1. **Problem 30:** A traveler goes 8 km upstream and then returns downstream. The stream speed is 4 km/h, and the upstream trip took 20 minutes longer than the downstream trip. Find the boat speed in still water. 2. **Formula and rules:** - Let the boat speed in still water be $v$ km/h. - Upstream speed = $v - 4$ - Downstream speed = $v + 4$ - Time = Distance / Speed - Given: Time upstream = Time downstream + $\frac{20}{60} = \frac{1}{3}$ hours 3. **Set up the equation:** $$\frac{8}{v-4} = \frac{8}{v+4} + \frac{1}{3}$$ 4. **Solve the equation:** Multiply both sides by $3(v-4)(v+4)$: $$3 \times 8 (v+4) = 3 \times 8 (v-4) + (v-4)(v+4)$$ $$24(v+4) = 24(v-4) + v^2 - 16$$ Expand: $$24v + 96 = 24v - 96 + v^2 - 16$$ Simplify: $$24v + 96 = 24v + v^2 - 112$$ Subtract $24v$ from both sides: $$96 = v^2 - 112$$ Add 112: $$v^2 = 208$$ Take square root: $$v = \sqrt{208} = 4\sqrt{13} \approx 14.42$$ 5. **Answer:** The boat speed in still water is approximately 14.42 km/h. --- 6. **Problem 31:** Travelers go 80 km upstream and return downstream on the Selenge river. Total time is 8 hours 20 minutes (8.33 hours). Stream speed is 4 km/h. Find the boat speed in still water. 7. **Set variables:** - Boat speed = $v$ - Upstream speed = $v - 4$ - Downstream speed = $v + 4$ - Total time = $$\frac{80}{v-4} + \frac{80}{v+4} = 8.33$$ 8. **Solve the equation:** Multiply both sides by $(v-4)(v+4)$: $$80(v+4) + 80(v-4) = 8.33 (v^2 - 16)$$ $$80v + 320 + 80v - 320 = 8.33 v^2 - 133.28$$ $$160v = 8.33 v^2 - 133.28$$ Rearranged: $$8.33 v^2 - 160 v - 133.28 = 0$$ 9. **Use quadratic formula:** $$v = \frac{160 \pm \sqrt{160^2 + 4 \times 8.33 \times 133.28}}{2 \times 8.33}$$ Calculate discriminant: $$160^2 = 25600$$ $$4 \times 8.33 \times 133.28 \approx 4440$$ $$\sqrt{25600 + 4440} = \sqrt{30040} \approx 173.3$$ 10. **Calculate roots:** $$v = \frac{160 \pm 173.3}{16.66}$$ Positive root: $$v = \frac{160 + 173.3}{16.66} = \frac{333.3}{16.66} \approx 20.0$$ Negative root is invalid. 11. **Answer:** The boat speed in still water is approximately 20 km/h. --- 12. **Problem 32:** A square metal sheet has a 3 cm wide rectangular strip cut out, leaving an area of 70 cm². Find the side length of the square. 13. **Set variables:** - Let the side of the square be $s$ cm. - The cut strip is 3 cm wide and length $s$ cm. - Area of strip = $3s$ - Remaining area = $s^2 - 3s = 70$ 14. **Form quadratic equation:** $$s^2 - 3s - 70 = 0$$ 15. **Solve quadratic:** $$s = \frac{3 \pm \sqrt{9 + 280}}{2} = \frac{3 \pm \sqrt{289}}{2} = \frac{3 \pm 17}{2}$$ 16. **Possible solutions:** $$s = \frac{3 + 17}{2} = 10$$ $$s = \frac{3 - 17}{2} = -7$$ (discard negative) 17. **Answer:** The side length of the square is 10 cm. --- 18. **Problem 134:** Aldar cycles 67 km in 4 hours. The last 27 km was cycled 2 km/h faster than the previous speed. Find the time taken for the last 27 km. 19. **Set variables:** - Let the speed for the first part be $v$ km/h. - Time for first part: $t_1 = \frac{67 - 27}{v} = \frac{40}{v}$ - Speed for last 27 km: $v + 2$ - Time for last part: $t_2 = \frac{27}{v+2}$ - Total time: $t_1 + t_2 = 4$ 20. **Equation:** $$\frac{40}{v} + \frac{27}{v+2} = 4$$ 21. **Solve equation:** Multiply both sides by $v(v+2)$: $$40(v+2) + 27v = 4v(v+2)$$ $$40v + 80 + 27v = 4v^2 + 8v$$ $$67v + 80 = 4v^2 + 8v$$ Rearranged: $$4v^2 + 8v - 67v - 80 = 0$$ $$4v^2 - 59v - 80 = 0$$ 22. **Use quadratic formula:** $$v = \frac{59 \pm \sqrt{59^2 + 4 \times 4 \times 80}}{8}$$ Calculate discriminant: $$59^2 = 3481$$ $$4 \times 4 \times 80 = 1280$$ $$\sqrt{3481 + 1280} = \sqrt{4761} = 69$$ 23. **Calculate roots:** $$v = \frac{59 \pm 69}{8}$$ Positive root: $$v = \frac{59 + 69}{8} = \frac{128}{8} = 16$$ Negative root: $$v = \frac{59 - 69}{8} = \frac{-10}{8} = -1.25$$ (discard) 24. **Find time for last 27 km:** $$t_2 = \frac{27}{v+2} = \frac{27}{16 + 2} = \frac{27}{18} = 1.5 \text{ hours}$$ 25. **Answer:** Aldar took 1.5 hours to cycle the last 27 km. --- 26. **Problem 135:** A group travels 260 km by bus. After 2 hours, they stop for 30 minutes. To arrive on time, they increase speed by 5 km/h. Find the original speed. 27. **Set variables:** - Original speed = $v$ km/h - Distance covered in first 2 hours: $2v$ - Remaining distance: $260 - 2v$ - Time for remaining distance at increased speed $(v+5)$: $\frac{260 - 2v}{v+5}$ - Total time with stop: $2 + 0.5 + \frac{260 - 2v}{v+5}$ - Total time without stop at original speed: $\frac{260}{v}$ 28. **Equation:** $$2 + 0.5 + \frac{260 - 2v}{v+5} = \frac{260}{v}$$ 29. **Simplify:** $$2.5 + \frac{260 - 2v}{v+5} = \frac{260}{v}$$ 30. **Multiply both sides by $v(v+5)$:** $$2.5 v (v+5) + v (260 - 2v) = 260 (v+5)$$ 31. **Expand:** $$2.5 v^2 + 12.5 v + 260 v - 2 v^2 = 260 v + 1300$$ Simplify left side: $$0.5 v^2 + 272.5 v = 260 v + 1300$$ 32. **Rearranged:** $$0.5 v^2 + 272.5 v - 260 v - 1300 = 0$$ $$0.5 v^2 + 12.5 v - 1300 = 0$$ Multiply by 2: $$v^2 + 25 v - 2600 = 0$$ 33. **Use quadratic formula:** $$v = \frac{-25 \pm \sqrt{25^2 + 4 \times 2600}}{2}$$ Calculate discriminant: $$25^2 = 625$$ $$4 \times 2600 = 10400$$ $$\sqrt{625 + 10400} = \sqrt{11025} = 105$$ 34. **Calculate roots:** $$v = \frac{-25 \pm 105}{2}$$ Positive root: $$v = \frac{-25 + 105}{2} = \frac{80}{2} = 40$$ Negative root: $$v = \frac{-25 - 105}{2} = -65$$ (discard) 35. **Answer:** The original speed of the bus was 40 km/h.