CAT 2018 Quant Questions

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

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

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

$\Rightarrow$ $x + y + z = 240$ ... (1)

Also, $\frac{{x + y + z + v + u}}{5} = 75$

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

$\Rightarrow$ $x + y + z + v + u = 375$

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.

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