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Bunuel
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 proportion
Trip 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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A bike travel 60 miles in an hour (60 mins)
So, 10 miles in 10 mins

It takes driver 5 mins more = 10+5= 15 mins
Bike Speed= d/t= 10 miles*60/15 = 40 mph.

Ans-B
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Distance = 10 miles.
Speed = 60mph.
1 hour = 60 miles
60 mins = 3600 seconds = 60 miles

if 3600 seconds = 60 miles then 10 miles =?

10 * 3600/60 = 600 seconds which is equal to 10 mins


since the question mentions that it takes 5 mins (300 seconds)longer
600+300=900 seconds


900 secs =10 miles
3600 secs = 1 mile

cross multiply

3600*10/900 = 40

So, ans is B




Please let me know if my method of understanding is correct
if not please correct me

Thank you in advance

And Good luck for your exam
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