Find two numbers whose A.M. exceeds their G.M. by 30 and their H.M. by 48.
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Answer:
15 & 135
Step-by-step explanation:
Find two numbers whose A.M. exceeds their G.M. by 30 and their H.M. by 48.
Let say two numbers are
a & b
AM = (a + b)/2 = x
GM = √ab
HM = 2ab/(a+b) = ab/x
AM = GM + 30
x= √ab+ 30
x-30 = √ab
Squaring both sides
x² - 60x + 900 = ab
AM = HM + 48
x= ab/x + 48
ab = x² - 48x
equating both
x² - 60x + 900 = x²- 48x
=> x = 75
a+b = 150
ab = 75² - 48*75 = 75(75-48) = 75*27
y² - 150y + 75*27 = 0
=> y² - 15y - 135y + 75*27 = 0
=> y(y-15) - 135(y-15) = 0
y = 15 & 135
a & b = 15 & 135
Two numbers are 15 & 135
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