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

Answer 1

Correct Answer: 105

Step 1: Use the Given Average of \( x, y, z \)

We are told:

\[ \frac{x + y + z}{3} = 80 \Rightarrow x + y + z = 240 \tag{1} \]

Step 2: Use the Given Average of \( x, y, z, u, v \)

It is also given:

\[ \frac{x + y + z + u + v}{5} = 75 \Rightarrow x + y + z + u + v = 375 \tag{2} \]

Step 3: Subtract Equation (1) from Equation (2)

\[ u + v = 375 - 240 = 135 \tag{3} \]

Step 4: Use Another Average Expression

It is given that:

\[ \frac{x + y}{2} + \frac{y + z}{2} = 135 \]

Simplify the left-hand side:

\[ \frac{x + y + y + z}{2} = \frac{x + 2y + z}{2} \Rightarrow \frac{x + 2y + z}{2} = 135 \Rightarrow x + 2y + z = 270 \tag{4} \]

Step 5: Subtract Equation (1) from Equation (4)

From equation (1): \( x + y + z = 240 \)

\[ x + 2y + z - (x + y + z) = 270 - 240 \Rightarrow y = 30 \]

Step 6: Plug Back to Find \( x + z \)

\[ x + y + z = 240 \quad \text{and} \quad y = 30 \Rightarrow x + z = 240 - 30 = 210 \]

Step 7: Minimize \( x \) Given \( x \geq z \)

Since \( x + z = 210 \), and \( x \geq z \), the minimum value of \( x \) occurs when \( x = z \). So:

\[ x = z = \frac{210}{2} = 105 \]

Final Answer:

\[ \boxed{x = 105} \]