The following 7 numbers have the same mean, median, and m ode. If 23 is not one of the numbers, find the value of x + y.
39, 25, 19, 19, 25, x, y
The "mode" exists.
Meaning there is a number which appears most often.
Thus, there are three possibilities:
1)Mode is 25 ($\displaystyle x=25)$.
2)Mode is 19 ($\displaystyle x=19)$.
3)Mode is 39 ($\displaystyle x=y=39$).
It cannot be case #3 because then the mean is 29.2 a violation of initial conditions of problem.
If #2 then the mean must be 19.
Meaning,
$\displaystyle \frac{39+3\cdot 19+2\cdot 25+y}{7}=19$
$\displaystyle \frac{146+y}{7}=19$
$\displaystyle y=-13$
Checking the mode we find that this works.
If #1 then the mean must be 25.
Meaning,
$\displaystyle \frac{39+3\cdot 25+2\cdot 19+y}{7}=25$
$\displaystyle \frac{152+y}{7}=25$
$\displaystyle y=23$
But the problem says it is not amongs the numbers.
Thus, we have determined the unique numbers the work.
The following 7 numbers have the same mean, median, and mode. If 23 is not one of the numbers, find the value of x + y.
39, 25, 19, 19, 25, x, y
For this question, I'll assume that the mode is infact unique, that is, you can't have more than one more, otherwise, you'll need 2 means, and that's impossible with a set of data
Rearrange to give 19, 19, 25, 25, 39... and x and y we don't know where to place yet. But no matter where we put it... the median has to br 19 or 25 (just move x and y around and use some logic, you'll see that this is true)
Mean = $\displaystyle \frac{39 + 25 + 19 + 19 + 25 + x + y}{7}$
We want this to be a nice whole number, so 39 + 25 + 19 + 19 + 25 + x + y must be a multiple of 7, 39 + 25 + 19 + 19 + 25 = 127, trial and error the next few mulitples of 7, i.e. 133, 140, 147, 154, 161, 168, 175, 182, 189, 196... we know x + y = 6, 13, 20, 27, 34, 41, 48, 55, 62, 69...
At this stage we still don't know what to expect x and y to be, but we're certainly getting closer... we want to achieve a unique mode, seeing we already have two of 19 and two or 25, we can safely say that one of x or y = 19 or 25. We've already narrowed it down to 2 cases, mode = 19 and mode = 25 (seeing the median can only be 19 or 25)
Test 1: If mode = 19, then mean = 19 (as the question suggests), so the sum of the scores must be $\displaystyle 19 \times 7 = 133$ suggesting that x + y = 6, and we want either x or y to be 19, so x = 19, y = -13, vice versa... I know this will raise a few questions in that y is negative, but it never said in the question that these numbers can't be negative, all they ask for is the sum. And I'll show you why in this following case ~
Test 2: If mode = 25, the mean = 25 (as the question suggests), so the sum of the scores must be $\displaystyle 25 \times 7 = 175$ suggesting that x + y = 48, to achieve a unique mode we need either x or y to be 25:
When x = 25, y = 23, and vice versa, this violates the rule that one of the numbers aren't 23.
So we know the mean = median = mode = 19 when the scores are:
-13, 19, 19, 19, 25, 25, 39
Check:
Mean = $\displaystyle \frac{-13 + 19 + 19 + 19 + 25 + 25 + 39}{7} = 19$
Mode = 19 (19 being the most frequent) and;
Median = 19, as 19 is the middle score.
So we can conclude that x + y = 6.