Boat and Stream

This is where the chapter starts. The question gives you two of the three quantities b, s and one effective speed, and asks for the third. Or it gives b and s and asks for the time or distance in one direction.

Form 1: b and s given, find D or U. Add or subtract. b = 15, s = 4 ⇒ D = 19 km/h, U = 11 km/h.

Form 2: b and one effective speed given, find s. If U is given, s = b − U. If D is given, s = D − b. For example, b = 15 and U = 11 ⇒ s = 4.

Form 3: time or distance in one direction. First get the right effective speed, then use TSD. b = 15, s = 4, distance 57 km downstream ⇒ time = 5719 = 3 hours.

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Sometimes the question hides b inside a still-water journey. A man rows 40 km in still water in 4 hours only tells you that b = 10 km/h. In the same way, covers 30 km downstream in 2 hours tells you D = 15 km/h.

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Two boats on the same river. The stream pushes both boats equally. So when you compare them (one chasing the other, or the gap between them), the stream speed cancels out. Relative speed between two boats on the same river does not depend on the stream.

Take Radhika and her friend, rowing downstream on the same river, one behind the other. Radhika's still-water speed is 9 km/h, her friend's is 6 km/h, and the river flows at 2 km/h. Their actual speeds are 11 and 8 km/h, but the gap between them changes at 11 − 8 = 3 km/h, which is just 9 − 6. The river played no part.

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Now the reverse question: you know how fast the boat goes downstream and upstream, and you have to find b and s.

We have two equations:
D = b + s
U = b − s

Add them: D + U = 2b
Subtract them: D − U = 2s

b = D + U2 and s = D − U2

So the boat's speed is the average of the two speeds, and the stream's speed is half their difference. It makes sense: the river adds s on one side and removes s on the other, so the boat's own speed sits exactly in the middle.

Radhika's boat goes downstream at 14 km/h and upstream at 10 km/h. Her speed in still water = 14 + 102 = 12 km/h, and the stream = 14 − 102 = 2 km/h. Check: 12 + 2 = 14 and 12 − 2 = 10.

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When journeys are given instead of speeds. Most questions say something like goes 30 km downstream in 2 hours and 21 km upstream in 3 hours. First change each journey into a speed: D = 302 = 15, U = 213 = 7. Then b = 11 and s = 4. Never average the distances or the times; average only the speeds.

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When the ratio of D and U is given with one more fact. If D : U = 3 : 2 and D + U = 20, then D = 12, U = 8, so b = 10 and s = 2. Since D + U = 2b and D − U = 2s, the sum of the two speeds really gives you b, and their difference gives you s.

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A swimmer going up and coming back. If a swimmer goes upstream for some minutes and then downstream for the same number of minutes, he ends up 2s × (time of each part) away from the start, in the direction of the flow. His own swimming cancels out; only the river's push remains.

Here the question does not stop at b or s. It first gives you enough to find them, and then asks about a different journey: in still water, in the opposite direction, or over a new distance. So the question has two steps, and you already know each of them.

Step 1: find the missing speed. Use Types 1 and 2. From D and U get b and s; or from one journey and the stream speed get b.

Step 2: answer the new journey. Pick the speed that suits it (b in still water, b + s downstream, b − s upstream) and use time = distancespeed.

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Three patterns keep coming back.

Pattern A: up and down journeys given, still-water journey asked. "Goes 4 km upstream in 1 hour and 2 km downstream in 15 minutes. How long for 18 km in still water?" U = 4, D = 8, so b = 6 and time = 186 = 3 hours. The stream is not even needed.

Pattern B: one direction and the stream given, the other direction asked. "Goes 40 km upstream in 5 hours; stream = 2 km/h; time for 60 km downstream?" U = 8 ⇒ b = 10 ⇒ D = 12 ⇒ time = 5 hours. Find b first; from b you can find the speed in either direction.

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Pattern C: times linked by a percentage. Downstream time is 20% less than upstream time means downstream time = 80% of upstream time. Find the other time, then both speeds.

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For example, Radhika rows 9 km upstream in 112 hours, and the river flows at 1.5 km/h. Her upstream speed = 6 km/h, so her own speed = 7.5 km/h. Downstream she will move at 9 km/h, and in still water 15 km will take her 2 hours.

When the distance is the same both ways, you do not need the distance at all. The ratio method finishes these questions in two lines.

The fact from TSD: for the same distance, time is inversely proportional to speed. So if the boat goes a distance downstream and comes back the same distance upstream:

t_up : t_down = D : U

If the upstream trip takes 5 hours and the downstream trip takes 3 hours, then D : U = 5 : 3.

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From D : U to b : s. Since b = D + U2 and s = D − U2, the halves cancel in a ratio:

b : s = (D + U) : (D − U)

For D : U = 5 : 3, b : s = 8 : 2 = 4 : 1. And because D : U = t_up : t_down, the same rule works directly on the times: b : s = (t_up + t_down) : (t_up − t_down).

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The "k times as long" form. It takes twice as long to row up as to row down means t_up : t_down = 2 : 1, so D : U = 2 : 1 and b : s = 3 : 1. In general, if upstream takes k times as long, then b : s = (k + 1) : (k − 1), that is s = b × k − 1k + 1.

• twice as long ⇒ b : s = 3 : 1
• thrice as long ⇒ b : s = 4 : 2 = 2 : 1
• four times as long ⇒ b : s = 5 : 3

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The "same time, different distances" form. If the boat covers d₁ km downstream in the same time as d₂ km upstream, then D : U = d₁ : d₂, and again b : s = (d₁ + d₂) : (d₁ − d₂).

Going backwards: b : s given, times asked. If b : s = 7 : 2, then D : U = 9 : 5, so t_up : t_down = 9 : 5. If the downstream trip took 50 minutes, the return trip takes 50 × 95 = 90 minutes.

Radhika rows to her grandmother's village downstream in 2 hours and comes back upstream in 3 hours. So D : U = 3 : 2, and b : s = 5 : 1. If her own speed is 10 km/h, the river flows at 2 km/h.

A round trip means the boat goes some distance d one way and comes back the same d the other way. One leg is downstream and the other is upstream, whichever direction it starts in. So:

Total time T = db + s + db − s

That is all you need. The question hides one of the four quantities (T, d, b, s) and asks for it.

1. Total time asked. Find D and U, divide d by each, add. b = 10, s = 2, d = 24 ⇒ T = 2412 + 248 = 2 + 3 = 5 hours.

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2. Distance asked. Take d common: T = d(1D + 1U). Or use the ready formula:

d = T(b² − s²)2b

It comes from adding the two fractions: db + s + db − s = 2bdb² − s².

3. Stream speed asked (b, d, T given). From the same formula, b² − s² = 2bdT, so s² = b² − 2bdT. With b = 10, d = 24, T = 5 ⇒ s² = 100 − 96 = 4 ⇒ s = 2. You can also try the options: put each value of s and see which one gives the total time.

4. Average speed of the round trip. The distances are equal and the speeds are D and U, so the average is the harmonic mean:

Average speed = 2DUD + U = (b + s)(b − s)b = b² − s²b

It is always less than b. A river never helps a round trip: the time it saves downstream is less than the time it wastes upstream. That is why a round trip on a river always takes longer than the same trip in still water. If the boat halts somewhere, add the halt to the total time before dividing.

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5. Ratio given with a round trip. If you know D : U (from "5 km down in the same time as 3 km up"), take D = 5k, U = 3k, write the total time in terms of k and find k.

Radhika rows at 5 km/h in still water on a river flowing at 1 km/h. She goes 12 km and comes back. Time = 126 + 124 = 2 + 3 = 5 hours, while in still water she would need only 245 = 4.8 hours. Her average speed = 25 − 15 = 4.8 km/h.

6. A floating object dropped from the boat. A bottle, hat or bag that falls into the river drifts at exactly the speed of the stream. So, relative to the water, the object is standing still and the boat simply goes away from it and comes back. If the boat goes away for t minutes and then turns, it takes the same t minutes to return, whatever the speeds. The object has therefore floated for 2t and its distance from the point of falling is s × 2t. The speed of the boat in still water does not matter at all.

In Type 5 the two times were added. Here they are subtracted. The question says something like for the same distance, the boat takes 3 hours more upstream than downstream. The distance is the same both ways, and the upstream time is the bigger one.

When b and s are known (distance asked). Find the time for 1 km each way and subtract: 1U − 1D. That is the extra time per km. Then:

d = extra time1U − 1D

Or use the ready formula. Simplifying the difference gives db − s − db + s = 2sdb² − s², so:

d = Δt × (b² − s²)2s

Compare it with the round-trip formula d = T(b² − s²)2b. Only one thing changes: 2b becomes 2s, because the difference of the two fractions has 2s on top, while the sum has 2b. Learn them as a pair.

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The LCM way. D and U are often such that their LCM is a neat number. If U = 4 and D = 12, then for 12 km the times are 3 h and 1 h, a gap of 2 hours. If the real gap is 5 hours, the distance is 12 × 52 = 30 km.

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When s is unknown. The equation then becomes a quadratic in s. Write it, clear the fractions and solve, or test the options, since only one of them will give the stated gap.

Take Radhika again. She rows at 7 km/h in still water and the river flows at 3 km/h, so D = 10 and U = 4. For 20 km (the LCM) she needs 2 h downstream and 5 h upstream, a gap of 3 hours. If a question says the gap is 6 hours, the distance is 40 km.

This is the hardest type of the chapter. The question gives two different trips, each with some distance upstream, some distance downstream and a total time. Neither b nor s is given. For example: "20 km upstream and 30 km downstream in 6 hours; 10 km upstream and 45 km downstream in 5 hours."

If you write 20b − s + 30b + s = 6, you get two ugly equations. Change the unknowns instead.

Take x = 1U and y = 1D. These are the times to cover 1 km upstream and 1 km downstream. Now each trip becomes a simple linear equation:
20x + 30y = 6
10x + 45y = 5

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Eliminate one unknown. Multiply the second by 2: 20x + 90y = 10. Subtract the first: 60y = 4 ⇒ y = 115, so D = 15. Put it back: 20x = 6 − 2 = 4 ⇒ x = 15, so U = 5. Now b = 10 and s = 5, exactly as in Type 2.

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Look for the multiple that makes the upstream or the downstream distances equal; that is the easiest elimination. When the two trips are mirror images (36 down + 24 up, and 24 down + 36 up), add and subtract the equations instead; you get x + y and x − y at once.

One trip with a relation. Sometimes there is only one trip, but b and s are tied, for example b is 4 km/h more than s. That tells you U = b − s = 4 directly. Use it to get the upstream time; the rest of the time belongs to the downstream part.

Radhika notices two things on her river. 6 km upstream and 10 km downstream take 3 hours; 12 km upstream and 10 km downstream take 5 hours. Subtracting, the extra 6 km upstream takes 2 hours, so U = 3. Then 10 km downstream takes 3 − 2 = 1 hour, so D = 10. Her speed = 6.5 km/h, the river = 3.5 km/h.

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Replace the river by the air and the boat by an aeroplane, and you get the same chapter. The aeroplane has a speed a in still air, and the wind blows at w. Everything you learnt for b and s carries over:

• Tail wind (wind in the direction of flight): speed over the ground = a + w, like downstream.
• Head wind (against the flight): speed over the ground = a − w, like upstream.

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A flight of 600 km at a = 300 km/h with a 50 km/h tail wind takes 600350 hours; on the way back with the same wind as a head wind it takes 600250 hours. The two ground speeds give the still-air speed and the wind just as D and U gave b and s:

a = (a + w) + (a − w)2, w = (a + w) − (a − w)2

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A moving walkway or escalator is the same thing. Radhika walking at v on a walkway moving at m covers the ground at v + m when both go the same way, and at v − m when they go opposite ways. Round trips, equal-distance average speed and the ratio method work exactly as in a river.

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