If speed of a aeroplane is reduced by 40km per hr,it takes 20 min more to cover 1200km.find the speed of the aeroplane
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We know Speed (S) = Distance (D) ÷ Time (T)
Hence S = 1200/T --------- (i)
We have also been given that:
S - 40 = 1200/(T + 20) -------------(ii)
Putting the value of S as obtained in (i) into (ii), we get
1200/T - 40 = 1200/(T + 20)
Simplifying this gives us a quadratic equation:
T² + 20T - 600 = 0
The formula for solving quadratic equation:
[-b -√(b² - 4ac)]/2a or [-b +√(b² - 4ac)]/2awhere a and are the coefficients of x² and x while c is the constant term in the quadratic equation ax² + bx + c
Hence we have T = [-20 + √(20² - 4(1)(-600)] /2
We have T = 16.457 hours
Since we have T now, we can calculate S from (i)
S = 1200/16.457 = 72.9 km/hr
Hence S = 1200/T --------- (i)
We have also been given that:
S - 40 = 1200/(T + 20) -------------(ii)
Putting the value of S as obtained in (i) into (ii), we get
1200/T - 40 = 1200/(T + 20)
Simplifying this gives us a quadratic equation:
T² + 20T - 600 = 0
The formula for solving quadratic equation:
[-b -√(b² - 4ac)]/2a or [-b +√(b² - 4ac)]/2awhere a and are the coefficients of x² and x while c is the constant term in the quadratic equation ax² + bx + c
Hence we have T = [-20 + √(20² - 4(1)(-600)] /2
We have T = 16.457 hours
Since we have T now, we can calculate S from (i)
S = 1200/16.457 = 72.9 km/hr
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