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Cars Y and Z travel side-by-side at the same rate of speed along paral

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Cars Y and Z travel side-by-side at the same rate of speed along parallel roads as shown above. When car Y reaches point P, it forks to the left at angle x°, changes speed, and continues to stay even with car Z as shown by the dotted line. The speed of car Y beyond point P is what percent of the speed of car Z?

(1) The speed of car Z is 50 miles per hour.
(2) x = 45

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2015-10-08_1848.png
2015-10-08_1848.png [ 4.22 KiB | Viewed 1477 times ]
[Reveal] Spoiler: OA

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Cars Y and Z travel side-by-side at the same rate of speed along paral [#permalink]

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New post 08 Oct 2015, 09:56
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Bunuel wrote:
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Cars Y and Z travel side-by-side at the same rate of speed along parallel roads as shown above. When car Y reaches point P, it forks to the left at angle x°, changes speed, and continues to stay even with car Z as shown by the dotted line. The speed of car Y beyond point P is what percent of the speed of car Z?

(1) The speed of car Z is 50 miles per hour.
(2) x = 45

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
2015-10-08_1848.png


(1) this info is not sufficient, we need info about X
(2) In this case we have an Isosceles Triangle with ratio of sides d:d:d\(\sqrt{2}\) add. we know that both need the same Time-t and we have a proportion for distance, let's say the distance d=5 -> Ratio \(5:5:5*\sqrt{2}\) ~7
we can setup an equtation Rate Y= 7/t, Rate Z =5/t Rate Y/ Rate Z = 7/t*t/5*100 ~140%
Answer (B)
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Re: Cars Y and Z travel side-by-side at the same rate of speed along paral [#permalink]

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New post 08 Oct 2015, 10:33
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1) The speed of car Z is 50 miles per hour....Insufficient
(2) x = 45deg, mean we can use 45:45:90=x:x:x√2 to calculate the distance. so assuming the the distance travelled by X be 'a', the distance travalled by Y becomes a√2. Let the speed of X be z, so the speed of X will be a/z. Since, X & Y both take the same time, the speed of Y will be a√2/z. So the ratio becomes √2 i.e. 140% approx.

So the answer is 'B'.

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Re: Cars Y and Z travel side-by-side at the same rate of speed along paral [#permalink]

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Bunuel wrote:
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Cars Y and Z travel side-by-side at the same rate of speed along parallel roads as shown above. When car Y reaches point P, it forks to the left at angle x°, changes speed, and continues to stay even with car Z as shown by the dotted line. The speed of car Y beyond point P is what percent of the speed of car Z?

(1) The speed of car Z is 50 miles per hour.
(2) x = 45

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
2015-10-08_1848.png


VERITAS PREP OFFICIAL SOLUTION:

Type of question: What Is the Value? The question asks for “the speed of car y (beyond point p) as a percent of the speed of car z.” Note: This is the same as asking for Distance/Time of car y as a percent of the Distance/Time of car z.

Given information in the question stem or diagram: The roads that car y and z are traveling on are parallel until point P. At point P the cars are even and then car y makes a turn of x degrees and continues to keep up with car z. The key information that you need to leverage from this given information is that the times will be the same for each car.

Statement 1: “The speed of car z = 50 mph.” This statement does not give any information about the speed of car y, nor about the extra distance that car y must travel. This statement is not sufficient on its own. Eliminate choices A and D, and note that this is a classic “Why Are You Here?” set-up. This statement is clearly insufficient on its own so you will want to consider whether it is important in relation to Statement 2.

Statement 2: “The measure of angle x = 45 degrees.” This statement may not appear sufficient until you realize that you do not actually need to know the speeds of cars y and z. Because the cars are keeping up with each other, the time that each car travels must be equal. That means that the increased percentage of distance that car y has traveled will be the same as the increased percentage of the rate of car y compared to car z. In other words, determining the ratio of the distance of y to the distance of z is enough to answer this question because the times are necessarily the same. While you do not need to do it, the actual mathematical solution is below. Also, note that this is a classic example of the “Why Are You Here?—Temptation” set-up that is used so often on harder problems. They have tempted you with a statement (the speed of car z) that is not necessary to determine sufficiency. Statement 2 is sufficient by itself.

The road that car y travels on after point P forms a triangle with the road that y was traveling on (that road is parallel to the road z is traveling on). This triangle is a right triangle with the right angle formed by the dotted line. Statement 2 gives the measure of x at 45 degrees. That means you have a 45–45–90 triangle with the distance that y has traveled as the hypotenuse and the distance z has traveled as one of the shorter sides. The hypotenuse is √2 bigger than the shorter sides, so car y is √2 faster than car z. As noted above, this is sufficient and the correct answer is B.
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Re: Cars Y and Z travel side-by-side at the same rate of speed along paral [#permalink]

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New post 26 Aug 2017, 06:06
how time is equal for Y and Z? I'm not able to understand

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Re: Cars Y and Z travel side-by-side at the same rate of speed along paral [#permalink]

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New post 26 Aug 2017, 06:21
r19 wrote:
how time is equal for Y and Z? I'm not able to understand



Hi..

read the following context

Quote:
continues to stay even with car Z as shown by the dotted line


this means both are moving in the same line when drawn perpendicular to the initial direction.
since the cars are moving continuously and none halts in between, so whenever you look at the distance at a certain point of time, the TIME would be same as both are moving till that point of time.
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Re: Cars Y and Z travel side-by-side at the same rate of speed along paral [#permalink]

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New post 29 Aug 2017, 07:00
Posting my way of approach by applying basic trigonometry. Assume the point of meeting to be Q. So distance traveled by Y is PQcosX and distance traveled by Z is PQ. Also assume speeds of Y and Z to be y and z respectively. We know that the time for both should be equal. So, PQcosX/y=PQ/z----cosX/y=1/z----y/z=cosX. Thus we can see that all we need is the value of x.

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Re: Cars Y and Z travel side-by-side at the same rate of speed along paral   [#permalink] 29 Aug 2017, 07:00
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