Minimum possible value of x

CAT 2018 Slot 2 · QA · Hard · Averages

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

Answer

105

Explanation

Given average of x,y,zx, y, z is 80: x+y+z=240x + y + z = 240

Average of x,y,z,u,vx, y, z, u, v is 75: x+y+z+u+v=375x + y + z + u + v = 375 Since x+y+z=240x + y + z = 240, we have u+v=375240=135u + v = 375 - 240 = 135.

Given u=x+y2u = \frac{x+y}{2} and v=y+z2v = \frac{y+z}{2}: u+v=x+2y+z2=135    x+2y+z=270u + v = \frac{x + 2y + z}{2} = 135 \implies x + 2y + z = 270

Subtracting x+y+z=240x + y + z = 240 gives: y=30y = 30

Then x+z=24030=210x + z = 240 - 30 = 210. Since xzx \ge z, to minimize xx, zz should be as large as possible. Since xzx \ge z, maximum z=105z = 105 (when x=z=105x = z = 105).

Therefore, the minimum possible value of xx is 105.

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