Question

A train covers a distance of 90 km at a uniform speed .Had the speed bee 15 km hour more, it would have taken 30 minutes less for the journey. Find the original speed of the train

Answer 1

AnswEr:

Let us Consider that the speed of the train be x km/hr.

Increased Speed = 15 km/hr.

Distance Covered = 90 km.

$\bigstar\large{\underline{\boxed{\sf{\pink{Time\:=\:\dfrac{Distance}{Speed}}}}}}$

$:\implies\sf\: \dfrac{90}{x}$ -------Eq(1)

When Speed Increased

$:\implies\sf\: \dfrac{90}{x\:+\: 15}$ -------Eq(2)

$\star\bold{\underline{\sf{According\: to\; Question\;Now,}}}$

$:\implies\sf\: \dfrac{90}{x} \:-\: \dfrac{90}{x\:+\;15} = \dfrac{1}{2}$

$:\implies\sf\: 90 = \dfrac{x\:+\;15\:-\;5}{x(x\:+\;15)} = \dfrac{1}{2}$

$:\implies\sf\: \dfrac{90 \times\;15}{x^2\:+\:15x} \:=\; \dfrac{1}{2}$

$:\implies\sf\: x^2 + 15x = 2700$

Now, Solving this Equation

$:\implies\sf\: x^2 + 15x = 2700=0$

$:\implies\sf\: x^2 + 16x - 14x - 2700 = 0$

$:\implies\sf\: x(x\:+\;16) -45(x\:+\: 60)=0$

$:\implies\sf\: (x\:+\:16) \:-\:(x\:-\:45) =0$

$:\implies\sf\: x + 16 = 0$

$:\implies\large{\underline{\boxed{\sf{\red{x\:=\: -16}}}}}$

$:\implies\sf\: x - 45 = 0$

$:\implies\large{\underline{\boxed{\sf{\red{x\:=\: 45}}}}}$

Speed Can't be Negative.

Therefore, Speed of the Train is 45 km/hr.

Answer 2

AnswEr:

Let us Consider that the speed of the train be x km/hr.

Increased Speed = 15 km/hr.

Distance Covered = 90 km.

time = distance / speed

:⟹ 90/x ------- equation 1

When Speed Increased

:⟹ 90/x+15 -------Eq(2)

AccordingtoQuestionNow,

:⟹ 90/x - 90/x+15 = 1/2

:⟹ 90 = x+15-5 / x(x+15) = 1/2

:⟹ 90*15 / x square + 15x = 1/2

:⟹ x square + 15x = 2700

Now, Solving this Equation

:⟹ x square + 15x = 2700 =0

:⟹ x square + 16x - 14x -2700 = 0

:⟹ x(x+15) - 45 (x+60) = 0

:⟹ (x +15) - (x-45) = 0

:⟹ x = (-16)

:⟹ x - 45 = 0

:⟹ x = 45

speed cannot be negative...

so the speed is 45 km/h...