Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 4)
4.
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
Answer: Option
Explanation:
Let the speeds of the two trains be x m/sec and y m/sec respectively.
Then, length of the first train = 27x metres,
and length of the second train = 17y metres.
![]() |
27x + 17y | = 23 |
x+ y |
27x + 17y = 23x + 23y
4x = 6y
![]() |
x | = | 3 | . |
y | 2 |
Discussion:
233 comments Page 6 of 24.
Shree said:
1 decade ago
Its simple .
T = D/S
Therefore total distance = 27x+17y and total speed = x+y
Total time = 23
23= 27x+17y/x+y
T = D/S
Therefore total distance = 27x+17y and total speed = x+y
Total time = 23
23= 27x+17y/x+y
Deepak said:
1 decade ago
@niveditha
Since both train crossing each other in 23s, so have to multiply 23 by sum of speed of both train.
Since both train crossing each other in 23s, so have to multiply 23 by sum of speed of both train.
GUFRAN said:
1 decade ago
Let, the ist train speed v1=X m/s and T1=27s
2nd trin speed V2=Y m/s and T2=17s
Then S1=V1*T1=27x
S2=V2*T2=17Y
Both are in opposite dir. Then speed is S=S1-S2=27X-17Y
They cross each other means time difference 23(X-Y)=23X-23Y
now...
27X-17Y=23X-23Y
4X=-6Y
X/Y=-3/2=3/2(BECZ speed is not -ve)
X:Y=3:2
2nd trin speed V2=Y m/s and T2=17s
Then S1=V1*T1=27x
S2=V2*T2=17Y
Both are in opposite dir. Then speed is S=S1-S2=27X-17Y
They cross each other means time difference 23(X-Y)=23X-23Y
now...
27X-17Y=23X-23Y
4X=-6Y
X/Y=-3/2=3/2(BECZ speed is not -ve)
X:Y=3:2
Mobarak said:
1 decade ago
If two trains running at the speed x & y respectively in opposit direction then their relative velocity is (x+y) m/s.
Now if two trains cross each other then they have to cross total length of two trains which is (27x+17y).
Now time taken for crssng the total length = (27x+17y)/(x+y) i.e 23.
Now if two trains cross each other then they have to cross total length of two trains which is (27x+17y).
Now time taken for crssng the total length = (27x+17y)/(x+y) i.e 23.
Narayan said:
1 decade ago
27 17
23
23-17=6 27-23=4
Then ratio become 3:2
Dheeraj said:
1 decade ago
Hi Lokesh,
Can you please explain why you've said 27x-24x=23y-17y.....?
I can't see figure of 24 ain't where???
Thanks in advance....
Can you please explain why you've said 27x-24x=23y-17y.....?
I can't see figure of 24 ain't where???
Thanks in advance....
Ankit said:
1 decade ago
Thanks Y.S.Rama Krishna
I understood the answer now. :)
I understood the answer now. :)
Shahul hameed.M said:
1 decade ago
Hi friends in smart way....
To find out the speed ratio so ,
formula is Length = speed*time
so,
speed = length /time
here 2 trains so two different length and speed so the formula
s1+s2=L1+L2/time
speed we don't know so we take X,Y for speed for 2 trains
respectively ....
use above formula:
X+Y=L1+L2/23
HERE
L1=speed*time (speed is not given so take it as X,time is given that is 27)
L2= speed*time (speed is not given so take it as Y,time is given that is 17)
SO L1=27X
L2=17Y
apply it,
X+Y=27X+17Y/23
SO SOLVE IT
23(X+Y)=27X+17Y
23X+23Y=27X+17Y
separate the X and Y terms so we get
23X-27X+23Y-17Y=0
WE GET:
-4X+6Y=0
6Y=4X
RATIO (means difference of X and Y)
SO FRIEND,
X/Y=6/4=3/2
SO THE RATIO IS 3:2
To find out the speed ratio so ,
formula is Length = speed*time
so,
speed = length /time
here 2 trains so two different length and speed so the formula
s1+s2=L1+L2/time
speed we don't know so we take X,Y for speed for 2 trains
respectively ....
use above formula:
X+Y=L1+L2/23
HERE
L1=speed*time (speed is not given so take it as X,time is given that is 27)
L2= speed*time (speed is not given so take it as Y,time is given that is 17)
SO L1=27X
L2=17Y
apply it,
X+Y=27X+17Y/23
SO SOLVE IT
23(X+Y)=27X+17Y
23X+23Y=27X+17Y
separate the X and Y terms so we get
23X-27X+23Y-17Y=0
WE GET:
-4X+6Y=0
6Y=4X
RATIO (means difference of X and Y)
SO FRIEND,
X/Y=6/4=3/2
SO THE RATIO IS 3:2
(1)
Ranjitha said:
1 decade ago
@ chanti
its length=speed*time
so length of first train=27*x mts
and length of 2nd train= 17*y mts
its length=speed*time
so length of first train=27*x mts
and length of 2nd train= 17*y mts
FATHIMA said:
1 decade ago
The 1st train sec 27-23=4.
The 2nd train sec 23-17=6.
So both are divide by 2.
4/2=2, and 6/3=3.
So the ratio is 3:2 its right method?
The 2nd train sec 23-17=6.
So both are divide by 2.
4/2=2, and 6/3=3.
So the ratio is 3:2 its right method?
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