CAT 2023 Slot 2 QA Question 1

The question says: Any non-zero real numbers x, y such that y ≠ 3 and (x/y) < (x + 3)/(y - 3). Which of the options will satisfy the condition.

The options are A, B, C, D, with various inequalities. The correct answer is supposed to be C, but We analyze why.

First, let me parse the given inequality: (x/y) < (x + 3)/(y - 3). I need to manipulate this inequality to find relationships between x and y that must hold true.

Since y ≠ 3, the denominator on the right side isn't zero, so that's Also, x and y are non-zero, so we don't have to worry about division by zero in the original inequality.

We have the inequality:

(x/y) < (x + 3)/(y - 3)

I want to solve this inequality for x and y. Maybe I can cross-multiply, but I have to be careful because the sign of y and (y - 3) will affect the direction of the inequality.

Alternatively, I can bring everything to one side and see what I get:

(x/y) - (x + 3)/(y - 3) < 0

Let me combine these two fractions. The common denominator would be y(y - 3). So:

[ x(y - 3) - (x + 3)y ] / [ y(y - 3) ] < 0

Expanding the numerator:

x(y - 3) - (x + 3)y = xy - 3x - xy - 3y = (-3x - 3y) = -3(x + y)

So the inequality becomes:

[ -3(x + y) ] / [ y(y - 3) ] < 0

Factoring out the -3:

-3(x + y) / [ y(y - 3) ] < 0

Since -3 is negative, I can multiply both sides by -1, which will reverse the inequality sign:

3(x + y) / [ y(y - 3) ] > 0

So now, the inequality is:

3(x + y) / [ y(y - 3) ] > 0

Since 3 is positive, I can ignore it for the inequality's sign:

(x + y) / [ y(y - 3) ] > 0

So, the sign of the expression (x + y) / [ y(y - 3) ] must be positive. That means the numerator and denominator must have the same sign.

So, either both (x + y) and y(y - 3) are positive, or both are negative.

Let me consider the denominator first: y(y - 3). Let's find when this is positive or negative.

y(y - 3) > 0 when y > 3 or y < 0. Because if y is positive and greater than 3, both y and (y - 3) are positive. If y is negative, both y and (y - 3) are negative, so their product is positive.

Similarly, y(y - 3) < 0 when 0 < y < 3. Because y is positive, but (y - 3) is negative, so their product is negative.

So, the denominator is positive when y > 3 or y < 0, and negative when 0 < y < 3.

Now, the numerator is (x + y). So, for the entire fraction to be positive, both numerator and denominator must be positive or both must be negative.

So, let's analyze each case.

Now, what each case implies.

Starting with Case 1:

Subcase 1a: y > 3 and x + y > 0.

This implies that x > -y. Since y is positive (greater than 3), x must be greater than a negative number. Not sure yet, but let's hold onto that.

Subcase 1b: y < 0 and x + y > 0.

Here, y is negative, so x must be greater than -y, which is positive. So x is positive and greater than |y|.

0 < y < 3 and x + y < 0.

Here, y is positive but less than 3, so x must be less than -y, which is negative. So x is negative and less than -y.

Now, how this relates to the options given.

Option A: (x/y) < (y/x)

Hmm, not sure yet. Let me think about this. Maybe I can test with numbers.

Option B: If y > 10, then -x > y.

Wait, from Case 1a, when y > 3, x > -y. If y > 10, then x > -y. So, -x < y. So, if y > 10, then -x < y. But Option B says -x > y, which would be the opposite. So B might not be correct.

Option C: If y < 0, then -x < y.

From Subcase 1b, when y < 0, x > -y. So, -x < y. Because if x > -y, then multiplying both sides by -1 (and reversing inequality) gives -x < y. So, if y < 0, then -x < y. That matches Option C.

Option D: If x < 0, then -x < y.

From Case 2, when 0 < y < 3, x + y < 0, so x < -y. Since y is positive, -y is negative. So x is less than a negative number, meaning x is negative. So, if x < 0, then x < -y, which implies that -x > y. Wait, that's the opposite of Option D, which says -x < y. So D might not be correct.

If x < 0, then from Case 2, x < -y. So, x is negative, and -y is negative because y is positive. So, x < -y implies that -x > y. Because if x is less than -y, then multiplying both sides by -1 (and flipping inequality) gives -x > y. So, if x < 0, then -x > y, which is the opposite of Option