某人骑车上学,速度为15km/h,则早到15min；速度9km/h 则迟到15min,现打算提前10min到达,

设路程为s，
15min = 1/4 hour
t1 = s/v1 = s/15
t2 = s/v2 = s/9
t2-t1 = (1/4)+(1/4) = 1/2
则 (s/9)-(s/15) = 1/2
s=11.25 km
则t1=0.75 h = 45 min
t2=1.25 h = 75 min
那么提前10min所用的时间t3= t1+5 = 45+5 = 50 min = 5/6 h
速度v3 = s/t3 = 11.25/(5/6) = 13.5 km/h

13.84km/h

设路程为x,正常时间为y,则:
x/15=y-15
x/9=y+15

=>x=675,y=60

=>车速=675/(60-10) =13.5

对的