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singmah1 · 05-28-2007, 01:41 AM

1. A, B and C run around a circular track of length 750m at speeds of 3 m/sec, 6 m/sec and 18 m/sec respectively. If all three start from the same point, simultaneously and run in the same direction, when will they meet for the first time after they start the race?

2. Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

mr_sriramesh · 05-28-2007, 04:54 PM

Is the answer 250sec

Nikoloz · 05-29-2007, 10:20 AM

Q2. After 5.25 meters.
J denotes Jane's speed and K - Karen's. First eqate time of the main distance 99.75/J=95/K
Then eqate time needed for Kane to reach Karen. X denotes the distance asked.
X/J=(X-0.25)/K
now devide one oqasion by the other so J and K cancell out.
X/99.75=(x-0.25)/95
I solve X=5.25

LMK2727 · 05-31-2007, 08:25 PM

Nikoloz · 06-01-2007, 10:50 AM

Quote:

Originally Posted by singmah1

1. A, B and C run around a circular track of length 750m at speeds of 3 m/sec, 6 m/sec and 18 m/sec respectively. If all three start from the same point, simultaneously and run in the same direction, when will they meet for the first time after they start the race?

lets first figure out when will A and B meet for the firt time.
3*t=6*t-750
t=250
so when A will make its once around the track. Now lets chack maybe C will be there too in 250 seconds. Vuala C will indeed be there!

LMK2727 · 06-05-2007, 12:23 AM

Why does 3t = 6t-750 and not the other way around?

How did we check to see that C was there?!

Thank you.

Nikoloz · 06-05-2007, 05:33 AM

3t=6t-750 because the distance A will cover is 3t and the distance B will cover will be the same plus one whole circle. so we have to substract the length of the track to end up with exactly the same distances.
C will cover in those 250 seconds 4500 meters, which is 6 lengths of the track. And indeed C is 6 times faster than A, so in time A makes one revolution, C will make 6 (and B will make two). I guess there is a shortcut to solving this problem in that 3, 6, and 18 are so closely tied to 3 (divisible by)

balaganesh · 07-20-2007, 05:46 PM

some useful formula...
if the number of men are 2, a,b speeds with a>b

time taken for them to meet first = L/(a-b) [both runs in same direction]
time taken for them to meet first = L/(a+b) [both runs in opposite direction]

time taken for them to meet first at starting point is
LCM of (L/a , L/b) for both same and opposite direction.

If the number of men are 3 ,a,b,c are speeds..a>b>c

Time taken for them to meet first time is
LCM of [L/(a-b),L/(b-c)]

Time taken for them to meet first time at the starting point is
LCM of [L/a,L/b,L/c]

Lorelai · 07-20-2007, 08:41 PM

If the number of men are 3 ,a,b,c are speeds..a>b>c

Time taken for them to meet first time is
LCM of [L/(a-b),L/(a-c)]

They will be together for the first time after the start at a time equal to the LCM of the time taken by the fastest to gain a complete round of each of the other two.

balaganesh · 07-21-2007, 07:43 AM

Hi Lorelai,

i think both LCM of [L/(a-b),L/(b-c)] and LCM of [L/(a-b),L/(a-c)] are same..

so both are correct formulas for calculation..
if wrong please correct..

05-28-2007, 01:41 AM	#1 (permalink)
singmah1 Eager! Join Date: Oct 2006 Posts: 85	Rate Problems 1. A, B and C run around a circular track of length 750m at speeds of 3 m/sec, 6 m/sec and 18 m/sec respectively. If all three start from the same point, simultaneously and run in the same direction, when will they meet for the first time after they start the race? 2. Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

05-28-2007, 04:54 PM	#2 (permalink)
mr_sriramesh Make everything fun... Join Date: Jan 2007 Location: Albuquerque Posts: 146	Is the answer 250sec _ _ _ _ SIG _ _ _ _ The difficult is what takes a little time; the impossible is what takes a little longer.

05-31-2007, 08:25 PM	#4 (permalink)
LMK2727 Eager! Join Date: Apr 2007 Posts: 51	How did we solve the 1st question?

06-05-2007, 12:23 AM	#6 (permalink)
LMK2727 Eager! Join Date: Apr 2007 Posts: 51	Quick question Why does 3t = 6t-750 and not the other way around? How did we check to see that C was there?! Thank you.

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05-29-2007, 10:20 AM	#3 (permalink)
Nikoloz Within my grasp! Join Date: Apr 2007 Location: Tbilisi, Georgia Posts: 108	Q2. After 5.25 meters. J denotes Jane's speed and K - Karen's. First eqate time of the main distance 99.75/J=95/K Then eqate time needed for Kane to reach Karen. X denotes the distance asked. X/J=(X-0.25)/K now devide one oqasion by the other so J and K cancell out. X/99.75=(x-0.25)/95 I solve X=5.25

06-05-2007, 05:33 AM	#7 (permalink)
Nikoloz Within my grasp! Join Date: Apr 2007 Location: Tbilisi, Georgia Posts: 108	3t=6t-750 because the distance A will cover is 3t and the distance B will cover will be the same plus one whole circle. so we have to substract the length of the track to end up with exactly the same distances. C will cover in those 250 seconds 4500 meters, which is 6 lengths of the track. And indeed C is 6 times faster than A, so in time A makes one revolution, C will make 6 (and B will make two). I guess there is a shortcut to solving this problem in that 3, 6, and 18 are so closely tied to 3 (divisible by)

07-20-2007, 05:46 PM	#8 (permalink)
balaganesh Trying to make mom and pop proud Join Date: Dec 2006 Posts: 13	some useful formula... if the number of men are 2, a,b speeds with a>b time taken for them to meet first = L/(a-b) [both runs in same direction] time taken for them to meet first = L/(a+b) [both runs in opposite direction] time taken for them to meet first at starting point is LCM of (L/a , L/b) for both same and opposite direction. If the number of men are 3 ,a,b,c are speeds..a>b>c Time taken for them to meet first time is LCM of [L/(a-b),L/(b-c)] Time taken for them to meet first time at the starting point is LCM of [L/a,L/b,L/c]

07-20-2007, 08:41 PM	#9 (permalink)
Lorelai Within my grasp! Join Date: Apr 2007 Posts: 154	If the number of men are 3 ,a,b,c are speeds..a>b>c Time taken for them to meet first time is LCM of [L/(a-b),L/(a-c)] They will be together for the first time after the start at a time equal to the LCM of the time taken by the fastest to gain a complete round of each of the other two.

07-21-2007, 07:43 AM	#10 (permalink)
balaganesh Trying to make mom and pop proud Join Date: Dec 2006 Posts: 13	Hi Lorelai, i think both LCM of [L/(a-b),L/(b-c)] and LCM of [L/(a-b),L/(a-c)] are same.. so both are correct formulas for calculation.. if wrong please correct..

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