Time, Speed and Distance

Speed1 = 6 km ÷ (5/3) hr = 3.6 km/hr (1 hr 40 min = 5/3 hr). Speed2 = 6 km ÷ (4/3) hr = 4.5 km/hr (1 hr 20 min = 4/3 hr). Difference = 4.5 - 3.6 = 0.9 km/hr. The difference is positive since the second speed is greater.

The sum of upstream and downstream speeds equals twice the boat's speed in still water (since (b-c)+(b+c)=2b). So boat speed = 36/2 = 18 km/hr. Downstream speed = 18+3 = 21 km/hr. Time = distance/speed = 52.5/21 = 2.5 hours.

Let the meeting point be at distance x from M and (x+200) from N (since second train traveled 200 km more). Time to meet is same: x/54 = (x+200)/63. Cross-multiplying: 63x = 54(x+200) = 54x + 10800, so 9x = 10800, x = 1200 km. Distance between M and N = 1200 + 1200 + 200 = 2600 km. Option C is correct.

When swimming against the current (upstream), effective speed = swimmer's speed in still water - river speed = 12 - 2.5 = 9.5 km/hr. Distance = Speed × Time = 9.5 × 3.5 = 33.25 km. Against the current means subtracting the river's speed from the swimmer's speed.

Aman cycles along the boundary of a square field from corner A to diagonally opposite corner C in 45 minutes at 12 km/hr. Distance covered = 12 × (45/60) = 9 km. For a square, traveling from one corner to the diagonally opposite corner along the boundary means going along two adjacent sides (half the perimeter). So two sides = 9 km, meaning one side = 4.5 km. Area of square = (4.5)² = 20.25 sq km.

Let distance = D, stream speed = s. Downstream: D/(15+s) = 11 min. Upstream: D/(15-s) = 22 min. Dividing: (15-s)/(15+s) = 11/22 = 1/2. So 2(15-s) = 15+s → 30-2s = 15+s → 15 = 3s → s = 5. The key insight is upstream time is exactly double downstream time, making the stream half the boat's speed. Options A, B, C are too small.

When the second train starts at 2 p.m., the first train has a 2-hour head start and has covered 60 × 2 = 120 km. The relative speed of the two trains is 75 - 60 = 15 kmph. Time to catch up = 120/15 = 8 hours. Distance from Delhi = 75 × 8 = 600 km (or 60 × 10 = 600 km). Both trains meet 600 km from Delhi.

Speed ratio 5:4 means relative speed = 1 unit. Gap = 8 km. Time to catch up = 8 / (5x - 4x) = 8/x hours. Fast train distance = 5x × (8/x) = 40 km. In ratio problems: distance = gap × (fast speed / relative speed).

Downstream speed (with current) = 20+1.5 = 21.5 km/hr. Upstream speed (against current) = 20-1.5 = 18.5 km/hr. Time for each leg of 3182 km: downstream = 3182/21.5 = 148 hours, upstream = 3182/18.5 = 172 hours. Total time = 148+172 = 320 hours. Options A (300), B (350), C (375), and E (250) are incorrect.

Let total distance be D. Time taken for 1/6 of D at 25 km/h: (D/6)/25 = D/150 hours. Time for 3/4 of D at 30 km/h: (3D/4)/30 = D/40 hours. Time for remaining (1 - 1/6 - 3/4) = 1/12 of D at 40 km/h: (D/12)/40 = D/480 hours. Total time = D/150 + D/40 + D/480 = D × (16/2400 + 60/2400 + 5/2400) = D × 81/2400 hours. Given total time = 162 minutes = 2.7 hours. So D × 81/2400 = 2.7, therefore D = 2.7 × 2400/81 = 6480/81 = 80 km.

Distance = (75 + 60) x (20/60) = 135 x 1/3 = 45 km. Using relative speed 135 km/h for 20 minutes gives 45 km. The bikers approach each other, so their speeds add. Time must be converted to hours.

For average speed when distance fractions are known: If 55% of distance is at 40 km/h and 45% at 60 km/h, the average speed is not the simple average. We use weighted harmonic mean: Avg = 1 / (0.55/40 + 0.45/60) = 1 / (0.01375 + 0.0075) = 1 / 0.02125 ≈ 47.06 km/h. Distance = Avg × Time = 47.06 × 17 ≈ 800 km. The answer rounds to 800 km.

Downstream:stream speed ratio is 17:2. Stream speed = 4 km/hr, so downstream speed = (17/2) × 4 = 34 km/hr. Let boat speed = b, stream speed = s. Downstream = b + s = 34, s = 4, so b = 30 km/hr. Upstream speed = b - s = 30 - 4 = 26 km/hr. Distance in 5.5 hours upstream = 26 × 5.5 = 143 km. Options A, B, D are incorrect.

Distance by car = 60 km/hr × 35/60 hr = 35 km. Distance by bus = 40 km/hr × 35/60 hr = 23.33 km. Total distance = 58.33 km. Total time = 70 min = 7/6 hr. Average speed = Total distance / Total time = 58.33 × 6/7 = 50 km/hr. Alternatively, average speed when distances are equal but speeds different is (2×s1×s2)/(s1+s2) = (2×60×40)/(60+40) = 4800/100 = 48, but this formula applies when TIME is equal, not distance. Since time IS equal (35 min each), we can use harmonic mean: 2/(1/60+1/40) = 2/((2+3)/120) = 240/5 = 48. Wait, let me recalculate: total distance = 35+23.33 = 58.33 km, time = 1.167 hr, speed = 58.33/1.167 = 50 km/hr.

Let boat speed = u, current speed = c. Upstream: 60/(u-c) = 8, so u-c = 7.5. Downstream: 60/(u+c) = 5, so u+c = 12. Solving: 2u = 19.5, so u = 9.75 km/h. This is the speed in still water.