One day Eesha started 30 min late from home and reached her office 50 min late while driving 25 slower

home---------------office(time taken =t,speed=u)
one day
time taken=t+20
speed=3u/4
t*u=(t+20)*(3u/4)
solving t=60min
10 years agoHelpfull: Yes(19) No(0)
25% decrease in the speed results in 33.33% increase in time due to inverse relationship of them.
so here 33.33%= 50-30 i.e 20 mins
then 100% = 20*3 =60mins
10 years agoHelpfull: Yes(5) No(1)
Let the distance be D and the usual speed S. Then the time taken to travel is D/S=T(1).The new speed is (S- 1/4 S) = ¾ S. The new time taken is D/3/4 S = T + 20 (50 – 30)(20) we get 60 as answer.
10 years agoHelpfull: Yes(4) No(6)
c.60 min
100/75=t+20/t
t=60
10 years agoHelpfull: Yes(3) No(3)
Speed is inversely proportional to time
While she drives 25% ,i.e 1/4S ..slower means she drove at 3/4(S)

When speed became 3/4(S) then Time taken should be 4/3(T)
i.e, She has taken 4/3(T) - T extra to cover the distance.
Extra Time = T/3 = 20 min (20 min late due to slow driving)
Actual time T = 60 Minutes
10 years agoHelpfull: Yes(3) No(0)
d=st s1*t1=s2*t2 100t=75(t+20) t=60min
10 years agoHelpfull: Yes(1) No(3)
speed is 3/4 then time is 4/3 which is 1+1/3
so 1/3 of actual speed =20
speed = 60
10 years agoHelpfull: Yes(1) No(0)
ans-b because
10 years agoHelpfull: Yes(0) No(3)
25%=20
thn 100=80
10 years agoHelpfull: Yes(0) No(1)

One day ,Eesha started 30 min late from home and reached her office 50 min late while driving 25% slower than her usual speed.how much time in min Eesha usually take to reach her office from home.

a. 80min
b. 50min
c. 60min
d. 70 min