Question

Three positive integers are all different they have a mean of 8 and a median of 11 find the values of the three integers

Answer 1

Answer:

1, 11, 12

Step-by-step explanation:

Let the integer x, y and z.

Given,

Median = 11

Since, the number of terms are odd

$Median = (\frac{n+1}{2} )^{th} = (\frac{3+1}{2} )^{th} = \frac{4}{2} ^{th} = 2^{nd} = y$

∴y = 11

Also given that,

Mean = 8

$Mean = \frac{x + y + z}{3}$

$\frac{x + 11 + z}{3} = 8$

x + z + 11 = 24

x + z = 24 - 11 = 13 ----(i)

for z to be more that y and less than (x + z),

z must be 12

and then, x becomes 1

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