1. **Stating the problem:** Given a frequency distribution of tire lifetimes (in 1000 km), we need to find: (1) The five modes of the distribution. (2) The average running distance (mean) of tires. (3) Estimate the annual cost of tires for 10 motor vehicles each running 25000 km/year. (4) The last two parts relate to drawing and tangent construction on a circle, which is unrelated to the tire data and incomplete; we focus on tire questions. 2. **Extracted data:** \begin{array}{|c|c|} \hline \text{Distance (1000 km)} & \text{Number of Tires} \\ \hline 18 - 24 & 5 \\ 24 - 30 & 6 \\ 30 - 36 & 3 \\ 36 - 42 & 10 \\ 42 - 48 & 15 \\ 48 - 54 & 1 \\ \hline \end{array} 3. **(1) Five Modes:** Mode is the value(s) with the highest frequency. Frequencies: 5, 6, 3, 10, 15, 1 The highest frequency is 15 for the class 42 - 48. So the mode class is $42 - 48$ (only one mode class). Since asked "five modes" likely means five modal class midpoints or a misunderstanding; since only one mode exists, only one mode class is $42-48$. We can list modes as the midpoints of top 5 frequencies in descending order: Frequencies descending: 15, 10, 6, 5, 3 Classes correspond to: 15: 42-48, midpoint $45$ 10: 36-42, midpoint $39$ 6: 24-30, midpoint $27$ 5: 18-24, midpoint $21$ 3: 30-36, midpoint $33$ Thus, the "five modes" by frequency descending midpoints: $45$, $39$, $27$, $21$, $33$ 4. **(2) Find the mean distance:** Calculate the mean using midpoints and frequencies. Midpoints $x_i$: $21$, $27$, $33$, $39$, $45$, $51$ Frequencies $f_i$: 5, 6, 3, 10, 15, 1 Calculate total frequency $N = 5+6+3+10+15+1 = 40$ Calculate sum $\sum f_i x_i$: $5\times21 + 6\times27 + 3\times33 + 10\times39 + 15\times45 + 1\times51$ $= 105 + 162 + 99 + 390 + 675 + 51 = 1482$ Mean $\bar{x} = \frac{1482}{40} = 37.05$ (1000 km) So average running distance $= 37050$ km. 5. **(3) Estimate annual tire cost:** Each vehicle driven 25000 km/year. One tire lasts average 37050 km. Number of tires for 10 vehicles: each has 4 wheels; total tires = 40. Tires needed per year per tire = $\frac{25000}{37050} \approx 0.675$ tires per year. For 40 tires, total tire replacements per year $= 40 \times 0.675 = 27$ tires. Cost per tire = 10000. Total annual cost = $27 \times 10000 = 270000$. **Final answers:** (1) Five modal midpoints: $45$, $39$, $27$, $21$, $33$ (2) Mean running distance = $37050$ km (3) Annual tire cost = 270000 6. **Note:** Parts (iv) and (v) about circle and tangent lack data and context and so are not addressed here.