Two ships

The distance from A to B is 300km. At 7 am started from A to B a ferry with speed higher by 20 km/h than a ship that leaves at 8 o'clock from B to A. Both met at 10h 24min. Determine how far they will meet from A and when they reach the destination.

Result

s1 =  204 km
t1 = 12:00 hh:mm Wrong answer
t2 = 15:30 hh:mm Wrong answer

Solution:

s=300 km s1+s2=s s1=(v+20)t s2=v(t1)  t=(10+24/60)7.00=175=3.4 h  (v+20)t+v(t1)=300  3.4 (v+20)+v 2.4=300 5.8v=232 295v=232 29v=1160 v=1160/29=40   s1=(v+20) t=(40+20) 3.4=204 km
t1=7.00+s/(v+20)=7.00+300/(40+20)=12=12:00 hh:mm
t2=8.00+s/v=8.00+300/40=312=15.5=15:30 hh:mm



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