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

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.

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

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.

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

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).

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:

GMAT assassins aren't born, they're made,
Rich

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?

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.

GMAT assassins aren't born, they're made,
Rich

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 ?

Hi utkarshrihand,

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

GMAT assassins aren't born, they're made,
Rich

\(? = x\)

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.

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

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

Hi, thanks for that quick approach. I'd love to understand this approach more. Could you please clarify how the the time becomes 6/7 the original? Appreciate it.

Speed Time Distance
x t 280
7/6x (t-1/2) 280

xt = 7/6x(t-1/2)
t= 7/2

put in the equation
x= 280*2/7 = 80

If I'm thinking about setting up an algebraic equation, I typically ask whether Plugging In The Answers (PITA) will be a better option. Let's try it.
I typically like testing B and D unless there's a reason to deviate from that. In this case, we are looking for something that will divide nicely into 280. Only C fits that description, so let's start there.

C: The bus goes 280km. It usually travels 80kph, so it usually takes 3.5 hours. Wait, we've gotta do something with either 6/7 or 7/6 and end up at 3 hours...I'm liking where this is going enough that I really might consider just clicking C and moving on. But I get it if you're less confident in how the dials turn, so we can finish the math. Anyway, today we drive (7*80)/6 kph. \(\frac{280}{(7*80)/6}=\frac{280*6}{7*80}=\frac{40*6}{80}=\frac{6}{2}=3\). Is that a half hour faster than 3.5 hours? Yep.

Answer choice C.


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