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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 paral  [#permalink]

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08 Oct 2015, 06:49
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Difficulty:

55% (hard)

Question Stats:

60% (01:43) correct 40% (02:03) wrong based on 264 sessions

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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.

Attachment:

2015-10-08_1848.png [ 4.22 KiB | Viewed 2451 times ]

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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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08 Oct 2015, 08:56
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Bunuel wrote:

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.

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%
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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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08 Oct 2015, 09:33
2
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.

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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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11 Oct 2015, 06:40
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Bunuel wrote:

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.

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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26 Aug 2017, 05:06
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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26 Aug 2017, 05:21
r19 wrote:
how time is equal for Y and Z? I'm not able to understand

Hi..

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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29 Aug 2017, 06:00
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]

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23 Sep 2018, 23:00
This asks for a specific number for the speed of car Y (beyond point P) as a percent of the speed of car Z. Notice that, while this is a "what is the value" question, it is asking for a ratio rather than a value for the speed of Y. This means that you may be able to find the relationship between the two speeds by leveraging your assets even if you cannot find a specific speed for each car. Remember also that complex situations like this one are often abstractions of fairly straightforward ideas. The test will reward you for breaking down that abstraction into something more concrete.
You are given that the roads that cars Y and Z are parallel until point P. At point P the cars are side-by-side. Then car Y makes a turn of x degrees, speeds up, and continues to keep up with car Z. So even though they are going at different speeds, they are travelling to the right (in the figure) at the same relative speed.

Statement (1) gives you the speed of car Z, 50 miles per hour. This statement does not give any information as to the speed of car Y, or the value of the angle x. It should be clear that this statement is not sufficient on its own, and you can prove it by thinking conceptually.

If the two cars had been travelling side by side on the same road before car Y veered off, you would be able to recognize that they seem to form two sides of a triangle with the angle x between them at point P. If angle x is very small, car Y will not have to go much faster than car Z in order to keep up, since it is travelling in almost the same direction.

However, as x increases, the speed of car Y has to increase as well. Thus, without knowing anything about the angle X, you can't make any deductions about the ratio of their speeds.

Eliminate choices A and D.

Statement (2) tells you that the measure of angle x is 45 degrees. While this statement may not appear sufficient, remember that you don't need to know the speeds of the two cars, only the ratio between them.
Because the cars are keeping up with each other, the amount of time that each car has had to travel is equal. That means that the increased percentage of distance that car y has traveled will be the same as the increased percentage or the rate of car Y compared to car Z. In other words: determine the ratio of the distance traveled by car Y to the distance traveled by car Z, and you have enough information to answer this question.

Again imagine that cars Y and Z were instead traveling side-by-side on the same road. As with statement (1), when car Y veers off at point P, it forms a triangle whose three sides are the distance between point P and car Y, the distance between point P and car Z, and the distance between cars Y and Z. Because you are told that cars Y and Z remain even, you should recognize that this means that this is a right triangle.
Going back to the information given in statement (2), you are told that x is 45 degrees. That means you have a 45-45-90 triangle with the distance that car Y has traveled as the hypotenuse and the distance car Z has traveled as one of the sides.

The ratio of the sides of a 45-45-90 triangle is x:x:x\sqrt{2}
where x is a length of one of the legs. The ratio of the distances traveled for Y to Z is the same as the ratio of the length of the hypotenuse to a side, x\sqrt{2}:x
which simplifies to \sqrt{2}:1
Since you have established that the ratio of the distances must be the same as the ratio of the speeds, you know that statement (2) is sufficient. The correct answer is B.

Veritas Prep Solution
Re: Cars Y and Z travel side-by-side at the same rate of speed along paral &nbs [#permalink] 23 Sep 2018, 23:00
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