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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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 $x \geq z$ , then the minimum possible value of $x$ is

Entered answer:

✅ Correct Answer: 105

We're told that the arithmetic mean of three numbers x, y, and z is 80.

Since the mean of x, y, z is 80:

$x + y + z = 80 \times 3 = 240$

The arithmetic mean of x, y, z, u, v is 75, so:

$x + y + z + u + v = 75 \times 5 = 375$

Since we know $x + y + z = 240$ :

$u + v = 375 - 240 = 135$

We're given that:

$u = \frac{x + y}{2}$

$v = \frac{y + z}{2}$

Substituting these into $u + v = 135$ :

$\frac{x + y}{2} + \frac{y + z}{2} = 135$

$\frac{x + y + y + z}{2} = \frac{x + 2y + z}{2} = 135$

$x + 2y + z = 270$

Now we have two equations:

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

$x + 2y + z = 270$ ... (2)

$(x + 2y + z) - (x + y + z) = 270 - 240$

$y = 30$

Since $y = 30$ and $x + y + z = 240$ :

$x + 30 + z = 240$

$x + z = 210$

We need to find the minimum value of x given that $x \geq z$ and $x + z = 210$ .

Since $x + z = 210$ is fixed, when x is at its minimum, z must be at its maximum.

Given the constraint $x \geq z$ , the minimum value of x occurs when $x = z$ .

When $x = z$ :

$x + x = 210$

$2x = 210$

$x = 105$

Therefore, the minimum possible value of x is 105.

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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 $x \geq z$ , then the minimum possible value of $x$ is