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

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Correct Answer: 105

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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CAT 2018 Quant Questions
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