Question

Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then the total time taken, in minutes, by train B to travel from station Y to station X is

• A. 15
• B. 12
• C. 6
• D. 10

Answer 1

The Correct Option is A

Given:
Train A reaches station Y in 10 minutes.
Train B takes 9 minutes to reach station X after meeting train A.
Let's assume that the meeting point of the two trains between stations X and Y is M.
Time taken by train A to travel from X to M = t minutes
Time taken by train B to travel from Y to M = t minutes
Time taken by train A to travel from M to Y = (10 - t) minutes
Time taken by train B to travel from M to X = 9 minutes
Since the two trains start simultaneously and travel equal distances, we can use the ratio of time taken by them:
\(\frac{t}{9}=\frac{(10 - t)} {t}\)
Now, let's solve for 't':
\(t^2 = 9(10 - t)\)
\(t^2 = 90 - 9t\)
\(t^2 + 9t - 90 = 0\)
(t + 15)(t - 6) = 0
t = 6 (since time cannot be negative)
So, the time taken by train B to travel from station Y to station X is 6 + 9 = 15 minutes.
Hence, the correct answer is 15 minutes.

Answer 2

Let the speed of train \(A = a\)
and the speed of train \(B = b\)
Then,
\(\frac xa=\frac {D-x}{b}\)
Given,
\(\frac Da=10\) and \(\frac xb=9\)

\(\frac {x}{\frac {D}{10}}=\frac{D-x}{\frac x9}\)

\(\frac {10x}{D}= \frac {9D-9x}{x}\)

\(10x^2= 9D^2-9Dx\)
\(10x^2+9Dx-9D^2=0\)
On solving,
\(x=\frac {3D}{5}\)
\(\frac xb=9\)

\(\frac {3D}{b \times 5}=9\)

\(\frac Db=15\)

Total time taken by train B from station Y to station X = 15 min.

So, the correct option is (A): 15