Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 25)
25.
A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from opposite direction in 6 seconds. The speed of the second train is:
Answer: Option
Explanation:
Let the speed of the second train be x km/hr.
| Relative speed | = (x + 50) km/hr | |||||||
|
||||||||
|
![]](/_files/images/aptitude/1-sym-cbracket-h1.gif)
Distance covered = (108 + 112) = 220 m.
 |
220 | = 6 | ||
|
250 + 5x = 660
x = 82 km/hr.
Discussion:
77 comments Page 4 of 8.
Rakshikasaravanan said:
5 years ago
First train speed 50sec.
Second train speed?
S = l1 + l2/t1 + t2.
Relative speed = x + 50s.
X + 50s = 108+112/6+t2.
X + 50s =792/6.
6x + 300=792.
6x = 792-300.
6x = 492.
X = 492/6.
X = 82.
Second train speed?
S = l1 + l2/t1 + t2.
Relative speed = x + 50s.
X + 50s = 108+112/6+t2.
X + 50s =792/6.
6x + 300=792.
6x = 792-300.
6x = 492.
X = 492/6.
X = 82.
Nileshshelar said:
7 years ago
(Total distance covered) ÷ (net speed) = time.
Mtrs/(mtr/sec) = sec.
220/(50+x) * 5/18=6,
So x=82.
But the unit is now mtr/sec because all we have converted into mtrs and seconds.
Mtrs/(mtr/sec) = sec.
220/(50+x) * 5/18=6,
So x=82.
But the unit is now mtr/sec because all we have converted into mtrs and seconds.
Prathap dha said:
1 decade ago
Formula : T=(x+y)/(u+v).
Where,
x,y -- Distance.
u,v -- Speed.
6 = (108+112)/((50*5/8)+V).
6v+250/3 = 220.
Therefore, v = 205/9 m/sec.
v = 205/9*(18/5) = 82km/hr.
Where,
x,y -- Distance.
u,v -- Speed.
6 = (108+112)/((50*5/8)+V).
6v+250/3 = 220.
Therefore, v = 205/9 m/sec.
v = 205/9*(18/5) = 82km/hr.
Thiyam surjakumar singh said:
1 decade ago
In the equation all the units meter, second get cancelled so the remaining is x only, whose units is km/hr as we have choosen at beginning of the problem. Hence the result is in km/hr.
Ram said:
1 decade ago
Easiest method is,
220*18 = 6(250+5x).
Multiply by 6(250+5x) you get 1500+30x.
3960 = 1500+30x.
Subtract the value of 3960-1500 = 2460.
Then divide by this form,
2460/30 = 82 kmph
220*18 = 6(250+5x).
Multiply by 6(250+5x) you get 1500+30x.
3960 = 1500+30x.
Subtract the value of 3960-1500 = 2460.
Then divide by this form,
2460/30 = 82 kmph
Rishav kumar said:
10 years ago
Yes answer is correct but checking through reverse process.
Let x = 82 km/h and we have find time to cross each other. Then then the answer will come 6 sec. So it is correct.
Let x = 82 km/h and we have find time to cross each other. Then then the answer will come 6 sec. So it is correct.
Ashok said:
1 decade ago
220/250+5x/18 = 6.
step 1: 18*220 = 3960.
step 2: 250+5x * 6 = 1500+30x
step 3: 3960 = 1500+30x.
step 4: 3960-1500 = 30x
step 5: 2460 = 30x.
step 6: x=2460/30.
step 7: x=82.
step 1: 18*220 = 3960.
step 2: 250+5x * 6 = 1500+30x
step 3: 3960 = 1500+30x.
step 4: 3960-1500 = 30x
step 5: 2460 = 30x.
step 6: x=2460/30.
step 7: x=82.
Roji said:
1 decade ago
Why should we convert to m/sec as in above problem (x+50)*5/18.
But in answer the required ans in km/hr again why we are taking 5/18.
Can anyone explain me please?
But in answer the required ans in km/hr again why we are taking 5/18.
Can anyone explain me please?
Sathish said:
1 decade ago
Hai I have another option.
D = 220.
S = 50 kmph.
T = 6.
Other speed = x.
So speed = (x+50)*5/18.
T = D/S. Find T subtract 50 that's answer = 82 kmph.
D = 220.
S = 50 kmph.
T = 6.
Other speed = x.
So speed = (x+50)*5/18.
T = D/S. Find T subtract 50 that's answer = 82 kmph.
Arunkumar said:
1 decade ago
108+112=220.
220/6 is the relative speed per sec.
(220/6)*(18/5)=132 is the relative speed per hour.
Speed of the 2nd train is 132-50 = 82kmph.
220/6 is the relative speed per sec.
(220/6)*(18/5)=132 is the relative speed per hour.
Speed of the 2nd train is 132-50 = 82kmph.
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