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:
236 comments Page 2 of 24.
Sourabh kumar said:
2 years ago
Let 1st train be x = 27.
2nd train be y = 17.
Two trains crossed each other=23(x+y).
27x + 17y = 23x + 23y.
How do you change the sign of x and y?
Anyone please explain to me.
2nd train be y = 17.
Two trains crossed each other=23(x+y).
27x + 17y = 23x + 23y.
How do you change the sign of x and y?
Anyone please explain to me.
(30)
Dhinesh said:
2 years ago
Well, I have something different, 1st train is 27 sec so take this time as a total as 100% minus the 2nd train which is 17 sec so 62. 96 % is the percent of 1st train slower than the second train or vice versa.
So, 62.
96 is a 3 ratio to 100 so the remaining 38.04 is 2.
So, it's 3:2 am I right?
So, 62.
96 is a 3 ratio to 100 so the remaining 38.04 is 2.
So, it's 3:2 am I right?
(29)
Thavapriya A said:
7 months ago
How to solve the following step?
27x+17y =23x+23y.
Could anyone tell me please?
27x+17y =23x+23y.
Could anyone tell me please?
(21)
Ishwarya said:
3 years ago
But why we can't use the formula of B:A?
Anyone, please explain me.
Anyone, please explain me.
(20)
Student said:
1 month ago
A train takes 27 sec to reach man and 23 sec to cross train B, so the difference is 4 sec
B train takes 17 sec to reach the man and 23 sec to cross train A, so the difference is 6 sec.
Speed is inversely proportional to time, so the ratio of speed is 6:4.
So 3:2 is the answer.
B train takes 17 sec to reach the man and 23 sec to cross train A, so the difference is 6 sec.
Speed is inversely proportional to time, so the ratio of speed is 6:4.
So 3:2 is the answer.
(19)
Adesh Katiya said:
2 years ago
u=27s
v=17s
Cross = 23s.
l1 = 27x.
l2 = 17y.
23 = l1 + l2/x + y
23 = 27x + 17y/x + y.
23x + 23y = 27x + 17y.
4x - 6y = 0.
4x = 6y.
x/y = 6/4.
x/y = 3/2.
Hence, option (b).
v=17s
Cross = 23s.
l1 = 27x.
l2 = 17y.
23 = l1 + l2/x + y
23 = 27x + 17y/x + y.
23x + 23y = 27x + 17y.
4x - 6y = 0.
4x = 6y.
x/y = 6/4.
x/y = 3/2.
Hence, option (b).
(14)
Ebe said:
4 years ago
I have a small doubt regarding the formula.
In the question, they have said to find the ratio of speed. Not the ratio of time.
So aren't we supposed to modify the formula accordingly? Please someone explain.
In the question, they have said to find the ratio of speed. Not the ratio of time.
So aren't we supposed to modify the formula accordingly? Please someone explain.
(11)
Dripta Majumdar said:
6 months ago
Why is the length of the platform not taken into consideration? Anyone, please explain to me.
(9)
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)
Anome said:
2 months ago
Very useful. Thanks all.
(7)
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