A train overtakes two persons who are walking in the same direction in which the train is going, at

shovan Manna   10 years AGO

here ,
9(v-2)=10(v-4)
v=22km/hr
so the length of the train is 9*(22-2)*5/18=50m

Like? Yes (62) | No (6)  

MRITUNJAY   10 years AGO

Speed and length of train is=x ,d respectively.
Relative speed of A. =x-2
relative speed of b=x-4
so equate with time
9/3600=d/x-2
10/3600=d/x-4
after solve d=50 metre

Like? Yes (15) | No (2)  

Surit Datta   10 years AGO

Let train speed be s
1st person's speed,s1 = 2km/hr =2*(5/18)=5/9 m/sec
2nd person's speed,s2 = 4km/h = 4*(5/18)=10/9 m/sec

Length of train,d = (s-s1)*t1 or (s-s2)*t2
so, (s-5/9)* 9 = (s-10/9)* 10 [t1=9sec, t2=10sec]
s=55/9 sec

Now,take 1st eq.
s-s1= d/t1 [bcz, Speed=Dis/Time]
(55/11)-(5/9)=d/9
Therefore we get d(length of train)=50m
option 4
That's it.

Like? Yes (12) | No (3)  

lakshmi   10 years AGO

Let the length of the train be x km and its speed y km/hr. Then, speed relative to first man = (y-2) km/hr.
Speed relative to second man = (y-4) km/hr.

x/y−2=9/60×60 and x/y−4=10/60×60

9y - 18=3600x or 10y - 40=3600x
So, 9y-18=10y-40 or y =22

x/22−2=93/600 or x = 20×9/3600=1/20

= (1/20×1000)m = 50 m

Like? Yes (13) | No (6)  

Sanu Panji   10 years AGO

1st person: rate = 2 kmph= 5/9 mps
2nd person: rate= 4 kmph= 10/9 mps
say length of the train is X
and speed of the train is Y mps

so, Y=(X/9)+(5/9)...............(i)
and Y=(X/10)+(10/9)..........(ii)

solving we get X= 50 metres

Like? Yes (7) | No   

Jaspalsingh Parmar   10 years AGO

speed=dist/tym
2 kmph = ( 2 x 5 ( m/sec = 5 m/sec.
18 9
4 kmph = ( 4 x 5 ( m/sec = 10 m/sec.
18 9
Let the length of the train be x metres and its speed by y m/sec.

Then, ( x ( = 9 and ( x ( = 10.
y - 5
9
y - 10
9
Therefore 9y - 5 = x and 10(9y - 10) = 9x

=> 9y - x = 5 and 90y - 9x = 100.

On solving, we get: x = 50.

Therefore Length of the train is 50 m.

Like? Yes (6) | No   

sravya   10 years AGO

let xmts for train and speed ym/sec..
1st man speed is 2*5/18=5/9m/sec
2nd man speed is 4*5/18=10/9m/sec
x/(y-5/9)=9sec-------------------1eqn
x/(y-10/9)=10sec----------------2ndeqn
solving bth x=50mts...ans...

Like? Yes (7) | No (2)  

Chirag Ranjan Lahiri   10 years AGO

9y - 5 = x and 10(9y - 10) = 9x
=> 9y - x = 5 and 90y - 9x = 100.
we get: x = 50.
Therefore Length of the train is 50 m

Like? Yes (7) | No (3)  

K.ARun   10 years AGO

2kmph=2*5/18 m/sec=> 5/9m/sec
4kmph=4*5/18 m/sec=>10/9m/sec

x/(y-(5/9))=9 and x/(y-(10/9))=10

on solving we get x=50m

Like? Yes (5) | No (2)  

Devendra Marghade   10 years AGO

If length of the train=x & its speed=y, then
Time to cross 1st person= x/(y-2) =9/3600 ---(i) &
Time to cross 2nd person=x/(y-4) =10/3600 ---(ii)
On solving (i) & (ii), x=2/40 Km =2*1000/40=50 meters

Like? Yes (4) | No (1)  

Shreya Banerjee   10 years AGO

First person speed = 2*(5/18) = 5/9 m/sec
Second person speed = 4*(5/18) = 10/9 m/sec

Let the length of train is x metre and speed is y m/sec

then,
x/(y−5/9)=9
= >9y−5=x
=>9y−x=5.....(i)
And,
x/(y−10/9) =10
=> 90y−9x=100.....(ii)
from (i) and (ii), we get, x=50.

So answer is 50 metres

Like? Yes (5) | No (2)  

sonali soni   10 years AGO

consider speed of train is s
and we know that d = s*t
d= (s-2)*9/60 ----------(1) relative speed
d= (s-4)*10/60----------(2) relative speed
(1) = (2) because distance is same
9 *s - 18 = 10*s - 40
s = 22 km/h

from eq (1)
d = (22-2)*5/18 *9
d = 50 meter

Like? Yes (3) | No   

Rats   10 years AGO

First person speed = 2*(5/18) = 5/9 m/sec
Second person speed = 4*(5/18) = 10/9 m/sec
Let the length of the train be x metres and its speed be y m/sec.

Therefore 9y - 5 = x and 10(9y - 10) = 9x

=> 9y - x = 5 and 90y - 9x = 100.

On solving, we get: x = 50.

Therefore Length of the train is 50 m.

Like? Yes (2) | No   

rupesh   10 years AGO

2kmph=5/9 m/sec,4kmph=10/9m/sec
length of train=x;speed of train=y
x/y-5/9=9,x/y-10/9=10
9y-x=5.....1st eq
90y-9x=100.... 2nd eq
frm 1 n 2
x=50

Like? Yes (2) | No   

purushothama   10 years AGO

First person speed = 2*(5/18) = 5/9 m/sec
Second person speed = 4*(5/18) = 10/9 m/sec

Let the length of train is x metre and speed is y m/sec

then,
xy−59=9=>9y−5=x=>9y−x=5.....(i)Also,xy−109=1090y−9x=100.....(ii)from (i) and (ii), we get,x=50

Like? Yes (2) | No (1)  

ITHA Naresh Babu   10 years AGO

Answer : Option 4

Explanation :

Let x is the length of the train in meter and v is its speed in kmph

x/9 = ( v-2)(10/36) ---(1)

x/10 =( v-4) (10/36) --- (2)


Dividing equation 1 with equation 2

10/9 = (v-2)/(v-4)

=> 10v - 40 = 9v - 18

=> v = 22

Substituting in equation 1, x/9 = 200/36 => x = 9×200/36 = 50 m

Like? Yes (2) | No (1)  

jayalakshmi   10 years AGO

Let the length of the train be x km and its speed y km/hr. Then, speed relative to first man = (y-2) km/hr.
Speed relative to second man = (y-4) km/hr.

x/y−2=9/60×60 and x/y−4=10/60×60

9y - 18=3600x or 10y - 40=3600x
So, 9y-18=10y-40 or y =22

x/22−2=93/600 or x = 20×9/3600=1/20

= (1/20×1000)m = 50 m

Like? Yes (1) | No   

veeranjaneyulu   10 years AGO

d=sxt

Like? Yes (1) | No   

Mahantesh Desai   10 years AGO

x/(y-5/9) = 9 and x/(y - 10 /9) = 10
9y - 5 =x 10(9y-10) = 9x => 9y - x = 5 and 90y - 9x =100
on solving we get x = 50
The length of train is 50m

Like? Yes (1) | No   

kumari jyoti   10 years AGO

let the length of the train is x and it's speed is s. then,
to pass the 1st person, the train will have to cover (x+2*9*5/18)m in given 9 secs. i.e.,{(x+2*9*5/18)/s}=9....(1).
again,to pass the 2nd person, the train will have to cover(x+4*10*5/18)m in given 10 secs. i.e., {(x+4*10*5/18)/s}=10....(2).
solving equation 1 &2, we get,x= 50 m.

Like? Yes (1) | No   

khushboo goenka   10 years AGO

let rain speedis x and 2kmph=5/9 mps
and 4kmph=10/9mtps
so,
thn x-5/9=lngth of train /9
x-10/9=lnth of train/10

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Rajat Rokade   10 years AGO

answer is 50 metres i.e. option 4.

Like? Yes (2) | No (1)  

Rahul Kingdon   10 years AGO

Explanation:

2 kmph = ( 2 x 5 ( m/sec = 5 m/sec.
18 9
4 kmph = ( 4 x 5 ( m/sec = 10 m/sec.
18 9
Let the length of the train be x metres and its speed by y m/sec.

Then, ( x ( = 9 and ( x ( = 10.
y - 5
9
y - 10
9
Therefore 9y - 5 = x and 10(9y - 10) = 9x

=> 9y - x = 5 and 90y - 9x = 100.

On solving, we get: x = 50.

Therefore Length of the train is 50 m.

Like? Yes (1) | No   

Raj   10 years AGO

2 kmph = (2*5/18)m/sec =5/9 m/sec
4 kmph = (4*5/18)m/sec =10/9 m/sec
Let the length of the train be 'x' metres & its speed by 'y' m/sec-

then,
(x / y-5/9)=9
=> 9y-5=x
=> 9y-x=5 .....(i)

(x / y-10/9)=10
=> y-9x= 100..... (ii)
now we get from (i) n (ii) is 50 of train length

Like? Yes (1) | No   

sreedhar   10 years AGO

2 kmph = ( 2 x 5 ( m/sec = 5 m/sec.
18 9
4 kmph = ( 4 x 5 ( m/sec = 10 m/sec.
18 9
Let the length of the train be x metres and its speed by y m/sec.

Then, ( x ( = 9 and ( x ( = 10.
y - 5
9
y - 10
9
Therefore 9y - 5 = x and 10(9y - 10) = 9x

=> 9y - x = 5 and 90y - 9x = 100.

On solving, we get: x = 50.

Therefore Length of the train is 50 m.

Like? Yes (1) | No   

Vinay Sipani    10 years AGO

let the length of train be 'L'm and speed be 's' m/s

The length of persons is negligible.Their speed are 2 kmph and 4 kmph.
In m/s,their speeds are 0.55 m/s and 1.11 m/s.

Time taken for train to overtake is given by
L/(s-x) = t where x represents speed of person.

We have,
L/(s-0.55)=9 => s = 0.55 + (L/9) and
L/(s-1.11)=10 => s = 1.11 + (L/10)

Equating both, we get L = 50 m

Like? Yes (1) | No (1)  

Ashwin Dinesh   10 years AGO

10l + 50 = 9l + 100
So l = 50
which is option 4)

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Supriya Mondal   10 years AGO

let length of the train is x m,
then the train goes in 9 sec is=(20/36)*9+x m
and it goes in 10 sec is=(40/36)*10+x m

so we can write,
((20/36)*9 + x)/9=((40/36)*10 + x)/10

by solving this equation we get x=50m
so option 4)50m is the correct answer

Like? Yes (1) | No (1)  

madhusudhanreddy   10 years AGO

First person speed = 2*(5/18) = 5/9 m/sec
Second person speed = 4*(5/18) = 10/9 m/sec

Let
the length of train is x metre and
speed is y m/sec
then,
xy−59=9
=>9y−5=x
=>9y−x=5.....(i)
Also,xy−109=1090y−9x=100.....(ii)
from (i) and (ii),
we get,x=50
so,the length of the train is 50m.

Like? Yes (1) | No (1)  

poornima    10 years AGO

2 kmph = (2 x 5 /18)m/sec =5/ 9m/sec
4 kmph =( 4 x 5 /18)m/sec = 10/9 m/sec.
Let the length of the train be x metres and its speed
by y m/sec.
Then,
9 y - 5 = x and 10(9 y - 10) = 9 x
ans X=50mtrs

Like? Yes (1) | No (1)  

POOJA SINGH   10 years AGO

2 kmph = 5/9m/s
4kmph=10/9m/s
Let the length of the train be x m and its speed be y m/sec.
x /(y-(5/9))=9 and x/(y-(10/9))=10
9y - 5 = x and 10(9y - 10) = 9x
=> 9y - x = 5 and 90y - 9x = 100.
On solving, we get: x = 50.
Therefore Length of the train is 50 m.

Like? Yes (1) | No (1)  

Tania Mondal   10 years AGO

Let the length of the train be x metres and its speed by y m/sec.
x/(y-5/9)=9
and
x/(y-10/9)=10

Therefore 9y - 5 = x and 10(9y - 10) = 9x

9y - x = 5 and 90y - 9x = 100.

So x = 50.

Therefore Length of the train is 50 m...

Like? Yes (1) | No (1)  

Vijaya Kumar Bathini   10 years AGO

Converting Km/hr to M/sec

(2*5)/18 = 5/9 m/sec
(4*5)/18 = 10/9 m/sec

Let Length of train be X M/sec and Speed is Y M/sec

Time = distance/speed

9 = x/(y-5/9) -> 9x/9y-5 =81 -> 9y-5 =X
10 = X/(y-10/9) ->10(9Y-10) = X

Solving these we get X=50 m

Like? Yes (1) | No (1)  

Narra Sirisha   10 years AGO

First person speed = 2*(5/18) = 5/9 m/sec
Second person speed = 4*(5/18) = 10/9 m/sec
9y-5 = x \
=> 9y-x = 5 .....(i)
90y-9x = 100 .....(ii)
from (i) and (ii), we get
x=50

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MRITUNJAY KUMAR   10 years AGO

Option)4

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sravani   10 years AGO

let length of the train be X
Speed be Y.
Relative speed with the first person in m/sec is y-5/9
Relative speed with the second person in m/sec is y-10/9
so,(y-5/9)*9 =X
and (y-10/9)*10=X
solving these
y=55/9
substitute in any of the eq
we get,
X=50meters...

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Pramod Gaikwad   10 years AGO

answer is option 4) 50m

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Nupur   10 years AGO

L/ (x-2) *5/18 = 9
L/ (x-4) *5/18 = 10

Solving the two we get that L = 50

Like? Yes (1) | No (2)  

shubham   10 years AGO

let speed of the train be s m/s and its length be=l.
speed of person 1=2kmph=5/9 m/s
speed of person 2=4kmph=10/9 m/s
so we have,
l/(s-5/9)=9--------(eq1)
and

l/(s-10/9)=10-------(eq2)
solving both of these we get,s=55/9 m/s.

substituting this value in eq1,we get,
l=50metres i.e.4th option

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sanjeev kumar   10 years AGO

50 m

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Rahul Rathod   10 years AGO

First person speed = 2*(5/18) = 5/9 m/sec
Second person speed = 4*(5/18) = 10/9 m/sec
Let the length of train is x metres and speed is y m/sec
then,
x/(y-(5/9))=9
and
x/(y-(10/9))=10
so
we get
9y-x=5
90y-9x=100
by solving we get x=50m
answer is 50m option 4)

Like? Yes (1) | No (3)