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 10 of 24.
DURAI said:
5 years ago
Short method is;
x person 27.
y person 17 add both distances and divide the total distance and multiple the same for both persons.
27+17/23 *2.
37/23 = 1.6 *2.
1.6*2.
Ans : 3.2.
x person 27.
y person 17 add both distances and divide the total distance and multiple the same for both persons.
27+17/23 *2.
37/23 = 1.6 *2.
1.6*2.
Ans : 3.2.
(2)
Sourabh kumar said:
1 year 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.
(29)
Vipul said:
1 decade ago
First we are doing KM Hr to MS.
Then Formula is a*5/18.
Then,
45*5/18=75/6 MS.
Distance = Speed*Time.
= 75/6*30.
= 375.
Distance = 375-130.
= 245.
Then Formula is a*5/18.
Then,
45*5/18=75/6 MS.
Distance = Speed*Time.
= 75/6*30.
= 375.
Distance = 375-130.
= 245.
Gokulakisan S said:
10 years ago
You can use this simple method:
Time to cross men = tm.
Time to cross each other = te.
So 27(tm)+23(te) = 60.
17(tm)+23(te) = 40.
Then 60/40 = 3/2.
Therefore ratio is 3:2.
Time to cross men = tm.
Time to cross each other = te.
So 27(tm)+23(te) = 60.
17(tm)+23(te) = 40.
Then 60/40 = 3/2.
Therefore ratio is 3:2.
NICK said:
1 decade ago
Can any one tell me how to calculate mod(23-17) : mod(23-27)
I forgot how to calculate mod and if we are calculating same mod(23-17) : mod(23-27) then how it comes to 6:4????
I forgot how to calculate mod and if we are calculating same mod(23-17) : mod(23-27) then how it comes to 6:4????
Adesh Katiya said:
1 year 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)
Carlo said:
1 decade ago
Let Train A distance = 27x Train A speed = x.
Train B distance = 17y Train B speed = y.
Time = distance/speed.
23 = 27x + 17y / x + y.
x/y = 3/2 ----> Answer.
Train B distance = 17y Train B speed = y.
Time = distance/speed.
23 = 27x + 17y / x + y.
x/y = 3/2 ----> Answer.
Prabhakaran said:
1 decade ago
Im not clear with above step can anyone give more explanation about this problem. I have doubt in (27x+17y) / (x+y) =23 what this statement mean by. Anyone help me please.
RAHUL tulskar said:
8 years ago
Look at this simplest ans.
Take both crossing time as an x y.
27as x 17 as y.
Then,
27x+17y/x+y=23.
27x+17y=23x+23y.
27x-23x=17y-23y.
4x =6y.
Or
XY=4/6or 2/3 get it.
Take both crossing time as an x y.
27as x 17 as y.
Then,
27x+17y/x+y=23.
27x+17y=23x+23y.
27x-23x=17y-23y.
4x =6y.
Or
XY=4/6or 2/3 get it.
Hemanth Jetti 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.
27x - 23x = 23y - 17y.
4x = 6y.
x = 6y/4.
x/y = 3/2.
= 3:2.
2nd train be y = 17.
Two trains crossed each other=23(x+y).
27x + 17y = 23x + 23y.
27x - 23x = 23y - 17y.
4x = 6y.
x = 6y/4.
x/y = 3/2.
= 3:2.
(157)
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