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nick1816
Asume length of track= LCM(12,15,20)= 60.

speed of A= 5m/min
speed of B= 4m/min
speed of C= 3m/min

Time taken by A and C to meet=\( \frac{60}{(5-3)}\)= 30min ( Both are running in same direction)

Time taken by A and B to meet= \(\frac{60}{(5+4)}= \frac{60}{9}\) min ( Both are running in opposite direction)

Time taken by B and C to meet=\( \frac{60}{(4+3)}= \frac{60}{7}\) min ( Both are running in opposite direction)

time after which all three A, B and C meet for the first time= LCM(30, 60/9,60/7)= \(\frac{LCM(30,60,60)}{HCF(1,9,7)}\)= 60mins


Bunuel
If A, B and C take 12 mins, 15 mins and 20 mins to complete one full round of a circular track, after how much time will all three A, B and C meet for the first time, if they start from the same point at the same time, with A and C running in clockwise direction and B running in anti-clockwise direction?

A. 1/2 hours
B. 1 hour
C. 2 hours
D. 5 hours
E. 30 hours


time after which all three A, B and C meet for the first time= LCM(30, 60/9,60/7)= \(\frac{LCM(30,60,60)}{HCF(1,9,7)}\)= 60mins

Can you explain the logic behind LCM(30, 60/9,60/7)= [m]\frac{LCM(30,60,60)}{HCF(1,9,7)} ? I was able to get the time but couldn't find to logic behind figuring out LCM (30, 60/9, 60/7).
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Bunuel
If A, B and C take 12 mins, 15 mins and 20 mins to complete one full round of a circular track, after how much time will all three A, B and C meet for the first time, if they start from the same point at the same time, with A and C running in clockwise direction and B running in anti-clockwise direction?

A. 1/2 hours
B. 1 hour
C. 2 hours
D. 5 hours
E. 30 hours

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Must be LCM ( 12 , 15 & 20 ) = 60, Hene Answer must be (B)
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metalhead2593


time after which all three A, B and C meet for the first time= LCM(30, 60/9,60/7)= \(\frac{LCM(30,60,60)}{HCF(1,9,7)}\)= 60mins

Can you explain the logic behind LCM(30, 60/9,60/7)= [m]\frac{LCM(30,60,60)}{HCF(1,9,7)} ? I was able to get the time but couldn't find to logic behind figuring out LCM (30, 60/9, 60/7).



Actually the answer is the LCM (30, 60/9, 60/7),

LCM (a/b, c/d) = LCM(a,c)/HCF (b,d)

refer URL for more details.

https://in.edugain.com/articles/6/LCM-of-Fractions/
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Hi, could you please explain the logic of these workings. I only understood that you found the LCM to simplify calculations. But why did you take the total lenght of track divided by the difference between rates of A,B and C

nick1816
Asume length of track= LCM(12,15,20)= 60.

speed of A= 5m/min
speed of B= 4m/min
speed of C= 3m/min

Time taken by A and C to meet=\( \frac{60}{(5-3)}\)= 30min ( Both are running in same direction)

Time taken by A and B to meet= \(\frac{60}{(5+4)}= \frac{60}{9}\) min ( Both are running in opposite direction)

Time taken by B and C to meet=\( \frac{60}{(4+3)}= \frac{60}{7}\) min ( Both are running in opposite direction)

time after which all three A, B and C meet for the first time= LCM(30, 60/9,60/7)= \(\frac{LCM(30,60,60)}{HCF(1,9,7)}\)= 60mins



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Sanath777
Hi, could you please explain the logic of these workings. I only understood that you found the LCM to simplify calculations. But why did you take the total lenght of track divided by the difference between rates of A,B and C


Because to meet, one runner has to make up one full lap relative to the other. So the required time is:

time = track length / relative speed

When they run in the same direction, relative speed is the difference of their speeds. When they run in opposite directions, relative speed is the sum.
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