Question

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

Answer 1

Heya frnd....


Here is your answer…




⭐QUADRATIC EQUATIONS ⭐

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

Time taken to cover 1500 km = 1500/x h

New speed = (x + 100) km/h

Time taken to cover 1500 km with new speed = 1500 / (x + 100) h

ATQ,

1500/x - 1500 / (x + 100) = 30/60

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

=> 1500 {x + 100 -x/x² + 100x} = 1/2

=> 1500 × 100 ×2 = x² + 100x

=> x² + 100x = 300000

=> x² + 100x - 300000 = 0

=> x² + 600x - 500x - 300000 = 0

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

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

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

=> x = 500 or x = -600

=> x = 500 (Neglecting -600)

Hence, the original speed of the plane is 500km/h.

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Hope this would help you.

Answer 2

Solution :-

Let the original speed of train be x km/hr

New speed = (x + 100) km/hr

We know that,

Time = Distance / Speed

Given : A plane left 30 minutes or 1/2 hours later than the scheduled time.

According to the question,

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

=> (1500x + 15000 - 1500x)/x(x + 100) = 1/2

=> 2(15000) = x(x + 100)

=> 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

∴ x ≠ - 600 (Because speed can't be negative)

Hence,

Its usual speed = 500 km/hr