PERCENTAGE

QUESTION 1 [3 marks] — Criterion A Medium

A shirt's price increases from \$40 to \$46.

a. Find the amount of increase.

[1]

Show Solution

$$46-40=\$6$$

b. Express this increase as a percentage of the ORIGINAL price.

[2]

Show Solution

$$\frac{6}{40}\times100=15\%$$

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]

Show Solution

Decrease $=900-720=\$180$. Percentage decrease $=\frac{180}{900}\times100=20\%$.

b. A second laptop, originally \$650, is also reduced by 20%. Find its new price.

[2]

Show Solution

$650 \times (1-0.20) = 650\times0.80=\$520$.

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]

Show Solution

After +10%: $200\times1.10=220$. After $-10\%$ of 220: $220\times0.90=198$.

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]

Show Solution

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. Explain algebraically why this happens, using $x$ for the original value.

[2]

Show Solution

$x \times 1.10 \times 0.90 = 0.99x$ — 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]

Show Solution

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. Explain why percentage increases and decreases of the SAME percentage never exactly cancel out (except when the percentage is 0%).

[2]

Show Solution

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]

Show Solution

$$240000 \times 1.05 = 252000$$

b. Find the population at the end of Year 2.

[2]

Show Solution

$$252000 \times 1.03 = 259560$$

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 Solution

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]

Show Solution

After 1 year: $42.8\times1.065\approx45.6$ million. After 2 years: $45.6\times1.065\approx48.5$ million.

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 Solution

Overall growth: $\frac{48.5-42.8}{42.8}\times100\approx13.4\%$. This exceeds the simple $13\%$ estimate because the SECOND year's 6.5% growth is calculated on the ALREADY-larger population from year 1, adding a small extra 'growth on growth' effect that simple addition of percentages misses (compound growth is always slightly more than simple linear addition of the same rate).

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]

Show Solution

Account A growth: $5000\times0.08=\$400$. Account B growth: $12000\times0.03=\$360$. Combined growth: $400+360=\$760$.

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 Solution

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]

Show Solution

$$\frac{156000}{1-0.085}=\frac{156000}{0.915}\approx\$170{,}492$$

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 Solution

Next year's cost (with efficient fleet, applying 12% price rise to this year's \$156,000): $156000\times1.12=\$174{,}720$. Without the fleet upgrade (12% price rise applied to the ORIGINAL \$170,492): $170492\times1.12\approx\$190{,}951$. Savings from the fleet upgrade next year: $190951-174720\approx\$16{,}231$ — the fleet upgrade continues to provide a substantial ongoing saving even as fuel prices rise, confirming its value beyond the initial year.

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]

Show Solution

$$\frac{4.88}{1.22}\approx4.00 \text{ tonnes/hectare}$$

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 Solution

Additional yield per hectare: $4.88-4.00=0.88$ tonnes/hectare. Total additional tonnes: $0.88\times340\approx299.2$ tonnes.

Percentage increase and decrease