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 =Â $$9\times\ 8$$
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 = 1Â $$\times\ $$ 85 or 5Â $$\times\ $$ 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.
Create a FREE account and get: