REAL NUMBERS AND RATIO

QUESTION 1 5 marks Criterion A

Medium

Order the fractions $\frac{3}{5}, \frac{7}{10}, \frac{2}{3}, \frac{5}{8}$ from smallest to largest.

a. Convert each fraction to an equivalent fraction with denominator 120.

[3]

b. Using your converted fractions, write the original fractions in order from smallest to largest.

[2]

Show complete worked solution

(a)

$\frac{3}{5}=\frac{72}{120}$, $\frac{7}{10}=\frac{84}{120}$, $\frac{2}{3}=\frac{80}{120}$, $\frac{5}{8}=\frac{75}{120}$

(b)

Comparing numerators (72, 84, 80, 75): smallest to largest gives $$\frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{7}{10}$$

QUESTION 2 3 marks Criterion A

Medium

Complete the following conversions, showing full working:

a. Write $\frac{47}{6}$ as a mixed number.

[1]

b. Write $5\frac{3}{8}$ as an improper fraction.

[1]

c. Find the fraction equivalent to $\frac{3}{4}$ with denominator 32.

[1]

Show complete worked solution

(a)

$47 \div 6 = 7$ remainder $5$, so $\frac{47}{6}=7\frac{5}{6}$

(b)

$$5\frac{3}{8} = \frac{5\times8+3}{8} = \frac{43}{8}$$

(c)

$$\frac{3}{4} = \frac{3\times8}{4\times8} = \frac{24}{32}$$

QUESTION 3 5 marks Criterion B

Medium

Investigate the sequence of fractions $\frac{n}{n+1}$ as $n$ increases: $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots$

a. Write each of $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}$ as a decimal, correct to 3 decimal places.

[3]

b. Describe what happens to the value of $\frac{n}{n+1}$ as $n$ gets larger.

[2]

Show complete worked solution

(a)

$\frac{1}{2}=0.500$, $\frac{2}{3}=0.667$, $\frac{3}{4}=0.750$, $\frac{4}{5}=0.800$, $\frac{5}{6}=0.833$

(b)

The value increases and gets closer and closer to 1, but never actually reaches 1 (since the numerator is always exactly 1 less than the denominator).

QUESTION 4 5 marks Criterion C

Medium

Explain, without using a calculator, how to determine which of $\frac{5}{7}$ and $\frac{7}{9}$ is larger.

a. Describe the cross-multiplication method for comparing two fractions.

[2]

b. Apply this method to determine which of $\frac{5}{7}$ and $\frac{7}{9}$ is larger, showing your working.

[3]

Show complete worked solution

(a)

To compare $\frac{a}{b}$ and $\frac{c}{d}$ (both positive denominators), multiply $a\times d$ and $c\times b$. Whichever product is larger corresponds to the larger fraction.

(b)

$5\times9=45$ and $7\times7=49$. Since $45<49$, this means $\frac{5}{7}<\frac{7}{9}$, so $\frac{7}{9}$ is larger.

QUESTION 5 5 marks Criterion D

Medium

Two pizza deals are advertised: Deal A gives $\frac{3}{8}$ of a large pizza for \$6. Deal B gives $\frac{5}{12}$ of a large pizza for \$9.

a. Find the cost of a WHOLE pizza under each deal (i.e. scale each deal up to a full pizza).

[3]

b. Which deal offers better value, and by how much (per whole pizza)?

[2]

Show complete worked solution

(a)

Deal A: $\$6 \div \frac{3}{8} = 6\times\frac{8}{3}=\$16$ per whole pizza. Deal B: $\$9 \div \frac{5}{12} = 9\times\frac{12}{5}=\$21.60$ per whole pizza.

(b)

Deal A is better value, since $\$16 < \$21.60$. The difference is $\$21.60-\$16=\$5.60$ per whole pizza.

QUESTION 1 7 marks Criterion B

Medium

Investigate the pattern formed when adding two consecutive unit fractions, $\frac{1}{n} + \frac{1}{n+1}$.

a. Calculate $\frac{1}{2}+\frac{1}{3}$, $\frac{1}{3}+\frac{1}{4}$, and $\frac{1}{4}+\frac{1}{5}$, leaving each answer as a single fraction (not simplified further unless needed).

[3]

b. Look at the numerators (5, 7, 9) and denominators (6, 12, 20) of your three answers. Describe the pattern in each.

[2]

c. Using your pattern, predict $\frac{1}{6}+\frac{1}{7}$ without doing the full calculation, then check your prediction by calculating it directly.

[2]

Show complete worked solution

(a)

$\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$. $\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$. $\frac{1}{4}+\frac{1}{5}=\frac{9}{20}$.

(b)

The numerators increase by 2 each time (5, 7, 9 — an arithmetic sequence). The denominators are each the product of the two consecutive numbers used ($2\times3=6$, $3\times4=12$, $4\times5=20$).

(c)

Prediction: numerator continues the pattern to 13, denominator $=6\times7=42$, so $\frac{13}{42}$. Checking directly: $\frac{1}{6}+\frac{1}{7} = \frac{7}{42}+\frac{6}{42} = \frac{13}{42}$ — matches the prediction.

QUESTION 2 6 marks Criterion C

Medium

A recipe requires $2\frac{1}{4}$ cups of flour. A baker is making $1\frac{1}{3}$ batches of the recipe.

a. Write a clear, step-by-step solution showing how much flour is needed in total, converting all mixed numbers to improper fractions first and showing every step.

[4]

b. Explain why converting to improper fractions is a more reliable method than trying to multiply mixed numbers directly.

[2]

Show complete worked solution

(a)

Convert to improper fractions: $2\frac{1}{4}=\frac{9}{4}$, $1\frac{1}{3}=\frac{4}{3}$. Multiply: $\frac{9}{4}\times\frac{4}{3}=\frac{36}{12}=3$. So exactly 3 cups of flour are needed.

(b)

Multiplying mixed numbers directly (e.g. multiplying whole number parts and fraction parts separately) does not give the correct product, since a mixed number represents an addition, not a simple combination — converting to a single improper fraction avoids this common error.

QUESTION 3 7 marks Criterion D

Medium

A water tank starts full. Over one day, $\frac{2}{5}$ of the water is used for irrigation in the morning, then $\frac{1}{3}$ of the remaining water is used for cleaning in the afternoon.

a. Find what fraction of the original full tank remains at the end of the day.

[3]

b. If the tank holds 4500 litres when full, how many litres remain at the end of the day?

[2]

c. The tank needs at least 1500 litres remaining to supply the household overnight. Based on your answer, is there enough water? Explain.

[2]

Show complete worked solution

(a)

After morning: remaining $=1-\frac{2}{5}=\frac{3}{5}$. Afternoon usage $=\frac{1}{3}\times\frac{3}{5}=\frac{1}{5}$. Remaining at end of day $=\frac{3}{5}-\frac{1}{5}=\frac{2}{5}$.

(b)

$$\frac{2}{5} \times 4500 = 1800 \text{ litres}$$

(c)

Yes — 1800 litres remain, which is more than the required 1500 litres, so there is enough water for the household overnight.

QUESTION 1 7 marks Criterion B

Medium

Investigate the decimal expansions of sevenths: $\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \ldots, \frac{6}{7}$.

a. Using a calculator, write out $\frac{1}{7}$ and $\frac{2}{7}$ as decimals, to at least 6 decimal places.

[2]

b. Both decimals repeat using the same 6 digits: 1, 4, 2, 8, 5, 7 — just starting at a different point in the cycle. Verify this is also true for $\frac{3}{7}$ and $\frac{4}{7}$.

[3]

c. Based on this pattern, predict the repeating decimal for $\frac{5}{7}$ without a calculator, then check your answer.

[2]

Show complete worked solution

(a)

$\frac{1}{7} = 0.142857142857\ldots$ (repeating). $\frac{2}{7} = 0.285714285714\ldots$ (repeating).

(b)

$\frac{3}{7}=0.428571\ldots$ and $\frac{4}{7}=0.571428\ldots$ — both use the same digit cycle $142857$, confirming the pattern.

(c)

Prediction (continuing the cyclic pattern): $0.714285\ldots$. Checking: $\frac{5}{7}=0.714285714285\ldots$ — matches the prediction.

QUESTION 2 6 marks Criterion C

Medium

A student needs to convert the recurring decimal $0.\dot{4}\dot{5}$ (meaning $0.454545\ldots$) into a fraction.

a. Let $x = 0.\dot{4}\dot{5}$. Explain why multiplying both sides by 100 is the correct first step (rather than multiplying by 10).

[2]

b. Complete the conversion, showing every step clearly: set up $100x$, subtract $x$, and solve for $x$ as a fraction in simplest form.

[4]

Show complete worked solution

(a)

The repeating block has 2 digits ("45"), so multiplying by $10^2=100$ shifts the decimal point exactly one full repeating cycle, meaning the decimal part after the shift lines up exactly with the original — this is essential for the subtraction step to eliminate the repeating part.

(b)

$100x = 45.454545\ldots$. Subtracting: $100x - x = 45.4545\ldots - 0.4545\ldots$, so $99x = 45$. Then $x = \frac{45}{99} = \frac{5}{11}$ (dividing by HCF 9).

QUESTION 3 5 marks Criterion D

Medium

A currency exchange booth converts Australian dollars (AUD) to Euros (EUR) at a rate of 1 AUD $=$ 0.61 EUR, and charges a flat \$5 AUD service fee, taken before conversion.

a. A tourist wants to exchange \$350 AUD. Calculate how many Euros they will receive.

[3]

b. The tourist was expecting approximately 213 EUR (based on the full \$350 without the fee). Explain the difference between their expectation and the actual amount.

[2]

Show complete worked solution

(a)

After fee: $350 - 5 = 345$ AUD. Converting: $345 \times 0.61 = 210.45$ EUR.

(b)

Without the fee, $350\times0.61=213.50$ EUR — close to the tourist's expectation. The actual amount (210.45 EUR) is lower because the flat \$5 service fee is subtracted before the exchange rate is applied, reducing the amount actually converted.

QUESTION 1 6 marks Criterion B

Medium

Investigate what happens when you square a decimal number between 0 and 1.

a. Calculate $(0.9)^2$, $(0.5)^2$, and $(0.3)^2$.

[2]

b. In each case, compare the squared result to the original number. What do you notice?

[2]

c. Explain why this happens, using the idea of multiplying a number by 'a fraction of itself'.

[2]

Show complete worked solution

(a)

$(0.9)^2=0.81$. $(0.5)^2=0.25$. $(0.3)^2=0.09$.

(b)

In every case, the squared result is smaller than the original number (e.g. $0.81<0.9$, $0.25<0.5$, $0.09<0.3$).

(c)

Squaring a number less than 1 means multiplying it by another number less than 1, which is the same as finding a fraction of the original number — and a fraction of something is always less than the whole thing (for positive numbers), so the result is always smaller than the original.

QUESTION 2 6 marks Criterion C

Medium

Calculate $$4.8 - 2.35 \times 1.6 \div 0.4$$

a. Show every step of your working, clearly stating which operation is performed at each stage, and give your final answer correct to 2 decimal places.

[4]

b. A classmate got a different answer by calculating $4.8-2.35$ first. Explain what error they made.

[2]

Show complete worked solution

(a)

Multiplication/division first (left to right): $2.35\times1.6=3.76$, then $3.76\div0.4=9.4$. Then subtraction: $4.8-9.4=-4.6$. Final answer: $-4.60$.

(b)

They performed the subtraction before the multiplication and division, which violates the order-of-operations convention — multiplication and division must be carried out before subtraction unless brackets say otherwise.

QUESTION 3 5 marks Criterion D

Medium

A phone plan costs \$45.90 per month, plus \$0.35 per minute for calls beyond the included minutes, plus \$0.12 per SMS beyond the included texts.

a. In one month, a customer used 40 extra minutes and sent 25 extra texts. Calculate their total bill for the month.

[3]

b. The customer budgeted \$60 for their phone bill this month. By how much have they gone over budget, and suggest one way they could reduce next month's bill.

[2]

Show complete worked solution

(a)

Extra minutes cost: $40\times0.35=14.00$. Extra texts cost: $25\times0.12=3.00$. Total: $45.90+14.00+3.00=62.90$.

(b)

They went over budget by $62.90-60.00=\$2.90$. They could reduce next month's bill by making fewer extra calls or sending fewer extra texts (e.g. using a messaging app over Wi-Fi instead of SMS).

QUESTION 1 4 marks Criterion A

Medium

For each number below, write it as a fraction $\frac{p}{q}$ (where $p, q$ are integers, $q\ne0$), showing that it is rational:

a. $0.75$

[1]

b. $-3$

[1]

c. $0.\dot{3}$ (i.e. $0.3333\ldots$)

[2]

Show complete worked solution

(a)

$$0.75 = \frac{75}{100} = \frac{3}{4}$$

(b)

$$-3 = \frac{-3}{1}$$

(c)

Let $x=0.\dot{3}$. Then $10x=3.333\ldots$, so $10x-x=3$, giving $9x=3$, so $$x=\frac{3}{9}=\frac{1}{3}$$

QUESTION 2 4 marks Criterion A

Medium

Order the rational numbers $-1.5, \frac{3}{4}, -\frac{2}{3}, 0.6$ from smallest to largest.

a. Convert each number to a decimal (correct to 3 decimal places where needed).

[2]

b. Write the four numbers in order from smallest to largest.

[2]

Show complete worked solution

(a)

$-1.5$, $\frac{3}{4}=0.750$, $-\frac{2}{3}=-0.667$, $0.6$

(b)

$$-1.5, \ -\frac{2}{3}, \ 0.6, \ \frac{3}{4}$$

QUESTION 3 5 marks Criterion B

Medium

Investigate whether adding or multiplying two rational numbers always gives a rational result (this is called 'closure').

a. Add $\frac{2}{3}$ and $\frac{1}{5}$, and multiply $\frac{2}{3}$ and $\frac{1}{5}$. Are both results rational?

[2]

b. Using general fractions $\frac{a}{b}$ and $\frac{c}{d}$, explain why the sum $\frac{a}{b}+\frac{c}{d}$ will always be rational.

[3]

Show complete worked solution

(a)

$\frac{2}{3}+\frac{1}{5}=\frac{10}{15}+\frac{3}{15}=\frac{13}{15}$ (rational). $\frac{2}{3}\times\frac{1}{5}=\frac{2}{15}$ (rational). Both results are rational.

(b)

$$\frac{a}{b}+\frac{c}{d} = \frac{ad+bc}{bd}$$ Since $a,b,c,d$ are integers, $ad+bc$ and $bd$ are also integers (sums and products of integers are integers), so the result is a ratio of two integers — by definition, rational.

QUESTION 4 5 marks Criterion C

Medium

Explain why every terminating decimal (a decimal that ends, like $0.125$) must be a rational number.

a. Using $0.125$ as an example, show how it can be written as a fraction with a power of 10 as the denominator.

[2]

b. Simplify this fraction, and then explain in general terms why ANY terminating decimal can always be written this way.

[3]

Show complete worked solution

(a)

$$0.125 = \frac{125}{1000}$$

(b)

$\frac{125}{1000}=\frac{1}{8}$ (dividing by HCF 125). In general, a terminating decimal with $n$ digits after the decimal point can always be written as (the whole number formed by removing the decimal point) $\div 10^n$ — both of these are integers, so the result is always a ratio of integers, i.e. rational.

QUESTION 5 6 marks Criterion D

Medium

A recipe requires ingredients measured as $0.75$ cups of sugar, $\frac{1}{3}$ cup of oil, and $1.2$ cups of flour, for one batch.

a. A baker is making $2\frac{1}{2}$ batches. Find the total amount of each ingredient needed, giving each answer as a fraction in simplest form.

[4]

b. The baker only has a $\frac{1}{4}$-cup measuring scoop. Explain whether the sugar amount can be measured exactly using only this scoop.

[2]

Show complete worked solution

(a)

Sugar: $0.75\times2.5=\frac{3}{4}\times\frac{5}{2}=\frac{15}{8}=1\frac{7}{8}$ cups. Oil: $\frac{1}{3}\times\frac{5}{2}=\frac{5}{6}$ cups. Flour: $1.2\times2.5=\frac{6}{5}\times\frac{5}{2}=3$ cups.

(b)

$1\frac{7}{8}$ cups $=\frac{15}{8}$ cups. Since $\frac{1}{4}=\frac{2}{8}$, and $15$ is not evenly divisible by $2$, the sugar amount cannot be measured using only whole $\frac{1}{4}$-cup scoops — a smaller measure would be needed for the extra $\frac{1}{8}$.

QUESTION 1 4 marks Criterion A

Medium

Classify each of the following as rational or irrational, giving a brief reason:

a. $\sqrt{16}$

[1]

b. $\sqrt{17}$

[1]

c. $\pi$

[1]

d. $\frac{22}{7}$

[1]

Show complete worked solution

(a)

Rational — $\sqrt{16}=4$, a whole number (which is rational).

(b)

Irrational — 17 is not a perfect square, so its square root cannot be written as a terminating or repeating decimal.

(c)

Irrational — $\pi$'s decimal expansion never terminates or repeats (a well-known proven result).

(d)

Rational — it is already expressed as a ratio of two integers, even though it is a common approximation for $\pi$.

QUESTION 2 4 marks Criterion A

Medium

Consider $\sqrt{30}$.

a. Between which two consecutive whole numbers does $\sqrt{30}$ lie? Justify your answer using perfect squares.

[2]

b. Estimate $\sqrt{30}$ correct to 1 decimal place, showing your reasoning (e.g. by testing values).

[2]

Show complete worked solution

(a)

$5^2=25$ and $6^2=36$. Since $25<30<36$, we know $5<\sqrt{30}<6$.

(b)

Since $5.4^2=29.16$ and $5.5^2=30.25$, $5.4<\sqrt{30}<5.5$. The rounding midpoint is $5.45$, and $$5.45^2=29.7025<30.$$ Hence $\sqrt{30}>5.45$, so it rounds upward to $$\boxed{\sqrt{30}\approx5.5}$$ correct to one decimal place.

QUESTION 3 6 marks Criterion B

Medium

Investigate which square roots of the whole numbers from 1 to 20 are rational, and which are irrational.

a. List the whole numbers from 1 to 20 whose square root is a whole number (i.e. a perfect square).

[2]

b. State which of these square roots are rational, and explain your reasoning for the rest of the numbers 1–20.

[3]

c. State a general rule: for which whole numbers $n$ is $\sqrt{n}$ rational?

[1]

Show complete worked solution

(a)

$1 (=1^2), 4 (=2^2), 9 (=3^2), 16 (=4^2)$ — these are the perfect squares between 1 and 20.

(b)

$\sqrt{1}=1$, $\sqrt{4}=2$, $\sqrt{9}=3$, $\sqrt{16}=4$ are all rational (whole numbers). Every other number from 1 to 20 (2, 3, 5, 6, 7, 8, 10–15, 17–20) is not a perfect square, so its square root is irrational.

(c)

$\sqrt{n}$ is rational exactly when $n$ is a perfect square (i.e. $n=k^2$ for some whole number $k$).

QUESTION 4 5 marks Criterion C

Medium

Explain why $\sqrt{2}$ cannot be written as a terminating or recurring decimal.

a. State what it would mean for $\sqrt{2}$ to be rational, in terms of being expressible as $\frac{p}{q}$.

[2]

b. It is a well-established mathematical result (proven by contradiction) that no such fraction exists for $\sqrt{2}$. Explain what this means for its decimal expansion.

[3]

Show complete worked solution

(a)

If $\sqrt{2}$ were rational, it could be written as a fraction $\frac{p}{q}$ in simplest form, where $p$ and $q$ are integers with no common factors.

(b)

Since $\sqrt{2}$ cannot be written as any fraction of integers, it cannot be rational — and every rational number has a decimal expansion that either terminates or eventually repeats. Since $\sqrt{2}$ is not rational, its decimal expansion must go on forever without ever repeating in a pattern.

QUESTION 5 5 marks Criterion D

Medium

A carpenter is building a square tabletop and wants the diagonal to measure exactly $2$ m.

a. Using the relationship (diagonal)$^2 = 2\times$(side length)$^2$ for a square, find the exact side length in the form $\sqrt{k}$, then as a decimal correct to 3 decimal places.

[3]

b. The carpenter's tape measure only shows millimetres (i.e. 3 decimal places in metres). Explain why the carpenter can never cut the side length with mathematically perfect accuracy, no matter how precise their tools are.

[2]

Show complete worked solution

(a)

$$2^2 = 2s^2 \Rightarrow s^2 = 2 \Rightarrow s = \sqrt{2} \approx 1.414 \text{ m}$$

(b)

Since $\sqrt{2}$ is irrational, its decimal expansion never terminates — so any real-world measurement (which must stop at some finite number of decimal places) can only ever be an approximation, never the mathematically exact value.

QUESTION 1 6 marks Criterion B

Medium

Investigate what happens to a ratio when both terms are multiplied by the same scale factor.

a. Starting with the ratio $3:5$, multiply both terms by 2, then by 3, then by 4. Record each resulting ratio.

[2]

b. Simplify each of your three new ratios back to simplest form. What do you notice?

[2]

c. Explain why this happens, and state a general rule for ratio $a:b$ scaled by a factor $k$.

[2]

Show complete worked solution

(a)

$\times2$: $6:10$. $\times3$: $9:15$. $\times4$: $12:20$.

(b)

$6:10$ simplifies to $3:5$. $9:15$ simplifies to $3:5$. $12:20$ simplifies to $3:5$. Every scaled ratio simplifies back to the original ratio.

(c)

Multiplying both terms of a ratio by the same factor doesn't change the underlying proportion between them — it's equivalent to multiplying a fraction's numerator and denominator by the same number, which doesn't change its value. General rule: $a:b$ is equivalent to $ka:kb$ for any $k>0$.

QUESTION 2 5 marks Criterion C

Medium

A garden bed is to have soil and compost mixed in the ratio $5:2$. The gardener has $35$ litres of soil available.

a. Show clear working to find how much compost is needed to maintain the correct ratio, and state the total volume of mix produced.

[3]

b. Explain, using the ratio, how you would check that your answer maintains the exact $5:2$ proportion.

[2]

Show complete worked solution

(a)

Since soil corresponds to the '5' part of the ratio: $1$ part $= 35\div5=7$ litres. Compost needed (2 parts) $=2\times7=14$ litres. Total mix $=35+14=49$ litres.

(b)

Checking: $35:14$. Dividing both by their HCF (7): $35\div7=5$, $14\div7=2$, giving $5:2$ — confirming the ratio is maintained.

QUESTION 3 6 marks Criterion D

Medium

A map has a scale of $1:25000$, meaning 1 cm on the map represents 25000 cm (250 m) in real life. The distance between two towns on the map measures 8.4 cm.

a. Find the real-life distance between the two towns, giving your answer in kilometres.

[3]

b. A hiker plans to walk between the towns at an average speed of 5 km/h, starting at 9:00 am, and wants to arrive before 10:00 am. Based on your answer to (a), will they arrive on time? Justify your answer.

[3]

Show complete worked solution

(a)

Real distance $= 8.4 \times 25000 = 210000$ cm $= 2100$ m $= 2.1$ km.

(b)

Time needed $= 2.1 \div 5 = 0.42$ hours $= 25.2$ minutes. Starting at 9:00 am, they would arrive at approximately 9:25 am, which is before 10:00 am — so yes, they will arrive on time, with about 35 minutes to spare.

Fractions 5 questions

Operations with fractions 3 questions

Decimal numbers 3 questions

Operations with decimal numbers 3 questions

Rational numbers 5 questions

Irrational numbers 5 questions

Ratio 3 questions