PERCENTAGE

QUESTION 1 [4 marks] — Criterion A Medium

Convert each of the following:

a. $35\%$ to a fraction in simplest form.

[1]

Show Solution

$$35\% = \frac{35}{100} = \frac{7}{20}$$

b. $\frac{9}{25}$ to a percentage.

[1]

Show Solution

$$\frac{9}{25} = \frac{36}{100} = 36\%$$

c. $0.625$ to a percentage.

[1]

Show Solution

$$0.625 = 62.5\%$$

d. $120\%$ to a decimal.

[1]

Show Solution

$$120\% = 1.2$$

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]

Show Solution

$\frac{3}{8}=37.5\%$. $42\%$ (already). $0.39=39\%$. $\frac{2}{5}=40\%$.

b. Write the original four values in order from smallest to largest.

[2]

Show Solution

$$\frac{3}{8}, \ 0.39, \ \frac{2}{5}, \ 42\%$$

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]

Show Solution

$17\%$, $53\%$, $8\%$.

b. State the rule connecting a fraction with denominator 100 and its percentage.

[1]

Show Solution

The numerator of the fraction (out of 100) IS the percentage number directly.

c. Using this rule, explain why $\frac{n}{100} \times 100\% = n\%$ for ANY fraction $\frac{n}{100}$, algebraically.

[2]

Show Solution

$\frac{n}{100}\times100 = \frac{100n}{100}=n$, 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]

Show Solution

'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. Show why the student's 'just use fractions like half and quarter' approach breaks down for a percentage like 37%.

[2]

Show Solution

$37\%=\frac{37}{100}$, 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]

Show Solution

Discount $=30\% \times 85 = 0.30\times85=\$25.50$. New price $=85-25.50=\$59.50$.

b. Was the shopper's mental estimate of '\$25' reasonable? Explain your reasoning.

[2]

Show Solution

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]

Show Solution

$1.40x=22 \Rightarrow x=22\div1.40\approx15.7\%$.

b. Express both efficiencies (22% now, ?15.7% a decade ago) as simplified fractions.

[2]

Show Solution

$22\%=\frac{22}{100}=\frac{11}{50}$. $15.7\%\approx\frac{157}{1000}$ (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]

Show Solution

After $+50\%$: $80\times1.5=120$. After $-50\%$ of 120: $120\times0.5=60$.

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 Solution

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]

Show Solution

After first 20%: $200\times0.8=160$. After second 20%: $160\times0.8=128$.

b. Compare this to applying a SINGLE 40% discount directly to \$200. Which gives the customer a better deal, and by how much?

[3]

Show Solution

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. 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 Solution

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]

Show Solution

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. 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 Solution

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]

Show Solution

On-time: $0.78\times2400=1872$. Delayed: $2400-1872=528$.

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 Solution

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.

Percentage