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
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
Updated On: Sep 30, 2024
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Correct Answer: 105
Solution and Explanation
Given x+y+z/3 = 80
⇒ x+y+z = 240....(1)
also x+y+z+u+v/5 = 75
x+y+z+u+v = 375…. (2)
From (1) and (2), u + v = 135…. (3)
x+y/2+y+z/2 = 135
x+2y+z = 270...(4)
From (1) & (4), y=30
⇒ x+z = 210
Since x ≥ z, x takes the minimum possible value at
x = 105
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