Question

A plane left 30 minute late than its schedule time and in order to reach destination 1500 km away in time, it had to increase its speed by 100 km/hr from its usual speed. find the usual speed.

Answer 1

Answer:

Step-by-step explanation:

Given,

• A plane left 30 minute late than its schedule time and in order to reach destination 1500 km away in time.
• It had to increase its speed by 100 km/hr from its usual speed.

To Find,

• The usual speed.

Solution :-

Let the usual speed of the plane be x km/hr.

Increased speed = 100 km/hr

Time taken to cover 1500 km = 1500/x hr.

Time taken to cover 1500 km with increased speed = 1500/x + 100 hr According to the Question,

⇒ 1500/x - 1500(x + 100) = 30/60

⇒ x² + 100x - 300000 = 0

⇒ x² + 600x - 500x - 300000 = 0

⇒ (x + 600)(x - 500) = 0

⇒ x = -600 or 500 (Neglecting negative sign as speed cannot be negative) ⇒ x = 500

Hence, the usual speed of plane is 500 km/hr.

Answer 2

Question :

A plane left 30 minute late than its schedule time and in order to reach destination 1500 km away in time, it had to increase its speed by 100 km/hr from its usual speed. find the usual speed.

To Find :

The usual speed of the plane

Let us assume :

The usual speed be a

Answer :

Increased speed of the plane be (a + 100) km/h

Time taken to reach the destination at normal speed→ t = 1500/a hrs.

Time taken to reach the destination at increased speed → 1500/a + 100 hr

Hence, the difference between the increased speed and normal speed is :

$\sf \leadsto \frac{1500}{a} - \frac{1500}{a + 100} = \frac{1}{2}$

$\sf\leadsto \frac{1500}{a} = \frac{1}{2} + \frac{1500}{a + 100}$

Now to make denominator same on the left hand side of the equation

$\sf \leadsto \frac{1500}{a} = \frac{a + 100 +3000}{2a + 200}$

By cross multiplication we get

$\sf \leadsto1500(2a + 200) = a(a + 100 + 3000)$

$\sf \leadsto3000a - 30000 = {a}^{2} + 100a +3000a$

$\sf \leadsto\cancel{3000a} - \cancel{3000a }- 100a = {a}^{2} +30000$

$\sf \leadsto{a}^{2} + 100a - 30000 = 0$

Now we need to apply middle term factorisation to split 100a

$\sf \leadsto {a}^{2} + (600 - 500)a - 30000 = 0$

$\sf \leadsto {a}^{2} + 600a - 500a - 30000 = 0$

As per the situation we need to factorise here

$\sf \leadsto a(a + 600) - 500(a + 600) = 0$

Taking (a + 600) as common we get

$\sf \leadsto(a + 600)(a - 500) = 0$

Hence,

a + 600 = 0

• a = - 600

a - 500 = 0

• a = 500

We know that speed cannot be negative so a = 500 km/h

∴ The usual speed of the plane is 500 km/h

Regards

# BeBrainly