Time, Speed and Distance

Let r be rickshaw speed and b be bus speed. 2/r + 12/b = 0.5. 4/r + 10/b = 0.5 + 9/60 = 0.65. Solving the system: 4/r + 24/b = 1. Subtracting: 14/b = 0.35, so b = 40. 2/r + 12/40 = 0.5 => 2/r = 0.5 - 0.3 = 0.2 => r = 10.

The distance is 16 km, which is 16000 meters. The time is 4/3 hours, which is (4/3) * 3600 = 4800 seconds. Speed = distance / time = 16000 / 4800 = 160 / 48 = 10 / 3 = 3 1/3 m/sec.

Let boat speed be v and stream speed be 1.5. Downstream: d = 5(v + 1.5). Upstream: d = 5.5(v - 1.5). Equating: 5v + 7.5 = 5.5v - 8.25 => 0.5v = 15.75 => v = 31.5.

Let car speed be v. Train speed is 1.5v. Time difference is 12.5 min = 12.5/60 = 5/24 hours. (75/v) - (75/1.5v) = 5/24. (75/v) - (50/v) = 5/24. 25/v = 5/24. v = 25 * 24 / 5 = 120 kmph.

Speed = Distance / Time. Distance = 10 km. Time = 5 min = 5/60 hr = 1/12 hr. Speed = 10 / (1/12) = 120 km/hr.

Distance = Speed * Time. 30 * 2.5 = 75 km.

Distance = 120 km = 120,000 m. Speed = 10 m/s. Time = 120,000 / 10 = 12,000 seconds. 12,000 / 3600 = 3.333 hours = 3 hours 20 minutes.

Let cruiser position be (0, y) and car be (x, 0). Distance D = sqrt(x^2 + y^2). D^2 = x^2 + y^2. 2D * dD/dt = 2x * dx/dt + 2y * dy/dt. D = sqrt(0.6^2 + 0.8^2) = 1.0. dD/dt = 20. dx/dt = v_car. dy/dt = -60 (cruiser moving towards intersection). 2 * 1 * 20 = 2 * 0.8 * v_car + 2 * 0.6 * (-60). 40 = 1.6 * v_car - 72. 112 = 1.6 * v_car. v_car = 112 / 1.6 = 70.

Let the correct travel time be T hours and the distance be D. The nine-minute difference gives D/70 - D/80 = 0.15, so D = 84 km, and the scheduled time is 84/70 - 12/60 = 1 hour.

Let speed be v. 300/v - 300/(v+10) = 1. 300(v+10-v) = v(v+10). 3000 = v^2 + 10v. v^2 + 10v - 3000 = 0. (v+60)(v-50) = 0. Speed = 50 kmph.

Let distance be D. Upstream speed = 8 - 4 = 4 km/h. Downstream speed = 8 + 4 = 12 km/h. Time = D/4 + D/12 = 20. (3D + D) / 12 = 20. 4D = 240. D = 60 km.

Time taken for one lap is distance/speed. A: 12/3 = 4 hrs, B: 12/7 hrs, C: 12/13 hrs. They meet at the LCM of (4, 12/7, 12/13) = LCM(4, 12, 12) / HCF(1, 7, 13) = 12 / 1 = 12 hours.

Circumference = pi * d = 1.4 * pi km = 1400 * pi meters. Relative speed = 165 - 110 = 55 m/min. Time to meet = Distance / Relative speed = 1400 * pi / 55. This calculation seems off; checking options. If they meet at the starting point, time must be a multiple of the time taken for one full lap. A takes 1400*pi / 165, B takes 1400*pi / 110. LCM of these times gives 1.33 hours.

Time = Distance / Speed. Given distance = 15, let speed = x. Time = 15 / x. The problem states time is two less than speed: 15 / x = x - 2. Multiplying by x gives 15 = x^2 - 2x, which rearranges to x^2 - 2x - 15 = 0.

Let car speed be v. Train speed = 1.5v. Time taken by car = 120/v. Time taken by train = 120/(1.5v) + 0.5. Since times are equal, 120/v = 80/v + 0.5. 40/v = 0.5, so v = 80 km/hr.