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

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Three friends A, B and C decide to run around a circular track. They  [#permalink]

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New post 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  [#permalink]

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New post 20 Oct 2019, 19:34
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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 = \(y:y-x:y-2x\) ...(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:y-x...(ii)

But the ratio from (i) has to be same so y:y-2x can be written as 3*y:3*(y-2x) or 3y:3y-6x. Comparing with (ii) gives\(y-x=3y-6x....2y=5x....y=\frac{5x}{2}\)
So \(y:y-x:y-2x\) 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:2y-x...(iii)
But the ratio from (i) has to be same so y:y-2x can be written as 3*y:3*(y-2x) or 3y:3y-6x. Comparing with (iii) gives\(2y-x=3y-6x....y=5x\)
So \(y:y-x:y-2x\) can be written as \(5x:5x-x:5x-2x=5x:4x:3x=5:4:3\)

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  [#permalink]

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New post 21 Oct 2019, 01:46
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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  [#permalink]

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New post 21 Oct 2019, 10:30
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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  [#permalink]

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New post 22 Oct 2019, 00:16
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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:(30-x):(30-2x).....(i)
V1/V3=30/(30-2x).....(ii)

SCENARIO 1: When A completes his 3rd lap (i.e. 90 meters), C is still on his 1st lap (i.e. he completes (30-x) meters):
V1/V3=90/(30-x).....(iii)
From (ii) and (iii), 30/(30-2x)=90/(30-x).....> 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+(30-x)] meters)
V1/V3=90/(60-x).....(iv)
From (ii) and (iv), we have 30/(30-2x)=90/(60-x).....> 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  [#permalink]

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New post 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 = \(y:y-x:y-2x\) ...(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:y-x...(ii)

But the ratio from (i) has to be same so y:y-2x can be written as 3*y:3*(y-2x) or 3y:3y-6x. Comparing with (ii) gives\(y-x=3y-6x....2y=5x....y=\frac{5x}{2}\)
So \(y:y-x:y-2x\) 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:2y-x...(iii)
But the ratio from (i) has to be same so y:y-2x can be written as 3*y:3*(y-2x) or 3y:3y-6x. Comparing with (iii) gives\(2y-x=3y-6x....y=5x\)
So \(y:y-x:y-2x\) can be written as \(5x:5x-x:5x-2x=5x:4x:3x=5:4:3\)

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?
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Re: Three friends A, B and C decide to run around a circular track. They   [#permalink] 31 Oct 2019, 12:43
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