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20 Oct 2019, 03:18
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Difficulty:
95%
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Question Stats:
29% (02:55) correct
71%
(02:27)
wrong
based on 203
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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
A. 5 : 4 : 2
B. 4 : 3 : 2
C. 5 : 4 : 3
D. 3 : 2 : 1
E. 3 : 2 : 2
22 Oct 2019, 00:16
Kudos
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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
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
General Discussion
20 Oct 2019, 19:34
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DisciplinedPrep
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 - "
C
