CAT 2018 Question Paper Shift 2 Solved Paper

Question : 94 of 100

Marks: +1, -0

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 ofx is

Your Answer:

Validate

Solution:

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+2 y+z=270

...........(4)
Since

x ≥ z, x

takes the minimum possible value at

{x}=105

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