Question

An aeroplane starts late by 30 minutes to reach the destination which is 1500 km away. The pilot increases the speed by 250km/hour just to reah on time. Find the original speed of the aeroplane.

Answer 1

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Step-by-step explanation:

Let the Original speed of the aeroplane be x km/h.

Time taken to cover 1500 km with the usual speed of x km/h = 1500 / x hrs

Time taken to cover 1500 km with the increase speed of (x+100) km/h = 1500 / (x+100) hrs

ATQ

1500 / x = 1500 / (x+100) + ½

1500 / x - 1500 / (x+100) = ½

1500(x+100) -1500x / x(x+100)= ½

1500x + 1500 × 100 -1500x / x² +100x = ½

1500 × 100 / x² +100x = ½

2(1500 × 100 ) = x² + 100x

300000 = x² +100x

x² +100x - 300000 =0

x² - 500 x + 600 x - 300000= 0

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

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

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

x = -600 or x = 500

Speed cannot be negative, so x = 500

Hence the usual speed of the plane is 500 km/h.

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Answer 2

Solution :-

Let the original speed of train be x km/hr

New speed = (x + 250) 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 + 250) = 1/2

=> (1500x + 37500 - 1500x)/x(x + 250) = 1/2

=> 2(37500) = x(x + 250)

=> 75000 = x² + 250x

=> x² + 250x - 75000 = 0

=> x² + 1000x - 750x - 75000 = 0

=> x(x + 1000) - 750(x + 1000) = 0

=> (x - 750) (x + 1000) = 0

=> x = 750 or x = - 1000

∴ x ≠ - 1000 (Because speed can't be negative)

Hence,

Its usual speed = 750 km/hr