Question

A train travels at a certain average speed for a distance of 63 km and then travels at a
distance of 72 km at an average speed of 6 km/hr more than its original speed. If it
takes 3 hours to complete total journey, what is the original average speed ?
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Answer 1

$\huge\ \pmb{ \red{«\: คꈤ \mathfrak Sฬєя \: » }}$

Let the original speed of train is x km/h

Time taken to cover 63 km with speed x km/h,

$\bold{T_1 = \frac{distance}{time}} \\ \\ \bold{T_1 =\frac{63}{x} hours}$

Again, question said , speed of train now (x + 6) km/h

Time taken to cover 72km with speed (x + 6) km/h ,

$\bold{T_2= \frac{72}{(x + 6)} \: km/h }$

Acc to question

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: T_1+T_2 = 3 \: hours \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:⇒ \frac{63}{x} + \frac{72}{(x + 6)} = 3 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:⇒ \frac{21}{x} + \frac{24}{(x + 6)}= 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:⇒ (21x + 126 + 24x) = \frac{x}{(x + 6)} \\ \\\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ⇒45x + 126 = x² + 6x \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:⇒ x² - 39x - 126 = 0 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:⇒x² - 42x + 3x - 126 = 0 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:⇒ x(x - 42) + 3(x - 42) = 0 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:⇒(x + 3)(x - 42) = 0$

∴x = 42 and -3 , but x ≠ -3 ∵ speed doesn't negative

Hence, original speed of train = 42 km/h

$\: \: \: \: \: \: \: \: \: \: \huge \bold{@WildCat7083 }$