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06 Jan 2020, 07:57
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48% (02:38) correct
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A, B and C can complete one full round of a circular track in 10 minutes, 30 minutes and 50 minutes respectively. If A runs clockwise and B and C run anticlockwise, find the time after which they will meet for the first time.
A. 20 mins
B. 30 mins
C. 60 mins
D. 75 mins
E. 150 mins
A. 20 mins
B. 30 mins
C. 60 mins
D. 75 mins
E. 150 mins
25 Sep 2020, 07:02
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When 3 people are running around a circular track, then the time taken to meet is LCM of the time taken by fastest person to meet each of the other 2 persons.
The time taken, is found using the concept of relative speed.
Let the length of the track be the LCM(10,30,50) = 150m
Speed of A = \(\frac{150}{10} = 15 \space m/min\)
Speed of B = \(\frac{150}{30} = 5 \space m/min\)
Speed of C = \(\frac{150}{50} = 3 \space m/min\)
A is the fastest, and he is moving in the opposite direction of the other 2. Therefore the Relative speed will be the sum of speeds.
Time taken for A to meet B for the 1st time = \(t_{AB} = \frac{150}{15 + 5} = \frac{150}{20} = \frac{15}{2}\)
Time taken for A to meet C for the 1st time = \(t_{AC} = \frac{150}{15 + 3} = \frac{150}{18}= \frac{25}{3}\)
Time taken for all 3 to meet for the first time is the LCM of \(t_{AB} \space and \space t_{AC}\) = LCM of \(\frac{15}{2} \space and \space \frac{25}{3}\)
LCM of Fractions = \(\frac{LCM \space of \space Numerators}{HCF \space of \space Denominators}= \frac{LCM(15,25)}{HCF(2,3)} = \frac{75}{1} = 75 \space minutes\)
Option D
Arun Kumar
The time taken, is found using the concept of relative speed.
Let the length of the track be the LCM(10,30,50) = 150m
Speed of A = \(\frac{150}{10} = 15 \space m/min\)
Speed of B = \(\frac{150}{30} = 5 \space m/min\)
Speed of C = \(\frac{150}{50} = 3 \space m/min\)
A is the fastest, and he is moving in the opposite direction of the other 2. Therefore the Relative speed will be the sum of speeds.
Time taken for A to meet B for the 1st time = \(t_{AB} = \frac{150}{15 + 5} = \frac{150}{20} = \frac{15}{2}\)
Time taken for A to meet C for the 1st time = \(t_{AC} = \frac{150}{15 + 3} = \frac{150}{18}= \frac{25}{3}\)
Time taken for all 3 to meet for the first time is the LCM of \(t_{AB} \space and \space t_{AC}\) = LCM of \(\frac{15}{2} \space and \space \frac{25}{3}\)
LCM of Fractions = \(\frac{LCM \space of \space Numerators}{HCF \space of \space Denominators}= \frac{LCM(15,25)}{HCF(2,3)} = \frac{75}{1} = 75 \space minutes\)
Option D
Arun Kumar
10 Jan 2020, 02:21
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Bunuel
Substitute the options and check
Number of rounds made by A, B & C after 't' mins = t/10, t/30 & t/50 respectively
No. of rounds -------- A ----- B ----- C
Time (mins)
-- 20 ----------------- 2 ---- 2/3 ---- 2/5 --> No
-- 30 ----------------- 3 ----- 1 ----- 3/5 --> No
-- 60 ----------------- 6 ----- 2 ----- 6/5 --> No
-- 75 ----------------- 7.5 --- 2.5 --- 1.5 --> Yes [Everyone met at mid way of track for the first time]
-- 150 ---------------- 15 ---- 5 ---- 3 --> No [Met for the second time]
IMO Option D
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