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The square of two digit number(ab) divided by half of itself resultant added by 18 then divided by 2 then we get reverse of original number(ba). how many combination of ab exist

(a) 9

(b) 8

(c) 7

(d) 6
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$\frac{(ab)^{2}}{\frac{1}{2}* ab} =2* ab$

2*ab+18

$\frac{2*ab+18}{2}$

ab+9 =ba

ba-ab=9

their are 8 possibilities

21-12 =9

32-23=9

43-34=9

54-45=9

65-56=9

76-67=9

87-78=9

98-89=9
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