Question

The speed of a boat in still water is 15 km/h. It can go 30 km upstream and return downstream to the original point in 4 h and 30 min. Find the speed of stream.

Answer 1

$\sf \: Please \: Note : -$

Here, we have to find out the speed of the stream. So, we assume it as x km/h and after that we use the condition that, if speed of boat in still water is a km/h (say), then speed of boat upstream = ( a – x ) km/h and speed of boat downstream = ( a + x ) km/h.

$\sf \: Now, \: we \: use \: Time = \frac{Distance}{Speed} \: and \: apply \: the \: condition$

$\sf \: given \: in \: the \: question \: to \: reach \: the \: quadratic \: equation.$

$\sf \: SOLUTION : -$

Let speed of the stream = x km/h

Given, speed of boat in still water = 15 km/h

Therefore, Speed of boat upstream = ( 15 – x ) km/h

and speed of boat downstream = ( 15 + x ) km/h

According to the question,

$\sf \: \frac{30}{15 - x} + \frac{30}{15 + x} = 4 \frac{1}{2}$

$\bigg [ \sf \: \because \: \: Time = \frac{Distance}{Speed} \: and \: distance = 30 \: km \: and \: also, \: 4h \: 30 \: min \bigg ]$

$\sf \: = \bigg(4 + \frac{30}{60} \bigg)h = 4 \frac{1}{2} h$

$\implies \: \sf \: \frac{30(15 + x) + 30(15 - x)}{(15 - x)(15 + x)} = \frac{9}{2}$

$\implies \: \sf \: \frac{450 + 30x + 450 - 30x}{(15) {}^{2} - {x}^{2} } = \frac{9}{2}$

$[\because \sf (A - B)(A + B) = {A}^{2} - B ^{2} ]$

$\implies \: \sf \: \frac{900}{225 - {x}^{2} } = \frac{9}{2}$

$\implies \: \sf \: \frac{900 \times 2}{9} = 225 - {x}^{2}$

$\implies \: \sf \: 200 = 225 - {x}^{2}$

$\implies \: \sf \: {x}^{2} = 25$

$\implies \: \sf \: x = \pm5$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: [taking \: square \: root \: on \: both \: sides]$

But speed cannot be negative.

$\therefore \: \sf \: x = 5$

Hence, speed of stream is 5 km/h.