The average (arithmetic mean) of x, y and z is 80, and the average (ar

Given
x>=z
x+y+z=80*3 (equation 1)
x+y+z+u+v=75*5 (equation 2)
u=(x+y)/2 (equation 3)
v=(y+z)/2 (equation 4)

Putting value of x+y+z from equation 1 into 2.
240+u+v=375
u+v=135
replace u and v in terms of x and y by using equation 3 and 4.

(x+y)/2+(y+z)/2=135
x+y+y+z=270 (equation 5)

putting value of x+y+z from equation 1 in equation 5.

240+y=270
y=30

putting value of y in equation 1.
x+z=210

Now, x has to be greater than or equal to z.
Average of x and z = 105
Condition is that x has to be minimum and yet greater than or equal to z.
This is only possible if x will be equal to z.
Hence, we are constrained to select a value of x which is equal to z.
Option B is the right choice.

Mean of X,Y, Z = 80
\(\frac{(X+Y+Z)}{3}\) = 80
therefore, X+Y+Z = 240

Mean of X,Y,Z,U,V = 75
\(\frac{(X+Y+Z+U+V)}{5}\) =75;
240 + U+V = 75*5
U+V = 375-240= 135

\(U= \frac{(X+Y)}{2}\) and \(V = \frac{(Y+Z)}{2}\)
U+V = \(\frac{(X+Y+Z+Y)}{2} = 135\)
240+Y = 135*2
Y= 30

X+Z = 240-30
X+Z = 210

Now its given that \(X>=Z\)

so X should be 105. Answer option B

Awesome response, there's just a small mistake at first when you subtracted the two equations you had "u+v= 135" but instead you wrote "u+y=135" kinda threw me off at first not gonna lie haha, thank you for your help !