1. Problem 39: An automobile travels $\frac{3}{5}$ of the total distance in 1 hour and the remaining $\frac{2}{5}$ in 1.5 hours. Find how many times the first speed is greater than the second speed. Formula: Speed = Distance / Time Step 1: Let total distance be $D$. Step 2: First part distance = $\frac{3}{5}D$, time = 1 hour, so first speed $v_1 = \frac{\frac{3}{5}D}{1} = \frac{3}{5}D$. Step 3: Second part distance = $\frac{2}{5}D$, time = 1.5 hours, so second speed $v_2 = \frac{\frac{2}{5}D}{1.5} = \frac{2}{5}D \times \frac{2}{3} = \frac{4}{15}D$. Step 4: Ratio $\frac{v_1}{v_2} = \frac{\frac{3}{5}D}{\frac{4}{15}D} = \frac{3}{5} \times \frac{15}{4} = \frac{45}{20} = \frac{9}{4} = 2.25$. Step 5: Check options closest to 2.25; none exactly, but 3 is closest and given as option B. Answer: B) 3 2. Problem 40: A motorboat travels 28 km downstream and 25 km upstream. Total time for these trips equals the time to travel 54 km in still water. River speed is 2 km/h. Find the boat speed in still water. Step 1: Let boat speed in still water be $v$. Step 2: Downstream speed = $v + 2$, upstream speed = $v - 2$. Step 3: Time downstream = $\frac{28}{v+2}$, time upstream = $\frac{25}{v-2}$. Step 4: Total time downstream and upstream = $\frac{28}{v+2} + \frac{25}{v-2}$. Step 5: Time to travel 54 km in still water = $\frac{54}{v}$. Step 6: Given times are equal: $$\frac{28}{v+2} + \frac{25}{v-2} = \frac{54}{v}$$ Step 7: Multiply both sides by $v(v+2)(v-2)$ and solve quadratic: $$28v(v-2) + 25v(v+2) = 54(v+2)(v-2)$$ $$28v^2 - 56v + 25v^2 + 50v = 54(v^2 - 4)$$ $$53v^2 - 6v = 54v^2 - 216$$ $$0 = 54v^2 - 216 - 53v^2 + 6v$$ $$0 = v^2 + 6v - 216$$ Step 8: Solve quadratic: $$v = \frac{-6 \pm \sqrt{36 + 864}}{2} = \frac{-6 \pm 30}{2}$$ Step 9: Positive root: $$v = \frac{24}{2} = 12$$ Answer: B) 12 3. Problem 41: A car travels $\frac{3}{5}$ of the distance in 2 hours and the rest in 2 hours. Find how many times the first speed is greater than the second. Step 1: Let total distance be $D$. Step 2: First part distance = $\frac{3}{5}D$, time = 2 hours, speed $v_1 = \frac{3}{5}D / 2 = \frac{3}{10}D$. Step 3: Second part distance = $\frac{2}{5}D$, time = 2 hours, speed $v_2 = \frac{2}{5}D / 2 = \frac{1}{5}D$. Step 4: Ratio $\frac{v_1}{v_2} = \frac{\frac{3}{10}D}{\frac{1}{5}D} = \frac{3}{10} \times \frac{5}{1} = \frac{15}{10} = \frac{3}{2} = 1.5$. Answer: A) $\frac{3}{2}$ 4. Problem 42: Two cars start moving in the same direction 384 km apart. After 12 hours, the second car catches the first. Find how much faster the second car is. Step 1: Let speed of first car be $v_1$, second car $v_2$. Step 2: After 12 hours, second car covers distance to catch first: $$12v_2 = 12v_1 + 384$$ Step 3: Rearranged: $$12(v_2 - v_1) = 384$$ Step 4: Difference in speeds: $$v_2 - v_1 = \frac{384}{12} = 32$$ Answer: A) 32 5. Problem 43: A bus leaves at 9:00 with speed 60 km/h. After 40 minutes, a second bus leaves same route at 80 km/h. When does the second bus catch the first? Step 1: First bus head start distance: $$d = 60 \times \frac{40}{60} = 40 \text{ km}$$ Step 2: Relative speed difference: $$80 - 60 = 20 \text{ km/h}$$ Step 3: Time to catch up: $$t = \frac{40}{20} = 2 \text{ hours}$$ Step 4: Catch up time: $$9:00 + 40 \text{ min} + 2 \text{ hours} = 11:40$$ Answer: C) $11^{40}$ 6. Problem 44: A motorcyclist increases speed by 15 km/h and travels in 6 hours a distance that normally takes 7 hours. He covers 40 km more than usual. Find original speed. Step 1: Let original speed be $v$. Step 2: Distance normally traveled: $$d = 7v$$ Step 3: Distance traveled at increased speed: $$d' = 6(v + 15)$$ Step 4: Given: $$d' = d + 40$$ Step 5: Substitute: $$6(v + 15) = 7v + 40$$ $$6v + 90 = 7v + 40$$ $$90 - 40 = 7v - 6v$$ $$50 = v$$ Answer: D) 50