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# a, b and c are integers such that a < b < c. Do they have a

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a, b and c are integers such that a < b < c. Do they have a [#permalink]  09 Jun 2012, 11:04
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Question Stats:

25% (04:39) correct 75% (01:19) wrong based on 4 sessions
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)

[Reveal] Spoiler: Solution
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

[Reveal] Spoiler: OA

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a, b and c are integers such that a < b < c. Do they have a   [#permalink] 09 Jun 2012, 11:04
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