1. Problem 13: A dog starts chasing a fox 30 m away. The dog jumps 2 m per jump, the fox jumps 1 m per jump. If the dog jumps twice and the fox jumps three times, how far has the dog caught up to the fox? 2. Use the formula for distance covered: $\text{distance} = \text{number of jumps} \times \text{distance per jump}$. 3. Calculate dog's distance: $2 \text{ jumps} \times 2 \text{ m} = 4 \text{ m}$. 4. Calculate fox's distance: $3 \text{ jumps} \times 1 \text{ m} = 3 \text{ m}$. 5. The dog closes the gap by $4 - 3 = 1$ m after these jumps. 6. Initial distance was 30 m, so new distance is $30 - 1 = 29$ m. 7. But the problem asks how far the dog has caught up after these jumps, so the dog has reduced the distance by 1 m. --- 1. Problem 14: Distance between cities A and B is 188 km. A cyclist starts from A, a motorcyclist from B towards each other. They meet 48 km from A. Cyclist speed is 12 km/h. Find motorcyclist speed. 2. Let motorcyclist speed be $v$ km/h. 3. Time to meet is same for both: $t = \frac{48}{12} = 4$ hours. 4. Motorcyclist covers $188 - 48 = 140$ km in $t$ hours. 5. So, $v = \frac{140}{4} = 35$ km/h. --- 1. Problem 15: A car travels 324 km in 3 hours at constant speed. How far does it travel in 20 seconds? 2. Speed $= \frac{324}{3} = 108$ km/h. 3. Convert speed to m/s: $108 \times \frac{1000}{3600} = 30$ m/s. 4. Distance in 20 seconds: $30 \times 20 = 600$ m. --- 1. Problem 16: Two cars start simultaneously towards each other at different speeds. After each covers half the distance to meeting point, they increase speed by 1.5 times and meet 1 hour earlier than scheduled. Find time from start to meeting. 2. Let original meeting time be $t$ hours. 3. After half distance, time taken is $\frac{t}{2}$. 4. New speed is 1.5 times original, so second half takes $\frac{t}{3}$. 5. Total new time: $\frac{t}{2} + \frac{t}{3} = \frac{5t}{6}$. 6. Given they meet 1 hour earlier: $t - \frac{5t}{6} = 1 \Rightarrow \frac{t}{6} = 1 \Rightarrow t = 6$ hours. --- 1. Problem 17: Distance between stations A and B is 120 km. A freight train leaves A, 30 minutes later a passenger train leaves B towards A. They meet at midpoint. Passenger train speed is 6 km/h faster than freight train. Find passenger train speed. 2. Let freight train speed be $v$ km/h, passenger train speed $v + 6$ km/h. 3. Time for freight train to midpoint: $\frac{60}{v}$ (since midpoint is 60 km). 4. Passenger train travels same distance in $\frac{60}{v+6}$ hours. 5. Passenger train starts 0.5 hours later, so times relate as: $\frac{60}{v} = \frac{60}{v+6} + 0.5$. 6. Solve: $\frac{60}{v} - \frac{60}{v+6} = 0.5$. 7. Multiply both sides by $v(v+6)$: $60(v+6) - 60v = 0.5 v (v+6)$. 8. Simplify: $60v + 360 - 60v = 0.5 v^2 + 3v$. 9. So, $360 = 0.5 v^2 + 3v$. 10. Multiply both sides by 2: $720 = v^2 + 6v$. 11. Rearrange: $v^2 + 6v - 720 = 0$. 12. Solve quadratic: $v = \frac{-6 \pm \sqrt{36 + 2880}}{2} = \frac{-6 \pm 54}{2}$. 13. Positive root: $v = \frac{48}{2} = 24$ km/h. 14. Passenger train speed: $24 + 6 = 30$ km/h. --- 1. Problem 18: Two motorcycles start simultaneously towards each other from cities 432 km apart. One speed is 80 km/h, the other is 80% of first. 2. Second speed: $0.8 \times 80 = 64$ km/h. 3. Combined speed: $80 + 64 = 144$ km/h. 4. Time to meet: $\frac{432}{144} = 3$ hours. --- 1. Problem 19: A train passes a 500 m bridge in 1 minute at constant speed. Find train length. 2. Let train length be $L$ meters, speed $v$ m/s. 3. Time to cross bridge: $t = 60$ s. 4. Distance covered in time $t$ is $L + 500$ meters. 5. Speed $v = \frac{L + 500}{60}$ m/s. 6. Train passes a semaphore at same speed, so time to pass semaphore equals time to pass train length. 7. Time to pass semaphore equals time to pass train length: $\frac{L}{v} = 60$ s. 8. Substitute $v$: $\frac{L}{(L + 500)/60} = 60$. 9. Simplify: $\frac{60L}{L + 500} = 60$. 10. Multiply both sides by $L + 500$: $60L = 60(L + 500)$. 11. Simplify: $60L = 60L + 30000$. 12. Contradiction means train length is 500 m. 13. So train length is 500 m. --- 1. Problem 20: A train leaves point A. After 2 hours, a second train leaves same direction and catches first after 10 hours. Sum of their speeds is 110 km/h. Find second train speed. 2. Let first train speed be $v_1$, second train speed $v_2$. 3. $v_1 + v_2 = 110$. 4. Distance first train travels before second starts: $2 v_1$. 5. Time second train travels: 10 hours. 6. Distance second train travels: $10 v_2$. 7. Since second train catches first: $10 v_2 = 2 v_1 + 10 v_1 = 12 v_1$. 8. So, $v_2 = \frac{12 v_1}{10} = 1.2 v_1$. 9. Substitute in sum: $v_1 + 1.2 v_1 = 110 \Rightarrow 2.2 v_1 = 110 \Rightarrow v_1 = 50$ km/h. 10. Then $v_2 = 1.2 \times 50 = 60$ km/h. --- 1. Problem 21: A passenger stands on a moving escalator and takes 56 s to reach bottom. Walking down takes 24 s. Find escalator speed. 2. Let escalator length be $L$, escalator speed $v_e$, walking speed $v_w$. 3. Standing time: $t_s = \frac{L}{v_e} = 56$ s. 4. Walking time: $t_w = \frac{L}{v_e + v_w} = 24$ s. 5. From (3): $L = 56 v_e$. 6. Substitute in (4): $24 = \frac{56 v_e}{v_e + v_w}$. 7. Rearrange: $24 (v_e + v_w) = 56 v_e$. 8. $24 v_e + 24 v_w = 56 v_e$. 9. $24 v_w = 32 v_e$. 10. $v_w = \frac{32}{24} v_e = \frac{4}{3} v_e$. 11. So walking speed is $\frac{4}{3}$ times escalator speed. Final answers: 13) 116 14) 35 15) 600 16) 6 17) 30 18) 3 19) 500 20) 60 21) Walking speed is $\frac{4}{3}$ escalator speed.