Time, Speed and Distance

Let boat speed be b and stream speed be s. Upstream speed = b - s = 24/6 = 4 km/hr. Downstream speed = b + s = 20/4 = 5 km/hr. Adding: (b - s) + (b + s) = 4 + 5, so 2b = 9, giving b = 4.5 km/hr in still water. Option A (4) and C (5) are the upstream and downstream speeds respectively, which are common mistakes.

Let distance = d and start time be t hours before 11 am. At 15 kmph, time taken = t hours. At 10 kmph, time taken = t + 2 hours (since 1 pm is 2 hours after 11 am). Distance is same, so 15t = 10(t + 2). Solving: 15t = 10t + 20, so 5t = 20, t = 4. Distance = 15 × 4 = 60 km. To reach at 12 noon (1 hour after 11 am), time needed = 4 + 1 = 5 hours. Speed = 60/5 = 12 kmph.

The round trip takes 6 hours: d/(4-s) + d/(4+s) = 6, where d is distance and s is stream speed. This is one equation with two unknowns, so multiple (d,s) pairs satisfy it. Without knowing stream speed, the distance cannot be uniquely determined.

Speed ratio 3:7 means time ratio is 7:3 (inverse). Let Puneet's time be 7x and Shashank's be 3x. Difference: 7x - 3x = 4x = 240 minutes, so x = 60. Shashank's time = 3x = 180 minutes = 3 hours. Speed and time are inversely proportional for fixed distance.

For equal distances, average speed = harmonic mean = 4/((1/10)+(1/20)+(1/30)+(1/60)). Simplifying: 4/((6+3+2+1)/60)=4/(12/60)=4×(60/12)=20 km/h. This is because at equal distances, more time is spent at lower speeds.

Let boat speed = v, stream speed = u. Downstream speed = v+u, upstream = v-u. From first trip: 18/(v+u) + 12/(v-u) = 3. From second: 36/(v-u) + 24/(v+u) = 6.5. Noting that 36/12 = 3 and 24/18 = 4/3, we can derive that 2×(first equation) gives 36/(v+u) + 24/(v-u) = 6. Comparing with second equation, we find v-u = 20/3 and v+u = 20, so v = 10 km/hr. Option B is correct.

Moving time = 20 km ÷ 10 km/hr = 2 hours = 120 minutes. The man stops after each kilometer EXCEPT after the 20th (final) kilometer. So he stops 19 times, taking 19 × 4 = 76 minutes. Total time = 120 + 76 = 196 minutes.

Downstream speed = 21/3 = 7 km/h, upstream speed = 15/3 = 5 km/h. Speed of boat = (downstream + upstream)/2 = 6 km/h. Speed of current = (downstream - upstream)/2 = 1 km/h. The current's speed is half the difference between downstream and upstream speeds.

Let time to meet be t hours. Distance by P = 20t, by Q = 30t. Given: 30t - 20t = 72, so 10t = 72, giving t = 7.2 hours. Total distance = 20t + 30t = 50t = 50 × 7.2 = 360 km. Option A is correct.

Distance = Speed × Time = 138 × 15 = 2070 km. To cover the same distance in 23 hours, new speed = 2070/23 = 90 km/hr. Decrease in speed = 138 - 90 = 48 km/hr.

Downstream speed = 60/6 = 10 km/hr. If total round trip time is 21 hours and downstream takes 6 hours, upstream takes 15 hours, giving upstream speed = 60/15 = 4 km/hr. Solving u+v=10 and u-v=4 gives u=7 km/hr.

Let A's speed = a km/hr, B's speed = b km/hr. A takes 2 hours more: 30/a = 30/b + 2. When A doubles speed: 30/(2a) = 30/b - 1. Solving: a = 5 km/hr, b = 6 km/hr. Check: 30/5 = 6 hrs, 30/6 = 4 hrs (difference 2 ✓). Doubled: 30/10 = 3 hrs, which is 1 less than 4 ✓. Option A is correct.

Speed = 90 km/hr = 90 × (1000/3600) m/s = 25 m/s. Distance = Speed × Time = 25 × 5 = 125 m. Option C is correct. Option A (100 m) would require speed of 20 m/s (72 km/hr). Option B (60 m) would require 12 m/s (43.2 km/hr).

In fixed time t, distance downstream = (v+3)t, distance upstream = (v-3)t. Given: (v+3)t = 2(v-3)t, so v+3 = 2v-6, v = 9. Downstream distance = 12t, upstream = 6t. Ratio is 2:1 as stated. Speed in still water is 9 km/hr.

Downstream speed = 3 + 2 = 5 km/h, upstream speed = 3 - 2 = 1 km/h. Time to go 5 km downstream and 5 km upstream: 5/5 + 5/1 = 1 + 5 = 6 hours. Note: the wording is unclear but 'with and against the stream' implies going with the stream and then returning against it.