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 ?
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.
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
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) = ?.
To solve this problem, we need to use the concept of average speed and the formula:
Speed = Distance / Time
Let the distance between the teacher's home and school be 'd' kilometers.
Let's calculate the time taken by the teacher if she had driven at her usual speed of 24 kmph.
Speed = 24 kmph
Time taken = Distance / Speed
= d / 24
Since she reached 5 minutes late, the time taken would be the scheduled time plus the delay:
Time taken = d / 24 + 5/60
Now, let's calculate the time taken if she had driven 25% faster.
25% faster than 24 kmph = 24 + (25% of 24) = 24 + 6 = 30 kmph
Time taken = Distance / Speed
= d / 30
Since she would reach 4 minutes earlier than the scheduled time, the time taken would be the scheduled time minus the early arrival:
Time taken = d / 30 - 4/60
Now we can equate the two time expressions and solve for 'd'.
d / 24 + 5/60 = d / 30 - 4/60
Multiply both sides by 120 to eliminate the fractions:
5d + 2 = 4d - 8
Rearrange the equation:
d = 2 + 8
d = 10
Therefore, the distance between the teacher's home and school is 10 kilometers.