M4maths Previous Puzzles

28 Oct 2015 Time and distance

A train running at 25 kmph takes 18 second to pass a platform. Next, it takes 27/2 second to pass a man walking at the rate of 5 kmph in the same direction. Find the length of the train and that of the platform.

OPtion

1) 65m and 50m
2) 75m and 65m
3) 75m and 50m
4) 85m and 55m
5) 95m and 75m
6) 95m and 50m
7) 55m and 80m
8) 75m and 85m
9) 75m and 70m
10)None of these

Solution

Speed of the train relative to man
= 25-5 kmph = 20 kmph = 20*5/18 m/sec = 50/9 m/sec
In passing a man, the train covers it own distance with relative speed.
Therefore length of train = relative speed * time = 50/9 * 27/2 m
= 75 m
Also, the distance traveled in 18 second at 25 kmph or 25*5/18 m/sec. is the sum of lengths of the train and the platform
Therefore Length of train+Length of platrom = 25* 5/18 * 18 m
= 125
Hence, the length of platfrom = 125-75 = 50 metre
Option 3) 75m and 50 m

Correct Option: 3 Best Solution (19)

RANJAN FADIA   9 years AGO

lenght of train = 75 m
combine length of train & platform = 125m
so length of platform =50 m

Like? Yes (4) | No   

JEGATHA.H   9 years AGO

train length=x;platform len =y;
speed of train relative to man = (25-5)=20kmph
=(20*5/18)m/sec
= 50/9 m/sec
leng of train =speed*time=50/9*27/2=75m
speed of train = 25*5/18=125/18m/sec
18(x+y)/125=18
x+y=125
y=125-75=50m

Like? Yes (2) | No   

Anubhav jain   9 years AGO

D=s*t
25*5/18*18=125 meter
option 3

Like? Yes (2) | No   

bhaskar   9 years AGO

13.5*(25-5)*1000/3600=75
(25*18*1000/3600)-75=50

Like? Yes (2) | No (1)  

SAGNIK DATTA   9 years AGO

train length = t platform =p
t+p = 25*18/3600
t = 27*20/7200
solving t=75
p=50

Like? Yes (2) | No (1)  

MRITUNJAY   9 years AGO

let train length be x m and platform length is y m

x/20*5/18=27/2

x=75 m

75+y/25*5/18=18

y= 50 m

option 3.

Like? Yes (1) | No   

Robert Yogenson   9 years AGO

length of train =(25-5)X5/18X27/2=75 m
and total length =25X5=125
platform =125-75=50 m

Like? Yes (1) | No   

jalaj   9 years AGO

I Calculated the length of the train first by using concept of relative velocity of the train and man.Since both are moving in same direction.So,speed=20 * 5/18m/s=
Time (given as)=27/2 sec From this we get Distance =75m
Now I use the equation. 75+Length of platform/18=25*5/18(I used the equation Distance/Time=Speed.From this,I get the Ans length of the platform as 50m

Like? Yes (1) | No   

Ruby kaushik   9 years AGO

suppose length of train is - L
length of bridge is - B
So L+B = time* speed
L+B = 18*(75/18) [change speed in m/sec - 25*5/18]
L+B = 125 -------- (1)

now the relative speed of train and man is 20 km (25-5) = 20*5/18 = 100/18 m/sec

So L = 100/18* 27/ 2 = 75 ------- (2)

from equation 1 and 2
L = 75m and B = 50m

Answer is 3

Like? Yes (1) | No   

Priya Verma   9 years AGO

(length of train + length of platform)=25*5/18*18=125m
length of train = 20*5/18*27/2=75m
and hence length of platform = (125-75)=50m

Like? Yes (1) | No   

Arifa   9 years AGO

speed of train=25 km/ph=(25*5/18)=125/18 m/s
speed of train relative to man=25-5=20 km/ph
=(20*5/18)=100/18 m/s
length of train=(27/2)*(100/18)=75 m
total distance =length of platform+length of train
total distance=18*125/18=125 m
length of platform =125-75=50 m
ans option 3) 75 m & 50 m

Like? Yes (1) | No   

Abhisheak Kumar Missu   9 years AGO

Length of platform+train= 25*5/18*18=125m
Relative speed of train=25-5=20km/h
Length of train=20*5/18*27/2=75m
so Length of platform=125-75=50m

Like? Yes (1) | No   

sanjeev kumar   9 years AGO

by changing 25 kmph to m/s multiply it by 5/18 = 125/18m/s
and in 18 sec it covers 125m.
and 25kmph-5kmph= 20 kmph = 20*5/18=100/18m/s
in 13.5 sec it will cover = 75m (which will be the length of the train as man will have a negligible length w.r.t train).
therfore, if train lenght =75m then 125-75= 50m will be platform length.

Like? Yes (1) | No   

nikhil kumar   9 years AGO

relative speed=25-4=20 km/hr=100/18m/s
length of train(x)=time taken to cross pltfrm x speed=27/2 x 100/18=75m
now total length of train and platfrom =x+y
then x+y=speed of train in m/s x time to pass platform=50m
so option 3 is correct

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ABHIJEET ANAND   9 years AGO

let lenth of train =x;
lenth of plateform=y;
now,
from question,
x/(20*5/18)=27/2;
x=75m;
again,
(y+75)/(25*5/18);
y=50

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Anuj Tiwari   9 years AGO

Realtive Speed=25-5 KMPH

Now, Let length of train be x meters.
S=D/T

20*(5/18) m/s =(x) /(27/2)

Therefore, x=75 meters (Length of train)

now let length of platform be y meters

25*(5/18)=(75+y)/18

y=50 meters (length of platform)

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ritu raj anand   9 years AGO

time taken to cross the platform= (Length of train + length of platform)/ speed of train
L of train+ L of platform= 125
again,
time taken to cross the man= L of train/ relative speed of train and man
therefore, L of train =75
hence, L of platform= 125-75 = 50.

Like? Yes (1) | No   

akash gupta   9 years AGO

let l train length and p platform length
25*5/18=(l+p)/18
l+p=125
check the option 75+50
s0 3 option is right

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vikash ranjan   9 years AGO

25*5/18 *18=125
(25-5)*5/18 * (27/2)=75
125-75=50

Like? Yes (1) | No (1)