Math, asked by ChristyCR7, 10 months ago

pls solve this question plss​

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Answered by anurag2305
1

Answer:

Answer is 750 kmph.

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Since, Speed = distance/ time.

s= d/t

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Now assume that, initially speed was 's' kmph and the time taken to cover a distance of d=1500 kms was 't' hours.

So, s= 1500/t. --------(i)

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Now, we know that, the new speed was 's+ 250' kmph and the time taken to cover the distance of d=1500 kms was 't- 0.5' hours, i.e. half an hour less time.

So, s+ 250= 1500/ (t- 0.5). ----------(ii)

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Now, we have 2 equations. On solving them, we get,

initial time= t= 2 hours. Initial speed= s= distance/time= 1500/2= 750kmph.

Answered by Siddharta7
0

Answer:

750 km/hr

Step-by-step explanation:

Let the usual speed of the plane = x km/hr

Distance to the destination = 1500 km

Increased speed = 250 km/hr

So, according to the question

1500/x - 1500(x+250) = 30/60

⇒ 1500 * 60 * (x + 250) - 1500 * 60 * x = 30 * x * (x + 250)

⇒ 90000(x + 250) - 90000x = 30x(x + 250)

⇒ 90000x + 22500000 - 90000x = 30x² + 7500x

⇒ 30x² + 7500x - 22500000 = 0

⇒ x² + 250x - 750000 = 0

⇒ x² + 1000x - 750x - 750000 = 0

⇒ x(x + 1000) - 750(x + 1000) = 0

⇒ (x - 750)(x + 1000) = 0

⇒ x = 750, -1000{Speed cannot be negative}

⇒ x = 750 km/hr

Therefore,

Usual speed is 750 km/hr.

Hope it helps!

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