SOLUTION: Respected sir, 1] A train travelling 25km an hour leaves Delhi at 9 a.m. & another train travelling 35 km an hour starts at 2 p.m. in the same direction . how many km from Delhi

Question 195507: Respected sir,
1] A train travelling 25km an hour leaves Delhi at 9 a.m. & another train travelling 35 km an hour starts at 2 p.m. in the same direction . how many km from Delhi are they together ?
2] Walking 3/4 of his usual rate,a man is 1 1/2 (one an half hour) hours too late.find his usual time.
plz sir solve my difficulty as soon as possible.
thank u.
Answer by ejazali_syed(7) (Show Source): You can put this solution on YOUR website!
1)A train travelling 25km an hour leaves Delhi at 9 a.m. & another train travelling 35 km an hour starts at 2 p.m. in the same direction . how many km from Delhi are they together ?
speed = distance/time => distance = speed *time
d = 25*5 = 125 kms. i.e. by the time the second train starts travelling which is 5 hrs after the first one the first train has covered a distance of 125kms.
for the first train => d1 = 25*t1;
for the second train => d2 = 35*t2
so the second train will catch the first train in t2 hrs.
relation of t1 in terms of t2 => t1 = t2+5 (since train 1 has started moving 5 hrs before t2)
we need the condition d1 = d2 => 25(t2+5) = 35(t2)
35t2-25t2 = 125 => t2 = 125/10
hence they will meet eachother after 12.5 hrs of train 2 leaving the station.
i.e. 2.30 am
------------------------------------------------
2)Walking 3/4 of his usual rate,a man is 1 1/2 (one an half hour) hours too late.find his usual time.
plz sir solve my difficulty as soon as possible.
same formula s = d/t --------- 1
let s be the usual rate. d be the distance travelled in t hrs
today he waliking at 3/4 of usual rate = 3s/4. and to cover the distance d it took 1.5 hrs more i.e. t+1.5

so 3s/4 = d/(t+1.5) ---------------2

usual time t = ?
from 1 d = st
from 2 d = 3s(t+1.5)/4
therefore st = s(3t+4.5)/4 ==> 4t = 3t +4.5 => t = 4.5 hrs is the usual time.
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