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In the figure above, car A and car B simultaneously begin travelling.. [#permalink]
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03 Feb 2015, 17:40
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In the figure above, car A and car B simultaneously begin travelling around a circular park with area of \(4\pi\) square miles. Both cars start form the same point, the START location shown in the figure, and travel with constant speed rates until they meet. Car A travels counterclockwise and car B travels clockwise. Where along the track will they meet? ( Note: figure not drawn to scale). (1) Car B travels 30 mph faster than car A. (2) Car A travels twice as slowly as car B. Could someone pleas explain the answer to this question. It will be much appreciated, thank you
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Last edited by Bunuel on 04 Feb 2015, 01:02, edited 1 time in total.
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Re: In the figure above, car A and car B simultaneously begin travelling.. [#permalink]
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03 Feb 2015, 18:11
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The key to this problem is understanding that the only information you need is the ratio of speeds. You don't need to solve for the actual answer, so the area and exact speeds are irrelevant.
Statement 1: This is insufficient. Let's consider two scenarios and image a different circular park with a circumference of 100m. If car A travels 10 mph, then car B travels 40 mph. In 2 hours, car B will have traveled 80% of the circumference when it intercepts car A. If car A travels 30 mph, then car B travels 60 mph. In a little over an hour, car B will have traveled about 60% of the distance when it intercepts car A.
Statement 2: This is sufficient. Regardless of the actual speeds, car A and B will meet at one place. In the time it takes car B to travel 2/3 of the distance, car A has traveled 1/3 of the distance. Together, they have covered the entire circumference of the park.



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In the figure above, car A and car B simultaneously begin traveling ar [#permalink]
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16 Jul 2015, 00:26



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Re: In the figure above, car A and car B simultaneously begin traveling ar [#permalink]
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16 Jul 2015, 02:40
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Bunuel wrote: In the figure above, car A and car B simultaneously begin traveling around a circular park with area of 4π square miles. Both cars start from the same point, the START location shown in the figure and travel with constant speed rates until they meet. Car A travels counterclockwise and car B travels clockwise. Where along the track will they meet? (1) Car B travels 30 mph faster than car A. (2) Car A travels twice slower than car B. Kudos for a correct solution.Attachment: Picture_circle.png Given: area of the circle = \(4\pi\) > radius of the circle = 2> circumference of the circle = \(4\pi\) Let Va and Vb be the speeds of A and B respectively. For 'where' do they meet, we have the following equation : \(Va*T+Vb*T =4\pi\)  (1), where T is the time after which they both meet. Per the question, Va.T = ? or Vb.T = ? Per statement 1, Vb = Va+30. This is not sufficient as this equation with 1 leaves us with 2 equations and 3 variables (Va , Vb, T) We can not isolate Vb.T or Va.T from this. Thus this statement is not sufficient. Per statement 2, Va = Vb /2 (twice slower = half the speed). Susbstituting this in equation (1) we get, \(Vb*T = 8\pi / 3\) . Thus this statement is sufficient. Hence B is the correct answer.



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Re: In the figure above, car A and car B simultaneously begin traveling ar [#permalink]
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16 Jul 2015, 09:57
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Bunuel wrote: In the figure above, car A and car B simultaneously begin traveling around a circular park with area of 4π square miles. Both cars start from the same point, the START location shown in the figure and travel with constant speed rates until they meet. Car A travels counterclockwise and car B travels clockwise. Where along the track will they meet? (1) Car B travels 30 mph faster than car A. (2) Car A travels twice slower than car B. Kudos for a correct solution.Attachment: Picture_circle.png From the property of Speed, Distance and Time For Constant Time Speed of A / Speed of B = Distance travelled by A / Distance travelled by B
Area of Track = 4π square miles = π*r^2 square miles i.e. radius = 2 We have the length of Track = 2*π*r miles = 4π So we only require the ratio of speeds to find the ratio of Distance travelled by them and find their meeting pointQuestion : Distance of Meeting point along the track = ?Statement 1: Car B travels 30 mph faster than car A.This statement provides the difference of speeds but not the ratio of speeds. Therefore, NOT SUFFICIENTStatement 2: Car A travels twice slower than car B.i.e. Speed of B = 2* Speed of A Ratio of speeds is 2/1 i.e. Ratio of distance covered till they meet = 2:1 i.e. Meeting Point = (2/3)4π = (8/3)π (in Clockwise direction) SUFFICIENTAnswer: Option B
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Re: In the figure above, car A and car B simultaneously begin traveling ar [#permalink]
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19 Jul 2015, 12:32
Bunuel wrote: In the figure above, car A and car B simultaneously begin traveling around a circular park with area of 4π square miles. Both cars start from the same point, the START location shown in the figure and travel with constant speed rates until they meet. Car A travels counterclockwise and car B travels clockwise. Where along the track will they meet? (1) Car B travels 30 mph faster than car A. (2) Car A travels twice slower than car B. Kudos for a correct solution.Attachment: Picture_circle.png 800score Official Solution:Knowing the area of the park, we can find its radius (πr² = 4π square miles, so r = 2 miles). Knowing the radius, we can find the length of the road (the circumference, it equals 2πr = 4π miles). To visualize this question, think about a circle as just a line segment joined at its two ends. Let’s first cut the circle at the starting point and make it into a straight line. Notice that cars A and B are at opposite ends of the line traveling toward each other. A [ ______________________ ] B Let’s denote the speed rate of car A by x mph and the speed rate of car B by y mph. When they meet, the whole distance will be the sum of the paths they have travelled. Since they’ve started simultaneously, then they both travelled for the same time. Let’s denote it by t hours. This gives us the following relation: x × t + y × t = 4π Statement (1) gives us the fact that y = x + 30. The relation transforms into xt + (x + 30)t = 4π. Let’s plug two values to see if we get different results. If x = 5, then t = π/10. So the cars would travel 5π/10 = π/2 and 35π/10 = 7π/2 miles respectively. If x = 10, then t = 4π/50. So the cars would travel 40π/50 = 4π/5 and 160π/50 = 16π/5 miles respectively. As wee see these results define different points along the track. Therefore statement (1) by itself is NOT sufficient. Statement (2) gives us the fact that y = 2x. The relation transforms into xt + 2xt = 4π. So xt = 4π/3. Therefore we know that car A travells 4π/3 miles and car B travels 8π/3 miles. That gives us the definite point along the track, no matter what the values of x and t are. So statement (2) by itself is sufficient. Statement (2) by itself is sufficient to answer the question, while statement (1) by itself is not. The correct answer is B.
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Re: In the figure above, car A and car B simultaneously begin travelling.. [#permalink]
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19 Jul 2015, 12:34



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Re: In the figure above, car A and car B simultaneously begin travelling.. [#permalink]
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15 Oct 2015, 06:25
Hello,
Even with Statement 2 in consideration, the cars will meet at three different points on the circle. Since the question does not ask us about the first meeting point, how can this be answered?
Thanks



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Re: In the figure above, car A and car B simultaneously begin travelling.. [#permalink]
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Re: In the figure above, car A and car B simultaneously begin travelling.. [#permalink]
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In the figure above, car A and car B simultaneously begin travelling around a circular park with area of 4π4π square miles. Both cars start form the same point, the START location shown in the figure, and travel with constant speed rates until they meet. Car A travels counterclockwise and car B travels clockwise. Where along the track will they meet? (Note: figure not drawn to scale). (1) Car B travels 30 mph faster than car A. (2) Car A travels twice as slowly as car B. It is DS ques. (1) says speed of A = x mph B = x + 30 mph x = 5 , x = 50 will yield diff results respectively. NS (2) it gives comparison of speed. we know dis = speed * time we know total dis. ratio of speed, time is same for both. Sufficient.
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