2 trains each of length 120 metres move in the same direction

Formula to find time to cross each other:
When 2 trains of lengths \(a\) \(meters\) and \(b\) \(meters\) are moving in same direction with speeds \(x\) \(m/s\) and \(y\) \(m/s\) respectively \((x > y)\), then the time taken to cross each other is \(= \frac{a + b}{x - y}\)

Given; 2 trains each of length \(120\) \(meters\).
Let the speed of faster train be \(x\) \(m/s\) and slower train be \(y\) \(m/s\).

Given, faster train completely overtakes the slower one in \(15\) \(seconds\). ie; time taken to cross each other = \(15 s\)
\(15 = \frac{120 + 120}{x - y}\) \(==>\) \(\frac{240}{x - y}\)
\(x - y = 16\)
\(y = x - 16\) -------------- (i)

If the slower train were to move at half its speed, faster train would overtake it in \(10\) \(seconds\).
Speed of slower train would become \(y/2\).
\(10 = \frac{240}{x - {y/2}}==>\) \(x - \frac{y}{2} = 24\) \(==>\) \(\frac{2x - y}{2}= 24\) \(==>\) \(2x - y = 48\) -------------------- (ii)

Substituting value of \(y\) from (i) in above equation (ii);
\(2x - (x - 16) = 48\) ==> \(2x - x + 16 = 48\)
\(x = 48 - 16 = 32\)

Therefore speed of faster train is \(32 m/s\)
Speed of slower train, \(y = x - 16\) \(==> 32 - 16 = 16\).
Speeds of trains are \(32\) \(m/s\) and \(16\) \(m/s\). Answer B...

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