The speed of a boat is 5 times the speed at which a river flows. What

\(? = {{{T_{{\rm{up}}}}} \over {{T_{{\rm{down}}}}}}\)

\({V_r} = {{1\,\,{\rm{m}}} \over {1\,\,\sec }}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{V_b} = {{5\,\,{\rm{m}}} \over {1\,\,\sec }}\)

\({\rm{distance}} = 12\,{\rm{m}}\)


\(\left. \matrix{\\
{\rm{down}}\,\,{\rm{:}}\,\,\,{\rm{12}}\,{\rm{m}}\,\,\left( {{{1\,\,\sec } \over {5 + 1\,\,{\rm{m}}}}} \right) = \,\,\,2\,\,\sec \,\,\, = \,\,\,{T_{{\rm{down}}}} \hfill \cr \\
{\rm{up}}\,\,{\rm{:}}\,\,\,{\rm{12}}\,{\rm{m}}\,\,\left( {{{1\,\,\sec } \over {5 - 1\,\,{\rm{m}}}}} \right) = \,\,\,3\,\,\sec \,\,\, = \,\,\,{T_{{\rm{up}}}} \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,? = \,\,{3 \over 2}\,\, = \,\,150\%\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Let x = the speed of the river (in miles per hour)
So, 5x = the speed of the boat in (miles per hour)
Let d = distance traveled (in miles)

This means the boat's speed going UPriver = 5x - x = 4x
And the boat's speed going DOWNriver = x + 5x = 6x

time = distance/speed
So, travel time going UPriver = d/4x
So, travel time going DOWNriver = d/6x

What percent of the time it takes for the boat to travel DOWNriver does it take for the boat to travel UPriver?
We must take the fraction (d/4x)/(d/6x), and convert it to a PERCENT

(d/4x)/(d/6x) = (d/4x)(6x/d)
= 6xd/4xd
= 6/4
= 3/2
= 150/100
= 150%

Answer: D

Cheers,
Brent

=>

Let \(d\) and \(v\) be the distance the boat travels and the speed of the river flow, respectively.
The time the boat takes to travel up the river is \(\frac{d}{( 5v – v )} = \frac{d}{4v}\).
The time the boat takes to travel down the river is \(\frac{d}{( 5v + v )} = \frac{d}{6v}\).
Let \(p\) be the percentage are looking for.
Using the IVY approach, we obtain \(\frac{d}{4v} = p(\frac{1}{100})(\frac{d}{6v})\) or \(p = (\frac{6}{4})*100 = 150(%).\)

Therefore, the answer is D.
Answer: D

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