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

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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.

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]

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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.

Please provide detailed explanations...

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.

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]

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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,\(t = \frac{300}{p+q}\)

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]

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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.

Please provide detailed explanations...

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..

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]

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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.

Please provide detailed explanations...

@Bunuel: Please move this question to Distance/Speed DS forum as it is not Work/Rate DS question. Thanks.

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]

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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.

Please provide detailed explanations...

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

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]

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14 Jan 2015, 00:00

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Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]

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19 Feb 2016, 11:56

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

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27 Dec 2016, 18:41

IMO B. It sounds logically clear without deducing actual equations.

From the question we are given total distance. Now Since both start simultaneously, time taken by both of them to reach X is same.When time is same, distance covered by them will be directly proportional to the their speed.

From 2: we get the speed ratio and thus their distance ratio. and we already know their sum i.e. 300. -- Sufficient.

to make it clear = s1/s2 = d1/d2 s2 = 4/3 s1 d1/d2 = 3/4 d2/d1 = 4/3 d2+d1/d1 = 7/3 300/d1= 7/3 d1 = 900/7
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Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]

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27 Dec 2016, 21:43

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Hi Fozzzy,

The following two points will serve you well when dealing with questions on speed, distance and time where two objects meet each other.

1. Always read the question and figure out the constant out of distance and time (there will always be one). If the distance is constant then we use the formula S1T1 = S2T2 where S1 and T1 are the speed and time of the first object and S2 and T2 are the speed and time of the second object.

If the time is constant then we use the formula D1/S1 = D2/S2 where D1 and S1 are the distance and speed of the first object and D2 and S2 are the distance and speed of the second object.

2. Once you figure out which of the two above formulae you need to use, always graphically represent the question.

So before getting into the statement let us first analyze the question and graphically represent the question.

Attachment:

Speed Distance and time.PNG [ 14.73 KiB | Viewed 265 times ]

The question asks us for a definite value of x. Let the speed of car P be SP and the speed of car Q be SQ

Since the time is constant DP/SP = DQ/SQ ------> x/SP = (300 - x)/SQ

Statement 1 : Up to location X, the average speed of car Q was 15 miles per hour faster than that of car P

SQ = SP + 15 ------> x/SP = (300 - x)/(SP + 15).

Now here since we have two unknowns 'x' and 'SP' we cannot find a definite value of x. Insufficient.

Statement 2 : Up to location X, the average speed of car Q was 4/3 times that of car P.

SQ = (4/3)SP ------> x/SP = (300 - x)/(4/3)SP.

Cancelling out SP on both sides we can solve for a definite value of x. Sufficient.

OA : B

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

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28 Dec 2016, 02:10

CrackVerbalGMAT wrote:

If the time is constant then we use the formula D1/S1 = D2/S2 where D1 and S1 are the distance and speed of the first object and D2 and S2 are the distance and speed of the second object.

Dear CrackVerbalGMAT,

Is the above statement true when solving questions with chasing cars? or only in cars in in different directions? If it is applicable, can you kindly use the above example to show it graphically??

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]

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28 Dec 2016, 02:44

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Mo2men wrote:

CrackVerbalGMAT wrote:

If the time is constant then we use the formula D1/S1 = D2/S2 where D1 and S1 are the distance and speed of the first object and D2 and S2 are the distance and speed of the second object.

Dear CrackVerbalGMAT,

Is the above statement true when solving questions with chasing cars? or only in cars in in different directions? If it is applicable, can you kindly use the above example to show it graphically??

thanks

Hi Mo2men,

This will always hold good whenever two objects meet each other. If they are moving in the same direction (two chasing cars) then again time will be a constant. Things become easier if the two objects start simultaneously. If the objects do not start simultaneously then the only extra working we need to do is to find the distance the first object traveled when the second object was at rest. This again now becomes a chasing scenario where the time will be a constant.

There are always two approaches of solving questions where two objects meet : Approach 1 : Always find the constant out of distance and time. There will always be one. Approach 2 : Use the concept of relative speed.

Hope this helps!

CrackVerbal Academics Team
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