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Re M1814
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16 Sep 2014, 01:03
Official Solution: (1) \(a + d = b + c\). That statement is true for ANY four consecutive even integers. Not sufficient. (2) \(b + c = d  a\). Let \(a=2k\), for some integer \(k\). Now, since \(a\), \(b\), \(c\) and \(d\) are consecutive even integers then \(b=2k+2\), \(c=2k+4\) and \(d=2k+6\). So, we have that \((2k+2)+(2k+4)=(2k+6)2k\). Solving gives \(k=0\), so \(a=0\), \(b=2\), \(c=4\) and \(d=6\). Hence, \(mean=3\). Sufficient. Answer: B
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Bunuel wrote: Official Solution:
(1) \(a + d = b + c\). That statement is true for ANY four consecutive even integers. Not sufficient. (2) \(b + c = d  a\). Let \(a=2k\), for some integer \(k\). Now, since \(a\), \(b\), \(c\) and \(d\) are consecutive even integers then \(b=2k+2\), \(c=2k+4\) and \(d=2k+6\). So, we have that \((2k+2)+(2k+4)=(2k+6)2k\). Solving gives \(k=0\), so \(a=0\), \(b=2\), \(c=4\) and \(d=6\). Hence, \(mean=3\). Sufficient.
Answer: B 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: M1814
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04 Oct 2015, 05:19
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 82, Why ? Because an even integer we need here is Zero. 0,2,4,6 2+4=60 > 2+4/2=3 Sufficient
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Re: M1814
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09 Jan 2017, 22:48
Bunuel wrote: Official Solution:
(1) \(a + d = b + c\). That statement is true for ANY four consecutive even integers. Not sufficient. (2) \(b + c = d  a\). Let \(a=2k\), for some integer \(k\). Now, since \(a\), \(b\), \(c\) and \(d\) are consecutive even integers then \(b=2k+2\), \(c=2k+4\) and \(d=2k+6\). So, we have that \((2k+2)+(2k+4)=(2k+6)2k\). Solving gives \(k=0\), so \(a=0\), \(b=2\), \(c=4\) and \(d=6\). Hence, \(mean=3\). Sufficient.
Answer: B Hi Bunuel I solved statement 1 like this: The 4 nos. are 2n, 2n+2, 2n+4, and 2n+6 a+d=b+c 2n+2n+6= 2n+2+2n+4 4n+6= 4n+6 We get no value from this as both the sides become zero. Therefore insufficient. The reason you explained for this result makes sense. Thanks.



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Re: M1814
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29 Apr 2017, 10:24
What about negative even integers?
What if set would be 2, 0, 2, 4?
Or does the question stem somehow rule negative numbers out?



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Re: M1814
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29 Apr 2017, 11:41



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What is the mean of four consecutive even integers \(a\), \(b\), \(c\) and \(d\), where \(a \lt b \lt c \lt d\)?
As number are Even Consecutive numbers
a, a+2, a+4, a+6
(1) \(a + d = b + c\)
a+a+6 = a+2+a+4
2a+6 = 2a +6
0=0..does not lead to get value of a
Insufficient
(2) \(b + c = d  a\)
a+2+a+4 = a+6a
2a +6 =6........2a = 0.......a=0...........Hence we can get the mean of the four numbers.
Sufficient
Answer: B



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Re: M1814
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20 Jun 2017, 05:01
The answer has to be option B. My take : The questions asks us to find the mean of four consecutive even numbers. If a,b,c,& d are four consecutive even numbers , a+d=b+c. We need to find the mean of the four numbers, the mean will be (b+c)/2. Statement 1  Does not give us the value of (b+c)/2 Statement 2  b+c=da. This implies d+a=da , or a=0. The series is 0,2,4,6. We can find the mean now. Hence statement 2 is sufficient. The answer has to be option B.
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Re: M1814
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20 Jun 2017, 05:30
reto wrote: Bunuel wrote: Official Solution:
(1) \(a + d = b + c\). That statement is true for ANY four consecutive even integers. Not sufficient. (2) \(b + c = d  a\). Let \(a=2k\), for some integer \(k\). Now, since \(a\), \(b\), \(c\) and \(d\) are consecutive even integers then \(b=2k+2\), \(c=2k+4\) and \(d=2k+6\). So, we have that \((2k+2)+(2k+4)=(2k+6)2k\). Solving gives \(k=0\), so \(a=0\), \(b=2\), \(c=4\) and \(d=6\). Hence, \(mean=3\). Sufficient.
Answer: B 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 I don't know if this will help, but I used to have the same issues. Anytime you see these problems where each number is related (consecutive ints, ratios etc.) you can often relate all the numbers to one variable For example, 4 consecutive even intergers are a, a+2, a +4 and a+6 In S2, we know that a +2 + a + 4 = a + 6  a solving for this we get: 2a + 6 = 6 2a = 0 Therefore a = 0 I like to double check just to make sure, so if we have 0 , 2, 4 and 6. Then 2 + 4 = 6  0 This satisfies the equation. Hope this helps.



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Re: M1814
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20 Jun 2017, 10:42
a) a+d=b+c this statement is obvious as if a=n,b =n+2,c=n+4 and d=n+6 then a+d=b+c b) b+c=ad if a =n then d=a+6(as a,b,c,d are consecutive even integers) b+c=6=a+d hence B



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Re: M1814
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21 Jun 2017, 23:16
I think it is easier than it seems to be... if a+d=b+c and da=b+c, then a=0, otherwise equations are not true. So, if a,b,c,d are consecutive even integers, then a=0, b=2, c=4, d=6 the mean is equal to (0+2+4+6)/4=3 the answer is 3.



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Re: M1814
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02 Sep 2018, 00:19
Quote: What is the mean of four consecutive even integers a, b, c and d, where a<b<c<d?
(1) a+d=b+c
(2) b+c=d−a Quote: Given a<b<c<d and they are even consecutive integers. STMT1: case 1: 2,4,6,8 mean =5 case 2: 8,6,4,2 mean = 5
2 solutions not sufficient STMT2: b+c=d−a b+a=dc now take 0,2,4,6 2=2 mean =3 Try another case : Not possible!! so B is the winner
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