An aeroplane left 40 minutes late due to heavy rains and in order to reach its destination, 1600km away, in time, it had to increase its speed by 400kmph from its original speed. Find the original speed of the aeroplane
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Let the original speed of the aeroplane be x kmph
Then,time taken to reach the destination=1600/x hr
On increasing the speed by 400 kmph,new velocity=x+400 kmph,then time taken to reach the destination=(1600/x)-(40/60)=1600/x-2/3 hr=4800-2x/3x hr
ATQ,
1600=(x+400)(4800-2x/3x)
1600=4000x-2x²+1920000/3x
4800x=4000x-2x²+1920000
4800x-4000x+2x²-1920000=0
2x²+800x-1920000=0
2(x²+400x-960000)=0⇒x²+400x-960000=0⇒x²+400x+(-960000)=0
x=[-400±√(400)²-4(1)(-960000)]/2(1) (quadratic equation formula)
=-400±√160000+3840000/2=-400±√4000000/2=(-400±2000)/2
⇒-400+2000/2,-400-2000/2
⇒1600/2,-2400/2⇒x=800,-1200
∴x=800 (neglecting -1200 as speed cannot be negative)
∴Hence,the original speed of the aeroplane is 800 kmph.
Then,time taken to reach the destination=1600/x hr
On increasing the speed by 400 kmph,new velocity=x+400 kmph,then time taken to reach the destination=(1600/x)-(40/60)=1600/x-2/3 hr=4800-2x/3x hr
ATQ,
1600=(x+400)(4800-2x/3x)
1600=4000x-2x²+1920000/3x
4800x=4000x-2x²+1920000
4800x-4000x+2x²-1920000=0
2x²+800x-1920000=0
2(x²+400x-960000)=0⇒x²+400x-960000=0⇒x²+400x+(-960000)=0
x=[-400±√(400)²-4(1)(-960000)]/2(1) (quadratic equation formula)
=-400±√160000+3840000/2=-400±√4000000/2=(-400±2000)/2
⇒-400+2000/2,-400-2000/2
⇒1600/2,-2400/2⇒x=800,-1200
∴x=800 (neglecting -1200 as speed cannot be negative)
∴Hence,the original speed of the aeroplane is 800 kmph.
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