A boat goes 30 km upstream in 3 hours and 36 km downstream in 3 hours — find speed of boat and stream

Question

A boat goes 30 km upstream in 3 hours and 36 km downstream in the same 3 hours. Find the speed of the boat in still water and the speed of the stream.

Solution — Step by Step

Let the speed of the boat in still water = $b$ km/h. Let the speed of the stream = $s$ km/h.

When the boat moves upstream (against the current), the effective speed = $b - s$ km/h. When the boat moves downstream (with the current), the effective speed = $b + s$ km/h.

This is the core concept: the stream helps you downstream and opposes you upstream.

Upstream: 30 km in 3 hours.

b - s = \frac{30}{3} = 10 \quad \text{...(1)}

Downstream: 36 km in 3 hours.

b + s = \frac{36}{3} = 12 \quad \text{...(2)}

Adding equations (1) and (2):

(b - s) + (b + s) = 10 + 12

2b = 22

b = 11 \text{ km/h}

Substituting back into equation (1):

11 - s = 10 \Rightarrow s = 1 \text{ km/h}

Speed of boat in still water = 11 km/h Speed of stream = 1 km/h

Verification:

Upstream speed = $11 - 1 = 10$ km/h. Distance in 3 hours = $10 \times 3 = 30$ km ✓
Downstream speed = $11 + 1 = 12$ km/h. Distance in 3 hours = $12 \times 3 = 36$ km ✓

Why This Works

This is a classic “two-unknowns, two-equations” problem. The upstream/downstream scenario naturally creates two linear equations in two variables — exactly what we need for a unique solution.

The key insight is that the stream speed $s$ adds or subtracts from the boat speed depending on direction. By adding the two equations, we eliminate $s$ entirely. By subtracting, we would eliminate $b$ .

This type of problem appears in every Class 8–10 CBSE exam under “Simultaneous Linear Equations.” The scenario changes (boats, trains, two workers, mixture problems) but the method is always the same: define variables, form two equations, solve.

Alternative Method

By inspection (since time is equal for both): The upstream and downstream speeds are directly:

Upstream speed = 30/3 = 10 km/h
Downstream speed = 36/3 = 12 km/h

Then: $b = \frac{\text{upstream} + \text{downstream}}{2} = \frac{10 + 12}{2} = 11$ km/h

And: $s = \frac{\text{downstream} - \text{upstream}}{2} = \frac{12 - 10}{2} = 1$ km/h

This is faster for cases where times are equal. When times differ (30 km in 2 hr upstream, 36 km in 3 hr downstream), always use the simultaneous equations method.

Common Mistake

Students often get confused between upstream and downstream and write “upstream speed = $b + s$ ” (adding the stream speed). Remember: upstream is AGAINST the current, so the stream opposes the boat — effective speed = $b - s$ . Downstream is WITH the current, so it helps the boat — effective speed = $b + s$ . If you confuse these, both equations are wrong and you get a negative speed for $s$ .

In exams where you’re given 4 quantities (two distances, two times) for upstream and downstream, first compute the two speeds directly (distance ÷ time), then use the half-sum and half-difference formulas to find $b$ and $s$ . It’s faster than writing out full equations.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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