Find number when median is given ...

[stevecowall]

Question:
Find all values of x such that the mean of 77, 137, and x is 1 more than their median.

My solution:
So (77+137+x)/3 = 137 + 1
So i solve it then I got x = 200

I'm not sure that is correct ...

 

[soroban]

Hello, stevecowall!

Find all values of \(\displaystyle x\) such that the mean of 77, 137, and \(\displaystyle x\) is 1 more than their median.

The mean of the numbers is: .\(\displaystyle \dfrac{77+137+x}{3} \:=\:\dfrac{x+214}{3}\)


There are three cases to consider.


(1) \(\displaystyle x\) is the smallest number.
. . .Then we have: .\(\displaystyle x\;--\;77\,--\,137\)
. . .The median is 77.

. . .We have: .\(\displaystyle \dfrac{x+214}{3} \:=\:78 \quad\Rightarrow\quad \boxed{x \,=\,20}\)


(2) \(\displaystyle x\) is the largest number.
. . .Then we have: .\(\displaystyle 77\;--\;137\;--\;x\)
. . .The median is 137.

. . .We have: .\(\displaystyle \dfrac{x+214}{3} \:=\:138 \quad\Rightarrow\quad \boxed{x \,=\,200}\)


(3) \(\displaystyle x\) is the median.
. . .Then we have: .\(\displaystyle 77\;--\;x\;--\;137\)

. . .We have: .\(\displaystyle \dfrac{x+214}{3} \:=\:x + 1 \quad\Rightarrow\quad \boxed{x \,=\,105.5}\)