### Question 25

SSC-CGL 2020 Mar 3 Shift 1

Speed of a train A is 20% more than that of train B. After meeting, they take x and 2-1/2 hours respectively to reach their destinations. What is x?

### Solution

Remember Shortcut: If two trains travelling at v1 and v2 meet, and after that take t1 and t2 to reach their destinations, then t1/t2 = (v2/v1)²Given $\displaystyle \frac{v_A}{v_B} = \frac{120}{100} = 1.2$

Times taken after that: x and 2.5

$\displaystyle \therefore \frac{x}{2.5} = \bigg(\frac{v_B}{v_A}\bigg)^2$

$\displaystyle \therefore \frac{x}{2.5} = \bigg(\frac{1}{1.2}\bigg)^2$

$\displaystyle \implies x = 2.5 \times \frac{1}{1.44} $

$\displaystyle = \frac{125}{72}$ hours Ans!

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