(a) Luis buys a season ticket to watch his local football team.
The season ticket costs $595.
(i) Luis buys the season ticket online and gets a 5% discount on the $595.
Work out how much Luis pays for the season ticket online.
(ii) A ticket to watch one match costs $38.
Luis watches 16 matches.
How much did Luis save by buying a season ticket online instead of 16 tickets at $38 each?
(b) The football stadium has 26272 seats.
The number of people who attend one match is 23854.
Calculate the percentage of the 26272 seats that are empty.
(c) The total number of people attending matches at the stadium last season was 506762.
Write 506762 in standard form, correct to 3 significant figures.
(d) Last season the team played a total of 45 matches.
The table shows the results of these matches.
| Result | Number of matches | Pie chart sector angle |
| Won | 20 | 160° |
| Drawn | 15 | |
| Lost | 10 |
(i) Complete the table.
(ii) Complete the pie chart.
(e) The table shows the total attendance figures for all the teams in the league for two seasons.
| Season | Total attendance |
| A | 9.76 × 106 |
| B | 1.36 × 107 |
Work out how much greater the attendance was in season B than in season A.
Give your answer in standard form.
▶️ Answer/Explanation
(a)(i) Ans: $565.25
5% discount on $595: \(595 \times 0.95 = 565.25\)
(a)(ii) Ans: $42.75
Cost of 16 single tickets: \(16 \times 38 = 608\)
Savings: \(608 – 565.25 = 42.75\)
(b) Ans: 9.2%
Empty seats: \(26272 – 23854 = 2418\)
Percentage empty: \(\frac{2418}{26272} \times 100 \approx 9.2\%\)
(c) Ans: \(5.07 \times 10^5\)
506762 rounded to 3 sig figs: 507000 = \(5.07 \times 10^5\)
(d)(i) Ans: 120°, 80°
Total matches: 45 (360°)
Drawn: \(\frac{15}{45} \times 360 = 120°\)
Lost: \(\frac{10}{45} \times 360 = 80°\)
(e) Ans: \(3.84 \times 10^6\)
Difference: \(1.36 \times 10^7 – 9.76 \times 10^6 = 3,840,000 = 3.84 \times 10^6\)
(a) Write in figures: six million three thousand and seventy six.
(b) (i) Find p when p = -0.6 ÷ 1.6
(ii) Find q when q = -0.6 – 1.6
(iii) Compare p and q using >, <, ≥, ≤, or =
(c) Mount Robson’s height is 3950 m (to nearest 10 m). Complete the inequality for h.
(d) Calculate \(2\frac{1}{12} ÷ 1\frac{1}{4}\) as a decimal (4 significant figures).
(e) (i) Find 80
(ii) Calculate 5-3 as a fraction
(iii) Simplify 8x5 × 3x4
▶️ Answer/Explanation
(a) Ans: 6,003,076
Breakdown:
– Six million = 6,000,000
– Three thousand = 3,000
– Seventy six = 76
Combined: 6,000,000 + 3,000 + 76 = 6,003,076
(b)(i) Ans: -0.375
Calculation: -0.6 ÷ 1.6 = -0.375
(b)(ii) Ans: -2.2
Calculation: -0.6 – 1.6 = -2.2
(b)(iii) Ans: >
Since -0.375 > -2.2 (negative numbers closer to zero are larger)
(c) Ans: 3945 ≤ h < 3955
When rounding to nearest 10m:
Lower bound: 3950 – 5 = 3945
Upper bound: 3950 + 5 = 3955
(d) Ans: 1.667
Convert to improper fractions:
\( \frac{25}{12} ÷ \frac{5}{4} = \frac{25}{12} × \frac{4}{5} = \frac{100}{60} = 1.\overline{6} \)
Rounded to 4 s.f.: 1.667
(e)(i) Ans: 1
Any non-zero number to power 0 equals 1: 80 = 1
(e)(ii) Ans: \(\frac{1}{125}\)
Negative exponent gives reciprocal: 5-3 = \(\frac{1}{5^3}\) = \(\frac{1}{125}\)
(e)(iii) Ans: 24x9
Multiply coefficients and add exponents:
8 × 3 = 24
x5 × x4 = x9