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Manager
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What is the mean of four consecutive even integers a , b , c [#permalink]
 03 Sep 2009, 16:35
Question Stats:
0% (00:00) correct
100% (01:48) wrong based on 0 sessions
What is the mean of four consecutive even integers a , b , c , d ?
1. a + d = b + c
2. b + c = d - a
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Manager
Joined: 25 Aug 2009
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a,b,c and d are 4 consecutive even integers.
Let a = 2n where n belongs to integers
b = 2n+ 2
c = 2n +4
d = 2n + 6
mean = \frac{(8n + 12)}{4}
=> mean = 2n + 3; we need value of n.
1.) a + d = b + c
=> 2n + 2n +6 = 2n + 2 + 2n + 4
=> 0 = 0 , we can not find value of n, hence, insufficient..
2.) b + c = d - a
=> 4n + 6 = 2n + 6 - 2n
=> n = 0
Hence, sufficient..
Hence, answer is B.
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CEO
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nikhilpoddar wrote: What is the mean of four consecutive even integers a , b , c , d ?
1. a + d = b + c
2. b + c = d - a Intresting question. B as OA is correct. The mean of 4 consecutive even integers can never be even. In fact it is always an odd integer. Using st. 2, if b + c = d - a, the only possible values for a, b, c, and d are 0, 2, 4, and 6 respectively. So mean is 3.
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Manager
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We can take four numbers as a, a+2(b),a+4(c), a+6(d)....
solve by given stmt..
B....mean is 3...
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