Aptitude - Problems on Trains - Discussion
Discussion Forum : Problems on Trains - General Questions (Q.No. 13)
13.
Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is:
Answer: Option
Explanation:
Let the speed of the slower train be x m/sec.
Then, speed of the faster train = 2x m/sec.
Relative speed = (x + 2x) m/sec = 3x m/sec.
 |
(100 + 100) | = 3x |
| 8 |
24x = 200
| x = |
25 | . |
| 3 |
| So, speed of the faster train = | 50 | m/sec |
| 3 |
| = |  |
50 | x | 18 |  km/hr |
| 3 | 5 |
= 60 km/hr.
Discussion:
74 comments Page 8 of 8.
Vaasudevarao said:
1 decade ago
Speed= d/t (100+100)=200*18/5=90km/h two train speed x+2x =3x 3x=90; x=90/3 x=30 so faster train speed is 2x =60
RAHAMAN said:
1 decade ago
SOLVING WITHOUT x
{(100+100)/8}*{(18/5)}=90K/H
NOW,
SPEED=2:1
SO,
90*2/3=60K/H
{(100+100)/8}*{(18/5)}=90K/H
NOW,
SPEED=2:1
SO,
90*2/3=60K/H
Sravan said:
1 decade ago
Actually,
if speed of the slower train be x m/sec.
Then, speed of the faster train y = 2x m/sec=> x=y/2
Relative speed =(x+y)= (y/2+ y) m/sec = 3y/2 m/sec.
(100 + 100)/8 = 3y/2
3y=50
y=50/3 m/s
y=(50/3)*(18/5)kmph
speed of the faster train = 60 kmph
if speed of the slower train be x m/sec.
Then, speed of the faster train y = 2x m/sec=> x=y/2
Relative speed =(x+y)= (y/2+ y) m/sec = 3y/2 m/sec.
(100 + 100)/8 = 3y/2
3y=50
y=50/3 m/s
y=(50/3)*(18/5)kmph
speed of the faster train = 60 kmph
Shiv said:
1 decade ago
Give me short cut.
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