M28-25

Hi Bunuel,

If D is must be true option then even C will also satisfy .

I tried with C and D with some random numbers and both seemed correct to me.

I have gone through the other explanations which you gave but didn't get.

Can you please help me to understand much better that how we can get MUST case and COULD BE in such questions.

As per the question, we know:
i. \(x\) and \(y\) are integers, and
ii. \(x + y = -12\)

From this, we get following five situations/ scenarios:
1) \(x < 0\), \(y < 0\) (eg. \(x = -5\), \(y = -7\) or \(x = -7\), \(y = -5\))
2) \(x < 0\), \(y > 0\) (eg. \(x = -13\), \(y = 1\))
3) \(x > 0\), \(y < 0\) (eg. \(x = 1\), \(y = -13\))
4) \(x = 0\), \(y < 0\) (\(x = 0\), \(y = -12\))
5) \(x < 0\), \(y = 0\) (\(x = -12\), \(y = 0\))

Let us now consider each of the options to determine whether they must be true for each of the above scenarios:
Option A:
Both \(x\) and \(y\) are negative
in other words: \(x <0\), \(y < 0\). This is true only for scenario 1) above. But this is not true for other 4 scenarios. Hence, this could be true, but not definitely true.

Option B:
\(xy>0\), this is only possible when \(x>0\) and \(y>0\) or when \(x<0 and y<0\). Again, this is true only for scenario 1) above. But this is not true for other 4 scenarios. Hence, this could be true, but not definitely true.

Option C:
If \(y<0\), then \(x>0\).
If we look at all the 5 scenarios, \(y<0\) in scenario 3 and 4. However, this statement is true for scenario 3 but not true for scenario 4. This is because, if \(y = -12\) (i.e. \(y<0\)), then \(x = 0\) (neither negative nor positive). Hence, this could be true, but not definitely true.

Option D:
If \(y>0\), then \(x<0\).
If we look at all the 5 scenarios, \(y<0\) only in scenario 2 and as per that scenario \(x>0\). So this must be true because whenever \(y\) is a negative integer, \(x\) has to be a positive integer. This is the correct answer.

Option E:
\(x-y>0\). This is true for scenario 4 but not true for scenario 5. Therefore, this could be true, but not definitely true. Remember, it is sufficient to consider 2 out of 5 scenarios, if we get different answers from them. We have to consider all the 5 scenarios only in cases where the statement is definitely true or must be true.

Hope this helps!!

Regards,
Nalin

A bit lengthy but a good detailed solution. Would add a couple of observations.

For any MUST BE TRUE question, you can only be sure of your answer once you find that examples that will negate the other options. Any correct option in a MUST BE TRUE question, will be true for ALL possible cases.

x+y=-12

A. Both x and y are negative . x=-12, y = 0 negate this condition. Eliminate.
B. xy>0. x=-12, y = 0 negate this condition. Eliminate.
C. If y<0, then x>0. x=0, y = -12 negate this condition. Eliminate.
D. If y>0, then x<0
E. x−y>0. x=-12, y = 0 negate this condition. Eliminate.

Thus, D is the correct answer. you only had to use (-12,0) or (0,-12) to arrive at the correct answer.

Hope this helps.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.