Question

The number of people living in $6$ houses is $3,8,4,x,y,z$ where $x<y<z$. What are $x,y,$ and $z$?
Given :
The median is $7.5$.
The mode is $8$.
The mean is $7$.

• A. $7,8,$ and $12$
• B. $8,9,$ and $12$
• C. $7,9,$ and $12$
• D. $7,8,$ and $9$

Answer 1

The correct option is A $7,8,$ and $12$

Given numbers: $3,8,4,x,y,z$

Case 1: Mode $=8$
$x<y<z$
Since $x,y,$ and $z$ all are different in values. And mode is given as 8. So, any one of them will be $8$
Now, the given numbers will be $3,4,8,8,?,?$.
$\Rightarrow ?$ used for unknown number
Case 2:
Median $=7.5$
There are $6$ set of points and median will be the average of two middle points.
Therefore, sum of two middle points must be $=15$
So one of the middle number will be $=8$
So, one unknow number will be placed in between $4$ and $8$
Ordered numbers will be $3,4,?,8,8,?$
So, $7.5=\frac{?+8}{2}$
$\Rightarrow ?=15-8=7$
So, the set of numbers are $3,4,7,8,8,?$

Case 3: Mean $=7$
Ordered numbers: $3,4,7,8,8,?$
$\frac{3+4+7+8+8+?}{6}=7$
$?=42-30$
$\Rightarrow ?=12$

Ordered numbers are : $3,4,7,8,8,12$
So,
The unknown numbers are $7,8$ and $12$.
Given that unknown numbers $x<y<z$
So,
$x=7$
$y=8$
$z=12$