Question

A faster train takes one hr less than the a slower train for a journey of 200km/hr . If the speed of slower train is 10km/hr less than that of the faster train . Find the speeds of the trains

Answer 1

$\boxed{\bold{\large{Answer:}}}$

$\text{Let\:the\:speed\:of\:the\:faster\:train\:be\:x Km/hr}$

$\text{So,the\:speed\:of\:the\:slower\:Train\:will\:be\:x\:-10\:Km/hr}$

$\text{According\:to\:the\:Question,}$

$\implies \frac{200}{x} = \frac{200}{x - 10} - 1 \\ \\ \implies \frac{200}{x} = 200 - \frac{x - 10}{x - 10} \\ \\ \implies200(x - 10) = 200x - x(x - 10) \\ \\ \implies200x - 2000 = 200x - {x}^{2} + 10x \\ \\ \implies {x}^{2} - 10x - 2000 = 0 \\ \\ \implies {x}^{2} - 50x + 40x - 2000 = 0 \\ \\ \implies(x - 50)(x + 40) = 0$

$\text{By\:zero\:product\:Rule,}$

$x = 50 \: and \: x = - 40$

$\text{Since\:speed\:can't\:be\:negative,}$

$\text{Speed\:of\:faster\:train\:=\:50Km/hr}$

$\text{Speed\:of\:The\:slower\:train\:will\:be\:=\:40Km/hr}$

$\boxed{\bold{\large{Hope\:it\:helps\:you!}}}$

Answer 2

let speed of fast train be x km/hr
speed of slower train = x - 10 km/hr

A/Q  fast train takes 1 hr less than slower one in covering 200 km

⇒ 200/ x = 200/x-10    -  1
⇒ 200/x = 200 - (x-10) / x-10
⇒ 200(x-10) = 200x - x (x-10)
⇒200x-2000 = 200x - x² + 10x
⇒x² - 10x - 2000 = 0
⇒ x² -50x + 40x -2000 = 0  ⇒ (x-50)(x+40) = 0
x = 50  and x= - 40

since x can't be negative , so x= speed of faster train = 50 km/hr
speed of slower train = 40 km/hr