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Let the mean of \(x, y^{-1}, z^{5}\) be 6 find the mean of \(10, y^{-1}, 12, x, z^{5}\)....


Question

Let the mean of \(x, y^{-1}, z^{5}\) be 6 find the mean of \(10, y^{-1}, 12, x, z^{5}\).

Options

A) 7

B) 8

C) 9

D) 10

The correct answer is B.

Explanation:

Since the mean of \(x, y^{-1}\) and \(z^5\) is \(= 6\)
\(\therefore \cfrac{x+ y^{-1} + z^5 }{3} = 6\)
\(\therefore x + y^{-1} + z^5 = 6 \times 3 = 18\)
With this information, the mean of \(10, y^{-1}, 12, x\) and \(z^5\)
\(= \cfrac{10+ y^{-1} + 12 + x + z^5}{5} = \text{M(mean)}\)
Substituting \(18\) for \(x+y^{-1}+z^5\)
We then have \(\cfrac{10+12+18}{5} = M\)
\(\therefore \text{M} = \cfrac{40}{5}\)
\(\text{M} = 8\)
\(\therefore\) the mean of \(10,y^{-1},12,x,z^5= 8\)

Explanation provided by Danny Babs


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Dicussion (1)

  • Since the mean of \(x, y^{-1}\) and \(z^5\) is \(= 6\)
    \(\therefore \cfrac{x+ y^{-1} + z^5 }{3} = 6\)
    \(\therefore x + y^{-1} + z^5 = 6 \times 3 = 18\)
    With this information, the mean of \(10, y^{-1}, 12, x\) and \(z^5\)
    \(= \cfrac{10+ y^{-1} + 12 + x + z^5}{5} = \text{M(mean)}\)
    Substituting \(18\) for \(x+y^{-1}+z^5\)
    We then have \(\cfrac{10+12+18}{5} = M\)
    \(\therefore \text{M} = \cfrac{40}{5}\)
    \(\text{M} = 8\)
    \(\therefore\) the mean of \(10,y^{-1},12,x,z^5= 8\)