Question

A bus covers a distance of 90 km at a uniform speed had the speed been 15 km/hr more it would have taken 30 minutes less for the journey. find the original speed of the bus

Answer 1

$\sf\underline{\pink{\:\:\: AnswEr:-\:\:\:}}$

Original speed of Bus is 45 km/hr.

$\sf\underline{\red{\:\:\: Given:-\:\:\:}}$

• Increased speed is 15 km/hr (x + 15).
• Distance is 90 km.

$\bf{\underline{\underline \blue{Explanation:-}}}$

We know that,

$:\implies\small{\underline{\boxed{\sf{\pink{Speed = \dfrac{Distance}{Time}}}}}}$

$:\implies\sf t_{1}= \dfrac{90}{x} \: hours$

After when speed increased -

$:\implies\sf t_{2} = \dfrac{90}{(x + 15)}$

Now, $\sf t_{2} - t_{1} = 30$

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

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

Taking 90 as common -

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

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

$\rule{150}2$

By using the splitting the middle term method -

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

$:\implies\sf\red{x^2 + 15x - 2700 = 0}$

$:\implies\sf x^2 + 60x - 45x - 2700 = 0$

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

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

Now, our factors are (x + 60) & (x - 45).

Equaling them with 0.

$:\implies\sf x + 60 = 0 \implies\sf x = -60$

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

Distance can't be negative. So, neglecting the -ve value - 60 & taking +ve value 45.

Thus, original speed of Bus is 45 km/hr.

Answer 2

Answer:

Let the original speed of train be y km/h.

then, increased speed : y + 15 km/h

Wkt, Speed = Distance/time

usual time = 90/y hrs

and, New time = 90/y+15 hrs

also, 30 min = 30/60 hr = 1/2 hr

According to question :-

$\frac{90}{y} - \frac{90}{y + 15} = \frac{1}{2}$

taking out 90 as common :

$90( \frac{1}{y} - \frac{1}{y + 15}) = \frac{1}{2}$

Taking lcm,

$90(\frac{ y + 15 - y}{y(y + 5)}) \: = \frac{1}{2}$

$\frac{90 \times 15}{ {y}^{2} + 15y } = \frac{1}{2}$

${y}^{2} + 15y \: = 2700$

${y}^{2} + 15y - 2700$

By factorisation :

${y}^{2} + 60y - 45y - 2700 = 0$

$y(y + 60) - 45(y + 60) = 0$

$(y - 45) \: (y + 60)$

y = 45 or -60

Rejecting the negative value we get ;

Y = 45

Hence,

usual speed of train is 45 km/hr