Question

Solve the ques in the attachment


Class 10

Quadratic Equation

Answer 1

$hope \: it \: help \: you \:$

Answer 2

Let the speed of the person while going be x km/hr.

Given that return speed is 10 km/hr more then the speed while going.

So, the return speed of the person = (x + 10) km/hr.

(i)

Total distance covered = 150 km.

We know that Time = Distance/Speed

⇒ Time taken for going = 150/x

⇒ Time taken for returning = 150/(x + 10).

Hence,

⇒ 150/x - 150/x + 10 = 2 1/2

⇒ 150/x - 150/x + 10 = 5/2

⇒ 5[30/x - 30/x + 10] = 5[1/2]

⇒ (30/x) - (30/x + 10) = 1/2

⇒ 30 * 2(x + 10) - 30 * 2x = x(x + 10)

⇒ 60x + 600 - 60x = x^2 + 10x

⇒ x^2 + 10x - 600 = 0

⇒ x^2 - 20x + 30x - 600 = 0

⇒ x(x - 20) + 30(x - 20) = 0

⇒ x = 20,-30{Speed cannot be negative}

⇒ x = 20.

Therefore,

⇒ Speed of the person while going = 20 km/hr.

⇒ Speed of the person while returning = x + 10 = 30 km/hr.

Hope it helps!