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Every day at noon, a bus leaves for Townville and travels at a speed

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Every day at noon, a bus leaves for Townville and travels at a speed [#permalink]

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New post 26 May 2016, 11:54
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Every day at noon, a bus leaves for Townville and travels at a speed of \(x\) kilometers per hour. Today, the bus left 30 minutes late. If the driver drives \(\frac{7}{6}\) times as fast as usual, she will arrive in Townville at the regular time. If the distance to Townville is 280 kilometers, what is the value of \(x\)?

A) 66
B) 72
C) 80
D) 84
E) 90
[Reveal] Spoiler: OA

Last edited by Sallyzodiac on 26 May 2016, 12:11, edited 1 time in total.

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Every day at noon, a bus leaves for Townville and travels at a speed [#permalink]

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New post 26 May 2016, 12:10
I did this algebraically, but I guess you could do it by plugging in numbers as well. My approach:

\(Time = \frac{Distance}{Speed}\); thus the required time to travel 280 km = \(\frac{280}{s}\).

However, this lady is running 30 minutes or \(\frac{1}{2}\) of an hour late, but is at the same time travelling \(\frac{7s}{6}\) as fast as usual and reaches Townville on time (as if she was travelling at her regular speed \(s\) and left at her regular time). Thus, we can set up the following equation:

\(\frac{280}{s}\) = \(\frac{280}{7s/6}\) + \(\frac{1}{2}\). Solving for \(s\), we get \(s = 80\), answer choice C.

Last edited by Sallyzodiac on 26 May 2016, 12:12, edited 1 time in total.

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Re: Every day at noon, a bus leaves for Townville and travels at a speed [#permalink]

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New post 26 May 2016, 12:11
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Sallyzodiac wrote:
Every day at noon, a bus leaves for Townville and travels at a speed of x kilometers per hour. Today, the bus left 30 minutes late. If the driver drives \(\frac{7}{6}\) times as fast as usual, she will arrive in Townville at the regular time. If the distance to Townville is 280 kilometers, what is the value of x?

A) 66
B) 72
C) 80
D) 84
E) 90


Distance = Time*Rate

\(280 = Usual \ Time*x\) --> \(Usual Time = \frac{280}{x}\);

\(280 = (Usual \ Time - \frac{1}{2})*(\frac{7}{6}*x)\) (30 minutes = 1/2 hours) --> \(Usual \ Time = \frac{40*6}{x} + \frac{1}{2}\)

\(\frac{280}{x} = \frac{40*6}{x} + \frac{1}{2}\);

\(\frac{40}{x} =\frac{1}{2}\);

\(x = 80\).

Answer: C.
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Every day at noon, a bus leaves for Townville and travels at a speed [#permalink]

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New post 26 May 2016, 14:51
x=normal speed of bus
t=normal time of trip
xt=280 km
(7x/6)(t-1/2)=280 km
xt=(7x/6)(t-1/2)
t=3.5 hours
280/3.5=80 kph

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Re: Every day at noon, a bus leaves for Townville and travels at a speed [#permalink]

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New post 26 May 2016, 20:16
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Sallyzodiac wrote:
Every day at noon, a bus leaves for Townville and travels at a speed of \(x\) kilometers per hour. Today, the bus left 30 minutes late. If the driver drives \(\frac{7}{6}\) times as fast as usual, she will arrive in Townville at the regular time. If the distance to Townville is 280 kilometers, what is the value of \(x\)?

A) 66
B) 72
C) 80
D) 84
E) 90


You can also use ratios.

If speed becomes 7/6 the original, time taken will become 6/7 the original (since same distance is traveled). The 1/7 th of the time taken is 30 mins so total time taken usually is 7*30 = 210 mins = 210/60 = 7/2 hrs

Usual Speed = 280/(7/2) = 80 mph

Answer (C)
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Re: Every day at noon, a bus leaves for Townville and travels at a speed [#permalink]

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New post 27 May 2016, 04:01
Total Distance: 280 Km.
Usual speed: x KM per hour.

Usual time: \(\frac{280}{x}\)

In the revised situation the bus covers the same distance by driving fast. So speed is more and Time taken is less

Our revised equation is:

\(\frac{280}{x}\) (-) \(\frac{1}{2}\) = 280/(\(\frac{7x}{6}\))

\(\frac{280}{x}\) (-) \(\frac{1}{2}\) =\(\frac{280*6}{7x}\)

\(\frac{280}{x}\) (-) \(\frac{280*6}{7x}\) =\(\frac{1}{2}\)

\(\frac{40}{x}\) =\(\frac{1}{2}\)

x = 80.

Option C is correct answer.

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Every day at noon, a bus leaves for Townville and travels at a speed [#permalink]

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New post 10 Jun 2016, 12:47
Sallyzodiac wrote:
Every day at noon, a bus leaves for Townville and travels at a speed of \(x\) kilometers per hour. Today, the bus left 30 minutes late. If the driver drives \(\frac{7}{6}\) times as fast as usual, she will arrive in Townville at the regular time. If the distance to Townville is 280 kilometers, what is the value of \(x\)?

A) 66
B) 72
C) 80
D) 84
E) 90


Algebra / plugging values are two approaches to solve this problem. There is one more approach and I took the road less travelled :wink:

The question stem mentions that "the driver drives \(\frac{7}{6}\) times as fast as usual". Before jumping to algebra / plugging values, take a look at the answer choices. All the options, except (C), are multiples of 6. The distance is 280 kms. Driving at 80 km/hr, it will take 3.5 hours to cover the total distance. To cover the distance in 3 hours, the speed should be \(\frac{280}{3}\). Multiplying 80 by \(\frac{7}{6}\), you get \(\frac{280}{3}\). Therefore, the answer is (C). :)

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Re: Every day at noon, a bus leaves for Townville and travels at a speed [#permalink]

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Re: Every day at noon, a bus leaves for Townville and travels at a speed   [#permalink] 13 Aug 2017, 10:51
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