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16 Sep 2014, 00:29



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01 Sep 2015, 23:56
given x+y=−12 and both x and y are integers. Consider few examples that give a sum of 12
8 + (4) = 12 4 + ( 16) = 12 Now check the given options. Only d holds good.



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02 Sep 2015, 02:25
please check this
d)16(x) + 4(y)=12> x<0, y>0
c)4(x)+(16)(y)=12>x>0,y<0
so there are two valid answer options in this question, correct me if i am wrong



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02 Sep 2015, 03:25



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02 Sep 2015, 04:31
ramarao443 wrote: please check this
d)16(x) + 4(y)=12> x<0, y>0
c)4(x)+(16)(y)=12>x>0,y<0
so there are two valid answer options in this question, correct me if i am wrong Option c (If y<0, then x>0) gets eliminated when we use these values x = 8 and y = 4 8 + (4) = 12



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01 Mar 2016, 13:29
this question is wrong I believe becuase what happens if
x=12 and y=0
x+y=12 and THE SAME HAPPENS viceversa
please explain us all



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02 Mar 2016, 07:33
Bunuel wrote: Official Solution:
If \(x\) and \(y\) are integers and \(x + y = 12\), which of the following must be true?
A. Both \(x\) and \(y\) are negative B. \(xy \gt 0\) C. If \(y \lt 0\), then \(x \gt 0\) D. If \(y \gt 0\), then \(x \lt 0\) E. \(xy \gt 0\)
Look at option D: if \(y\) is positive, then \(x\) must be negative in order the sum of \(x\) and \(y\) to be negative.
Answer: D 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.



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02 Mar 2016, 20:35
msk0657 wrote: Bunuel wrote: Official Solution:
If \(x\) and \(y\) are integers and \(x + y = 12\), which of the following must be true?
A. Both \(x\) and \(y\) are negative B. \(xy \gt 0\) C. If \(y \lt 0\), then \(x \gt 0\) D. If \(y \gt 0\), then \(x \lt 0\) E. \(xy \gt 0\)
Look at option D: if \(y\) is positive, then \(x\) must be negative in order the sum of \(x\) and \(y\) to be negative.
Answer: D 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: \(xy>0\). 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



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nalinnair wrote: msk0657 wrote: Bunuel wrote: Official Solution:
If \(x\) and \(y\) are integers and \(x + y = 12\), which of the following must be true?
A. Both \(x\) and \(y\) are negative B. \(xy \gt 0\) C. If \(y \lt 0\), then \(x \gt 0\) D. If \(y \gt 0\), then \(x \lt 0\) E. \(xy \gt 0\)
Look at option D: if \(y\) is positive, then \(x\) must be negative in order the sum of \(x\) and \(y\) to be negative.
Answer: D 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: \(xy>0\). 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<0E. 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.



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Re: M2825
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12 Nov 2018, 08:04
If x and y are integers and x+y=−12, which of the following must be true?
Use logic and test numbers
X = 22, y = 10 22 + 10 = 12
A. Both x and y are negative, testing at least the solution above proves for this to not be true. B. x>0 y>0, testing at least the solution above proves for this to not be true. C. If y<0, then x>, testing at least the solution above proves for this to not be true. D. If y>0, then x<0, do not need to prove any further that of the given answer choices this is the only one that can be correct. E. x−y>0[/color], testing at least the solution above proves for this to not be true.



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12 Nov 2018, 08:33
As per the question we are given x+y=12 and as per the option D , we have If y >0 then x <0 i followed the following approach instead of putting some random values, i put the value of y in the inequality (y>0). What we get after putting the value y from the given equation Quote: 12x>0 .......(1) On simplifying the equation 1 we get Quote: x<12 means x will always be negative (or x<0). this is our required answer (D)
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03 Jan 2019, 18:12
I don't agree with the explanation. If x=24 and y=12 then option D is correct If x=12 and y=24, then option C is correct... How are you judging that y has to be > 0



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04 Jan 2019, 00:09










