Math, asked by BrainlyHelper, 1 year ago

While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/hr. Find the original speed/hour of the plane.

Answers

Answered by nikitasingh79
12

SOLUTION :  

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

A.T.Q

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

-  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 original speed of the plane is 500 km/h.

HOPE THIS  ANSWER WILL HELP YOU..


Anand230703: x =500
Answered by KnowMore
12
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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