Question

A motor boat whose speed is 24km/hr takes 1 hr more to go 32km upstream than to return to downstream to the same spot. Find speed of streem​

Answer 1

QuestioN

• A motor boat whose speed is 24km/hr takes 1 hr more to go 32km upstream than return to downstream to the same spot. Find speed of stream?

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${\normalsize{\mathcal{\pmb{\underline{Given\: parameters- }}}}}$

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$\tt \:\:\:\:\:\:\:\:\:\:\succ\:Total \:Distance \: = \: 32\:km$

$\tt \:\:\:\:\succ\:Speed\: in\: Still \:Water\: = \:24\:km/h$

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${\normalsize{\mathcal{\pmb{\underline{Now, }}}}}$

$\:\:\:\frak{\red{Let}}\begin{cases}\sf{\:\:\:\:\:\:\:\:\:\:The\:speed\:of\:stream\:be\:x} &\\ \sf {\:\:\:\:\:Speed_{(boat)}\:in\:Upstream\:is\:24-x}&\\ \sf{Speed_{(boat)}\:in\: downstream\:is\:24+x}\end{cases}$

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${\large{\frak{\pmb{\underline{ According\:to\: question,}}}}}$

Time taken in the upstream journey - Time taken in the downstream journey = 1 hour

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✎ $\textit Hence \:the\: equation\: becomes$

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$\sf \:\:\:\:\:\implies\:\:\bf\dfrac{32}{24-x}\:-\:\bf\dfrac{32}{24+x}\:=\:1$

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$\sf \:\:\:\:\:\:\:\:\:\:\: \looparrowright\:\:\bf\dfrac{\cancel{24}+x\:\cancel{-24}+x}{(24)^2-x^2}\:=\:\bf\dfrac{1}{32}$

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$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\looparrowright\:\:\bf\dfrac{2x}{576-x^2}\:=\:\bf\dfrac{1}{32}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sf |By\:Cross\: Multiplication|$

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$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\looparrowright\:x^2+64x-576\:=\:0$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sf |By\:splitting\: mid\:term|$

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$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\rightsquigarrow\:x^2+(72-8)x-576\:=\:0$

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$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\rightsquigarrow\:x^2+72x-8x-576\:=\:0$

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$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\rightsquigarrow\:x(x+72)-8(x+72)\:=\:0$

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$\:\:\Bigg|\begin{array}{|c c|} \sf x\: - \:8\:=\:0\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sf x\:+\:72\:=\:0 \\\\ \sf \:x\: =\: 8\:\:(\checkmark)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sf x\:=\:-72\end{array}\Bigg|$

$\tt \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \:\big|Since,\:the\:speed\\ \: \tt \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:can't\:be\: negative\big|$

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$\:\:\:\:\:\ddagger\small{\underline{\boxed{\mathcal{\underline{\purple{\therefore \:The \:speed \:of \:the \:stream\: is \:8\: km/hr}}}}}}$

Answer 2

Answer:

Speed of stream = 8 km/hr. HOPE IT HELPS YOU

Step-by-step explanation:

Speed of boat = 24km/hr

Let speed of stream is x

TIME = Distance/Speed

Speed in upstream = 24 - x

Speed in Downstream = 24 + x

Time taken in upstream is 1 hr more. So,

So, ATQ

time taken in upstream = time taken in downstream + 1 hour

Distance of upstream/Speed = Distance of downstream/Speed + 1 hour

$\frac{32}{24-x} = \frac{32}{24+x} + 1 hr\\\frac{32}{24-x} = \frac{32+24+x}{24+x}\\\frac{32}{24-x} = \frac{56+x}{24+x}\\Cross Multiply$

32(24+x) = (24-x)(56+x)

768 + 32x = 1344 + 24x - 56x - x^2

768 = 1344 + 24x - 56x - 32x - x^2

768 = 1344 - 64x - x^2

x^2 + 64x + 768 - 1344 = 0

x^2 + 64x - 576 = 0

Now factorize:

$x^{2} + 64x - 576 = 0\\x^{2} + (72-8)x - 576 = 0\\x^{2} + 72x - 8x - 576 = 0$

taking common

x(x + 72) - 8(x+72) = 0

(x-8)(x+72) = 0

x-8 = 0,, x + 72 = 0

x = 8 ,, x = (-72)

Speed can't be negative so neglecting negative.

So,

speed of stream is 8 km/hr