Statement I: Given arithmetic mean of (7,8,9,a,b,10,8,7) is 9 and mode is 8 7+8+9+a+b+10+8+7=9×8 49+a+b=72a+b=23It is mentioned that mode is 8,8 occurred twice and 7 also occurred twice. Therefore, a or b should be 8 If a=8,b=15 If b=8,a=15 ab=120 Statement I is true Statement II: a+b+ab=84 1+a+b+ab=85 (1+a)(1+b)=85 85=1×85or5×17 It cannot be 1×85, as it is mentioned both a and b are positive integers If 1+a=5,1+b=17 a=4 and b=16 If 1+a=17,1+b=5 a=16 and b=4 In both cases, a+b=20 Statement II is true Answer is option A.