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(C) Let the speed of train $A$ be $v_A = 45\, 	ext{km/h}$ and the speed of train $B$ be $v_B$.Let $t_1$ be the time taken by train $A$ to reach $Y$ a

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(C) Let the speed of train $A$ be $v_A = 45\, 	ext{km/h}$ and the speed of train $B$ be $v_B$.Let $t_1$ be the time taken by train $A$ to reach $Y$ after passing $B$,and $t_2$ be the time taken by train $B$ to reach $X$ after passing $A$.Given: $t_1 = 4\, 	ext{hours } 48\, 	ext{minutes} = 4 + \frac{48}{60} = 4 + \frac{4}{5} = \frac{24}{5}\, 	ext{hours}$.Given: $t_2 = 3\, 	ext{hours } 20\, 	ext{minutes} = 3 + \frac{20}{60} = 3 + \frac{1}{3} = \frac{10}{3}\, 	ext{hours}$.The formula for the ratio of speeds when two objects meet and then travel to their respective destinations is $\frac{v_A}{v_B} = \sqrt{\frac{t_2}{t_1}}$.Substituting the values: $\frac{45}{v_B} = \sqrt{\frac{10/3}{24/5}} = \sqrt{\frac{10}{3} 	imes \frac{5}{24}} = \sqrt{\frac{50}{72}} = \sqrt{\frac{25}{36}} = \frac{5}{6}$.Therefore,$5 	imes v_B = 45 	imes 6$.$v_B = \frac{45 	imes 6}{5} = 9 	imes 6 = 54\, 	ext{km/h}$.

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