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Question

The arithmetic mean of two positive numbers is 2 more than its geometric mean. If their difference is 8, then the numbers are

A
10,2
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B
9,1
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C
8,3
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D
12,4
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Solution

The correct option is B 9,1
Let the numbers be a and b.
According to the given condition,
|ab|=8(a+b)24ab=64(a+b)2=4ab+64 (1)

Given, AMGM=2
a+b2ab=2a+b=2ab+4(a+b)2=4ab+16+16ab
Using equation (1), we get
4ab+64=4ab+16+16abab=3

Now, a+b=10
and |ab|=8
a=9,b=1 or a=1,b=9

Therefore, the required numbers are 9,1

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