Question

A plane taken 30 min lease for journey of 1500 km if speed is increased by 100km/h from the usual speed . find the it speed​

Answer 1

Answer:

Step-by-step explanation:

Let the original time be X and original speed be y

Distance = 1500 km

Distance = Speed x Time

1500 km = xy                                 x = 1500 / y

If time taken is 30 min less and speed is 100 km/hr more

time = (x - 1/2) hrs              speed = (y + 100) km/hr

Distance = Speed x Time

1500 km  = (x - 1/2)(y + 100)

1500 = xy + 100x - y/2  - 50

1500 = 1500 + 100x - y/2 - 50

100x - y/2 - 50 = 0

200x - y - 100 = 0           (Multiplying the equation by 2)

200 * 1500/ y - y - 100 = 0

300000 - y^2 - 100 y = 0

y^2 - 100y - 300000 = 0

y^2 - 600y + 500 y - 300000 = 0

y(y - 600) + 500 ( y - 600 ) = 0

(y - 600)(y + 500) = 0

Therefore , y = 600 or y = - 500

Since speed cannot be negative , y = 600 km/hr

Answer 2

Given:-

• A plane taken 30 min lease for journey of 1500 km.
• If speed is increased by 100km/h from the usual speed.

To find:-

• find the it speed..?

Solutions:-

• Let the actual speed of plane be x km/hr.

The journey of 1500km, the time taken.

=> 1500/x = t1

Speed is increased by 100km/hr from the usual speed

=> 1500/x + 100 = t2

According to the questions;-

=> t1 - t2 = 30/60

=> 1500/x - 1500/x + 100 = 1/2

=> 1500(x + 10) - 1500x/x(x + 100) = 1/2

=> 1500x + 15000 - 1500x/x² + 100x = 1/2

=> 15000 × 2 = x² + 100x

=> 30000 = x² + 100x

=> x² + 100x - 30000 = 0

=> x² + 600x - 500x - 30000 = 0

=> x(x + 600) - 500(x +600) = 0

=> (x - 500) (x + 600) = 0

=> x = 500 or x = -600 (can't be in negative)

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