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

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A bike traveling at a certain constant speed takes 5 minutes longer to  [#permalink]

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New post 26 Sep 2018, 04:03
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

84% (01:49) correct 16% (01:29) wrong based on 70 sessions

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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

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Re: A bike traveling at a certain constant speed takes 5 minutes longer to  [#permalink]

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New post 28 Sep 2018, 13:34
1
D = 10
V1 = 60
T1 = 10/60 = 1/6 * 60 = 10 mins

Now T2 = 10 + 5 = 15 mins = 15/60 = 1/4 hour.

Find V2 = D/T2 = 10/(1/4) = 10 * 4 = 40

Answer choice B.

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A bike traveling at a certain constant speed takes 5 minutes longer to  [#permalink]

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New post 29 Sep 2018, 05:50
Distance = 10 miles
Speed = 60 miles/hour
Time = Distance/speed = 10/60=1/6=0.16 = 9.60 minutes(0.16x60)

Distance = 10 miles
Time = 9.6+5=14.6
Speed =Distance/time = 10/14.6 = 100/146 = 0.68 = 40.8 minutes(0.68x60)

Hence 40 minutes. Answer is B
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A bike traveling at a certain constant speed takes 5 minutes longer to  [#permalink]

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New post 29 Sep 2018, 11:57
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 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 traveling at a certain constant speed takes 5 minutes longer to  [#permalink]

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New post 13 Oct 2018, 05:41
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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A bike traveling at a certain constant speed takes 5 minutes longer to &nbs [#permalink] 13 Oct 2018, 05:41
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