Latest Elitmus Aptitude Question SOLUTION: a train met with an accident 60 km away from Amsterdam station.It completed the remaining journey at 5/6 th of the previous speed and reached Trent Bridge station 1 hour 12 in late

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(#M40018076) ELITMUS Numerical Ability question elitmus Keep an EYE

a train met with an accident 60 km away from Amsterdam station.It completed the remaining journey at 5/6 th of the previous speed and reached Trent Bridge station 1 hour 12 in late.Had accident taken place 60 km further ,it woud have been only 1 hour late a What is th normal speed of train what ist he distance between Amsterdam and Trent Bridge ?
explanation is needed

Asked In Elitmus SHRESTH (11 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

Mohit Agarwal  (11 years ago)

LET THE NORMAL SPEED OF TRAIN BE "x km/hr"
LET THE TOTAL DISTANCE BE "y km"

-Accident occur after 60km from Amsterdam i.e; train has travelled 60km with normal speed. hence time taken at normal speed till accident= (60/x) hr.
-Rest distance from accident to Trent Bridge= (y-60) km.
-Speed for rest of distance= (5/6)x = (5x/6).
-time taken to travel rest distance= (y-60)/(5x/6) = ((6(y-60))/5x)
-if accident would not have occurred den time taken = y/x
-train late by 1hr 12min= 1+(12/60)= 6/5 hr
- Hence 1st equation is =
(60/x)+ ((6(y-60))/5x) = y/x + 6/5 ------eqn 1
- similarly eqn 2 can be written as
(120/x) + ((6(y-120))/5x) = y/x + 1 ------eqn 2

simplify and solve both the equations...
ans will be x = 60 km/hr and y = 420 km
Thanks.

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Shalu Jain  (11 years ago)

normal speed of the train 60km/hr.
&
distance between Amsterdam and Trent Bridge is 420 km


explanation
let normal speed of the train be x km/hr and distance between Amsterdam and Trent Bridge be y km.
normally train takes y/x hrs to reach Trent Bridge.
in case 1
time taken till accident=60/x hr
time taken 2 cover remaining distance= 6 * (y-60)/5x
according 2 question,
60/x + 6 * (y-60)/5x= y/x + 1.2 (converting minute into hour)
on solving this equation we get,
y-6x=60 .................(1)

now in case 2
time taken till accident=(60+60)/x hr=120/x hr
time taken 2 cover remaining distance= 6 * (y-120)/5x
according 2 question,
120/x + 6 * (y-120)/5x= y/x + 1
on solving this equation we get,
y-5x=120 .................(2)

on solving (1) and (2) we get
x=60
y=420

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Deepak kumar Dubey  (11 years ago)

speed of train is :60kmhr
and distance bw amsterdam and trent bridge is :50km

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Sumit Rawal  (11 years ago)

5/6 of the speed means 6/5 of the time
6/5T=5/5T+1/5T
here it is given that car is late by 1 hour 12 minutes means6/5 hours
-1/5T=6/5hours
T=6 hours
this implies car took 6 hours at normal speed from accident point to trent bridge
2nd condition
If accidnent took place 60Km further car goes late by 1 hour
1/5T=1 hour
T= 5 hours
Distance travelled in 1 hour at normal speed =120-60=60
Speed = 60Km/hour
Total distance = 60*6 +60(distance from amsterdam to accident point) =420 Km/h

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RAHUL RAJ  (11 years ago)

Normal speed of train =60km/hr.
Distance between Amsterdam & Trent Bridge=420km
Exp:
Let normal speed be a;
Actual distance be b;
Actual time = a/b;
Time taken for 60km=60/a;
time taken for remaining distance= (b-60)/(5a/6)= 6*(b-60)/5a = (6b-360)/5a;
total time = (6b-360)/5a+60/a= 6b-60/5a;
time difference = 6b-60/5a-b/a= 1hr 12 min= 6/5;
=> b=6a+60.........(i);
similarly
6b-120/5a-b/a=1;
=> b=5a+120........(ii);
solving i and ii
a=60 and b=420

Is this solution Helpfull? Yes (3) | No (5)  

Akshay Mandlik  (11 years ago)

No need to find speed in this one.
Let the total Distance bet Amster and TB be x
The delay at 5/6th of the original speed for a distance of (x-60)kms is 1hr 12mins
The delay at 5/6th of the original speed for a distance of (x-120)kms (if the accident took place 60 kms further i.e 60+60 kms from Amster) is 1hr
Therefore, at 5/6th of the original speed the delay is 12 mins per 60kms
Therefore, for a delay of 1hr 12 mins the distance covered is 60*(72mins/12mins)
= 60*6 =360kms
total distance bet stns is 360+60kms (60kms travelled before accident)
=420kms

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Akshay Mandlik  (11 years ago)

Didn't know speed was asked,
Let T1,S1 & D1 be the original time, speed and distance
T2, S2& D2 be the time, speed and distance after accident
D1=D2=60kms
T2=T1+12mins
S2=5S1/6
s=d/t
60=S1*T1=S2*T2
S1*T1=(T1+12)*5S1/6
T1=5T1/6 + 10
T1/6=10
T1=60MINS=1hr
S1=D1/T1=60/1
=60kmph

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