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# 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

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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

Dec 4, 2015

#2
+96171
+5

The distance to be closed in each case is the same.........=  k   km

And it will obviously take more time for the faster rider to catch the slower when they go in the same direction than when they close this distance traveling in the opposite direction

So......let S be the rate of the faster cyclist and s  be the rate of the slower

When they are going in the same direction, the effective rate rate of closure = S - s........when they are closing the distance from the opposite direction, the effective rate of closure  = S + s

And we have

k  km       =     k km

And Rate * Time   = Distance......... so.......

[S  - s] r    =   [ S + s] t

Sr - sr =  St + st

Sr - St  =  st + sr

S[r - t]  = s[r + t]

S/s   = [r + t ] / [ r - t ]

Dec 4, 2015

#1
+97561
+10

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.

---------------------------------

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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Dec 4, 2015
edited by Melody  Dec 4, 2015
#2
+96171
+5

The distance to be closed in each case is the same.........=  k   km

And it will obviously take more time for the faster rider to catch the slower when they go in the same direction than when they close this distance traveling in the opposite direction

So......let S be the rate of the faster cyclist and s  be the rate of the slower

When they are going in the same direction, the effective rate rate of closure = S - s........when they are closing the distance from the opposite direction, the effective rate of closure  = S + s

And we have

k  km       =     k km

And Rate * Time   = Distance......... so.......

[S  - s] r    =   [ S + s] t

Sr - sr =  St + st

Sr - St  =  st + sr

S[r - t]  = s[r + t]

S/s   = [r + t ] / [ r - t ]

CPhill Dec 4, 2015