Rashmi walks to her bus stop at 5 kmph and reaches there late by 10 minutes On the next day

let distance x km and time to reach bus stand t hr
distance = speed*time
first day x= 5*(t+10/60)
second day x= 6*(t-10/60)
on equating t=11/6 hr
then x= 5*(11/6+10/60) =10 km
11 years agoHelpfull: Yes(6) No(1)
Let the distance to the bus stop is "d"km
Time taken by her(at 5kmph)=d/5hour, and
Time taken by her(at 6kmph)=d/6 hour
Now the equation is
d/6-d/5=1/3( total saving in time is 2o minutes equivalent to 1/3hour)
Solving this, we get d=10km(option d)
11 years agoHelpfull: Yes(2) No(2)
12d=t+10
10d=t-10
solving d=10
11 years agoHelpfull: Yes(2) No(1)
distance = time * speed

s=d/t

=d/6 - d/5
=1/3 hour

solving we get
d=10
11 years agoHelpfull: Yes(1) No(2)
We can use the direct formula for these type of questions :
Total time diff = late-(-early) = 10-(-10) = 10 + 10 =20
direct formula =
Distance =( Product_of_Speeds * (time_diff) )/ (Diff_of_speeds )
=((5*6)*20)/(6-5)= 10 [ANS]
7 years agoHelpfull: Yes(0) No(2)
The answer is 10 km
3 years agoHelpfull: Yes(0) No(0)

Rashmi walks to her bus stop at 5 kmph and reaches there late by 10 minutes. On the next day,she increases her speed to 6 kmph and reaches the bus stop 10 minutes early.How far is the bus stop ?

(a)13
(b)12
(c)11
(d)10