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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 Updated on: 26 May 2016, 12:11
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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

Originally posted by Sallyzodiac on 26 May 2016, 11:54.
Last edited by Sallyzodiac on 26 May 2016, 12:11, 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, 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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Every day at noon, a bus leaves for Townville and travels at a speed  [#permalink]

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New post Updated on: 26 May 2016, 12:12
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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.

Originally posted by Sallyzodiac on 26 May 2016, 12:10.
Last edited by Sallyzodiac on 26 May 2016, 12:12, 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, 12:11
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

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

\(280 = (Usual \ Time - \frac{1}{2})*(\frac{7}{6}*x)\) (because 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 27 May 2016, 04:01
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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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New post 16 Mar 2018, 11:08
Hi All,

This question CAN be solved with either algebra or TESTing THE ANSWERS. Either way, this question involves the Distance Formula:

Distance = Rate x Time

We're told that the distance = 280 km, so

280 = R x T

The question also mentions an exact difference of 30 minutes, which is a "round number" (relative to time). It makes me think that the question is probably designed around other round numbers. While I would normally start with answers B or D when TESTing the Answers, here I'm going to start with 80 (since it's a round number)…

So if X the original speed and X = 80, we'd have.

280 = 80 x T
280/80 = T
T = 3.5 hours

Now let's see what happens when we subtract .5 hours (since the bus left 30 minutes late) and increase the speed by 7/6…

80(7/6) x (3) = ???

560/6 x 3

1680/6 = 280

This MATCHES the original distance, so it MUST be the correct answer.

Final Answer:

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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 19 Mar 2018, 11:18
Bunuel wrote:
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.


My question may sound silly to you since its been a long time I studied math, My question is that, In the second equation, why are you subtracting 30 minutes from usual time? As the question says that she reaches 30 minutes late shouldn't it be " usual time+30 minutes " since she is late or am i missing something?
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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 19 Mar 2018, 11:27
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Hi utkarshrihand,

The prompt states that "the bus LEFT 30 minutes LATE...." and that the bus driver drives FASTER so that "she will arrive in Townville at the REGULAR TIME."

Thus, the bus ride took 1/2 an hour LESS, and we have to subtract that half-hour from the total time.

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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 19 Mar 2018, 11:50
EMPOWERgmatRichC wrote:
Hi utkarshrihand,

The prompt states that "the bus LEFT 30 minutes LATE...." and that the bus driver drives FASTER so that "she will arrive in Townville at the REGULAR TIME."

Thus, the bus ride took 1/2 an hour LESS, and we have to subtract that half-hour from the total time.

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Thank you for clarifying. I have one more question. In general if a question states that a person arrive say y minutes late, and usual time is a x minutes, then we would add y with x i.e total time equals x+y minutes, right ?
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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 20 Mar 2018, 10:50
Hi utkarshrihand,

Yes - if you 'arrive LATE', then you would ADD time to the total time traveled.

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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 09 Nov 2018, 11:32
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

\(? = x\)

Let´s use UNITS CONTROL, one of the most powerful tools of our method!

\(\frac{{30\,\,\,{\text{minutes}}\,\,{\text{saved}}}}{{280\,\,{\text{km}}}} = \,\,\frac{{\,\,\frac{3}{{28}}\,\,\,{\text{minutes}}\,\,{\text{saved}}\,}}{{1\,\,\,{\text{km}}}}\,\,\,\,\,\left( * \right)\)

\(\left. \begin{gathered}
x\,\,\frac{{{\text{km}}}}{{\text{h}}}\,\,\,\,::\,\,\,1\,{\text{km}}\,\,\left( {\frac{{1\,\,{\text{hour}}}}{{\,x\,\,{\text{km}}\,}}} \right)\left( {\frac{{60\,\,{\text{minutes}}\,}}{{\,1\,\,{\text{hour}}\,}}} \right)\,\,\,\, = \,\,\,\,\frac{{60}}{x}\,\,\,{\text{minutes}} \hfill \\
\frac{{7x}}{6}\,\,\frac{{{\text{km}}}}{{\text{h}}}\,\,\,:\,:\,\,\,1\,{\text{km}}\,\,\left( {\frac{{6\,\,{\text{hour}}}}{{\,7x\,\,{\text{km}}\,}}} \right)\left( {\frac{{60\,\,{\text{minutes}}\,}}{{\,1\,\,{\text{hour}}\,}}} \right)\,\,\,\, = \,\,\,\,\frac{{6 \cdot 60}}{{7x}}\,\,\,{\text{minutes}}\,\,\, \hfill \\
\end{gathered} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{60}}{x} - \frac{{6 \cdot 60}}{{7x}}\,\,\,\mathop = \limits^{\left( * \right)} \,\,\frac{3}{{28}}\)

\(\frac{{60 \cdot \boxed7}}{{x \cdot \boxed7}} - \frac{{6 \cdot 60}}{{7x}}\, = \frac{3}{{28}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\frac{{60}}{{7x}} = \frac{3}{{28}}\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,? = x = \frac{{28 \cdot 60}}{{7 \cdot 3}} = 80\,\,\,\,\,\,\left[ {\,\frac{{{\text{km}}}}{{\text{h}}}\,} \right]\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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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 12 Nov 2018, 08:07
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


Let’s let the normal speed = x. The time for today is:

280/(7x/6) = (280 * 6)/7x = (40 * 6)/x = 240/x

The regular time is 280/x.

Since today the bus left 30 minutes, or ½ hour, late, we can create the equation:

240/x + 1/2 = 280/x

Multiplying by 2x we have:

480 + x = 560

x = 80

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 02 Oct 2019, 19:11
time taken by bus usually ; 280/x
today time ; 280/7x/6 ; 240/7x
given
240/7x+1/2 = 280/x
solve for x = 80
IMO C


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
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Re: Every day at noon, a bus leaves for Townville and travels at a speed   [#permalink] 02 Oct 2019, 19:11
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