walking at 5 7 of his usual rate a man reaches the market 16 minutes late find his usual time

normal speed (s1) = s,
reduced speed (s2) = 5s/7,
t2-t1=16
(d/s2)-(d/s1)=16
(7d/5s)-(d/s)=16
(d/s)(2/5)=16
t1=(16*5)/2
ti=40
11 years agoHelpfull: Yes(11) No(1)
ans 40 min.
Let distance to be traveled=x and Usual speed=s, Usual time taken=x/s
According to condition x/(5s/7) - x/s = 16/60 -> 3x=2s
Therefore usual time= x/s=2/3 Hr=40 minutes
11 years agoHelpfull: Yes(3) No(1)
usual time=40min ans
11 years agoHelpfull: Yes(1) No(1)
let x and t be d actual speed n reaching time of d man res.
from ques if he walk by speed of 5x/7 he reach at t-16 minutes.
so,
x speed t minutes
1 " t/x "
5x/7 " 5t/7 "

so,5t/7=t-16
2t=7*16
t=56

so,56 m is d usal time to reach d market.
10 years agoHelpfull: Yes(1) No(2)
(5/7)is usual speed then 7/5 is the usual time.
so 7/5 of usual time - usual time(i.e. 1)=16 min.
2/5 of the usual time=16 min
(16*5)/2=40
9 years agoHelpfull: Yes(1) No(0)
usual time taken is 112mins
11 years agoHelpfull: Yes(0) No(12)
Speed rate is 5/7
And given time 16 min late
Late time for remaining 2/7 Th of total rate
Here 1/7 is 8 min
Total 5/7 for 40 min
Answer is 40 min
7 years agoHelpfull: Yes(0) No(0)
S=D/T-----------
here
(5/7)S=D/(T+16)
using
solve and get T=40 minutes
5 years agoHelpfull: Yes(0) No(0)

walking at 5/7 of his usual rate.a man reaches the market 16 minutes late.find his usual time to reach the market.