Question

# The arithemetic mean and the geometric mean of two distinct 2-digit numbers x and y are two integers one of which can be obtained by reserving the digits of the other(in base 10 representation) Then x+y equals

A
82
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B
116
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C
130
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D
148
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Solution

## The correct option is C 130Given, AM and GM of two number x and y are integer1<a<90<b<9x+y2=10a+b⇒x+y=2(10a+b)(1)√xy=10b+a⇒xy=(10b+a)2(2)Squaring equation (1) we getx2+y2+2xy=400a2+80ab+4b2(3)On multiplying (2) by 4 we get4xy=400b2+4a2+80ab(4)Subtracting equation (4) from (3) we get(x−y)2=396(a2−b2)(x−y)=6√11(a2−b2)For 11(a2−b2) to be a perfect square (a2−b2) should be equal to 11n(a2−b2)=11,44,...(a+b)(a−b)=11⇒(x−y)=66and (a+b)(a−b)=44⇒|(x−y)|=132but a can be atmost 9 and b can be atmost 9a+b≤18(a+b)(a−b)≤44a+b=11,a−b=4 but there is no interger value of aHence (a+b)(a−b)=11(x−y)=6662−52=11a=6,b=5(a,b)(6,5)x+y=2(10a+b)=2(10×6+5)=130

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