If x and y are negative numbers, is x<y?

Dear solitaryreaper,
I'm happy to help.

Statement #2 is not sufficient by itself. For most of the region 2x−3<3y−4, it's true that x < y.

For example, if (x, y) = (0, 5), then that is consistent with the statement #2 inequality, and it gives a "yes" answer to the prompt question, "is x < y?"

By contrast, if (x, y) = (9, 7), then this is also consistent with the statement #2 inequality, but it gives a "no" answer to the prompt question.

We can pick different values consistent with the statement #2 inequality that give different answers to the prompt question, so Statement #2 by itself is most certainly not sufficient.

Part of the problem is that whoever posted the question posted the WRONG OA! The correct OA for this question is (C), as the graphs I posted above show. I don't have the ability to made that edit, but perhaps Bunuel can make that change.

Does this answer your questions?
Mike

Hi Mike,

The stem says that x and y are negative numbers, so the OA is correct.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Since we have 2 variables and 0 equation, C is most likely to be the answer.

Condition 1) & 2)
\(3x + 4 < 2y + 3\)
\(2x - 3 < 3y - 4\)
If we add both inequalities, we have \(5x + 1 < 5y - 1\).
\(5x + 1 < 5y - 1 < 5y + 1 ⇔ 5x < 5y or x < y\).
Both conditions together are sufficient.

Since this question is one of key questions, we need to check A or B by CMT(Common Mistake Type) 4 (A).

Condition 1)
\(3x + 4 < 2y + 3\)

\(x = -2\), \(y = -1\) : Yes
\(x = -20\), \(y = -21\) : No.

This is not sufficient.

Condition 2)

\(2x - 3 < 3y - 4 ⇒ 3x - 4 < 3x - 3 < 2x - 3 < 3y - 4\) since \(x < 0\).
We have \(3x < 3y\) or\(x < y\).

This is sufficient.

Therefore, the answer is B.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Hi @mathsrevolution VeritasKarishma Bunuel

for first statement 1, I tested values and concluded that st 1 is enough. My question here would be how do we know when we have checked "sufficient" values. I checked extreme values (x=-100 y= -1) and close values (x= -4 and y =-3) both cases gave me a YES. therefore I concluded with st 1 as sufficient.

I will not use testing numbers to establish sufficiency. To establish insufficiency, yes, that is fine. If I am quickly able to arrive at two values which give a yes and a no, I can say that the statement is not sufficient and move on. But if all values I try give the same answer, I will need to establish sufficiency using logic.


(1) 3x + 4 < 2y + 3
3x < 2y−1
x < 2y/3 - 1/3
Add and subtract y/3 from RHS
x < y - y/3 - 1/3

x < y - (y+1)/3

Now, (y+1)/3 can be a positive number or a negative number say
when y = -1/2, (y+1)/3 is positive
when y = -2, (y+1)/3 is negative

So this inequality could give us e.g.
x < y - 1/6 (we don't know if x is less than y)
or x < y + 1/3 (yes, x is less than y)

Hence, not sufficient.

If \(x\) and \(y\) are negative numbers, is \(x < y\)?

(1) \(3x + 4 < 2y + 3\).

Re-arrange to: \(3x < 2y-1\).

Express \(2y\) as \(3y -y\) to get: \(3x < 3y - y - 1\), which can also be written as: \(3x < 3y - (y + 1)\).

Divide each term by 3: \(x < y -\frac{y+1}{3}\).

Now, \(\frac{y+1}{3}\) can be positive if \(y > -1\), and negative if \(y < -1\).

When \(\frac{y+1}{3}\) is positive, \(y -\frac{y+1}{3}\) will be less than \(y\), so we'd have: \(x < y -\frac{y+1}{3} < y\).

When \(\frac{y+1}{3}\) is negative, \(y -\frac{y+1}{3}\) will be greater than y, so we'd have: \(y < y -\frac{y+1}{3}\). In this case, \(x\) could be between \(y\) and \(y -\frac{y+1}{3}\), thus being more than \(y\): \(y < x < y -\frac{y+1}{3}\), as well as to the left of \(y\), thus being less than \(y\): \(x < y < y -\frac{y+1}{3}\).

Not sufficient.

Alternatively, after getting \(3x < 2y-1\), we could plug in numbers. If \(x\) is a very small number, for instance -100, and \(y\) is -1, then \(x < y\) and the answer is YES. However, if \(x=-2\) and \(y=-2\), then \(x=y\) and the answer is NO.

(2) \(2x - 3 < 3y - 4\).

Re-arrange to: \(2x < 3y - 1\).

Express \(3y\) as \(2y + y\) to get: \(2x < 2y + y - 1\).

Divide each term by 2: \(x < y + \frac{y-1}{2}\).

Since \(y\) is negative, \(\frac{y-1}{2}\) will also be negative, thus \(y + \frac{y-1}{2}\) will be less than \(y\). Therefore, we have \(x < y + \frac{y-1}{2} < y\).

Sufficient.


Answer: B