Given below are two statements:

Statement I : The set of numbers (7,8,9,a,b,10,8,7) has an arithmetic mean of 9 and mode(most frequently occurring number) as 8, Then a x b = 120.
Statement II : Let a and b be two positive integers such that a + b + a x b = 84, then a + b =20.
In the light of the above statements, choose the most appropriate answer from the options given below

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 = $$9\times\ 8$$

49+a+b = 72

a+b = 23

It 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 $$\times\ $$ 85 or 5 $$\times\ $$ 17

It cannot be 1$$\times\ $$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.