Question

When Kunal drives his car 25% faster than his usual speed, he reaches his office 30 minutes earlier than the usual time. By what percent (approximate) of his usual speed he should reduce his speed, so that he reaches his office 20 minutes later than the usual time?

Answer 1

Answer:

11.76 % reduction in speed

Step-by-step explanation:

Let say Usual Speed   =  S km/Hr

Usual Time Taken  = T hr

Distance of Office = ST

Speed with 25% faster  = S + (25/100)S  = 1.25S  km/hr

Time taken = T - 30/60  =  T - 0.5  Hrs

Distance = 1.25S(T - 0.5)

ST = 1.25ST  - 1.25S*0.5

=> 0.25ST =  1.25S*0.5

=> T =2.5 hr = 5/2 hrs

Let say reduced Speed = R km/hr

Time Taken = 5/2 + 20/60  = 17/6 hr

Distance = R(17/6)  km

ST = R(T + 1/3)

=> S (5/2)  = R(17/6)

=> R = 15S/17

Reduction in speed = S - R   =  S - 15S/17 =  2S/17

% reduction in speed = ((2S/17)/S) * 100

= 11.76 %

11.76 % reduction in speed