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I am trying to follow Walker's approach on nailing down the Inequalities under DS. It sure can be solved under 2 min mark with 100% accuracy, provided you are comfortable drawing the graphs from the Inequalities. So attached is my approach graphically.

well......Walker is one of the forum members just like us. But his Quant approaches are wonderful. I have attached the document on how I solved the DS problem. But I wanted to give credit to Walker for his approach on this.

Search the forum on Inequalities. You might get various approaches on how to solve them. Of them I picked this approach.

Even though it is time consuming to understand the concepts on how to draw graphs from the Linear (In)equalities , once you get it right, you should be comfortable solving the DS questions.

well......Walker is one of the forum members just like us. But his Quant approaches are wonderful. I have attached the document on how I solved the DS problem. But I wanted to give credit to Walker for his approach on this.

Search the forum on Inequalities. You might get various approaches on how to solve them. Of them I picked this approach.

Even though it is time consuming to understand the concepts on how to draw graphs from the Linear (In)equalities , once you get it right, you should be comfortable solving the DS questions.

hi mrsmarthi, you explanation is good. but please highlight the intersection or proof you are trying to prove in the diagram, so that the diagram would be more expressive. Walker - please suggest us the name of the software you are using to draw the graph.

1. 2x-2y-=1 2x = 1 + 2y x = (1+2y)/2=0.5 +y not suff.

2. x/y>1 multiply by y => x>y not suff.

combining them both, it's sufficient: x=0.5+y and x>y => 0.5+y>y this makes sense only if y<0

So, it's C.

X/y > 1 ==> x > y This is is not true in Ineqalities. You cannot multiple denominator value on both sides.

Here is an ex Let x = -3 and y = -2 ==? -3 / -2 > 1. This does not mean that -3 > -2. So your approach on Second clue is incorrect. I think you need to revisit the logic.

1. 2x-2y-=1 2x = 1 + 2y x = (1+2y)/2=0.5 +y not suff.

2. x/y>1 multiply by y => x>y not suff.

combining them both, it's sufficient: x=0.5+y and x>y => 0.5+y>y this makes sense only if y<0

So, it's C.

X/y > 1 ==> x > y This is is not true in Ineqalities. You cannot multiple denominator value on both sides.

Here is an ex Let x = -3 and y = -2 ==? -3 / -2 > 1. This does not mean that -3 > -2. So your approach on Second clue is incorrect. I think you need to revisit the logic.

Thank you so much, I can't believe I still make this stupid mistake. You are correct, I cannot do that.

Here's the (2) and the solution, then:

2. x/y>1, so then x and y have the same signs, either -, or +. Furthermore, x>y if their signs are positive and x<y if their signs are negative. insuff.

Let's combine (1) and (2).

if x and y are negative, then -x=0.5+(-y) or -x+y=0.5 - this can never be possible, because from (2) we know that x<y if their signs are negative. E.g. if x=-3, and y=-2, then -3+2 cannot equal 0.5.

if x and y are positive, then x=0.5+y or x-y=0.5 - this is possible, because from (2) we know that x>y if their signs are positive.

So, C is sufficient, x and y are both positive. I hope it makes sense.

1.) Is a subset of the solution a SUFFICIENT condition? Or should I make sure that ALL points of the condition case statement meets the original statement? In his example there is only a subset and looks like it is SUFFICIENT.

Hi mrsmarthi , Could you please explain how will you draw x/y >1 using your technique? using same technique how are you getting solution space between x-axis and the line x/y>1?..your equation is x-y>0; according to your technique, solution space will be right side of the line x-y=0; Thanks in advance

Was surprised that I got couple of questions to be answered. So thought I have a separate thread to document my preparation on solving Inequalities in DS. Here is the link.