Question:
The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x ≥ z, then the minimum possible value of x is
Correct Answer: 105
Given $\frac{x+y+z}{3}=80$
$\Rightarrow x+y+z=240$ …. (1)
also $\frac{x+y+z+u+v}{5}=75$
$x+y+z+u+v=375$ …. (2)
From (1) and (2), $\mathrm{u}+\mathrm{v}=135$ …. (3)
$\frac{\mathrm{x}+\mathrm{y}}{2}+\frac{\mathrm{y}+\mathrm{z}}{2}=135$
$x+2 y+z=270$ …. (4)
From (1) & (4), y=30
$\Rightarrow x+z=210$
Since x ≥ z, x takes the minimum possible value at
x = 105
Given $\frac{x+y+z}{3}=80$
$\Rightarrow x+y+z=240$ …. (1)
also $\frac{x+y+z+u+v}{5}=75$
$x+y+z+u+v=375$ …. (2)
From (1) and (2), $\mathrm{u}+\mathrm{v}=135$ …. (3)
$\frac{\mathrm{x}+\mathrm{y}}{2}+\frac{\mathrm{y}+\mathrm{z}}{2}=135$
$x+2 y+z=270$ …. (4)
From (1) & (4), y=30
$\Rightarrow x+z=210$
Since x ≥ z, x takes the minimum possible value at
x = 105
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