Question

A and B are two railway stations 90 km apart. A train leaves A at 9:00 am, heading towards B at a speed of 40 km/hr. Another train leaves B at 10:30 am, heading towards A at a speed of 20 km/hr. The trains meet each other at

• A. 11 : 45 am
• B. 10 : 45 am
• C. 11 : 20 am
• D. 11 : 00 am

Answer 1

The Correct Option is D

The correct answer is (D): \(11 : 00 am\)

The distance travelled by \(A\) before \(B\) starts his journey = \(40×1.5 = 60\)

The time taken by them to meet each other = \(\frac{90-60}{40+20} = \frac{30}{60} = 0.5\) hours

Required answer = \(10:30 \;a.m. + 30 \;min = 11:00 \;a.m.\)

Answer 2

The distance covered by \(A\) from \(9:00 \ AM\) to \(10:30\ AM\) is,
\(=\frac 32 \times 40\)
\(= 60\ km\)
Now, they are \(30\ km\) apart and let the time taken to meet be \(t\).
Distance covered by \(A\) in time \(t\) \(+\) Distance travelled by \(B\) in time \(t = 30\)
\(40t+20t=30\)
\(⇒60t= 30\)
\(⇒t= \frac {30}{60}\)

\(⇒t= \frac {1}{2}\) hour

\(⇒t = 30\ min\)
Therefore, they meet at \(11:00\ AM\).

So, the correct option is (D): \(11:00\ AM\)