750$. Total: $18000+6750=\$24750$.
(b)
Monthly instalment: $24750\div60=\$412.50\approx\$413$ (rounding to nearest dollar). Whether this is 'reasonable' depends heavily on the borrower's income and other expenses — for many households, a \$413 monthly car payment represents a significant but often manageable portion of a typical budget, though it would need to be weighed against other essential costs (housing, food, utilities) to determine true affordability for a specific borrower.
QUESTION 15
7 marks
Criterion D
Hard
A retiree has \$120,000 in savings and wants to withdraw a FIXED amount each year as 'income', while the remaining balance earns 3.2% simple interest annually on the ORIGINAL \$120,000 (assume withdrawals don't reduce the interest-earning principal, i.e. a simplified model where interest is always calculated on the original amount).
a.
Find the annual interest earned on the original \$120,000.
[2] b.
If the retiree withdraws exactly this \$3,840 interest each year for 15 years (never touching the \$120,000 principal), find the total amount withdrawn over 15 years, and explain why this simplified model (interest always on the ORIGINAL amount) might not perfectly reflect a REAL bank account, where the principal itself might be required to also be drawn down eventually.
[5] Show complete worked solution
(a)
$$I=\frac{120000\times3.2\times1}{100}=\$3840 \text{ per year}$$
(b)
Total withdrawn over 15 years: $3840\times15=\$57{,}600$. This simplified model assumes the \$120,000 stays completely untouched and keeps earning interest forever at a fixed rate — in reality, banks might not offer guaranteed fixed simple interest indefinitely, inflation could erode the real value of a fixed \$3,840 annual withdrawal over 15 years, and if the retiree ever needed to withdraw MORE than just the interest (e.g. for an emergency), the principal would shrink, reducing future interest earned — making this a simplified, optimistic model rather than a fully realistic retirement income plan.
QUESTION 16
3 marks
Criterion C
Medium
A student calculates simple interest correctly but writes the working in the WRONG order: '3=T, 5500=P, 0.055=R... 5500×0.055×3=907.50'.
a.
Rewrite this using standard, clearly labelled mathematical convention (defining $P$, $R$, $T$ BEFORE the calculation, using the standard formula $I=\frac{PRT}{100}$ with $R$ as a percentage, not a decimal), and explain why writing $R$ as 0.055 instead of 5.5 in the formula, while mathematically equivalent here, deviates from the standard convention.
[3] Show complete worked solution
(a)
Standard form: Let $P=5500$, $R=5.5$ (as a percentage), $T=3$ years. $$I=\frac{PRT}{100}=\frac{5500\times5.5\times3}{100}=\$907.50$$ The student's version used $R=0.055$ (a decimal) with no division by 100 — this happens to give the same numeric answer, but deviates from the conventional formula structure, which could cause confusion or errors if applied inconsistently in other problems.
QUESTION 17
6 marks
Criterion D
Medium
A farmer takes out a \$25,000 equipment loan at 5.8% simple interest, to be repaid in full after 3 years, expecting the new equipment to increase annual crop revenue by \$8,500 per year starting immediately.
a.
Find the total amount owed after 3 years.
[2] b.
Determine whether the farmer's projected extra revenue over the SAME 3 years is enough to cover the total loan repayment, and calculate the net financial position (surplus or shortfall) at the end of year 3.
[4] Show complete worked solution
(a)
$I=\frac{25000\times5.8\times3}{100}=\$4350$. Total: $25000+4350=\$29350$.
(b)
3 years of extra revenue: $8500\times3=\$25{,}500$. Since $25500<29350$, the extra revenue is NOT enough to fully cover the loan repayment. Shortfall: $29350-25500=\$3850$ — the farmer would need additional funds beyond the projected equipment-driven revenue increase to fully repay the loan within 3 years.