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# Cars P and Q started simultaneously from opposite ends of a

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Cars P and Q started simultaneously from opposite ends of a [#permalink]  07 Sep 2013, 23:41
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Cars P and Q started simultaneously from opposite ends of a straight 300-mile express way and travelled towards each other, without stopping, until they passed at location X. To the nearest mile, how many miles of the express way had car P travelled when the two cars passed each other?

1) Up to location X, the average speed of car Q was 15 miles per hour faster than that of car P.
2) Up to location X, the average speed of car Q was $$1 \frac{1}{3}$$times that of car P.

[Reveal] Spoiler: OA

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Last edited by fozzzy on 07 Sep 2013, 23:50, edited 2 times in total.
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Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]  07 Sep 2013, 23:49
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fozzzy wrote:
Cars P and Q started simultaneously from opposite ends of a straight 300-mile express way and travelled towards each other, without stopping, until they passed at location X. To the nearest mile, how many miles of the express way had car P travelled when the two cars passed each other?

1) Up to location X, the average speed of car Q was 15 miles per hour faster than that of car P.
2) Up to location X, the average speed of car Q was 1 1/3 times that of car P.

The time taken for the 2 cars to meet at point X = $$\frac{300}{p+q}$$ , where p and q are the respective speeds.
Thus, the distance travelled by car P = $$p*\frac{300}{p+q}$$

F.S 1 states that q = p+15. Thus, substituting this above, we get $$p*\frac{300}{2p+15}$$. Clearly depends on the value of p. Insufficient.

F.S 2 states that $$q =p* \frac{4}{3}$$, and this yields = $$p*\frac{300}{p+q}$$ = $$\frac{900}{7}$$. Sufficient.

B.
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Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]  07 Sep 2013, 23:53
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The time taken for the 2 cars to meet at point X = $$\frac{300}{p+q}$$ , where p and q are the respective speeds.

Can you explain how you formed that equation so quick?
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Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]  08 Sep 2013, 00:01
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fozzzy wrote:
The time taken for the 2 cars to meet at point X = $$\frac{300}{p+q}$$ , where p and q are the respective speeds.

Can you explain how you formed that equation so quick?

This is nothing but applied concept of relative velocity.

For two objects moving in the opposite direction, always add their speeds. For objects moving in the same direction, subtract their speeds.
For example, just as this example, as they are moving in opposite directions, treat their relative velocity to be a sum total of their individual speeds, and this is their relative speed. Thus, the time taken, as we know is $$\frac{Distance}{Speed}$$, thus, $$t = \frac{300}{p+q}$$.

To understand this using algebra, Imagine this : Let the time taken for both to meet be t.

Now, in time t, car P covers say distance x. Thus,$$x = p*t$$. Again, car Q must have covered a distance of (300-x). Thus, $$300-x = q*t$$
Replace the value of x from the first equation, we get $$300-p*t = q*t$$
Thus,$$t = \frac{300}{p+q}$$

Hope this helps.
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Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]  12 Sep 2013, 07:51
fozzzy wrote:
Cars P and Q started simultaneously from opposite ends of a straight 300-mile express way and travelled towards each other, without stopping, until they passed at location X. To the nearest mile, how many miles of the express way had car P travelled when the two cars passed each other?

1) Up to location X, the average speed of car Q was 15 miles per hour faster than that of car P.
2) Up to location X, the average speed of car Q was $$1 \frac{1}{3}$$times that of car P.

Time taken for P to meet Q at a common point is same for both P, and Q.

Time for P, = Distance / speed = 'x'(assume) / P(assume)
Time for Q, = 300-x / (P+20) from A

Equate both as they are equal. so x / P = 300-x / (p+20)
cannot solve..

from B

Time for Q = 300-x/1.33P

Equate now

x / P = 300-x/ 1.33P,

Solve for x, Hence B is suff.
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Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]  14 Oct 2013, 15:56
fozzzy wrote:
Cars P and Q started simultaneously from opposite ends of a straight 300-mile express way and travelled towards each other, without stopping, until they passed at location X. To the nearest mile, how many miles of the express way had car P travelled when the two cars passed each other?

1) Up to location X, the average speed of car Q was 15 miles per hour faster than that of car P.
2) Up to location X, the average speed of car Q was $$1 \frac{1}{3}$$times that of car P.

@Bunuel: Please move this question to Distance/Speed DS forum as it is not Work/Rate DS question. Thanks.
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Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]  31 Dec 2013, 07:13
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fozzzy wrote:
Cars P and Q started simultaneously from opposite ends of a straight 300-mile express way and travelled towards each other, without stopping, until they passed at location X. To the nearest mile, how many miles of the express way had car P travelled when the two cars passed each other?

1) Up to location X, the average speed of car Q was 15 miles per hour faster than that of car P.
2) Up to location X, the average speed of car Q was $$1 \frac{1}{3}$$times that of car P.

Same-o, I've seen couple of questions such as this one

When one has a constant distance, all we need to know what the problem is asking is the relative rate, that is the RATIO between both rates. Not the difference

So B is clearly the way to go here

Hope it helps
Cheers!

J
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Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]  14 Jan 2015, 00:00
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Re: Cars P and Q started simultaneously from opposite ends of a   [#permalink] 14 Jan 2015, 00:00
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