CAT 2018

Slot 2

QA

Question & Solution

Given that the arithmetic mean of x, y and z is 80.

$\Rightarrow$ $\dfrac{x+y+z}{3} = 80$

$\Rightarrow$ $x+y+z = 240$  ... (1)

Also,  $\dfrac{x+y+z+v+u}{5} = 75$

$\Rightarrow$ $\dfrac{x+y+z+v+u}{5} = 75$

$\Rightarrow$ $x+y+z+v+u = 375$

Substituting values from equation (1),

$\Rightarrow$ $v+u = 135$

It is given that u=(x+y)/2 and v=(y+z)/2.

$\Rightarrow$ $(x+y)/2+(y+z)/2 = 135$

$\Rightarrow$ $x+2y+z = 270$

$\Rightarrow$ $y = 30$   (Since $x+y+z = 240$) 

Therefore, we can say that $x+z = 240 - y = 210$. We are also given that x ≥ z, 

Hence, $x_{min}$ = 210/2 = 105.