A train speeds past a pole in 15 second and a platform 100 m long in 25 second. Its length in metres

Let the length of the train is x meter and Speed of the train is y meter/second.
Then x/y = 15
=> y = x/15(speed of train)
a/q
(100+x)/25=x/15
=>x=150 m

Let the length of the train be x metres and its speed by y m/sec.

Then, x/y= 15 => y = x /15

Therefore x + 100/25 = x/15

=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

Let the length of the train be x metres and its speed by y m/sec.

Then, x/y= 15 => y = x /15

Therefore x + 100/25 = x/15

=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

length distance covered by train to cross a pole=length of the train=x,
If speed of the train=y m/s ,then
x/y= 15 ---(i) &
(x+100)/y= 25 ----(ii)
On solving (i) & (ii), y=10, x=150

length(train) = x m
speed(train) = y m/s.
Then,
speed=distance / time
y = x/15
train takes 25 seconds to travel 100 m platform ditance + it\'s Length.
(x + 100)/25 = x/15
= 15(x + 100) = 25x
= 15x + 1500 = 25x
= 1500 = 10x
= x = 150 m.


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Previous Day Puzzle and users best solution . 11 SEP 2014 : Discount A owes B, Rs. 1120 payable 2 years hence and B owes A, Rs. 1081.50 payable 6 months hence. If they decide to settle their accounts forthwith by payment of ready money and the rate of interest be 6% per annum, then who should pay and how much.
OPTION 1) A, Rs.50
2) A, Rs.70
3) B, Rs.80
4) B, Rs.70
5) A, Rs.60
6) B, Rs.100
7) B, Rs.50
8) A, Rs.90
9) B, Rs.25
10)None of these
SolutionP.W. of Rs. 1120 due 2 years hence at 6%
= Rs. [100*1120 / 100+(6*2)] = Rs.1000
P.W. of Rs. 1081.50 due 6 month hence at 6%
= Rs. [100*1081.50 / 100+(6/2)]
= Rs. [100*1081.50 / 103]
= Rs. 1050
So A owes B, Rs. 1000 cash and B owes a Rs. 1050 cash.
therefore
B must pay Rs.50 to A.

Option 7) B, Rs.50 Correct Option: 7 Best Solution (21) Best user's solutionDEVENDRA M
Devendra MIf today's A's amount payable to B =x, then
x+ (x*2*6/100)=1120 , x=1000
Similarly,if today's B's amount payable to A= y ,then
y+(y*(1/2)*6/100)= 1081.5, y=1050
So B should pay 1050-1000= 50 Rs. to A
(Option 7)Is this solution helpfull? Yes (3) | No (2) JITENDRA VARSHNEY
jitendra varshneyp.w of rs. 1120 due 2 years hence at 6%=rs[(100*1120)/(100+(6*2))]
=rs.1000
p.w of rs. 1081.50 due 6 monts hence at 6%=rs[(100*1081.50)/(100+(6*1/2))]
=rs.[(1000*1081.50)/103]
=rs. 1050
therefore ,A owes B,rs.1000 cash and B owes A rs. 1050 cash

so,B must pay rs. 50 to A
so.,option 7 is correct B,50 Is this solution helpfull? Yes (2) | No (1) KUSHAGRA TIWARI
Kushagra TiwariPw of 1120 due 2 years and hence 6%
={(100*1120)/100+(6*2)} = rs 1000

PW of rs 1081.50 due 6 month and hence 6%

={(100*1081.50)/(100+(6/2))}
={(100*1081.50)/103}
= rs 1050

so a owes b rs 1000 and B owes a rs 1050

so b pays rs 50 to aIs this solution helpfull? Yes (1) | No (1) DIPAK TIWARI
DIPAK TIWARILet A owes B X₹
Hence,(1120-X)=(X*6*2)/100
=>X=1000₹
Again let B owes A Y₹
Hence, (1081.50-Y)=(Y*1/2 *6)/100
=>Y=1050₹
Hence B will give 50₹ to A
Option 7Is this solution helpfull? Yes (1) | No (2) PAWAN KUMAR
PAWAN KUMARLet the amount to be paid to B by A be Rs. x and the amount to be paid to A by B be Rs. y.

Now, A's debt after 2 years = x + (x*6*2/100) = 1120
=> x= Rs. 1000
Also, B's debt after 6 months= y + (y*6*1/2*100) = 1081.50
=> y= Rs. 1050
Since y>x so, B will pay Rs. 50 to A .Is this solution helpfull? Yes (1) | No (2) PURUSHOTTAM SAH
purushottam sahPW of 1120 due 2 yr at 6%=1000
PW of 1081.5 due 6 month at 6%=1050
so A owes Rs 1000
B owes 1050
B pay RS 50
Is this solution helpfull? Yes | No (1) ANANDHAKUMAR
anandhakumaranswer is rs.50 by B
ans is option 7Is this solution helpfull? Yes | No (1) SUDHAKAR
sudhakarans is rs.50Is this solution helpfull? Yes | No (1) SELVAKUMAR
selvakumarb by rs.50Is this solution helpfull? Yes | No (1) ENUKONDA RAMAKRISHNA
Enukonda Ramakrishnagot the answerIs this solution helpfull? Yes | No (1) UDAYBHAN VISWKARMA
Udaybhan ViswkarmaA owes=P.W of Rs.1120 due 2 years hence
=Rs.[(1120*100)/(100+(6*2))]=Rs.(1120*100)/(112)=Rs.1000

Similarly

B owes=P.W of Rs.1081.50 due 6 months hence,
=Rs.[(1081.50*100)/(100+(6*0.5))] =Rs.(1081.50*100)/(103) =Rs.1050

Therefore B pay Rs 50 to AIs this solution helpfull? Yes | No (1) SHAIK JAHANGEER BASHA
Shaik Jahangeer BashaA owes=P.w. of Rs.1120 due 2 years hence
=Rs[1120*100/(100+(6*2))]
=1000
B owes=P.w. of Rs.1081.50 due 1/2 years hence
=Rs[1081.50*100/(100+(6*(1/2)))]
=1050
B must pay Rs.50 to AIs this solution helpfull? Yes | No (1) SHIKHA PANWAR
shikha panwarA owes=1120 due 2 years hence
=Rs[1120*100/(100+(6*2))]
=1000
B owes=1081.50 due 1/2 years hence
=Rs[1081.50*100/(100+(6*(1/2)))]
=1050
B must pay Rs.50 to AIs this solution helpfull? Yes | No (1) ABHISHEK VISHWAKARMA
Abhishek VishwakarmaPw of 1120 due 2 years and hence 6%
={(100*1120)/100+(12)} = rs 1000

PW of rs 1081.50 due 6 month and hence 6%

={(100*1081.50)/(100+(6/2))}
={(100*1081.50)/103}
= rs 1050

so a owes b 1000 and
B owes a 1050

so b pays rs 50 to aIs this solution helpfull? Yes | No (1) ASHISH SHENDE
Ashish ShendePw of 1120 due 2 years and hence 6%
={(100*1120)/100+(12)} = rs 1000

PW of rs 1081.50 due 6 month and hence 6%

={(100*1081.50)/(100+(6/2))}
={(100*1081.50)/103}
= rs 1050

so a owes b 1000 and
B owes a 1050

so b pays rs 50 to aIs this solution helpfull? Yes | No (1) MAHESH
Mahesh1081.50/1.03=1050
1120/1.12=1000
so, B oes 50 rupees to AIs this solution helpfull? Yes (1) | No (2) VAIBHAV SONI
vaibhav soniowes=P.w. of Rs.1120 due 2 years hence
=Rs[1120*100/(100+(6*2))]
=1000
B owes=P.w. of Rs.1081.50 due 1/2 years hence
=Rs[1081.50*100/(100+(6*(1/2)))]
=1050
B must pay Rs.50 to AIs this solution helpfull? Yes | No (1) CHIRAG RANJAN LAHIRI
Chirag Ranjan LahiriPW of Rs 1112 due 2 years hence at 6%
= (100*1120)/(100+6*2) = Rs 1000

and similarly in the next case it is Rs 1050

So A owes B 1000 and B owes A 1050
So B must pay Rs 50 to A, hence option "7"Is this solution helpfull? Yes (1) | No (2) SUMIT KUMAR
sumit kumarLet original amounts are A and B
interest for A=1120-A
and from formula
interest=A*6*2/100
so, 1120-A=(A*6*2)/100
A=1000 Rs.
same as
1081.5-B=(1081.5*6*.5)/100
B=1050 Rs.
B-A=50 Rs. option(7) ans.Is this solution helpfull? Yes (1) | No (3) SANJIT KUMAR SHARMA
Sanjit Kumar SharmaA owes=P.W of Rs. 1120 due 2 years hence
=Rs. [(1120*100)/(100+(6*2))]= Rs. 1000
B owes =P.W of Rs. 1081.50 due 1/2 years hence
=Rs.[(1081.50*100)/(100+(6*1/2))]=Rs. 1050
therefore, B must pay Rs. (1050-1000)= Rs. 50 to A.Is this solution helpfull? Yes | No (2) LAKSHMI NALLURI
lakshmi nallurifor rs.1120,pw=100*1120/112=1000
p.w of 1081.50,6mnths,6%=100*1081.5/(100+6/2)
=100*1081.5/103=1050
a to b ,1000rs
b to a,1050 rs
hence,b pay to a 50/-Is this solution helpfull? Yes | No (2)
Advertisements (X)

Train crosses the platform in (25-15) = 10 sec. hence Speed is (100/10) = 10 m/sec.
length of the train = 10 * 15 = 150 m

Let x is the length of the trn..
Speed=(x/15) m/s.
Agn the speed of the trn when it passd the platform..
Speed=(x+100)/25.
x/15=(x+100)/25
After solving..x=150..
Ans.150 m..

D/V=15
(D+100)/25=D/15
D=150m

Let the length of the train be x meters and its speed be y m/sec.
Then, (x+100)/25=x/15
By solving ,
x=150 m

Consider the length of train = x m.

in 25 sec. the train cover (100+x) m. distance and
in 15 sec. the train cover x m. distance.

so, x*25/15 = 100 + x

x= 150 m.

So, it's length 150 m.

Let the length of the train be x metres and its speed by y m/sec.

Then, x = 15 => y = x .
y 15
Therefore x + 100 = x
25 15
=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

let the length of train be l, then
l/15=s ............(1)
and (l+100)/25=s ...........(2)
substituting value of l from 1 in 2 we get
l=150m

(100+X)25=X/25
X=150

150 m.

Let x is length of train; we know that s=d/t, so
s=x/15
s=(x+100)/25
on equating both:
25x=15x+1500
=>10x=1500
=>x=150m

let the speed of train be x m/s.It crosses a pole in 15 secs. then its lenght=15x.now time taken to cross 100m long platform = 25secs. (100+15x)/x=25.so x=10 m/s. so lenght of train=15*10=150m

Let length of train is l;
then according to problem

l/15=(l+100)/25
l=150

Let the length of the train be x metres and its speed by y m/sec.

Then, x = 15 => y = x .
y 15
Therefore x + 100 = x
25 15
=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

so, option 9

as to cover extra 100 meter it takes extra 10 second so its speed is 10m/s
therefore in 15 second 10*15-150 m

let the speed of the train be y and distance be x
speed=distance/time
y=x/15...(1)
y=x+100/25
x/15=x+100/25
x=150
option9

Let length of train be x m.
Therefore, x*25/15= x+100
Solving, x=150

10=(100+l)/25
l=150m

150 meter

let the distance of train is x mtr
therefore
s=d/t=x/15--------(i)
and also
s=(x+100)/25----------(ii)

since the speed of the train is same
therefore x/15 = (x+100)/25

x=150mtr

let speed of train be x.
(100+x)/25=x/15
=> x= 150m.

x/15=(x+100)/25
=>x=150

Ans 9

Let the length of the train is X m.
So, the train covers X m in 15 sec => in 1 sec it goes (X/15)m
and it covers (X+100)m in 25 sec => in 1 sec it goes (X+100)/25 m.
so , X/15= (x+100)/25
solving these we get x=150.
so length of the train is 150 m.

The answer is 150m.
Let the length of the train be x metres and its speed by y
m/sec.

Then,
x/y = 15 => y = x/15 .
Therefore
(x + 100)/25 = x/15
=> 15(x + 100) = 25x
=> 15x + 1500 = 25x
=> 1500 = 10x
=> x = 150 m.

Let the length of the train is x meter and Speed of the train is y meter/second
Then x/y = 15 [because distance/speed = time]
=> y = 15/x

=>x+10025=x15x=150 meters

if train length x
100+x/25=x/15
x=150

150 is the answer.

let the length be x.
then X/15=(X+100)/25

let train's distance=xmtr
so speed of train during crossing pole=x/15
speed of train during cross platfoam =x+100/25
so x/15 = x+100/25 so x = 150

Let the length of the train be x metres and its speed by y m/sec.

Then, x = 15 => y = x .
y 15
Therefore x + 100 = x
25 15
=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

Let the length of the train be x metres and its speed by y m/sec.

Then, x = 15 => y = x .
y 15
Therefore x + 100 = x
25 15
=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

Let the length of the train be x metres and its speed by y m/sec.

Then, x = 15 => y = x .
y 15
Therefore x + 100 = x
25 15
=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

d/15=100+d/25
25d=15d+1500

150m

L/15 = (L+100)/25
so 2L= 300
L=150

Let the length of train be x metres.
Then, x/15 = (x+100)/25
=> 5x = 300 + 2x
=> x = 150 m Ans.

l=15v.....1
when cross the platform
(l+100)=25v.........2
from 1 and 2
we get l as 150

length of train is x metre
speed of train= (pole length+ train lenght)*time taken to cross pole by train
x*25=(100+x)*15
x=150metre

let x is length of train
then speed is x/15 or (100+x)/25
so x/15=(100+x)/25
25x=15x+1500
25x-15x=1500
x=1500/10.
x=150 meters
length of train is 150 meters

100+x/25=x/25=150

L/15=L+100/25
25L=15l+1500
L=150m

lets length=x meter,speed = vmeter/second

then x/v=15
again x+100/v=25
100/v=10
v=10
so x=150

length of train=x meter
a/c problem, x/15=(x+100)/25
x=150

x+100/25=x/15

150m

speed of the train is constant
say the train is x m
so,
x/15=x+100/25
=>x=150

150m is the ans

length of train= speed of train{x} * time {15}
Length of train + length of platform{100m} = x * 25
LT+100=25x
15x +100 =25x
10x=100
x=10m/s
length of train= speed of train*time
LT= 10*15
LT=150m Ans

let the length of the train be x
speed=distance/time
so 100+x/25=x/15
therefore x=150m

Let the length of the train be x metres and its speed by y m/sec.

Then, x = 15 => y = x .
y 15
Therefore x + 100 = x
25 15
=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

Let the length of the train be x metres and its speed by y m/sec.

Then, x/y=15=>y=x/ 15
Therefore x + 100/25=x/ 15
=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

Let the length of the train be x metres and its speed by y m/sec.

Then x/y=15
or y=x/15

(x+100)/25=x/15
25x=15x+1500
10x=1500
x=150m.

Suppose the length of train be x,

to cross a pole means train covers x meters in 15 secs.

and to cross plateform train will cover 100 + x meters in 25 secs

train covers 100 meters in:

(100 + x) - x = 25 - 15 = 10 secs

i.e. it covers 10 meters in 1 sec.

so value of x = 15 * 10 = 150

x+100 * 15/x=25
x=150

(l+100)/25=l/15
solve for length

first condition is of pole,from which we get speed=l/15(l:length of train in meters)
and from second condition,we have speed =(l+100)/25.
now equating both the speed , we have l=150 m.

let x length of rain and y be the speed of train.
So for first case: x=y*15
2nd case: 100+x=25*25
solving above two we get x =150 meter answer

let, the length of train is x
then,
speed of train = (x+100)/25 .................equation(1)

speed of train w.r.t pole = x/15........................equation(2)

equation(1)= equation(2)

(x+100)/25 = x/15

after solving this equation
we get,

x = 150m

answer (9) : 150 m
s=d/t
let length of train = L
A/q
s=L/15 ---------(1)
s=(L+100)/25 --------(2)
on solving (1) & (2)
L= 150 m

train passes a pole in 15 sec.
let the length be x and speed be y
therefore, x/y=15-->y=x/15
x+100/25=x/15
=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

let the length of the train is x
and speed of the train is y
so according to the question
x/(y*5/18)=15
and
(x+100)/(y*5/18)=25
solving above equations we get
x=150
so length of the train is 150 m..
thats it...

D=length of the train
v=speed of the train
D=v*15----(1)
100+v*15=25*v
v=10;
D=10*15=150

Let the length of the train is x meter and Speed of the train is y meter/second
Then x/y = 15 [because distance/speed = time]
=> y = 15/x

=>x+10025=x15x=150 meters

let x is length of train
then speed is x/15 or (100+x)/25
so x/15=(100+x)/25
25x=15x+1500
25x-15x=1500
x=1500/10.
x=150 meters
length of train is 150 meters

length of train is 15*s
100+15*s=25*s
s=10
15*10=150

speed= distance / time
let train length be x,
train speed to cross pole,
s= x/15 -------(1)
similarly to cross the platform length 100m is,
s=(x+100)/25 ------(2)
equating (1) and (2),
x/15=(x+100)/25
by solving,
1500+15x=25x
1500 =10x
therefore train length x= 150m

the speed of the train remains the same.
so let the length of the train be x
so,
(x+100)/25 = x/15
x=150

let the length of train be x metre
is passes the pole in 15 sec
therefore distance travelled = x(i.e length of train)
as dist=speed*time
so x=(speed of train) s *15
x=15s
now train crosses a platform of 100 m
therefoe total dist now is x+100
so x+100=s*25 now solving the 2 equations obtained we get x=150m

Length of train be L, Speed be S.
L/S=15 ;
L+100/S = 25 ;
Solving S=10 m/s ;
L= 150 m

10 sec=100m
So 15 sec=150m.

x/15=x+100/25
x/3=x+100/5
5x/3=x+100
x=150

x/15= (100+x)/25
which satisfied by x = 150

X/15=(X+100)/25;
x=150;

the lenght of the train is X mtrs and its speed y m/sec
x/y=15 then y=x/15
therefore x+100/25=x/15 then ans is x=150m..

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

we know that speed = distance/time
= x/y

for pole, y = x/15
for platform, y = (x+100)/25

now. x/15 = (x+100)/25
solving we get, x = 150 metre.

option 9) 150 m

(x+100)/25=x/15
=>x=150m

answer 9)

let Length of Train be x metres and its speed be y m/s
x/y=15
or,y=x/15
or,(x+100)/25=x/15
or,x=150m

time taken to cover 100m is 10 sec
so speed=10m/s
so length of train =10*15=150m

Let the length of the train be x metres and its speed by y m/sec.

Then, x = 15 => y = x .
y 15
Therefore x + 100 = x
25 15
=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

speed is same
so, x/15=(x+100)/25
x=150m

let the length and speed of the train be d & s resp.
then according to the question,
d=s*15
& (d+100)=s*25
solving these two equations we get d=150m

Let the length of the train be x metres and its speed by y m/sec.

Then, x = 15 => y = x .
y 15

Therefore x + 100 = x
25 15

=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

15=x/y
so x=15y
now
25=(15y+100)/y
so y =10
then length of train is 10* 15=150 m

consider train lenth is X than X/15=100+X/25 solve this eq. than X=150.

x/15=(100+x)/25
x=150m

let the speed of train is x, and it\'s length is L

then x*15/3600=L;

also, x*25/3600=(L+100/1000)

after solving these two equations we find L=150 m

If L is the length of the train
and S is the speed of the train
and t is time

d = S t
to cross the pole
L = 15 S

To cross the platform
L + 100 = 25 S
15 S + 100 = 25 S
10 S = 100
S = 10
L = 15 (10) = 150 meters

length of train be x
speed of train while passing pole=speed while passing platform
i.e x/15=(x+100)/25
x=150

length be x, velocity=x/15
so, (100+x)/25=x/15
=>x=150 (9).

x/15 =v;
(100+x)/(x/15)=25
x=150 m

let the length of the train be l and its speed be x.
so with the pole condition we have l/15=x;

and the platform condition is
(100+l)/25=x;

solving we get x=10 and l=150 m

x/15=(x+100)/25
25x=15x+1500
25x-15x=1500
10x=1500
x=150
the length of train is 150m

train moves his own length in 15 secs so he takes 10 sec to cross the plathform of length 100m
so in each second train travels 100/10 m = 10m.
in 15 seconds it travels 150 m which is his own length ..

Let the length of the train = x m
Time taken to cross the pole = 15 s
Speed of the train =x/15 m/s
Total distance of train and length of the platform = x + 100 m
Time taken to cross the platform =25 s
Hence speed of the train =(x + 100)/25 m/s
Hence x/15 =(x + 100)/25
or 25x = 15x +1500
10x = 1500
or x = 150 m

when train across pole
speed =train length(l)/15=l/15

speed when train across the platform speed=(l+100)/25
by equating
l=100m

let the length of the train be L,
L/15=(L+100)/25
L=150m

let d be the length of train in m
x be its speed in m/sec
d/x=15
d+100/x=25
solving d=150m

150 is correct answer

x/15= (100+x)/25
10x=1500
x=150
so the length of train is 150 mtr

D/15=S and D+100/25=S
therefore 25D=15D+100
D=150..option 9..

(X/15)=((x+100)/25) . . .by solving we get length x=150m

speeed constant, take it in both case and equate them

let the length of train be x m
then speed of the train is x/15 m/s
so,it takes 25 second to cover a distance of (x+100) m
now speed will be (x+100)/25 m/s
since the speed will be same
x/15 =(x+100)/25
we get, x=150

speed of train=distance/time

s=m/15 while crossing pole,
s=m+100/25 while crossing platform

solving we get
m=150

let x be speed
15x+100=25x
x=10
lenth of train 15x=150

(x+100)*15/x=25
so, x=150m.

Let the length of the train is x meter and Speed of the train is y meter/second
Then x/y = 15 [because distance/speed = time]
=> y = 15/x

=>x+10025=x15x=150 meters


So length of the train is 150 meters

(100+x)*3=x*5
=> x=150

N\A

speed=x/15
{(100+x)*15}/x=25

if length is L meters and speed is v m/sec

then time taken to pass a pole = L / v = 15 ....... (i)
similarly time taken to cross the platform = (L+100) / v = 25...... (ii)
now writing both eq in terms of and equating both
L / 15 = (L + 100) / 25

solving this we get L = 150

let the speed of the train be x. so if it speeds past a pole in 15 seconds, then speed is x/15. Again if it covers a platform of length 100m in 25 seconds then speed is (x+100)/25. equating the two speeds we get x=150. So length of the train is 150 meters

(100+x)/25=x/15
->15(100+x)=25x
->1500=10x
x=150
where is length of train

x/(x+100)=15/25
or,x=150

Let train Length be X
so (X+100)/25= X/15
now solve X we get,
X=150

length = 150 m

150...................

speed(s) = distance(d)/time(t)
s = d/15=(d+100)/25
d*5 = d*3 +300
2d = 300
d = 150m

For covering the platform the train is taking 10 seconds more compared to covering the pole.
This is because of the length of the platform, so train takes 10 seconds to cover that 100m. Its speed is 10 meters per second.
Now in the first case the train takes 15 seconds to cover pole, so it covers a distance of 150 meters in 15 seconds. the distance it covers is nothing but the length of the train.
Answwer is 150 meters

Let the length of train be Xmetre
X/15 = (X+100)/25
=>X=100metre
Option 9

Let the length of the train be x metres and its speed by y m/sec.

Then, x = 15 y = x .
y 15
x + 100 = x
25 15
15(x + 100) = 25x

15x + 1500 = 25x

1500 = 10x

x = 150 m.

let length of the train = x m
train passes a pole in 15 sec, therefore speed = x/15
now, train passes a platform of 100 m in 25 sec,
therefore speed= (x+100)/25
now, x/15=(x+100)/25
x=150 (option 9)

l/15=(l+100)/25
so,l=150m

we know that,speed=time/distance,
lets take train speed =x,
and train speed=y
so x/y=15,
so y=x/15
than total distance=x+100 and time is 25 second
by the formula of s=d/t
x+100/25=x/15
by solving this equation we get x=150 meters so answer is 9 option

let length of the train= x m
speed of train = y m/sec

so x/y=15
y=x/15

therefore x+100/25=x/15
by solving we get x=150m i.e length of the train

so option (9) is the answer

let length of train be x
then x/15=(x+100)/25
hence x=150m

from 1st
length of train(d)=s*15

from 2nd
100+d=s*25

after putting value of d in 2nd
s=10 and d=150

Let length of the train = X meter
speed of the train = S meter/sec
then x/s = 15
or X=15*S--------- (i)

x+100/S=25
s = 10
hence X = 150 meter

Let the length of the train be x m.
then,speed to cross a pole in 15 seconds=x/15
and speed to cross 100mplatform in 25 seconds=x+100/25
so, x/15=x+100/25
=>25x=15x+1500
=>10x=1500
=>x=150m

Length of the train=150 m.

x/s=15
(x+100)/s=25
solving we get x=150

Let the length of the train = x metres

Length of the platform=100 m(given)

Length of the pole=>negligible==0

Speed=distance/time
Speed of the train is same so equate speed in crossing pole and platform

Speed crossing pole= Speed crossing platform

(0+x)/15=(100+x)/25
25x=15x+1500
10x=1500
x=150
x

Let, speed of train=x m/s
so, in 15 sec= 15x m and in 25 sec= 25x m
25 x-15x=100 So, x=10 m/s
hence, length of train= 150 m

train length = x m
100+x cross 25 Sec
x cross 15 sec
100 m cross 10sec (10m=1sec)
x=150 m cross 15sec

Let the length of the train be x metres and its speed by y m/sec.

Then, x = 15 => y = x .
y 15
Therefore x + 100 = x
25 15
=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.