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M18-14

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Bunuel
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M18-14 [#permalink] New post   16 Sep 2014, 01:03
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60% (01:55) correct 40% (01:58) wrong based on 237 sessions
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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) \(b + c = d - a\)
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Bunuel
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Re M18-14 [#permalink] New post   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) \(a + d = b + c\).

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
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reto
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M18-14 [#permalink] New post   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
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Re: M18-14 [#permalink] New post   04 Oct 2015, 05:19
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We can solve this one with number picking
1) Doesn't give us any additional information - Not Suff
2) Just pick 4 even integers (it doesn't matter which .. the second is always 2 more than the previous) 2 4 6 8 and check this expression b+c=d−a
4+6 is not equal to 8-2, Why ? Because an even integer we need here is Zero. 0,2,4,6 2+4=6-0 -> 2+4/2=3 Sufficient
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Re: M18-14 [#permalink] New post   28 Aug 2023, 13:19
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re M18-14 [#permalink] New post   04 Apr 2024, 23:34
I think this is a high-quality question and I agree with explanation. Beautiful question
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Re M18-14 [#permalink]
04 Apr 2024, 23:34
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