The geometric mean of three positive numbers a b c is 3 and the geometric mean of another three posi...........

Quantitative Aptitude (CDS) Statistics

The geometric mean of three positive numbers a, b, c is 3 and the geometric mean of another three positive numbers d, e, f is 4. Also, at least three elements in the set {a, b, c, d, e, f} are distinct. Which one of the following inequalities gives the best information about M, the arithmetic mean of the six numbers?

$M>2\sqrt{3}$

M > 3.5

M $\geq$ 3.5

It is not possible to set any precise lower limit for M

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Explanation:

The least arithmetic mean M is obtained when a = b = c = 3 and d = e = f = 4.
Therefore, the least value of M= (3+3+3+4+4+4)/6

=21/6=3.5

Also, M ≠ 3.5 as at least three elements among a, b, c, d, e, f are distinct, which will give the value of M greater than M (one such set is a = 3, b = 3, c = 3, d = 2, e = 4, f = 8).
∴M>3.5