Question

A boat running upstream takes $8$ hours $48$ minutes to cover a certain distance while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and the speed of the water current respectively?

A

2:1

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B

3:2

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C

8:3

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D

3:8

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Solution

The correct option is C 8:3Step 1: Given dataThe distance covered upstream in ${T}_{u}=8hrs48min$ is equal to the distance covered downstream in ${T}_{d}=4hrs$.Step 2: Formula usedIf the speed downstream is $ykm/hr$and the speed upstream is $xkm/hr$, then:The speed of the boat is given by ${s}_{b}=\frac{1}{2}\left(downstreamspeed+upstreamspeed\right)=\frac{x+y}{2}$The speed of the water stream is given by ${s}_{w}=\frac{1}{2}\left(downstreamspeed-upstreamspeed\right)=\frac{y-x}{2}$[ where $x=$ speed of the boat upstream and $y=$ the speed of the boat downstream ]Step 3: Calculating the speed of the boat upstream and the speed of the downstreamSuppose the speed of the boat upstream is x km/hr and the speed for the downstream is y km/hr.Now, by using the given data we get the equation as multiplication of speed for downstream to its time and multiplication of upstream speed with its time as: $x×{T}_{u}=y×{T}_{d}\phantom{\rule{0ex}{0ex}}x×8\frac{48}{60}=y×4$Then, we will solve the above expression to get the value of the ratio x:y as:$8\frac{4}{5}x=4y\phantom{\rule{0ex}{0ex}}\frac{44}{5}x=4y\phantom{\rule{0ex}{0ex}}\frac{x}{y}=\frac{4×5}{44}\phantom{\rule{0ex}{0ex}}\frac{x}{y}=\frac{5}{11}$Step 4: Calculating the ratio between the speed of the boat and the speed of the waterDividing the speed of the boat with the speed of the water current to get the ratio as:$⇒\frac{{s}_{b}}{{s}_{w}}\phantom{\rule{0ex}{0ex}}⇒\frac{\frac{x+y}{2}}{\frac{y-x}{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{x+y}{y-x}$Divide the numerator and denominator by y to get the required ratio as:$\frac{\frac{x}{y}+1}{1-\frac{x}{y}}$Substituting the value of $\frac{x}{y}=\frac{5}{11}$$⇒\frac{\frac{5}{11}+1}{1-\frac{5}{11}}\phantom{\rule{0ex}{0ex}}⇒\frac{\frac{5+11}{11}}{\frac{11-5}{11}}\phantom{\rule{0ex}{0ex}}⇒\frac{16}{6}\phantom{\rule{0ex}{0ex}}⇒\frac{8}{3}$So, we get the ratio between the speed of the boat and the speed of the water current respectively as $8:3$.Hence, option (c) is the correct answer.

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