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(1) \(3x + 4 \lt 2y + 3\). Re-arrange: \(3x \lt 2y-1\). \(x\) can be some very small number, for instance -100 and \(y\) some large enough number for instance -3 and the answer would be YES, \(x \lt y\) BUT if \(x=-2\) and \(y=-2.1\), then the answer would be NO, \(x \gt y\). Not sufficient.

(2) \(2x - 3 \lt 3y - 4\). Re-arrange: \(x \lt 1.5y-\frac{1}{2}\). Re-write as \(x \lt y+(0.5y-\frac{1}{2}) = y + \text{negative}\). So, \(x \lt y\) (as \(y+\text{negative}\) is "more negative" than \(y\)). Sufficient.

Bunuel why can we apply same logic, which you applied in option B, in option A. We have x<2/3 y -1/3. please tell me if I am missing anything. y is negative and we are subtracting and -ve number.

Bunuel why can we apply same logic, which you applied in option B, in option A. We have x<2/3 y -1/3. please tell me if I am missing anything. y is negative and we are subtracting and -ve number.

Hi, I'll try to answer this for you ..

We know x and y are negative numbers ..

and the answer to the Q lies in

1)knowing that if a -ive number is multiplied with a bigger positive number/integer, The product will become even smaller.. meaning 3x<2x etc 2) role of < and > signs so if we want to check on the relative values of x and y, they have to brought in some way to same Coeff..

Can you spot the difference in two statements given keeping the above info in mind..

the statements are..

(1) 3x+4<2y+3 So, 3x+1<2y.. 3x<2y-1 now x and y are negative integers.. so 3x< 2x.. but we cannot make a relation between 2x and 2y-1.. Insuff

(2) 2x−3<3y−4 2x<3y-1.. now 3x<2x.. so we can write 3x<2x<3y-1.. or 3x<3y-1 x<y-1/3.. since both x and y are negative numbers and x is lesser than sum of y, a -ive number, and another -ive number.. x<y Suff

If 1.5 times x (i.e. a -ve term) and 0.5 added to it is still less than y, then x should be less that y..isn't it?

what am I doing wrong?

(2) 2x−3<3y−4

2x-3+4<3y

2x+1<3y

.66x+.33<y

I used the same logic as stated above....X being -ve, .66 times x is also -ve and 0.33 added to it is still less than y. Hence x should be less than y.

Graphically One easy way to do this is to see then the lines intersect by graphing them. In the attached, red is (1), blue is (2), and yellow is the prompt. You can see that (1) intersects with y>x when x, y negative, and thus you can pick options that allow the red line to be true, but can literally go on either side of y > x. For the blue line, that is also possible, but only when y and x are positive. Thus, we know that (2) is sufficient for all x,y negative.

Algebraically Another easy way to do this is to see when the lines intersect, but by using algebra.

(1) 2y > 3x+1 Pretend this is 2y=3x+1. Set this equal to what we are comparing it to, y=x, or, 2y=2x. 2y=3x+1=2x They intersect at x = -1. Either side of this, you will a Y or N answer when seeing if (1) is sufficient.

(2) 3y > 2x+1 Pretend this is 3y=2x+1. Set this equal to what we are comparing it to, y=x, or, 3y=3x. 3y=2x+1=3x They intersect at x=1. If less than 1, you will always get the same answer when seeing if (2) is sufficient. Since x < 0 < 1, (2) is sufficient.

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Why are we using picking numbers for statement 1 and solving the second statement using inequalities and algebra? Both inequalities are of the same nature. Can someone please explain the sufficiency of the first statement using algebra?

(1) \(3x + 4 \lt 2y + 3\). Re-arrange: \(3x \lt 2y-1\). \(x\) can be some very small number, for instance -100 and \(y\) some large enough number for instance -3 and the answer would be YES, \(x \lt y\) BUT if \(x=-2\) and \(y=-2.1\), then the answer would be NO, \(x \gt y\). Not sufficient.

(2) \(2x - 3 \lt 3y - 4\). Re-arrange: \(x \lt 1.5y-\frac{1}{2}\). Re-write as \(x \lt y+(0.5y-\frac{1}{2}) = y + \text{negative}\). So, \(x \lt y\) (as \(y+\text{negative}\) is "more negative" than \(y\)). Sufficient.

Answer: B

Question: Bunuel case-1) 3x < 2y -1 y=0 => 3x < -1 => x < -0.33 x=0 => y > 0.5 (but y < 0 given). Therefore, we can have both x < y (-0.33 < -0.1) and x > y (-0.33 > -0.4). Thus 1 NS

case-2) x < (3/2)y - 1/2 y = 0 => x < -1/2 x = 0 => y > 1/3 (but y < 0 given). Therefore, we can have both x < y (-0.5 < -0.1) and x > y (-0.5 > -0.6). Thus 2 NS

Case-1+2) x < -0.33 & x < -0.5 => x < -0.33 & y < 0. Thus 1+2 NS => E

What am i doing wrong? Can you please clarify Bunuel...

(1) \(3x + 4 \lt 2y + 3\). Re-arrange: \(3x \lt 2y-1\). \(x\) can be some very small number, for instance -100 and \(y\) some large enough number for instance -3 and the answer would be YES, \(x \lt y\) BUT if \(x=-2\) and \(y=-2.1\), then the answer would be NO, \(x \gt y\). Not sufficient.

(2) \(2x - 3 \lt 3y - 4\). Re-arrange: \(x \lt 1.5y-\frac{1}{2}\). Re-write as \(x \lt y+(0.5y-\frac{1}{2}) = y + \text{negative}\). So, \(x \lt y\) (as \(y+\text{negative}\) is "more negative" than \(y\)). Sufficient.

Answer: B

Question: Bunuel case-1) 3x < 2y -1 y=0 => 3x < -1 => x < -0.666 x=0 => y > 0.5 (but y < 0 given). Therefore, we can have both x < y (-0.33 < -0.1) and x > y (-0.33 > 0.4). Thus 1 NS

case-2) x < (3/2)y - 1/2 y = 0 => x < -1/2 x = 0 => y > 1/3 (but y < 0 given). Therefore, we can have both x < y (-0.33 < -0.1) and x > y (-0.33 > 0.4). Thus 2 NS

What am i doing wrong? Can you please clarify Bunuel...

We are told that BOTH x and y are negative numbers. Can you give an example when that condition and \(2x - 3 \lt 3y - 4\) are satisfied and x is not less than y?
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