Question

Find two numbers whose am exceed their gm by 30 & their hm by 48

Answer 1

Answer:

15 & 135

Step-by-step explanation:

let say two numbers are a b

AM =( a+b)/2 = x

x = sqrt(ab) + 30

x = 2ab/(a+b) + 48

2ab/(a+b)= x-48

ab = x(x-48)

sqrt(ab) = x-30

squaring both sides

ab = x^2 + 900 -60x

x^2 -48x = x^2 +900 -60x

12x = 900

x = 75

a + b = 150

ab = 45^2 = 75*27

y^2 -150y + 2025 = 0

roots of these will be solution

y-^2 -15y -135y +2025 = 0

(y-15)(y-135) = 0

15 & 135 are numbers

am = 75

Gm = 45

HM = 27