1. **State the problem:** Two cyclists start simultaneously from two cities 120 km apart and ride towards each other. They meet after 2 hours and 40 minutes. One cyclist covers 1 km in 2 minutes less than the other. We need to find each cyclist's speed and the distance each traveled until they met. 2. **Convert time to hours:** 2 hours and 40 minutes = 2 + \frac{40}{60} = 2 + \frac{2}{3} = \frac{8}{3} hours. 3. **Define variables:** Let the speed of the slower cyclist be $v$ km/h. 4. **Relate speeds:** Since one cyclist covers 1 km in 2 minutes less, convert 2 minutes to hours: $\frac{2}{60} = \frac{1}{30}$ hours. The time to cover 1 km for the slower cyclist is $\frac{1}{v}$ hours. The faster cyclist covers 1 km in $\frac{1}{v} - \frac{1}{30}$ hours. 5. **Find the speed of the faster cyclist:** Speed is inverse of time per km, so $$v_{fast} = \frac{1}{\frac{1}{v} - \frac{1}{30}} = \frac{1}{\frac{30 - v}{30v}} = \frac{30v}{30 - v}$$ 6. **Use the meeting condition:** They meet after $\frac{8}{3}$ hours, so the sum of distances equals 120 km: $$v \times \frac{8}{3} + v_{fast} \times \frac{8}{3} = 120$$ Substitute $v_{fast}$: $$\frac{8}{3} v + \frac{8}{3} \times \frac{30v}{30 - v} = 120$$ Divide both sides by $\frac{8}{3}$: $$v + \frac{30v}{30 - v} = \cancel{\frac{120}{\frac{8}{3}}} = 45$$ 7. **Solve for $v$:** Multiply both sides by $30 - v$: $$v(30 - v) + 30v = 45(30 - v)$$ Expand: $$30v - v^2 + 30v = 1350 - 45v$$ Combine like terms: $$60v - v^2 = 1350 - 45v$$ Bring all terms to one side: $$-v^2 + 60v + 45v - 1350 = 0$$ $$-v^2 + 105v - 1350 = 0$$ Multiply both sides by $-1$: $$v^2 - 105v + 1350 = 0$$ 8. **Factor the quadratic:** $$v^2 - 105v + 1350 = (v - 30)(v - 45) = 0$$ So, $v = 30$ or $v = 45$. 9. **Check which is slower cyclist:** If $v=30$, then $$v_{fast} = \frac{30 \times 30}{30 - 30}$$ Denominator zero, invalid. If $v=45$, then $$v_{fast} = \frac{30 \times 45}{30 - 45} = \frac{1350}{-15} = -90$$ Negative speed invalid. Re-examine step 4: The faster cyclist covers 1 km in 2 minutes less, so the slower cyclist takes $t$ hours per km, the faster cyclist takes $t - \frac{1}{30}$ hours per km. If the slower cyclist speed is $v = \frac{1}{t}$, then the faster cyclist speed is $\frac{1}{t - \frac{1}{30}}$. Let $t$ be the time per km for the slower cyclist. Then speeds: $$v = \frac{1}{t}, \quad v_{fast} = \frac{1}{t - \frac{1}{30}}$$ Sum of distances: $$(v + v_{fast}) \times \frac{8}{3} = 120$$ Substitute speeds: $$\left(\frac{1}{t} + \frac{1}{t - \frac{1}{30}}\right) \times \frac{8}{3} = 120$$ Divide both sides by $\frac{8}{3}$: $$\frac{1}{t} + \frac{1}{t - \frac{1}{30}} = 45$$ Find common denominator: $$\frac{(t - \frac{1}{30}) + t}{t(t - \frac{1}{30})} = 45$$ Simplify numerator: $$\frac{2t - \frac{1}{30}}{t^2 - \frac{t}{30}} = 45$$ Multiply both sides by denominator: $$2t - \frac{1}{30} = 45 \left(t^2 - \frac{t}{30}\right)$$ Expand right side: $$2t - \frac{1}{30} = 45 t^2 - \frac{45 t}{30} = 45 t^2 - 1.5 t$$ Bring all terms to one side: $$0 = 45 t^2 - 1.5 t - 2 t + \frac{1}{30} = 45 t^2 - 3.5 t + \frac{1}{30}$$ Multiply entire equation by 30 to clear fraction: $$0 = 1350 t^2 - 105 t + 1$$ Use quadratic formula: $$t = \frac{105 \pm \sqrt{105^2 - 4 \times 1350 \times 1}}{2 \times 1350}$$ Calculate discriminant: $$105^2 = 11025, \quad 4 \times 1350 = 5400$$ $$\sqrt{11025 - 5400} = \sqrt{5625} = 75$$ So, $$t = \frac{105 \pm 75}{2700}$$ Two solutions: $$t_1 = \frac{180}{2700} = \frac{1}{15} = 0.0667 \text{ hours}$$ $$t_2 = \frac{30}{2700} = \frac{1}{90} = 0.0111 \text{ hours}$$ Choose $t_1 = \frac{1}{15}$ hours per km (slower cyclist). Speeds: $$v = \frac{1}{t} = 15 \text{ km/h}$$ $$v_{fast} = \frac{1}{t - \frac{1}{30}} = \frac{1}{\frac{1}{15} - \frac{1}{30}} = \frac{1}{\frac{2}{30} - \frac{1}{30}} = \frac{1}{\frac{1}{30}} = 30 \text{ km/h}$$ 10. **Calculate distances traveled:** $$d_{slow} = v \times \frac{8}{3} = 15 \times \frac{8}{3} = 40 \text{ km}$$ $$d_{fast} = 30 \times \frac{8}{3} = 80 \text{ km}$$ 11. **Final answer:** - Slower cyclist speed: 15 km/h, distance traveled: 40 km. - Faster cyclist speed: 30 km/h, distance traveled: 80 km. This matches the given solution.