Two cyclists, k km apart, and starting at the same time, would be together in r hours if they travelled in the same direction, but would pass each other in t hours if they travelled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is11:19 AM
do this if you call yourself master of maths
I do not call myself anything except for Melody. Your 'challenge' sounds a little impolite but your question is interesting.
What is 11:19 AM ???
What am I meant to be finding ??? I have found the ratio of the speeds.
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Let x be the speed of the faster one and let y be the speed of the slower one. Find x/y
Let the faster one be a 0 on the number line and let the slower one be at +k on the number line.
As they approach each other:
x km/hour = xt km in t hours
y km/hour = yt km in t hours
xt+yt=k
(x+y)t=k
AS they move in the same direction. Both towards th poitive end of the number line
x km/hour = xr km in r hours
y km/hour = yr km in r hours NEW position will be k+yr
\(k+yr-xr=0\\ xr-yr=k\\ r(x-y)=k\\ so \\ (x+y)t=(x-y)r\\ xt+yt=xr-yr\\ yt+yr=xr-xt\\ y(t+r)=x(r-t)\\ \frac{r+t}{r-t}=\frac{x}{y}\\ \frac{x}{y}=\frac{r+t}{r-t}\)
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