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In order to prove that option B is the right answer we are multiplying the inequality by the LCM (6) without knowing if x+y is positive or negative.

Is there any reason for x+y to be assumed to be positive?
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pratikshetty123
In order to prove that option B is the right answer we are multiplying the inequality by the LCM (6) without knowing if x+y is positive or negative.

Is there any reason for x+y to be assumed to be positive?

We are multiplying by 6, which is positive. We care about the sign of the value by which we are multiplying an inequality.
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Bunuel

Tough and Tricky questions: Statistics.



If set S = {7, y, 12, 8, x, 9}, is x + y less than 18?

(1) The range of set S is less than 9.
(2) The average of x and y is less than the average of set S.

Statement 1. If x=4 and y=5 then answer is Yes. But if x=6 and y=14 then answer is No. Insufficient
Statement 2. We have (x+y)/2<(x+y+36)/6 Multiply both sides by 6 to get 3x+3Y<x+y+36 Hence x+y<18 Sufficient
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Raisakatyal
If set S= (7,y,12,8,x,9), is x+y less than 18?
(1) The range of set S is less than 9.
(2) The average of x and y is less than the average of set S.

S = {7, y, 12, 8, x, 9}. Is x + y < 18?

Statement 1:
Range < 9
Case 1: Let x = y = 4
Range = 12 - 4 = 8 < 9
x + y = 8 < 18
Case 2: Let x = y = 15
Range = 15 - 7 = 8 < 9
x + y = 30 > 18
Statement 1 is Not Sufficient.

Statement 2:
\(\frac{x + y}{2} < \frac{7 + y + 12 + 8 + 9 + x}{6}\)
=> 3 * (x + y) < 36 + x + y
=> 2 * (x + y) < 36
=> x + y < 18
Statement 2 is Sufficient.

So, correct answer is option B.
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