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Alleya
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Tyro
Asked: 2021-04-08T09:10:48+00:00 In: ,

Rational Algebraic Equations

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A speed boat travels at a rate of 20 kmph in still water. It travels 55 km upstream in the same time that it travels 90 km downstream. What is the rate pf the current?

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  2. acedstud
    This answer was edited.

    First, recall the relationship between distance, time, and speed. That is, \text{speed}=\dfrac{\text{distance}}{\text{time}}.

    Let x be the speed of the current and let t be the time at which the boat travelled in uptstreem and in the downstream, then

    20-x=\dfrac{55}{t} . . . . . (1)
    20+x=\dfrac{90}{t} . . . . . (2)

    Dividing (1) by (2) gives \dfrac{20-x}{20+x}=\dfrac{\frac{55}{t}}{\frac{90}{t}}=\dfrac{55}{t}\cdot\dfrac{t}{90}=\dfrac{55}{90}.

    Multiplying both sides by the LCD, 90(20+x), (croo-multiplying), we have 90(20-x)=55(20+x)\Rightarrow1800-90x=1100+55x\Rightarrow1800-1100=55x+90x

    \Rightarrow700=145x\Rightarrow\dfrac{700}{145}=\dfrac{145x}{145}\Rightarrow x\approx4.83.

    Therefore, the rate of the current is about 4.83 kmph.

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