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MySolution Using (1) \((a+b+c+4)/4 > (a +b+c)/3\) or \(a+b+c<12\) No clue about a, b, c. Insufficient
Using (2) four possibilities: (a, b, c, 4), median (b+c)/2 < b or c < b, not possible (4, a, b, c), median (b+a)/2 < b or b > a, already known (a, 4, b, c), median (b+4)/2 < b or b > 4 (a, b, 4, c), median (b+c)/2 < b or b > 4 Considering all the above cases: \(b > 4\) But still no clue about a & c. Insufficient.
Using (1) & (2), \(b_m_i_n = 5\), (a, b, c are integers) assuming all are in A.P. which means, a = b - k c = b + k from (1), \(a+b+c < 12\) or \((b-k)+b+(b+k) < 12\) or\(3b < 12\) or \(b < 4\) but \(b > 4\) (from (2)), thus the assumption is wrong. and a, b, c cannot be in AP
Re: a, b and c are integers such that a < b < c. Do they have a [#permalink]
10 Jun 2015, 10:52
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Re: a, b and c are integers such that a < b < c. Do they have a [#permalink]
18 Jun 2015, 04:08
Hi, I see that no one responded to this one. I cannot figure it out muself, but will share my thoughts.
[1] a+b+c+4 / 4 > a+b+c/3 I tested the means for these numbers: 2-4-6-4 (if a,b,c did have a common difference). It turns out that the means are the same. I tested the means for these numbers: 2-5-6-4 (if a,b,c did not have a common difference). It turns out that the first means is greater than the second.
However, I have no idea what this means, or if, in the condition when they d have a common difference, they should also have a common difference with 4 (the extra number).
[2] Let's say that these are the numbers in increasing order: a-b-c-4 or 4-a-b-c (and I am not sure here is 4 is supposed to be greatersmaller than the rest or could even be in between).
Then bc (or ab) > b. If the numbers are: 1,2,3,4, then the median is 2.5. For 1,2,3 the median is 3. So, the second median is greater, which should not happen. If the numbers are: 1,3,4,4, then the median is 3.5. For 1,3,4 the median is 3. The first median is greater, which is what should happen.
Re: a, b and c are integers such that a < b < c. Do they have a [#permalink]
22 Jun 2015, 13:06
pacifist85 wrote:
Hi, I see that no one responded to this one. I cannot figure it out muself, but will share my thoughts.
[1] a+b+c+4 / 4 > a+b+c/3 I tested the means for these numbers: 2-4-6-4 (if a,b,c did have a common difference). It turns out that the means are the same. I tested the means for these numbers: 2-5-6-4 (if a,b,c did not have a common difference). It turns out that the first means is greater than the second.
However, I have no idea what this means, or if, in the condition when they d have a common difference, they should also have a common difference with 4 (the extra number).
[2] Let's say that these are the numbers in increasing order: a-b-c-4 or 4-a-b-c (and I am not sure here is 4 is supposed to be greatersmaller than the rest or could even be in between).
Then bc (or ab) > b. If the numbers are: 1,2,3,4, then the median is 2.5. For 1,2,3 the median is 3. So, the second median is greater, which should not happen. If the numbers are: 1,3,4,4, then the median is 3.5. For 1,3,4 the median is 3. The first median is greater, which is what should happen.
But again, I have no idea what this means...
We need to find out whether is a,b,c - progression
1.s (a+b+c+4)/4>(a+b+c)/3 , thus a+b+c<12 or in other words a+b+c<=11 as a,b,c - integers. Let's test it 2,3,4=9 ok 2,3,5=10 also satisfy. Thus s1 is unsuf. 2. median of a,b,c is b as a<b<c. Median in a,b,c,4 could be 4 versions (4,a,b,c) , (a,4,b,c), (a,b,4,c), (a,b,c,4) (4,a,b,c) means b>(a+b)/2 if a>4 . b>a>4 (a,4,b,c) means b>(4+b)2 if a<4<b . b>4 - (a,b,4,c) means b>(b+4)/2 if b<4<c. b>4 - doesn't work as 4>b condition (a,b,c,4) means b>(b+c)/2 if b<c<4. b>c - doesn't work as c>b condition Conclusion: b>a>4 for sure Clearly statement 2 is not suf, as a,b,c could be progression or could be not. Simply input numbers
Let's combine 1 and 2 statements. We know that a+b+c<12, and b>a>4. Let's input numbers a=5, b=6,c=7 total>12. the answer is no, thus sufficient. Answer C.
gmatclubot
Re: a, b and c are integers such that a < b < c. Do they have a
[#permalink]
22 Jun 2015, 13:06
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