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7.5 Fractions & Percentages

Pack A — Réponses

# Question Réponse
1 Find a fraction equivalent to $\dfrac{1}{2}$ with a denominator greater than 100. Justify that it is equivalent. E.g. $\dfrac{51}{102}$; equivalent because $\frac{1}{2} = \frac{1 \times 51}{2 \times 51} = \frac{51}{102}$
2 Simplify $\dfrac{6}{9}$ fully. State the HCF you used and why dividing by it gives the simplest form. $\dfrac{2}{3}$; HCF = 3; no further simplification possible because 2 and 3 share no common factor other than 1
3 Which fraction is larger: $\dfrac{5}{8}$ or $\dfrac{3}{8}$? Justify by explaining what the numerator tells you. $\dfrac{5}{8}$; same denominator means same-size pieces, so 5 pieces is more than 3 pieces
4 Calculate $\dfrac{1}{4} + \dfrac{1}{4}$ and simplify your answer. Explain why the denominator does not change when adding. $\dfrac{1}{2}$; the denominator stays 4 because the pieces are the same size — you only count up the pieces
5 Calculate $\dfrac{3}{4} - \dfrac{1}{4}$ in its simplest form. Check your answer by adding back. $\dfrac{1}{2}$; check: $\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$ ✓
6 Find $\dfrac{1}{3}$ of 24. Explain why you divide by the denominator to find a unit fraction of an amount. 8; dividing by 3 splits the amount into 3 equal parts, and one part is the answer
7 Find 50% of 80 in two different ways. Show both methods give the same answer. 40; Method 1: 50% = ½, so 80 ÷ 2 = 40. Method 2: 50% = 50/100, so 80 × 50 ÷ 100 = 40.
8 Write 1$\,\dfrac{2}{3}$ as an improper fraction. Justify by explaining what each whole number contributes. $\dfrac{5}{3}$; the whole number 1 contributes $\frac{3}{3}$, plus $\frac{2}{3}$ gives $\frac{5}{3}$
9 Write $\dfrac{7}{2}$ as a mixed number. Check by converting back to an improper fraction. $3\,\dfrac{1}{2}$; check: $3 \times 2 + 1 = 7$, so $\frac{7}{2}$ ✓
10 Write $\dfrac{1}{2}$ as a decimal. Explain the connection between the fraction and the decimal. 0.5; $\frac{1}{2}$ means 1 divided by 2, and $1 \div 2 = 0.5$
11 Calculate $\dfrac{1}{2} + \dfrac{1}{3}$. Give your answer in its simplest form. $\dfrac{5}{6}$
12 Calculate $\dfrac{3}{4} - \dfrac{1}{3}$. Give your answer in its simplest form. $\dfrac{5}{12}$
13 Calculate $\dfrac{2}{3} \times \dfrac{3}{4}$. Give your answer in its simplest form. $\dfrac{1}{2}$
14 Calculate $\dfrac{3}{4} \div \dfrac{1}{2}$. Give your answer in its simplest form. $1\,\dfrac{1}{2}$
15 Find $\dfrac{3}{4}$ of 48. 36
16 Write 35% as (a) a decimal and (b) a fraction in its simplest form. (a) 0.35 (b) $\dfrac{7}{20}$
17 Find 25% of 64. 16
18 Calculate $1\,\dfrac{1}{2} + 2\,\dfrac{1}{4}$. $3\,\dfrac{3}{4}$
19 Write $\dfrac{2}{3}$, $\dfrac{3}{5}$, and $\dfrac{7}{10}$ in order from smallest to largest. $\dfrac{3}{5} < \dfrac{2}{3} < \dfrac{7}{10}$
20 Increase £40 by 20%. £48
21 Find the HCF of 18 and 24, then use it to simplify $\dfrac{18}{24}$ fully. HCF = 6; $\dfrac{3}{4}$
22 What is 30% of −£40? −£12
23 Calculate $1\,\dfrac{1}{2} \times 2\,\dfrac{2}{3}$. 4
24 Calculate $3\,\dfrac{1}{2} \div 1\,\dfrac{3}{4}$. 2
25 Decrease £120 by 35%. £78
26 Write $\dfrac{1}{3}$ as a decimal. State whether it is terminating or recurring. $0.\overline{3}$ (recurring)
27 Write $-\dfrac{1}{2}$ and $-\dfrac{2}{3}$ in order from smallest to largest. $-\dfrac{2}{3} < -\dfrac{1}{2}$
28 Find the value of $3a + 1$ when $a = \dfrac{1}{2}$. $\dfrac{5}{2}$ (or 2.5)
29 Find $\dfrac{3}{8}$ of 2 m. Give your answer in centimetres. 75 cm
30 40% of a number is 48. What is the number? 120
31 A jacket costs £80. It is reduced by 25% and then by a further 10%. What is the final price? £54
32 After a 20% increase, a price is £72. What was the original price? £60
33 Solve $\dfrac{2}{3} \times x = 14$. $x = 21$
34 Find the value of $a^2 + b$ when $a = \dfrac{3}{4}$ and $b = \dfrac{1}{2}$. $\dfrac{17}{16}$
35 Calculate $2\,\dfrac{3}{4} + 1\,\dfrac{1}{6} - \dfrac{3}{8}$. $3\,\dfrac{13}{24}$
36 Arrange in order from smallest to largest: $-\dfrac{3}{4},\; \dfrac{1}{3},\; -\dfrac{1}{2},\; \dfrac{2}{5}$. $-\dfrac{3}{4} < -\dfrac{1}{2} < \dfrac{1}{3} < \dfrac{2}{5}$
37 Convert the recurring decimal $0.\overline{3}$ to a fraction in its simplest form. $\dfrac{1}{3}$
38 A price changes from £40 to £52. Find the percentage change and state whether it is an increase or decrease. 30% increase
39 $\dfrac{3}{4}$ of a class of 32 students passed a test. Of those who passed, $\dfrac{2}{3}$ also completed the homework. How many students both passed and completed the homework? 16
40 Shop A sells 400g of cereal for £1.20. Shop B sells the same cereal: 600g for £1.65. Which is the better value? Show your working. Shop B (27.5p per 100g vs 30p per 100g)

Pack B — Réponses

# Question Réponse
1 Find a fraction equivalent to $\dfrac{2}{3}$ with a denominator greater than 50. Justify that it is equivalent. E.g. $\dfrac{34}{51}$; equivalent because $\frac{2}{3} = \frac{2 \times 17}{3 \times 17} = \frac{34}{51}$
2 Simplify $\dfrac{8}{12}$ fully. State the HCF you used and why dividing by it gives the simplest form. $\dfrac{2}{3}$; HCF = 4; 2 and 3 share no common factor other than 1
3 Which fraction is larger: $\dfrac{7}{10}$ or $\dfrac{4}{10}$? Justify by explaining what the numerator tells you. $\dfrac{7}{10}$; same denominator, so 7 pieces out of 10 is more than 4 pieces out of 10
4 Calculate $\dfrac{2}{5} + \dfrac{1}{5}$ and simplify your answer. Explain why the denominator does not change when adding. $\dfrac{3}{5}$; the denominator stays 5 because the piece-size (fifths) does not change
5 Calculate $\dfrac{5}{6} - \dfrac{1}{6}$ in its simplest form. Check your answer by adding back. $\dfrac{2}{3}$; check: $\frac{2}{3} + \frac{1}{6} = \frac{5}{6}$ ✓
6 Find $\dfrac{1}{4}$ of 32. Explain why you divide by the denominator to find a unit fraction of an amount. 8; dividing by 4 splits the amount into 4 equal parts
7 Find 50% of 60 in two different ways. Show both methods give the same answer. 30; Method 1: 50% = ½, so 60 ÷ 2 = 30. Method 2: 50% = 50/100, so 60 × 50 ÷ 100 = 30.
8 Write 2$\,\dfrac{3}{4}$ as an improper fraction. Justify by explaining what each whole number contributes. $\dfrac{11}{4}$; each whole is $\frac{4}{4}$; two wholes give $\frac{8}{4}$, plus $\frac{3}{4}$ gives $\frac{11}{4}$
9 Write $\dfrac{9}{4}$ as a mixed number. Check by converting back to an improper fraction. $2\,\dfrac{1}{4}$; check: $2 \times 4 + 1 = 9$, so $\frac{9}{4}$ ✓
10 Write $\dfrac{1}{4}$ as a decimal. Explain the connection between the fraction and the decimal. 0.25; $\frac{1}{4}$ means 1 divided by 4, and $1 \div 4 = 0.25$
11 Calculate $\dfrac{1}{4} + \dfrac{1}{3}$. Give your answer in its simplest form. $\dfrac{7}{12}$
12 Calculate $\dfrac{5}{6} - \dfrac{1}{4}$. Give your answer in its simplest form. $\dfrac{7}{12}$
13 Calculate $\dfrac{3}{5} \times \dfrac{5}{6}$. Give your answer in its simplest form. $\dfrac{1}{2}$
14 Calculate $\dfrac{2}{3} \div \dfrac{1}{4}$. Give your answer in its simplest form. $2\,\dfrac{2}{3}$
15 Find $\dfrac{2}{5}$ of 60. 24
16 Write 60% as (a) a decimal and (b) a fraction in its simplest form. (a) 0.60 (b) $\dfrac{3}{5}$
17 Find 30% of 90. 27
18 Calculate $1\,\dfrac{1}{3} + 2\,\dfrac{1}{6}$. $3\,\dfrac{1}{2}$
19 Write $\dfrac{1}{2}$, $\dfrac{3}{8}$, and $\dfrac{5}{12}$ in order from smallest to largest. $\dfrac{3}{8} < \dfrac{5}{12} < \dfrac{1}{2}$
20 Increase £60 by 15%. £69
21 Find the HCF of 30 and 42, then use it to simplify $\dfrac{30}{42}$ fully. HCF = 6; $\dfrac{5}{7}$
22 What is 15% of −£60? −£9
23 Calculate $1\,\dfrac{1}{4} \times 2\,\dfrac{1}{3}$. $2\,\dfrac{11}{12}$
24 Calculate $3\,\dfrac{2}{3} \div 1\,\dfrac{1}{3}$. $2\,\dfrac{3}{4}$
25 Decrease £200 by 45%. £110
26 Write $\dfrac{2}{9}$ as a decimal. State whether it is terminating or recurring. $0.\overline{2}$ (recurring)
27 Write $-\dfrac{3}{4}$ and $-\dfrac{5}{8}$ in order from smallest to largest. $-\dfrac{3}{4} < -\dfrac{5}{8}$
28 Find the value of $3a + 1$ when $a = \dfrac{2}{3}$. 3
29 Find $\dfrac{5}{6}$ of 3 m. Give your answer in centimetres. 250 cm
30 35% of a number is 63. What is the number? 180
31 A jacket costs £120. It is reduced by 20% and then by a further 15%. What is the final price? £81.60
32 After a 25% increase, a price is £90. What was the original price? £72
33 Solve $\dfrac{3}{4} \times x = 18$. $x = 24$
34 Find the value of $a^2 + b$ when $a = \dfrac{2}{3}$ and $b = \dfrac{1}{4}$. $\dfrac{25}{36}$
35 Calculate $2\,\dfrac{2}{3} + 1\,\dfrac{3}{4} - \dfrac{5}{6}$. $3\,\dfrac{7}{12}$
36 Arrange in order from smallest to largest: $-\dfrac{3}{4},\; \dfrac{1}{3},\; -\dfrac{1}{2},\; \dfrac{2}{5}$. $-\dfrac{5}{6} < -\dfrac{2}{3} < \dfrac{1}{4} < \dfrac{3}{8}$
37 Convert the recurring decimal $0.\overline{63}$ to a fraction in its simplest form. $\dfrac{7}{11}$
38 A price changes from £60 to £45. Find the percentage change and state whether it is an increase or decrease. 25% decrease
39 $\dfrac{5}{8}$ of a class of 48 students passed a test. Of those who passed, $\dfrac{3}{5}$ also completed the homework. How many students both passed and completed the homework? 18
40 Shop A sells 250g of cereal for £0.90. Shop B sells the same cereal: 400g for £1.32. Which is the better value? Show your working. Shop B (33p per 100g vs 36p per 100g)

Problèmes — Solutions détaillées

1

**Compound percentage.** A jacket originally costs £80. It is reduced by 25% in a sale. The following week, the sale price is reduced by a further 10%. What is the final price? Is the total reduction 35%? Explain why or why not.

Réponse

£54; total reduction is 32.5%, not 35%.

After 25% off: $\pounds 80 \times 0.75 = \pounds 60$. After a further 10% off: $\pounds 60 \times 0.90 = \pounds 54$. Total reduction from original: $\pounds 80 - \pounds 54 = \pounds 26$. As a percentage: $\frac{26}{80} \times 100 = 32.5\%$. The reductions are **not** simply added (35%) because the second 10% is taken off the already-reduced price, not the original £80. Sequential percentage changes multiply: $0.75 \times 0.90 = 0.675$, giving a single reduction of $1 - 0.675 = 32.5\%$.
2

**Fraction wall.** In a school of 360 students: - $\dfrac{3}{8}$ study French. - $\dfrac{1}{4}$ of those French students also study Spanish. - The remaining French students study French only. How many students study French only?

Réponse

101 (or 102 depending on rounding — see working)

$\frac{3}{8}$ of 360 = $360 \div 8 \times 3 = 135$ students study French. $\frac{1}{4}$ of 135 = $135 \div 4 = 33.75$. Since we need a whole number of students, round to 34 who also study Spanish. French-only students: $135 - 34 = \mathbf{101}$. Note: the problem's fractions do not combine to give whole numbers with 360 students — a useful teaching point about mathematical modelling. If we round down (33 bilingual), French-only = 102.
3

**Pipe filling.** Pipe A fills a tank in 4 hours. Pipe B fills the same tank in 6 hours. If both pipes are open together, how long does it take to fill the tank?

Réponse

$2\,\dfrac{2}{5}$ hours (2 hours 24 minutes)

In one hour, Pipe A fills $\frac{1}{4}$ of the tank and Pipe B fills $\frac{1}{6}$. Together they fill $\frac{1}{4} + \frac{1}{6}$ per hour. LCD = 12: $\frac{3}{12} + \frac{2}{12} = \frac{5}{12}$ per hour. Time to fill whole tank = $1 \div \frac{5}{12} = \frac{12}{5} = 2\frac{2}{5}$ hours = **2 hours 24 minutes**.
4

**Fractions that sum to 1.** The ancient Egyptians only wrote fractions with numerator 1 (called "unit fractions"), such as $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$. (a) Write $\dfrac{5}{6}$ as a sum of **two different** unit fractions. (b) Write $\dfrac{7}{12}$ as a sum of **two different** unit fractions. (c) Show that $\dfrac{2}{n} = \dfrac{1}{n} + \dfrac{1}{n}$ does not count (same fraction twice). Instead, find unit fractions $\dfrac{1}{a}$ and $\dfrac{1}{b}$ with $a < b$ such that $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{2}{9}$. (d) Is it always possible to write any proper fraction as a sum of unit fractions? Try $\dfrac{3}{7}$.

Réponse

(a) $\frac{1}{2} + \frac{1}{3}$ (b) $\frac{1}{3} + \frac{1}{4}$ (c) $\frac{1}{5} + \frac{1}{45}$ (d) Yes: $\frac{3}{7} = \frac{1}{3} + \frac{1}{11} + \frac{1}{231}$ (one way)

(a) $\frac{5}{6} = \frac{1}{2} + \frac{1}{3}$. Check: $\frac{3}{6} + \frac{2}{6} = \frac{5}{6}$ ✓. (b) $\frac{7}{12} = \frac{1}{3} + \frac{1}{4}$. Check: $\frac{4}{12} + \frac{3}{12} = \frac{7}{12}$ ✓. (c) We need $\frac{1}{a} + \frac{1}{b} = \frac{2}{9}$ with $a < b$, $a \neq b$. Since $\frac{1}{a} > \frac{1}{9}$ (as it's the larger part), we need $a < 9$. Also $\frac{1}{a} < \frac{2}{9}$ means $a > \frac{9}{2} = 4.5$, so $a \geq 5$. Try $a = 5$: $\frac{1}{b} = \frac{2}{9} - \frac{1}{5} = \frac{10}{45} - \frac{9}{45} = \frac{1}{45}$. So $b = 45$. Check: $\frac{1}{5} + \frac{1}{45} = \frac{9}{45} + \frac{1}{45} = \frac{10}{45} = \frac{2}{9}$ ✓. Answer: $\frac{1}{5} + \frac{1}{45}$. (d) $\frac{3}{7}$: try $a = 3$ (largest unit fraction less than $\frac{3}{7}$ since $\frac{1}{3} \approx 0.333 < \frac{3}{7} \approx 0.429$... actually $\frac{1}{3} < \frac{3}{7}$ since $7 < 9$). $\frac{3}{7} - \frac{1}{3} = \frac{9}{21} - \frac{7}{21} = \frac{2}{21}$. Now write $\frac{2}{21}$ as a unit fraction sum: $a = 11$ gives $\frac{1}{b} = \frac{2}{21} - \frac{1}{11} = \frac{22}{231} - \frac{21}{231} = \frac{1}{231}$. So $\frac{3}{7} = \frac{1}{3} + \frac{1}{11} + \frac{1}{231}$. It is always possible (Fibonacci/Sylvester's sequence guarantees this).
5

**Percentage chain.** A shop increases all prices by 20% in January. In July, it reduces all prices by 20%. A customer claims "the prices are back to where they started." Is she correct? Give a numerical example to support your answer.

Réponse

She is wrong. The final price is 96% of the original.

Take an example: original price £100. After 20% increase: $\pounds 100 \times 1.20 = \pounds 120$. After 20% decrease: $\pounds 120 \times 0.80 = \pounds 96$. The price is now **£96**, which is **less** than the original £100. Multiplying the multipliers: $1.20 \times 0.80 = 0.96$, a 4% overall **decrease**. The 20% increase and 20% decrease do not cancel because the decrease is calculated on the higher (post-increase) price.
6

**Fraction of the way.** A road is 24 km long. A cyclist has completed $\dfrac{5}{8}$ of the journey. (a) How far has she cycled? (b) What fraction of the journey remains? (c) If the remaining distance takes 45 minutes, what is her average speed in km/h for that section?

Réponse

(a) 15 km (b) $\dfrac{3}{8}$ (c) 12 km/h

(a) $\frac{5}{8}$ of 24 km: $24 \div 8 \times 5 = 15$ km. (b) Remaining fraction: $1 - \frac{5}{8} = \frac{3}{8}$. Remaining distance: $\frac{3}{8} \times 24 = 9$ km. (c) 45 minutes = $\frac{3}{4}$ hour. Speed = distance ÷ time = $9 \div \frac{3}{4} = 9 \times \frac{4}{3} = 12$ km/h.
7

**Average with constraints.** Five **different** positive integers have mean 6 and median 5. (a) What is their sum? (b) The smallest is 1. Find all possible sets of five integers satisfying every condition. (c) If the smallest must be 2 instead of 1, does a valid set still exist? Explain.

Réponse

(a) 30 (b) Many valid sets, e.g. {1, 2, 5, 8, 14}, {1, 3, 5, 7, 14} — see working for all (c) Yes — e.g. {2, 3, 5, 6, 14}.

(a) Mean = 6, five values: sum = $6 \times 5 = \mathbf{30}$. (b) The median is the 3rd value (when ordered), so the 3rd value = 5. We have: $a < b < 5 < d < e$ with $a + b + 5 + d + e = 30$, so $a + b + d + e = 25$. Smallest = 1 (given): $a = 1$. So $b + d + e = 24$ with $1 < b < 5$, $5 < d < e$, all different integers. $b \in \{2, 3, 4\}$. - $b = 2$: $d + e = 22$, $d \geq 6$. Try $d = 6, e = 16$; $d = 7, e = 15$; $d = 8, e = 14$; $d = 9, e = 13$; $d = 10, e = 12$; $d = 11, e = 11$ (equal — invalid). Valid: $\{1,2,5,6,16\}$, $\{1,2,5,7,15\}$, $\{1,2,5,8,14\}$, $\{1,2,5,9,13\}$, $\{1,2,5,10,12\}$. - $b = 3$: $d + e = 21$, $d \geq 6$. Valid: $\{1,3,5,6,15\}$, $\{1,3,5,7,14\}$, $\{1,3,5,8,13\}$, $\{1,3,5,9,12\}$, $\{1,3,5,10,11\}$. - $b = 4$: $d + e = 20$, $d \geq 6$. Valid: $\{1,4,5,6,14\}$, $\{1,4,5,7,13\}$, $\{1,4,5,8,12\}$, $\{1,4,5,9,11\}$. Many valid sets exist (the problem asks to "find all possible sets" — students can list them). (c) If smallest = 2: $a = 2$, so $b + d + e = 23$, $2 < b < 5$ means $b \in \{3, 4\}$. - $b = 3$: $d + e = 20$, $d \geq 6$. Gives $\{2,3,5,6,14\}$ etc. — **valid** ✓. So a valid set **does** still exist with smallest = 2. (The answer "No" above was incorrect — corrected here: **Yes**, e.g. $\{2,3,5,6,14\}$.)
8

**Simplifying with prime factors.** (a) Write 126 and 210 each as a product of prime factors. (b) Hence find the HCF of 126 and 210. (c) Use the HCF to simplify $\dfrac{126}{210}$ fully.

Réponse

(a) $126 = 2 \times 3^2 \times 7$; $210 = 2 \times 3 \times 5 \times 7$ (b) HCF = 42 (c) $\dfrac{3}{5}$

(a) $126 \div 2 = 63$; $63 \div 3 = 21$; $21 \div 3 = 7$; 7 prime. $126 = 2 \times 3^2 \times 7$. $210 \div 2 = 105$; $105 \div 3 = 35$; $35 \div 5 = 7$; 7 prime. $210 = 2 \times 3 \times 5 \times 7$. (b) HCF: take lowest powers of shared primes ($2^1, 3^1, 7^1$): $2 \times 3 \times 7 = 42$. (c) $\frac{126}{210} = \frac{126 \div 42}{210 \div 42} = \frac{3}{5}$.
9

**Fractions and algebra.** The perimeter of a rectangle is 15 cm. One side has length $2\dfrac{1}{4}$ cm. Find the length of the other side as a mixed number.

Réponse

$5\,\dfrac{1}{4}$ cm

Perimeter = $2(l + w) = 15$, so $l + w = 7.5 = 7\frac{1}{2}$. One side $w = 2\frac{1}{4}$. Other side $l = 7\frac{1}{2} - 2\frac{1}{4}$. Convert: $7\frac{1}{2} = 7\frac{2}{4}$. So $l = 7\frac{2}{4} - 2\frac{1}{4} = 5\frac{1}{4}$ cm. Check: $5\frac{1}{4} + 2\frac{1}{4} = 7\frac{1}{2}$; perimeter = $2 \times 7.5 = 15$ ✓.
10

**Farey sequence investigation.** A Farey sequence $F_n$ contains all fractions between 0 and 1 (inclusive) with denominators at most $n$, written in ascending order. $F_3$: $\dfrac{0}{1},\ \dfrac{1}{3},\ \dfrac{1}{2},\ \dfrac{2}{3},\ \dfrac{1}{1}$ (a) Write out $F_4$ in full (all fractions with denominators 1, 2, 3, or 4, in order). (b) Pick any two adjacent fractions in $F_4$, say $\dfrac{a}{b}$ and $\dfrac{c}{d}$. Calculate $bc - ad$. What do you notice? (c) For two adjacent Farey fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$, the mediant is $\dfrac{a+c}{b+d}$. Find the mediant of $\dfrac{1}{3}$ and $\dfrac{1}{2}$. Is it between them? (d) Verify that the mediant of two adjacent Farey fractions always lies strictly between them.

Réponse

(a) $\frac{0}{1}, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{1}{1}$ (b) $bc - ad = 1$ always (c) $\frac{2}{5}$, yes (d) See working.

(a) Fractions with denominator ≤ 4, in order: $\dfrac{0}{1},\ \dfrac{1}{4},\ \dfrac{1}{3},\ \dfrac{1}{2},\ \dfrac{2}{3},\ \dfrac{3}{4},\ \dfrac{1}{1}$. (b) Check adjacent pairs: $\frac{0}{1}$ and $\frac{1}{4}$: $1 \times 1 - 0 \times 4 = 1$. $\frac{1}{4}$ and $\frac{1}{3}$: $1 \times 4 - 1 \times 3 = 1$. $\frac{1}{3}$ and $\frac{1}{2}$: $1 \times 3 - 1 \times 2 = 1$. The cross-product $bc - ad = \mathbf{1}$ for every adjacent pair. This is a remarkable property of Farey sequences. (c) Mediant of $\frac{1}{3}$ and $\frac{1}{2}$: $\frac{1+1}{3+2} = \frac{2}{5}$. Is $\frac{1}{3} < \frac{2}{5} < \frac{1}{2}$? $\frac{1}{3} \approx 0.333$, $\frac{2}{5} = 0.4$, $\frac{1}{2} = 0.5$. Yes, $\frac{2}{5}$ lies strictly between them. (d) If $\frac{a}{b} < \frac{c}{d}$ are adjacent Farey fractions, their mediant is $\frac{a+c}{b+d}$. To show $\frac{a}{b} < \frac{a+c}{b+d}$: cross-multiply: $a(b+d) < b(a+c) \Leftrightarrow ab + ad < ab + bc \Leftrightarrow ad < bc \Leftrightarrow bc - ad > 0$, which holds since $bc - ad = 1 > 0$. Similarly $\frac{a+c}{b+d} < \frac{c}{d}$ since $d(a+c) < c(b+d) \Leftrightarrow cd - ad < bc + cd - dc \Leftrightarrow$ same condition. So the mediant always lies strictly between adjacent Farey fractions.
11

**Mixed units.** A recipe needs $\dfrac{3}{4}$ kg of flour for 12 biscuits. (a) How much flour is needed for 20 biscuits? (b) A bag holds 1.5 kg. What fraction of the bag is used for 20 biscuits? (c) What percentage of the bag is left over?

Réponse

(a) $1\dfrac{1}{4}$ kg (b) $\dfrac{5}{6}$ (c) $16.\overline{6}\%$

(a) Flour per biscuit: $\frac{3}{4} \div 12 = \frac{3}{48} = \frac{1}{16}$ kg. For 20 biscuits: $\frac{1}{16} \times 20 = \frac{20}{16} = \frac{5}{4} = 1\frac{1}{4}$ kg. (b) Fraction of bag: $\frac{5/4}{3/2} = \frac{5}{4} \times \frac{2}{3} = \frac{10}{12} = \frac{5}{6}$. (c) Fraction remaining: $1 - \frac{5}{6} = \frac{1}{6}$. As a percentage: $\frac{1}{6} \times 100 = 16.\overline{6}\%$.
12

**Jamie's monthly budget.** Jamie earns £960 per month. - He saves $\dfrac{1}{4}$ of his earnings. - He spends 35% of the remainder on rent. - He spends $\dfrac{2}{5}$ of what is left on food and bills. How much does Jamie have left each month for discretionary spending?

Réponse

£280.80

Savings: $\frac{1}{4}$ of £960 = £240. Remainder after saving: $\pounds 960 - \pounds 240 = \pounds 720$. Rent: 35% of £720 = $\pounds 720 \times 0.35 = \pounds 252$. After rent: $\pounds 720 - \pounds 252 = \pounds 468$. Food and bills: $\frac{2}{5}$ of £468 = $\pounds 468 \div 5 \times 2 = \pounds 93.60 \times 2 = \pounds 187.20$. Discretionary: $\pounds 468 - \pounds 187.20 = \mathbf{\pounds 280.80}$.