Question

PLEASE ANSWER IT. ...........

Answer 1

HEYA!!!!

Here is your answer :
______________________________

Let the orginal speed of train be x

distance travelled by train = 63 and 72 KMS

Time = Distance / Speed

If total journey taken in 3 hours then,

$\frac{63}{x} + \frac{72}{x + 6} = 3 \\ \\ \frac{63(x + 6) + 72x}{ {x}^{2} + 6x } = 3 \\ \\ \frac{63x + 378 + 72x}{ {x}^{2} + 6x } = 3 \\ \\ 63x + 378 + 72x = 3( {x}^{2} + 6x) \\ \\ 135x + 378 = 3 {x}^{2} + 18x \\ \\ 3 {x}^{2} + 18x - 135x - 378 = 0 \\ \\ 3 {x}^{2} - 117x - 378 = 0 \\ \\ 3( {x}^{2} - 39x - 126) = 0 \\ \\ {x}^{2} - 39x - 126 = 0 \\ \\ {x }^{2} - 42x + 3x - 126 = 0 \\ \\ x(x - 42) - 3(x - 42) = 0 \\ \\ x = 42 \: \: \: \: x = 3$


Therefore, the original speed is 42kmph..


HOPE THIS HELPS U...