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09 Jun 2012, 12:04
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
36% (02:32) correct
64%
(02:24)
wrong
based on 55
sessions
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Not Attempted Yet
Found the question in some training material:
a, b and c are integers such that a < b < c. Do they have a common difference (or Is a, b, c an Arithmetic progression?)
(1) Mean of (a, b, c, 4) is greater than mean of (a, b, c)
(2) Median of (a, b, c, 4) is less than median of (a, b, c)
a, b and c are integers such that a < b < c. Do they have a common difference (or Is a, b, c an Arithmetic progression?)
(1) Mean of (a, b, c, 4) is greater than mean of (a, b, c)
(2) Median of (a, b, c, 4) is less than median of (a, b, c)
Show SpoilerSolution
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
Answer (C).
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
Answer (C).
18 Jun 2015, 05:08
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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...
[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...
22 Jun 2015, 14:06
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pacifist85
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.
