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Need help for solving this by plugging in the answers

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by [email protected] » Sun Mar 30, 2014 9:10 pm
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 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?

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by [email protected] » Sun Mar 30, 2014 9:55 pm
Hi shibsriz,

This question CAN be solved with either algebra or TESTing THE ANSWERS. Since you asked about how to solve by using the answers, here's how you would do it....

This question involves the Distance Formula

Distance = Rate x Time

We're told that the distance = 280km, 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: C

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by hlobprise » Tue Apr 15, 2014 12:15 am
Rich, how about the algebraic way? The answer starts with solving it as 280/x=(280/(7/6)x) +(1/2), but my brain has trouble getting into that position....what would be another way to set the problem up?
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by Tushar14 » Tue Apr 15, 2014 12:56 am
First equation becomes 280 = XT ( X is speed in km/hr & T is time)
Second becomes 7X/6 = 280/(T-.5)

Now solving these X & T from two equations becomes easy and gets X=80.
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by GMATGuruNY » Tue Apr 15, 2014 2:46 am
[email protected] 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 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?

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Alternate approach:

Since time and rate are RECIPROCALS, 7/6 the usual rate implies 6/7 the usual time.
Thus, the 1/2 hour saved by traveling at the faster rate represents 1/7 of the usual time:
1/2 = (1/7)t
t = 7/2 hours.
Since the usual time is 7/2 hours, the usual rate = d/t = 280/(7/2) = 80.

The correct answer is C.
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by [email protected] » Tue Apr 15, 2014 3:02 pm
Hi hlobprise,

Tushar14 has properly translated the information into 2 equations, which could then be used to answer the question. I'm going to add a few more details to his explanation.

This prompt is based on the Distance Formula:

Distance = Rate x Time

The original prompt tells us that the distance is 280 kilometers and refers to the rate as X kph. Using this information, the original equation can be written as....

280km = (X kph) x (T)
280 = XT

The next sentence changes the information: the bus leaves 30 minutes (1/2 an hour) late and drives 7/6 the speed. This can be written as....

280km = (7X/6 kph) x (T - 1/2)
280 = (7X/6)(T - 1/2)

With these two equations, you can now solve algebraically.

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