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 10:24. Find how far they will meet from A and when they will reach the destination.


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

Step-by-step explanation:

s=300 km s1 + s2 = s s1 = (v+20)t s2 = v(t1)  t=(10+24/60)7.00=517=3.4 h  (v+20)t + v(t1) = 300  3.4 (v+20)+v 2.4=300 5.8v=232 529v=232 29v=1160 v=1160/29=40  s1=(v+20) t=(40+20) 3.4=204 km

3.4 · (v+20) + v·2.4 = 300

5.8v = 232

v = 232/5.8 = 40

v = 40

Our simple equation calculator calculates it.
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=231=15.5=15:30 hh:mm

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