# Speed

• May 14th 2005, 02:29 PM
Speed
Two cars on Crete left the towns of Knossos and Phaistos at the same time to drive to the other town, passing each other at Gortyns and both travelling at different constant speeds. The car from Knossos travels 20km/h faster than the car from Phaistos. The distance from Knossos to Gortyns is 10km more then the distance from Phaistos to Gortyns. The car from Phaistos completed the journey from Gortyns to Knossos in 35mins. Find the distance from Knossos to Phaistos.
• May 16th 2005, 12:18 AM
paultwang
I do not see any question from you.
• Jun 5th 2005, 02:24 PM
Quote:

Two cars on Crete left the towns of Knossos and Phaistos at the same time to drive to the other town, passing each other at Gortyns and both travelling at different constant speeds. The car from Knossos travels 20km/h faster than the car from Phaistos. The distance from Knossos to Gortyns is 10km more then the distance from Phaistos to Gortyns. The car from Phaistos completed the journey from Gortyns to Knossos in 35mins.

:confused:
What is the distance from Knossos to Phaistos?
(any help is greatly appreciated)
• Jun 6th 2005, 05:22 AM
ticbol
As posted, your question here is not clear. Are the 3 mentioned towns in a straight line, and that Grotyns is in between Knossos and Phaistos?
The given information seem to show that. I assume that is so.

Draw on paper or imagine the figure.
It is a straight line segment PGK. A car from point P is to go to point K; a car from point K is to go to point P. The two cars have different constant speeds. They start on the same time, and they pass each other at point G.

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Let:
PG = a = distance from Phaistos to Gortyns, in km.
KG = b = distance from Knossos to Gortyns, in km.
PK = KP = distance from Knossos to Phaistos, in km.

"The distance from Knossos to Gortyns is 10km more then the distance from Phaistos to Gortyns."
So,
b = a+10 ----(i)

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Let s = speed of car from Phaistos, in km/hr.
And r = speed of car from Knossos, in km/hr.

"The car from Knossos travels 20km/h faster than the car from Phaistos."
So,
r = s+20 ----(ii)

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Distance = rate*time = speed*time

"The car from Phaistos completed the journey from Gortyns to Knossos in 35mins."
Our speeds are in km/hr, so we convert 35 minutes into hr.
Hence,
b = s*(35/60) = s(7/12) ---(iii)

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Distance = rate*time
So, time = distance / rate

"...left...at the same time..., passing each other at Gortyns..."
That means they spent the same time when they met at Gortyns.
So,
a/s = b/r ----(iv)

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We have four unknowns (a,b,r,s) and four independent equations.
Solvable.

We eliminate b and r from (iv).
Substitute the b from (i), and the r from (ii), into (iv),
a/s = (a+10)/(s+20)
Cross multiply,
a(s+20) = s(a+10)
as +20a = as +10s
20a = 10s
2a = s ----(v)

b from (i) = b from (iii),
a+10 = s(7/12)
s = (a+10)/(7/12)
s = 12(a+10)/7 ----(vi)

s from (v) = s from (vi),
2a = 12(a+10)/7
7*2a = 12(a+10)
14a = 12a +120
14a -12a = 120
2a = 120
a = 120/2 = 60 km ---***

Substitute that into (i),
b = a+10 = 60 +10 = 70 km ---***

Therefore, distance from Knossos to Phaistos = a+b = 60+70 = 130 km. ----answer.

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On second thought, Knossos, Gortyns and Phaistos don't have to be in a straight line.
• Jun 10th 2005, 02:37 PM