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16 Oct 2023, 09:44
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45% (02:11) correct
55%
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wrong
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If \(m\) and \(n\) are consecutive integers, is \(m\) greater than \(n\)?
(1) \(m-1\) and \(n+1\) are consecutive integers
(2) \(\frac{m}{n}\) is an even integer
(1) \(m-1\) and \(n+1\) are consecutive integers
(2) \(\frac{m}{n}\) is an even integer
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16 Oct 2023, 13:21
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Bunuel
Given
\(m\) and \(n\) are consecutive integers
Inference:
On a number line \(m\) and \(n\) can have one of the following positions with respect to each other.
Attachment:
Screenshot 2023-10-17 014316.png [ 45.56 KiB | Viewed 1317 times ]
Question: is \(m\) greater than \(n\)
Inference: Does \(m\) lie to the right of \(n\)
Statement 1
(1) \(m-1\) and \(n+1\) are consecutive integers
\(m-1\) denotes a position one unit to the left of \(m\)
\(n+1\) denotes a positon one unit to the right of \(n\)
Attachment:
Screenshot 2023-10-17 014653.png [ 72.27 KiB | Viewed 1304 times ]
We see that Case 2, Case 4, and Case 6 satisfy the criteria. In all cases \(m > n\). Hence, this statement is sufficient to conclude. We can eliminate B, C, and E.
Statement 2
(2) \(\frac{m}{n}\) is an even integer
This statement alone is not sufficient. For example, in case 2, m = 2 and n = 1. In this case \(m > n\).
However in case 5, m = 0 and n = 1. In this case \(m < n\)
Hence this statement alone is not sufficient.
Option A
24 Oct 2023, 08:45
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If \(m\) and \(n\) are consecutive integers, is \(m\) greater than \(n\)?
(1) \(m-1\) and \(n+1\) are consecutive integers.
If \(m\) were less than \(n\), then \(m-1\) (an integer less than \(m\)) and \(n+1\) (an integer greater than \(n\)) wouldn't be consecutive. Hence, \(m\) must be greater than \(n\). Sufficient.
Or look at this in another way: the stem says that the distance between \(m\) and \(n\) is 1. Now, if \(m < n\), then the distance between \(m-1\) and \(n+1\) would be 3, and they couldn't be consecutive as (1) states. Thus, it must be true that \(m > n\).
(2) \(\frac{m}{n}\) is an even integer.
Could \(m\) be greater than \(n\)? Certainly. For instance, if \(m = 2\) and \(n = 1\), or if \(m = 0\) and \(n = -1\). However, \(m\) can also be less than \(n\), such as when \(m = -2\) and \(n = -1\), or if \(m = 0\) and \(n = 1\). Not sufficient.
Answer: A
(1) \(m-1\) and \(n+1\) are consecutive integers.
If \(m\) were less than \(n\), then \(m-1\) (an integer less than \(m\)) and \(n+1\) (an integer greater than \(n\)) wouldn't be consecutive. Hence, \(m\) must be greater than \(n\). Sufficient.
Or look at this in another way: the stem says that the distance between \(m\) and \(n\) is 1. Now, if \(m < n\), then the distance between \(m-1\) and \(n+1\) would be 3, and they couldn't be consecutive as (1) states. Thus, it must be true that \(m > n\).
(2) \(\frac{m}{n}\) is an even integer.
Could \(m\) be greater than \(n\)? Certainly. For instance, if \(m = 2\) and \(n = 1\), or if \(m = 0\) and \(n = -1\). However, \(m\) can also be less than \(n\), such as when \(m = -2\) and \(n = -1\), or if \(m = 0\) and \(n = 1\). Not sufficient.
Answer: A
