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Three friends A, B and C decide to run around a circular track. They
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20 Oct 2019, 03:18
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Three friends A, B and C decide to run around a circular track. They start at the same time and run in the same direction. A is the quickest and when A finishes a lap, it is seen that C is as much behind B as B is behind A. When A completes 3 laps, C is the exact same position on the circular track as B was when A finished 1 lap. Find the ratio of the speeds of A, B, and C? A. 5 : 4 : 2 B. 4 : 3 : 2 C. 5 : 4 : 3 D. 3 : 2 : 1 E. 3 : 2 : 2
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Three friends A, B and C decide to run around a circular track. They
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20 Oct 2019, 19:34
DisciplinedPrep wrote: Three friends A, B and C decide to run around a circular track. They start at the same time and run in the same direction. A is the quickest and when A finishes a lap, it is seen that C is as much behind B as B is behind A. When A completes 3 laps, C is the exact same position on the circular track as B was when A finished 1 lap. Find the ratio of the speeds of A, B, and C?
A. 5 : 4 : 2 B. 4 : 3 : 2 C. 5 : 4 : 3 D. 3 : 2 : 1 E. 3 : 2 : 2 Let the circular track be y units and the distance of B from A when A finishes the lap be x behind A.. They start at the same time and run in the same direction.  So the time is the same for distance covered and distance covered should be in the same ratio as the speeds.. A is the quickest and when A finishes a lap, it is seen that C is as much behind B as B is behind A.  SO distance covered by A:B:C = \(y:yx:y2x\) ...(i) When A completes 3 laps, C is the exact same position on the circular track as B was when A finished 1 lap. Here we do not know that C is still on first lap or 2nd lap, but can be sure that he cannot be on 3rd lap as he cannot be just x units behind A in 3rd lap as C was already 2x behind in first lap itself. Two scenarios (I) C is in first lap, then distance covered by A:C is in 3y:yx...(ii)But the ratio from (i) has to be same so y:y2x can be written as 3*y:3*(y2x) or 3y:3y6x. Comparing with (ii) gives\(yx=3y6x....2y=5x....y=\frac{5x}{2}\) So \(y:yx:y2x\) can be written as \(\frac{5x}{2}:\frac{5x}{2}x:\frac{5x}{2}2x = 5x:3x:x=5:3:1\) (II) C is in second lap, then distance covered by A:C is in 3y:2yx...(iii)But the ratio from (i) has to be same so y:y2x can be written as 3*y:3*(y2x) or 3y:3y6x. Comparing with (iii) gives\(2yx=3y6x....y=5x\) So \(y:yx:y2x\) can be written as \(5x:5xx:5x2x=5x:4x:3x=5:4:3\) Thus two ratios 5:4:3 and 5:3:1, so the question should be  " Find What could be the ratio of the speeds of A, B, and C?"C
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Re: Three friends A, B and C decide to run around a circular track. They
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21 Oct 2019, 01:46
Good analysis, chetan2u. However, there is a typo under 'Two scenarios': should be "(II) C is in SECOND lap.....".



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Re: Three friends A, B and C decide to run around a circular track. They
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21 Oct 2019, 10:30
DisciplinedPrep wrote: Three friends A, B and C decide to run around a circular track. They start at the same time and run in the same direction. A is the quickest and when A finishes a lap, it is seen that C is as much behind B as B is behind A. When A completes 3 laps, C is the exact same position on the circular track as B was when A finished 1 lap. Find the ratio of the speeds of A, B, and C?
A. 5 : 4 : 2 B. 4 : 3 : 2 C. 5 : 4 : 3 D. 3 : 2 : 1 E. 3 : 2 : 2 We can start from the sentence "it is seen that C is as much behind B as B is behind A". This means the difference between A and B is the same as that of B and C. Therefore A  B = B  C. So we can eliminate A and E first since they don't have the same differences. The important thing to take away here is we can represent the speeds with only 2 variables. If A = x, then B = x  y and C = x  2y. Furthermore, we can pretend each lap has a distance of x and the time A needs to complete a lap is 1. All we care about is proportions of the speeds. Then for each lap A runs, B runs a distance of x  y and C runs a distance of x  2y. Next is "when A completes 3 laps ... when A finished 1 lap". The focus is on what happened between these laps. Either C is three times as slow as B, or C ran {1 lap + B's position after one lap of A which is x  2y} during the 3 laps of A. The former gives \((x  2y) * 3 = x  y\), \(x = 2.5y\). A = 2.5y, B = 1.5y, C = 0.5y. So A:B:C = 5:3:1, which isn't a choice. The latter sets up an equation, again using x as the distance for one lap: One lap + (B's position after one lap from A) = (C's position after 3 laps from A) \(x + (x  y) = 3*(x  2y)\) \(x = 5y.\) So A = 5y, B = 4y, C = 3y. A:B:C = 5:4:3. The answer is C
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Re: Three friends A, B and C decide to run around a circular track. They
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22 Oct 2019, 00:16
A simpler and possibly faster way is by assuming a numerical value for the length of the track:
ASSUMPTIONS: (1) Track length: 30 meters (2) V1, V2 and V3 are the speeds of A, B and C respectively (3) B and C are 'x' and '2x' meters behind A when A completes the 1st lap
CALCULATION: From Assumption (3) we have V1:V2:V3=30:(30x):(302x).....(i) V1/V3=30/(302x).....(ii)
SCENARIO 1: When A completes his 3rd lap (i.e. 90 meters), C is still on his 1st lap (i.e. he completes (30x) meters): V1/V3=90/(30x).....(iii) From (ii) and (iii), 30/(302x)=90/(30x).....> x=12 Substituting this value of 'x' in (i), we have V1:V2:V3=30:18:6=5:3:1
SCENARIO 2: When A completes his 3rd lap, C is on his 2nd lap (i.e. he completes [30+(30x)] meters) V1/V3=90/(60x).....(iv) From (ii) and (iv), we have 30/(302x)=90/(60x).....> x=6 Substituting this value of 'x' in (i), we have V1:V2:V3=30:24:18=5:4:3



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Re: Three friends A, B and C decide to run around a circular track. They
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31 Oct 2019, 12:43
chetan2u wrote: DisciplinedPrep wrote: Three friends A, B and C decide to run around a circular track. They start at the same time and run in the same direction. A is the quickest and when A finishes a lap, it is seen that C is as much behind B as B is behind A. When A completes 3 laps, C is the exact same position on the circular track as B was when A finished 1 lap. Find the ratio of the speeds of A, B, and C?
A. 5 : 4 : 2 B. 4 : 3 : 2 C. 5 : 4 : 3 D. 3 : 2 : 1 E. 3 : 2 : 2 Let the circular track be y units and the distance of B from A when A finishes the lap be x behind A.. They start at the same time and run in the same direction.  So the time is the same for distance covered and distance covered should be in the same ratio as the speeds.. A is the quickest and when A finishes a lap, it is seen that C is as much behind B as B is behind A.  SO distance covered by A:B:C = \(y:yx:y2x\) ...(i) When A completes 3 laps, C is the exact same position on the circular track as B was when A finished 1 lap. Here we do not know that C is still on first lap or 2nd lap, but can be sure that he cannot be on 3rd lap as he cannot be just x units behind A in 3rd lap as C was already 2x behind in first lap itself. Two scenarios (I) C is in first lap, then distance covered by A:C is in 3y:yx...(ii)But the ratio from (i) has to be same so y:y2x can be written as 3*y:3*(y2x) or 3y:3y6x. Comparing with (ii) gives\(yx=3y6x....2y=5x....y=\frac{5x}{2}\) So \(y:yx:y2x\) can be written as \(\frac{5x}{2}:\frac{5x}{2}x:\frac{5x}{2}2x = 5x:3x:x=5:3:1\) (II) C is in second lap, then distance covered by A:C is in 3y:2yx...(iii)But the ratio from (i) has to be same so y:y2x can be written as 3*y:3*(y2x) or 3y:3y6x. Comparing with (iii) gives\(2yx=3y6x....y=5x\) So \(y:yx:y2x\) can be written as \(5x:5xx:5x2x=5x:4x:3x=5:4:3\) Thus two ratios 5:4:3 and 5:3:1, so the question should be  " Find What could be the ratio of the speeds of A, B, and C?"C Are you assuming that they all start from the same point of the circle? Or that doesn't matter?




Re: Three friends A, B and C decide to run around a circular track. They
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