Question

the speed of a boat in still water is 15km/hr. it can not go 30km upstream and return downstream to the original point 4hours 30 min. find the speed of the stream .​

Answer 1

Appropriate Question:

• The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and returns downstream to the original point in 4 hours 30 minutes. Find the speed of the stream.

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Given: The speed of a Boat in still water is 15 km/hr. And, the boat goes 30 km upstream & returns downstream to the Original point in 4 hours & 30 minutes.

Need to find: The Original speed of the stream?

❍ Let's say, that the speed of the stream be x km/hr.

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$\underline{\bigstar\:\boldsymbol{According\; to\; the \;Question :}}$

• As it is given that, The boat can go 30 km upstream and returns downstream to the Original point in 4 hours 30 minutes.

Therefore,

• Speed of the boat in upstream (–ve) = (15 – x)
• Speed of the boat in downstream (+ve) = (15 + x)
• Time taken to go upstream = 30/(15 – x)
• Time taken to go downstream = 30/(15 + x)
• Time to reach original point = 4 hours and 30 minutes = 4 + ½.

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As we know that,

$\bigstar\;{\underline{\boxed{\frak{Time = \dfrac{Distance}{Speed}}}}}$

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Therefore,

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$:\implies\sf \bigg\{\dfrac{30}{15 - x}\bigg\} + \bigg\{\dfrac{30}{15 + x}\bigg\} = 4 + \dfrac{1}{2} \\\\\\:\implies\sf 30 \times \bigg\{\dfrac{1}{15 - x} + \dfrac{1}{15 + x}\bigg\} = \dfrac{9}{2} \\\\\\:\implies\sf \dfrac{15 + \cancel{x}\;+ 15 -\cancel{ x}\;}{(15^2 - x^2)} = \dfrac{9}{30 \times 2}\\\\\\:\implies\sf \dfrac{30}{225 - x^2} =\cancel\dfrac{9}{60}\\\\\\:\implies\sf \dfrac{30}{225 - x^2 } = \dfrac{3}{20}\\\\\\:\implies\sf 225 - x^2 = \dfrac{3}{20 \times 30}\\\\\\:\implies\sf 225 - x^2 = \cancel\dfrac{3}{600}\\\\\\:\implies\sf 225 - x^2 = 200\\\\\\:\implies\sf x^2 = 225 - 200\\\\\\:\implies\sf x^2 = 25\\\\\\:\implies\sf x = \sqrt{25}\\\\\\:\implies\underline{\boxed{\frak{\purple{x = \pm\;5}}}}\;\bigstar$

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Here,

• Speed can't be –ve. Therefore, ignoring negative value. Hence, x = 5.

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$\therefore{\underline{\textsf{Hence, the speed of the stream is \textbf{5 km/hr}.}}}$

Answer 2

Correct question:-

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

Given:-

• Speed of a boat in still water = 15km/hr

• It can go 30km upstream and return downstream to the original point 4hours 30 min.

To Find:-

• Speed of the stream

Solution:-

Let the Speed of Stream be x.

• Total Distance = 30 km

• Speed of the boat in still water = 15 km/hr

• Speed of the boat upstream = (15 - x) km/hr

• Speed of the boat downstream = (15 + x) km/hr

As we know,

${\boxed{\implies\:Speed\:=\dfrac{Distance}{Time}}}$

So,

• Time taken for upstream T = $\dfrac{30}{15-x}$

• Time taken for downstream T' = $\dfrac{30}{15+x}$

According to question,

$\implies\:T+T'=4\dfrac{1}{2}$

$\implies\dfrac{30}{15-x}+\dfrac{30}{15+x}=\dfrac{9}{2}$

Taking LCM,

$\implies\dfrac{30(15+x)+30(15-x)}{(15-x)(15+x)}=\dfrac{9}{2}$

$\implies\dfrac{450+30x+450-30x}{15^2-x^2}=\dfrac{9}{2}$

$\implies\dfrac{450+450}{225-x^2}=\dfrac{9}{2}$

$\implies\dfrac{900}{225-x^2}=\dfrac{9}{2}$

$\implies\dfrac{100}{225-x^2}=\dfrac{1}{2}$

$\implies\:225-x^2=100×2$

$\implies\:x^2=225-200$

$\implies\:x^2=25$

$\implies\:x=\sqrt{25}$

$\implies\:x=±5$

Speed can't be negative So we will ignore negative term, Then

${\boxed{\implies\:x=+5}}$

Hence, The speed of stream is 5 km/hr.

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