Question

A teacher driving his vehicle at 24 kmph reaches her school 5 minutes late.If She had driven the vehicle 25% faster on an average she would have reached 4 minutes earlier than the scheduled time. How far is her school ?

Answer 1

Let d be the distance she drives to school, in km.
Let t be the difference between the time school starts and the time the teacher leaves for school, in hours.

5 minutes = 1/12 hours
4 minutes = 1/15 hours

d=s*t

d=24 * (t+1/12).....eq1
d=(1.25)(24) * (t-1/15).....eq2

Solve the system of equations to find t and d.

Answer 2

d = 24*(T+5/60) = 1.25*24*(T-4/60).
24T + 2 = 30T - 2,
T = 4/6 = 2/3 h.

d = 24*(2/3+5/60) = ?.

Answer 3

D=(S1*S)(/(S2-S1)*T1+T2(in hour)
S1=24kmph
S2=25%of 24= 24*25/100=6 ie, 24+6=30kmph
T1=5min
T2=4min
T1+T2=9min to hour= 9/60
D=(24*30)/6 * 9/60 =18km