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
SuffStmt2: y-x<2; or x>y-2. Given (1), LT2y>y-2 => LTy>-2
InsuffSo, the answer is A.
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