A boat covers 32 km upstream and 36 km downstream in 7 hours. Also, it covers 40 km upstream and 48 km downstream in 9 hours. The speed of the boat in still water is:

Let speed of boat in still water be $b$ km/hr and speed of stream be $s$ km/hr.

Upstream speed = $(b - s)$ km/hr

Downstream speed = $(b + s)$ km/hr

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For the first trip, 32 km upstream and 36 km downstream takes 7 hours:

$\frac{32}{b-s} + \frac{36}{b+s} = 7$ ...(1)

For the second trip, 40 km upstream and 48 km downstream takes 9 hours:

$\frac{40}{b-s} + \frac{48}{b+s} = 9$ ...(2)

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Let $x = \frac{1}{b-s}$ and $y = \frac{1}{b+s}$

The equations become:

$32x + 36y = 7$ ...(1)

$40x + 48y = 9$ ...(2)

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Multiply equation (1) by 5:

$160x + 180y = 35$

Multiply equation (2) by 4:

$160x + 192y = 36$

Subtracting:

$12y = 1$

$y = \frac{1}{12}$

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Since $y = \frac{1}{b+s} = \frac{1}{12}$:

$b + s = 12$ ...(3)

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Substitute $y = \frac{1}{12}$ into equation (1):

$32x + 36 \times \frac{1}{12} = 7$

$32x + 3 = 7$

$32x = 4$

$x = \frac{1}{8}$

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Since $x = \frac{1}{b-s} = \frac{1}{8}$:

$b - s = 8$ ...(4)

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From equations (3) and (4):

$b + s = 12$

$b - s = 8$

Adding both equations:

$2b = 20$

$b = 10$

Therefore, the speed of the boat in still water is 10 km/hr.