Weekend Speed-Time Puzzle 2?

if T = time for A for the distance town-airport

elapsed time to meet B = T-10

T-10 = 17 + T'-20

T = T'+7

with T' = time for B for the distance airport-town

10x + 20y = D

with D = distance town-airport

x = D/T | y = D/T'

20D/T' + 10D/T = D

20/T' + 10/T = 1

20/T' + 10/(T'+7) = 1

==>T' = 28 and T = 35

T-10 = 25 minutes

at time : 9:25 am ..........answer

++++++++++++++++++++++++++++++++++++++

if car A leaves town at 9:00 am and it takes it 10 minutes to get to the airport then it will be at the airport at 9:10.

since car B doesn't leave the airport until 9:17, they won't really be passing each other.

They will be sitting next to each other from 9:10 , when car A arrives, until 9:17 when car B leaves.

oops, thank you M3 for pointing that out. My students do it all the time - they misread a problem and solve the problem they read instead of the one I made up.

OK, let's do it again. The distance traveled in time t by A equals the distance traveled in 20 minutes by B.

also, the distance traveled by A in 10 minutes equals the distance traveled by B in t-17 minutes.

so xt = 20y and 10x = (t-17)y

so x = 20y/t and x = (t-17)y/10

so 20/t = (t-17)/10

so 200 = t2 - 17t

and t = 25

so they meet at 9:25

never mind, I see J already got the answer.

Julius