Rate Problems

[Rangya]

1) All three will meet at same point at time which will be the LCM of time taken by A,B and C to cover the entire track...

i.e. LCM of ( 250,125,125/3) which is 250 sec.

[CrackXam]

Hi,

Do u suppose it should be 250*3 = 750 seconds?
I agree with the approach though.

[Lorelai]

Balaganesh, i think both the formulas give the same answer in this case. But this may not be the case always.

[way2fast]

I dare to disagree with the solution offered. While it may work in this case, it will not work in others.

Who says that they must cover the entire track before they meet? Suppose A and B run together at speeds of 5 m/s, and that C runs at speed of 50 m/s. They will meet again long before A and B have managed to cover the entire track.

[RamN]

Find below my take on the two questions,

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?

A - Takes 250 sec to finish one round (750/3 = 250 sec)
B - Takes 125 sec to finish one round - This means he completes 2 rounds in 250 sec
C - Takes 41.67 sec to finish one round - so he does 6 rounds in 250 seconds

A completes his first round and by then B & C would have completed 2 & 6 rounds respectively. Hence they meet after 250 sec


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?

Jane covers 99.75 mtrs in the same time as Karen covers 95 mtrs.

This means finding the number of mtrs covered by Karen in a sec and comparing it with the same by Jane should give us the answer

99.75/95 = 1.05

The above implies for every 1 mtr run by Karen, Jane covers 1.05 mtr.

To cover up for the remaining .25 mtrs, an additional 5 mtrs would be required (0.25/0.05 = 5)