Question

Think of 5 positive integers that have a mode of 4 and 6, a median of 6 and a mean of 6.

Answer 1

For the mode to be 6, there has to be at least two 6s.

For the median to be 6, with two 6s, there must be at least one number above 6.

For the mean to be 6, the sum is 30, so the three remaining numbers must total 18.

Let x, y & z be the remaining numbers in order of value.

So x + y + z = 30,

and x < y < z as if any are equal, we’d have two modes

For a range of 6, z = x + 6, so we’re left with

2x + y + 6 = 18

or 2x + y = 12

or y = 12 - 2x

If x = y, then y = 12 - 2y or 3y = 12 or y = 4

so x < 4 and z < 10

so x, y , z is one of {1, 10, 7}, {2, 8, 8}, {3, 6, 9}

and the last of these is the only one with no repeats and y < z

So the answer is: 3, 6, 6, 6, 9

hope it helps you ❤️...