Question 825020: In a 500 m race, Ravi beats Ramesh by 5 seconds or 100 m. They decide to run another race and this time Ravi gives Ramesh a head start of 200 m. If Ravi’s speed is twice his previous speed and Ramesh’s speed is one and half times his previous speed, how far from the starting point should the winning post be so that they finish at the same time?
Answer by y6n3o9p4(20)   (Show Source):
You can put this solution on YOUR website! Since Ravi ran 100m in 5 seconds in the first race, that means he ran 500m in 25 seconds. Ramesh was 100m behind, so he ran 400m in 25 seconds.
In the second race, Ravi's speed is doubled so he runs at a rate of 1000m in 25 seconds and Ramesh runs 50% faster: a rate of 600m in 25 seconds.
If Ramesh gets a 200m head start then he runs 800m in 25 seconds.
But this means Ravi still beats him by 200m if they run for 25 seconds so we need to set the winning post closer than 1000m.
Let's try decreasing the time by 5 seconds: Ravi runs 800m in 20 seconds and Ramesh runs 480m + 200m (680m) in 20 seconds.
Ravi still wins by 120m, so the winning post still needs to be moved closer than 800m. However now we can see that decreasing the time by 5 seconds decreases Ravi's lead by 80m. Divide Ravi's remaining lead (120m) by 80, then multiply it by 5 to find out how many more seconds we need to decrease the time by: 7.5 seconds.
Now, if they run for 7.5 seconds less, Ravi runs 500m in 12.5 seconds and Ramesh runs 300m + 200m (500m) in 12.5 seconds. They both reach the winning post at the same time.
The winning post should be set at 500m.
|
|
|
