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Re: DS: Inequalities [#permalink]
25 Aug 2008, 00:03

Given x/y>2.....implies both x and y can be positive or both of them are negative If both of them are negative, 3x+2Y<0. If both of them are positive we have x>2y..............(1)

Statement1: x-y<2 so x<2+y......(2) From (1) and (2) 2y<x<2+y So 2y<2+y Hence y<2.......(3) From (2) and (3), x<4 --------(4) So from (3) and (4), it is clear that 3x+2y<16. Hence statement (1) alone is sufficient

Coming to statement 2, we can think along the same lines and easily find two contradicting examples.

Let x=100 and y=1 Clearly x/y >2, and y-x<2 and the value of 3x+2y=302 (>18)

Let x=3 and y=1 Clearly x/y>2 and y-x<2 and the value of 3x+2y=11(<18) So statement(2) is not sufficient

Hence A.

Hi, Kevin........I am happy to come back again with your question

Re: DS: Inequalities [#permalink]
27 Dec 2008, 18:50

1

This post received KUDOS

I tried solving this problem graphically and got A as well..

Solution: Following the steps below

1. Draw 3x + 2y = 18 (x intercept 6, y intercept 9)..The question asks if region BELOW 3x + 2y = 18 is satisfied by given data 2. Draw y = 1/2 x and shade the region BELOW the line. Find intersection with last line (intersects at (4.5, 2.25)). 3. Draw x - y = 2 4. condition A: x - y < 2 --> shade the region above x - y = 2 and find intersection with shaded region belonging to y = 1/2 x --> shaded intersection is BELOW 3x + 2y = 18 --> sufficient 5. condition B: x - y > 2 --> shade region below x - y = 2 and find intersection with shaded region belonging to y = 1/2 x --> shaded intersection is below as well as above 3x + 2y = 18 --> insufficient

Although this method appears complicated when written out, it actually is quite fast if you're comfortable with plotting lines / curves

Re: DS: Inequalities [#permalink]
29 Dec 2008, 04:24

xyz21 wrote:

I tried solving this problem graphically and got A as well..

Solution: Following the steps below

1. Draw 3x + 2y = 18 (x intercept 6, y intercept 9)..The question asks if region BELOW 3x + 2y = 18 is satisfied by given data 2. Draw y = 1/2 x and shade the region BELOW the line. Find intersection with last line (intersects at (4.5, 2.25)). 3. Draw x - y = 2 4. condition A: x - y < 2 --> shade the region above x - y = 2 and find intersection with shaded region belonging to y = 1/2 x --> shaded intersection is BELOW 3x + 2y = 18 --> sufficient 5. condition B: x - y > 2 --> shade region below x - y = 2 and find intersection with shaded region belonging to y = 1/2 x --> shaded intersection is below as well as above 3x + 2y = 18 --> insufficient

Although this method appears complicated when written out, it actually is quite fast if you're comfortable with plotting lines / curves

Solving graphically can be very easy - thanks for sharing your method xyz. Would you mind to post plotted graph that illustrates this? Below I have tried to solve the problem algebraically.

Question stem: To start with, we have x/y>2; is 3x+2y<18? x/y>2 => x>0, y>0; or x<0, y<0. When x<0, y<0 =>the answer to "3x+2y<18?" will be "Yes", so we are really looking at the case when x>0, y>0. Let me know if someone disagrees with my thinking here.

Let's denote "Greater Than" as "GT" and "Less Than" as "LT" Given x>0; y>0, x/y>2 =>(x-2y)/y>0 => x>2y and y>0. We can depict x>2y as x=LT2y. ----- (1) Substituting (1) for x in 3x+2y<18 => 3(LT2y)+2y<18 => LT6y+2y<18 => LT8y<18 => LTy<9/4.

Given the above, the question is really asking is 0<y<9/4 true?

Stmt1: x-y<2; or given (1), LT2y-y<2 => LTy<2. Given y>0, stmt 1 => 0<y<2 Suff

Stmt2: y-x<2; or x>y-2. Given (1), LT2y>y-2 => LTy>-2 Insuff