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16 Sep 2014, 01:03
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What is the average (arithmetic mean) of four consecutive even integers \(a\), \(b\), \(c\) and \(d\), where \(a \lt b \lt c \lt d\)?
(1) \(a + d = b + c\)
(2) \(b + c = d - a\)
(1) \(a + d = b + c\)
(2) \(b + c = d - a\)
16 Sep 2014, 01:03
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Official Solution:
What is the average (arithmetic mean) of four consecutive even integers \(a\), \(b\), \(c\) and \(d\), where \(a \lt b \lt c \lt d\)?
(1) \(a + d = b + c\).
This statement holds true for all sets of four consecutive even integers \(a\), \(b\), \(c\), and \(d\), where \(a \lt b \lt c \lt d\). Not sufficient.
(2) \(b + c = d - a\).
Let \(a = 2k\), where \(k\) is an integer. Given that \(a\), \(b\), \(c\), and \(d\) are consecutive even integers, we can deduce that \(b = 2k + 2\), \(c = 2k + 4\), and \(d = 2k + 6\). This statement implies that \((2k + 2) + (2k + 4) = (2k + 6) - 2k\). Solving for \(k\), we find \(k = 0\). Hence, \(a = 0\), \(b = 2\), \(c = 4\), and \(d = 6\). This makes the average 3. Sufficient.
Answer: B
What is the average (arithmetic mean) of four consecutive even integers \(a\), \(b\), \(c\) and \(d\), where \(a \lt b \lt c \lt d\)?
(1) \(a + d = b + c\).
This statement holds true for all sets of four consecutive even integers \(a\), \(b\), \(c\), and \(d\), where \(a \lt b \lt c \lt d\). Not sufficient.
(2) \(b + c = d - a\).
Let \(a = 2k\), where \(k\) is an integer. Given that \(a\), \(b\), \(c\), and \(d\) are consecutive even integers, we can deduce that \(b = 2k + 2\), \(c = 2k + 4\), and \(d = 2k + 6\). This statement implies that \((2k + 2) + (2k + 4) = (2k + 6) - 2k\). Solving for \(k\), we find \(k = 0\). Hence, \(a = 0\), \(b = 2\), \(c = 4\), and \(d = 6\). This makes the average 3. Sufficient.
Answer: B
General Discussion
17 Jun 2015, 11:32
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Wow, this approach is just brilliant. Once you see it, it's so easy. The big question is, how do you ever come to this idea to plug in 2k for a? Of course 2k, 2x whatever stands for an even integer. What's really bothering me is the logic behind. I stare at this task and never in my life would have come up with this genious approach ... 
Please direct me to this thinking... Any help is very very very appreciated.
Thanks
Please direct me to this thinking... Any help is very very very appreciated.
Thanks
