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If the arithmetic and geometric means of two numbers are 10 , 8 respectively, then one number exceeds the other number by
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12
Let a and $b$ be two numbers such that
A.M. $=\frac{a+b}{2}=10$
$\Rightarrow \mathrm{a}+\mathrm{b}=20$ ...(i)
and G.M. $=\sqrt{a b}=8 \Rightarrow \mathrm{ab}=64 \quad$...(2)
From (1) and (2), we have $a^{2}-20 a+64=0 \Rightarrow(a-4)(a-16)=0$
$\Rightarrow \mathrm{a}=4,16$
Thus, when $\mathrm{a}=4, \mathrm{~b}=16$ and when $\mathrm{a}=16, \mathrm{~b}=4$
Hence, one number exceeds the other number by $12 .$
A.M. $=\frac{a+b}{2}=10$
$\Rightarrow \mathrm{a}+\mathrm{b}=20$ ...(i)
and G.M. $=\sqrt{a b}=8 \Rightarrow \mathrm{ab}=64 \quad$...(2)
From (1) and (2), we have $a^{2}-20 a+64=0 \Rightarrow(a-4)(a-16)=0$
$\Rightarrow \mathrm{a}=4,16$
Thus, when $\mathrm{a}=4, \mathrm{~b}=16$ and when $\mathrm{a}=16, \mathrm{~b}=4$
Hence, one number exceeds the other number by $12 .$
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