Hussain15 wrote:

If (x/y)>2, is 3x+2y<18?

(1) x-y is less than 2

(2) y-x is less than 2

It will be great to see how do you guys approach this lethal one.

I would solve this question with graphic approach, by drawing the lines. With this approach you will "see" that the answer is A. But we can do it with algebra as well.

\(\frac{x}{y}>2\) tells us that \(x\) and \(y\) are either both positive or both negative, which means that all points \((x,y)\) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or

when both \(x\) and \(y\) are positive.

In I quadrant, as \(x\) and \(y\) are both positive, we can rewrite \(\frac{x}{y}>2\) as \(x>2y>0\) (remember \(x>0\) and \(y>0\)).

So basically question becomes: If \(x>0\) and \(y>0\) and \(x>2y>0\), is \(3x+2y<18\)?

(1) \(x-y<2\).

Subtract inequalities \(x>2y\) and \(x-y<2\) (we can do this as signs are in opposite direction) --> \(x-(x-y)>2y-2\) --> \(y<2\).

Now add inequalities \(x-y<2\) and \(y<2\) (we can do this as signs are in the same direction) --> \(x-y+y<2+2\) --> \(x<4\).

We got \(y<2\) and \(x<4\). If we take maximum values \(x=4\) and \(y=2\) and substitute in \(3x+2y<18\), we'll get \(12+4=16<18\).

Sufficient.

(2) \(y-x<2\) and \(x>2y\):

\(x=3\) and \(y=1\) --> \(3x+2y=11<18\) true.

\(x=11\) and \(y=5\) --> \(3x+2y=43<18\) false.

Not sufficient.

Answer: A.

In statement 1, you got y<2 and x<4 but when x=2 & y=1 or even your point x=4 & y=2 so x/y>2 is not satisfied because 2/1 or 4/2 is not bigger 2. How come still statement 1 sufficient?