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 4 of 24.
Harish said:
1 decade ago
I have solved it in yet simple(crude) way.
Don't worry about their crossing each other in 23 seconds or whatever.
On the face of it, one can make out that he has asked the simplified ration for 27:17.
Simply take them to next higher level so that both of them are divisible by common highest divisor.
i.e. 30/20 = 3/2.
(but recommended not to follow it, as it doesn't apply to every problem?).
Don't worry about their crossing each other in 23 seconds or whatever.
On the face of it, one can make out that he has asked the simplified ration for 27:17.
Simply take them to next higher level so that both of them are divisible by common highest divisor.
i.e. 30/20 = 3/2.
(but recommended not to follow it, as it doesn't apply to every problem?).
Prem said:
2 years ago
Let's assume the distance is constant.
1st train has 27 sec time and X speed
2nd train has 17 sec time and Y speed.
Know that, two trains cross a platform in 23 sec.
The relative seed in opposite directions is (s1+s2).
s1--------x.
s2----------y.
Distance = speed * time.
total distance = Distance 1 + distance 2.
D = d1 + d2
(x+y) 23 = [x * 27] + [y * 17]
23x + 23y = x27 + y17
x/y = 3/2.
1st train has 27 sec time and X speed
2nd train has 17 sec time and Y speed.
Know that, two trains cross a platform in 23 sec.
The relative seed in opposite directions is (s1+s2).
s1--------x.
s2----------y.
Distance = speed * time.
total distance = Distance 1 + distance 2.
D = d1 + d2
(x+y) 23 = [x * 27] + [y * 17]
23x + 23y = x27 + y17
x/y = 3/2.
(45)
Jason valentine daniel said:
1 year ago
Let the speed of the two trains be x m/sec and y m/sec.
Then, the length of the first train = 27x metres,
And length of the second train = 17y metres,
Time taken by the train to cross each other is 23 sec.
Time taken by the train to cross each other = Total Distance of both trains/ Total of Both trains.
So, 23 = (27x + 17y) / (x + y).
23x + 23y = 27x + 17y.
6y = 4x.
x/y = 6/4 => 3:2.
Then, the length of the first train = 27x metres,
And length of the second train = 17y metres,
Time taken by the train to cross each other is 23 sec.
Time taken by the train to cross each other = Total Distance of both trains/ Total of Both trains.
So, 23 = (27x + 17y) / (x + y).
23x + 23y = 27x + 17y.
6y = 4x.
x/y = 6/4 => 3:2.
(56)
Nishitha said:
4 years ago
Given data:
t1=27 sec t2=17 sec
S=d/t,d=s * t.
d is consider as l,
l1=s1 * t1 =17 s1,l2 = s2* t2 =27 s2.
In question ,it clearly move in a opposite direction,so we have to use this formula->(l1+l2) /s1+s2.
Total time = (17s1+27s2)/s1 + s2,
23(s1+s2)= 17s1+27s2.
Rearrange it,
23s1-17s1=27s2-23s2,
6s1 = 4s2,
s1/s2 = 6/4=3/2.
So,the answer is 3:2
t1=27 sec t2=17 sec
S=d/t,d=s * t.
d is consider as l,
l1=s1 * t1 =17 s1,l2 = s2* t2 =27 s2.
In question ,it clearly move in a opposite direction,so we have to use this formula->(l1+l2) /s1+s2.
Total time = (17s1+27s2)/s1 + s2,
23(s1+s2)= 17s1+27s2.
Rearrange it,
23s1-17s1=27s2-23s2,
6s1 = 4s2,
s1/s2 = 6/4=3/2.
So,the answer is 3:2
(8)
Sikandar said:
10 years ago
Two train of length a meter and b meter are moving in opposite direction at you m/s and we m/s then time taken by the faster train to cross the slower train = a+b/u+v.
Or in same direction = a+b/u-v.
In above problem train are in opposite direction.
So a = 27x & b = 17y (x = speed of fastest train y = speed of slow train).
= 27x+17y/x+y = 23.
4x = 6y.
x:y = 3:2.
Or in same direction = a+b/u-v.
In above problem train are in opposite direction.
So a = 27x & b = 17y (x = speed of fastest train y = speed of slow train).
= 27x+17y/x+y = 23.
4x = 6y.
x:y = 3:2.
Arun said:
1 decade ago
Based on the below formula -- >
If two trains of length a(i.e. 27x) metres and b(i.e. 17y) metres are moving in opposite directions at x m/s and y m/s, then:
The time taken by the trains to cross each other (i.e.23 sec) =
(27x + 17y)
```````` secs
(x + y)
Therefore, the above solution is provided based on this derivation. Its given in the formulas section.
If two trains of length a(i.e. 27x) metres and b(i.e. 17y) metres are moving in opposite directions at x m/s and y m/s, then:
The time taken by the trains to cross each other (i.e.23 sec) =
(27x + 17y)
```````` secs
(x + y)
Therefore, the above solution is provided based on this derivation. Its given in the formulas section.
Amit Jana said:
1 decade ago
#Vivek kumar.
Here the questions is : two train of length a and b are run in opposite directions at u m/sec. and v m/sec., then the time taken by the trains to cross each other = (a+b)/(u+v).
And if two train of length a and b are run in same directions at u m/sec. and v m/sec., then the time taken by the faster trains to cross the slower train = (a+b)/(u-v)
Here the questions is : two train of length a and b are run in opposite directions at u m/sec. and v m/sec., then the time taken by the trains to cross each other = (a+b)/(u+v).
And if two train of length a and b are run in same directions at u m/sec. and v m/sec., then the time taken by the faster trains to cross the slower train = (a+b)/(u-v)
Prasanna Kumar said:
6 years ago
If we think let x and y be the length of the trains then, x/27,by/17 will be the speed of the trains respectively and the equation becomes x+y/23b=bx/27+y/17.
Solving we get x/y=81/34.
x+y = total distance.
23 = total time.
Speed = x/27+y/27.
Speed = distance/time.
Therefore x+y/23 = x/27+y/17 is the equation,
Can you tell me why this equation is wrong?
Solving we get x/y=81/34.
x+y = total distance.
23 = total time.
Speed = x/27+y/27.
Speed = distance/time.
Therefore x+y/23 = x/27+y/17 is the equation,
Can you tell me why this equation is wrong?
Jyotirmoy said:
1 decade ago
Let the speed of the two trains be a=x m/sec and b=y m/sec
Length of a=27x and b=17y
So (27x+17y)=23(x+y).
So 4x=6y
So x/y=3/2
[In First sight I thought as the 1st train is taking more time to cross the man , so it will be the slower train, but as the lenght of the 1st train longer so it has taken longer time althought it is the faster train]
Length of a=27x and b=17y
So (27x+17y)=23(x+y).
So 4x=6y
So x/y=3/2
[In First sight I thought as the 1st train is taking more time to cross the man , so it will be the slower train, but as the lenght of the 1st train longer so it has taken longer time althought it is the faster train]
AraShan said:
1 decade ago
By the formula time=distance/speed.
but, we have two trains with two diferent distances and speeds.
then we use the formula,
total time = [(distance1)+(Distance2)]/[speed1+speed2]
for finding distance= speed*time.[so,distance1= 27*x....like wise.
finally we get, 27x+17y/(27+17)= 23 (given totaltime.)
at last the answer is... x:y (3:2)
but, we have two trains with two diferent distances and speeds.
then we use the formula,
total time = [(distance1)+(Distance2)]/[speed1+speed2]
for finding distance= speed*time.[so,distance1= 27*x....like wise.
finally we get, 27x+17y/(27+17)= 23 (given totaltime.)
at last the answer is... x:y (3:2)
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