A bike traveling at a certain constant speed takes 5 minutes longer to

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

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

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