Bunuel wrote:

A bike traveling at a certain constant speed takes 5 minutes longer to travel 10 miles than it would take to travel 10 miles at 60 miles per hour. At what speed, in miles per hour, is the bike traveling?

(A) 36

(B) 40

(C) 42

(D) 48

(E) 50

R*T=D - just change time from hours to minutes and back

Scenario 1 - find time \(R*T=D\)

\(60mph * T_{hrs}=10mi\)

\(T_1=\frac{10mi}{60mph}=\frac{1}{6}\) of an hour

\(\frac{1}{6}hr*\frac{60min}{1hr}==10\) minutes

Scenario 2 find rate

Same distance.

5 minutes longer: \((10+5)=15\) minutes

Time in hours?

\((15mins *\frac{1 hr}{60mins}=

\frac{1}{4}\) hour

Rate\(_2\)? R*T = D

Rate, \(R_2*\frac{1}{4}hr=10mi\)

Rate, \(R_2\)

\(=\frac{10mi}{(\frac{1}{4}hr)}=(10*\frac{4}{1})mph=40\) mph

Answer B

Inverse proportionTrip 1, time in minutes:

\(\frac{10mi}{60mph}=\frac{1}{6}\) hour

\(\frac{1}{6}hr*60mins=10\) minutes

Trip 2, R is 5 minutes longer = 15 minutes

\(\frac{T_2}{T_1}=\frac{15}{10}=\frac{3}{2}\)

Same distance. Rate and time are inversely proportional. Flip the time ratio:\(\frac{2}{3}\)

Requiring \(\frac{3}{2}\) the time of Trip 1, the bike in Trip 2 will travel at \(\frac{2}{3}\) its former rate in Trip 1.

\(60mph*\frac{2}{3}=40\) mph

Answer B

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