Lily asked in Science & MathematicsMathematics · 3 months ago

# Can you find 6 numbers with a range of 7 a mean of 9 a median of 9 and a mode of 6 ?

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• 3 months ago

6 numbers with a range of 7.

That means the difference of the max and min is 7.

So  let's come up with variables for your 6 numbers, then an expression for the last one:

a, b, c, d, e, f

We know that f is 7 more than a, so:

f = a + 7

Leaving us with these 6 numbers:

a, b, c, d, e, (a + 7)

The mean and median are both 9.

The median of an even number of values is the mean of the middle to (c and d).  So:

(c + d) / 2 = 9

c + d = 18

And we know the mean of all 6 values is 9, so:

(a + b + c + d + e + a + 7) / 6 = 9

2a + b + c + d + e + 7 = 54

2a + b + c + d + e = 47

We know that c + d = 18, so:

2a + b + (c + d) + e = 472a + b + 18 + e = 472a + b + e = 29

We know the mode is 6.  So two of these three numbers has to be a 6.  Since "e" has to be greater than 9, that leaves "a" and "b" to be 6.  So now we can solve for "e":

2(6) + 6 + e = 2912 + 6 + e = 2918 + e = 29e = 11Going back to our numbers we now have:

6, 6, c, d, 11, (a + 7)

We know "a" so we know (a + 7):

6, 6, c, d, 11, 13The only thing left now is that we know the mean of c and d must be 9.  So back to this:

(c + d) / 2 = 9

c + d = 18

So like the last one, we have a few possible answers.  If we keep these to only integers and know the must be 18 and the two values must be greater than 6 and less than 11 and not equal to each other, there is only one possible set this can be:

6, 6, 8, 10, 11, 13

If c was 6 (which would still make 6 a mode), d would have to be 12, which makes the median equal to 10, which breaks that rule.

If c was 7, d would have to be 11, which doesn't make 6 the only mode.

If c was 9, d would be 9, which also doesn't ake 6 the only mode.

Those are why those values are excluded to leave the only set left to be correct.

• 3 months ago

try 6 6 6 11 12 13.

Right mean (9), right range (7), right mode (6),

wrong median (8.5).

try 6 6 x y z 13, requiring that

x + y = 18, x + y + z = 29, and z > y.

Then z must be 11, so x and y could be 8 and 10.

6 6 8 10 11 13.

Right mean, right median, right mode, right range.