Math, asked by amarpriya3006, 8 months ago

please answer this question and give me full explanation. I'll follow you

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Answers

Answered by kjha95830
1

Step-by-step explanation:

Let the usual speed of the plane is=s km/ hr

and the usual time it takes to reach the destination=t hr

Distance= 1500 km

Now, speed=distance/time

s=1500/t

t=1500/s-(I)

The plane ✈ got delayed by half an hour and speed was increased by 250 km/ hr to reach destination on time.

now, speed of plane=s+250 km/ hr

t = t -1/2hr

Now, speed × time = distance

(s+250) (t-1/2) =1500

(s+250) (1500/s-1/2) =1500 { from(I) we know t= 1500/t}

{s+250}{(3000-s)/2s}=1500

3000s-s^2+250×3000-250s = 3000s

-s^2+250×3000-250s=0

s^2+250s-250×3000=0

s^2+1000s-250s-250×3000=0

(s+1000) (s-750) =O

therefore,

s = -1000, s = +750

speed can't be negative.

so, usual speed = 750 km

or usual time = 1500/750

= 2 hr

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