Speed, distance, trains and boats for IBPS PO

Every speed question is one relationship wearing a different costume: distance = speed × time. Trains add their own length to the distance. Boats add or subtract the stream from the speed. Two moving objects combine their speeds. Once you see which costume the question is wearing, the arithmetic is short.

In IBPS PO, speed problems appear as standalone arithmetic in the prelims and as statements in quantity comparison and data sufficiency. Mains caselets sometimes build a small story around a journey. The numbers are designed to be friendly, so a clean setup is worth more than speed of calculation.

The core relationships

Why you cannot average the two speeds directly: on a 120 km round trip at 60 km/h and 40 km/h, the outward leg takes 1 hour and the return takes 1.5 hours. Total 120 km in 2.5 hours is 48 km/h, not 50. The slower leg lasts longer, so it pulls the average down.

Worked examples

Example 1 (average speed): A car goes to a city at 60 km/h and returns along the same road at 40 km/h. Find the average speed.

• Equal distances, so average = 2 × 60 × 40 ÷ (60 + 40) = 4,800 ÷ 100 = 48 km/h.

Example 2 (trains crossing): Trains of length 150 m and 200 m run at 54 km/h and 72 km/h. How long do they take to cross each other when moving in opposite directions? And in the same direction?

• Distance in both cases = 150 + 200 = 350 m.
• Opposite: 54 + 72 = 126 km/h = 126 × 5/18 = 35 m/s. Time = 350 ÷ 35 = 10 seconds.
• Same direction: 72 − 54 = 18 km/h = 5 m/s. Time = 350 ÷ 5 = 70 seconds.

Example 3 (pole and platform): A train crosses a pole in 12 seconds and a 300 m platform in 27 seconds. Find its length and speed.

• The extra 15 seconds are spent covering the platform's 300 m, so speed = 300 ÷ 15 = 20 m/s = 72 km/h.
• Length = 20 × 12 = 240 m.

Example 4 (boats): A boat goes 36 km downstream in 3 hours and returns in 4.5 hours. Find the speeds of the boat in still water and of the stream.

• Downstream = 36 ÷ 3 = 12 km/h. Upstream = 36 ÷ 4.5 = 8 km/h.
• Boat = (12 + 8) ÷ 2 = 10 km/h. Stream = (12 − 8) ÷ 2 = 2 km/h.

Example 5 (late and early): Walking at 5 km/h, a person reaches the station 7 minutes late. Walking at 6 km/h, they reach 5 minutes early. Find the distance.

• The two times differ by 7 + 5 = 12 minutes = 1/5 hour.
• d/5 − d/6 = 1/5, so d/30 = 1/5 and d = 6 km.
• Check: 6 km at 5 km/h takes 72 minutes; at 6 km/h, 60 minutes. The gap is 12 minutes.

Example 6 (after meeting): Two trains start at the same time from P and Q towards each other. After they meet, they take 9 hours and 16 hours to reach Q and P respectively. Find the ratio of their speeds.

• Let the speeds be a and b, and let them meet after t hours.
• After meeting, the first train covers what the second covered before meeting: 9a = bt. Likewise 16b = at.
• Dividing, 9a ÷ 16b = b ÷ a, so a² ÷ b² = 16 ÷ 9 and a : b = 4 : 3.
• The shortcut: speed ratio = √(time of second ÷ time of first) = √(16/9).

The boat equation trick

Some mains-level questions give two trips and two total times, with the boat and stream speeds unknown. Write 1/U = x and 1/D = y, and the question becomes two linear equations. Practice question 8 is this type.

Common mistakes

• Mixing units. Convert to m/s whenever lengths are in metres and times are in seconds.
• Forgetting the train's own length when it crosses a platform, a bridge or another train.
• Adding speeds for trains in the same direction. Same direction subtracts.
• Averaging speeds directly for a round trip. Use 2ab ÷ (a + b) for equal distances, or total distance ÷ total time.
• Swapping downstream and upstream. Downstream is always the faster one.
• Treating "late" and "early" as the same side. Add them when one is late and the other is early; subtract when both are late.

Practice set

1. Find the average speed for equal distances covered at 45 km/h and 30 km/h.
2. A 240 m train running at 72 km/h crosses a 360 m platform. How long does it take?
3. Trains of 120 m and 180 m run at 60 km/h and 48 km/h in the same direction. How long does the faster one take to pass the slower one?
4. A boat's downstream speed is 16 km/h and its upstream speed is 10 km/h. Find the speed of the stream.
5. At 4 km/h a person is 10 minutes late; at 5 km/h they are 8 minutes early. Find the distance.
6. A person drives to a town at 40 km/h and returns at 60 km/h. The round trip takes 5 hours. How far is the town?
7. A train running at 54 km/h passes a person walking at 6 km/h in the same direction in 18 seconds. Find the train's length.
8. A boat goes 30 km upstream and 44 km downstream in 10 hours. It goes 40 km upstream and 55 km downstream in 13 hours. Find the speed of the boat in still water and of the stream.

Answers:

1. 36 km/h. 2 × 45 × 30 ÷ 75 = 2,700 ÷ 75.
2. 30 seconds. 72 km/h = 20 m/s; distance 240 + 360 = 600 m.
3. 90 seconds. Relative speed 12 km/h = 10/3 m/s; distance 300 m; 300 ÷ 10/3 = 90.
4. 3 km/h. (16 − 10) ÷ 2. The boat's speed is 13 km/h.
5. 6 km. Gap = 18 minutes = 0.3 hour; d/4 − d/5 = d/20 = 0.3.
6. 120 km. d/40 + d/60 = 5, so 5d/120 = 5 and d = 120.
7. 240 m. Relative speed 48 km/h = 40/3 m/s; 40/3 × 18 = 240.
8. Boat 8 km/h, stream 3 km/h. With x = 1/U and y = 1/D: 30x + 44y = 10 and 40x + 55y = 13. Multiply the first by 4 and the second by 3 and subtract: 11y = 1, so D = 11. Then 30x = 6, so U = 5. Boat = (11 + 5) ÷ 2; stream = (11 − 5) ÷ 2.

What to do next

• Write the table above from memory once, then solve eight speed questions a day for five days.
• For every train question, draw two short lines for the trains and mark what has to pass what. It takes three seconds and prevents the length mistake.
• Pair this with time and work; both are rate × time and are often tested together.