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Bunuel
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The average (arithmetic mean) of x, y and z is 80, and the average (ar 08 Apr 2020, 10:29
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The average (arithmetic mean) of x, y and z is 80, and the average (arithmetic mean) of x, y, z, u and v is 75. If \(u = \frac{x+y}{2}\), \(v = \frac{y+z}{2}\) and \(x ≥ z\), then what is the minimum possible value of x ?

A. 104
B. 105
C. 106
D. 107
E. 106

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B
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The average (arithmetic mean) of x, y and z is 80, and the average (ar 08 Apr 2020, 10:49
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    \(x + y + z = 80*3 = 240\)…(I)
    \(x + y + z + u + v = 75*5 = 375\)
      \(u + y = 375 – 240 = 135\)…(II)
    \(u = \frac{(x + y)}{2}\)
    \(v = \frac{(y + z)}{2}\)
      \(u + v = \frac{(x + 2y + z)}{2}\)
      \(2*135 = x + 2y + z\)
      \(x + 2y + z = 270\)…(III)
EQ (III) – EQ (II)
    \(y = 270 – 240 = 30\)
      \(x + z = 240-30 = 210\)
    Minimum possible value of x is when x = z, all other values of x will be greater than z
      \(2x = 210\)
      \(x = 105\)
Answer: Option B.
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The average (arithmetic mean) of x, y and z is 80, and the average (ar 10 Apr 2020, 08:37
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The average (arithmetic mean) of x, y and z is 80, and the average (arithmetic mean) of x, y, z, u and v is 75. If \(u = \frac{x+y}{2}\), \(v = \frac{y+z}{2}\) and \(x ≥ z\), then what is the minimum possible value of x ?

A. 104
B. 105
C. 106
D. 107
E. 106

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Given: The average (arithmetic mean) of x, y and z is 80, and the average (arithmetic mean) of x, y, z, u and v is 75.

Asked: If \(u = \frac{x+y}{2}\), \(v = \frac{y+z}{2}\) and \(x ≥ z\), then what is the minimum possible value of x ?

x + y + z = 240

x + y + z + u + v = 75*5 = 375

u + v = 135

x + y = 2u
y + z = 2v

x - z = 2 (u-v) >= 0

x + 2y + z = 2*135 = 270

y = 30
x+z = 210

2x = 210 + 2(u-v)
x = 105 + (u-v)

Minimum possible value of x = 105 ; Since u-v>=0

IMO B
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The average (arithmetic mean) of x, y and z is 80, and the average (ar 10 Apr 2020, 11:13
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Given
x>=z
x+y+z=80*3 (equation 1)
x+y+z+u+v=75*5 (equation 2)
u=(x+y)/2 (equation 3)
v=(y+z)/2 (equation 4)

Putting value of x+y+z from equation 1 into 2.
240+u+v=375
u+v=135
replace u and v in terms of x and y by using equation 3 and 4.

(x+y)/2+(y+z)/2=135
x+y+y+z=270 (equation 5)

putting value of x+y+z from equation 1 in equation 5.

240+y=270
y=30

putting value of y in equation 1.
x+z=210

Now, x has to be greater than or equal to z.
Average of x and z = 105
Condition is that x has to be minimum and yet greater than or equal to z.
This is only possible if x will be equal to z.
Hence, we are constrained to select a value of x which is equal to z.
Option B is the right choice.
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The average (arithmetic mean) of x, y and z is 80, and the average (ar 11 Apr 2020, 04:26
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Mean of X,Y, Z = 80
\(\frac{(X+Y+Z)}{3}\) = 80
therefore, X+Y+Z = 240

Mean of X,Y,Z,U,V = 75
\(\frac{(X+Y+Z+U+V)}{5}\) =75;
240 + U+V = 75*5
U+V = 375-240= 135

\(U= \frac{(X+Y)}{2}\) and \(V = \frac{(Y+Z)}{2}\)
U+V = \(\frac{(X+Y+Z+Y)}{2} = 135\)
240+Y = 135*2
Y= 30

X+Z = 240-30
X+Z = 210

Now its given that \(X>=Z\)

so X should be 105. Answer option B
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The average (arithmetic mean) of x, y and z is 80, and the average (ar 27 Dec 2022, 10:40
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Enchanting
    \(x + y + z = 80*3 = 240\)…(I)
    \(x + y + z + u + v = 75*5 = 375\)
      \(u + y = 375 – 240 = 135\)…(II)
    \(u = \frac{(x + y)}{2}\)
    \(v = \frac{(y + z)}{2}\)
      \(u + v = \frac{(x + 2y + z)}{2}\)
      \(2*135 = x + 2y + z\)
      \(x + 2y + z = 270\)…(III)
EQ (III) – EQ (II)
    \(y = 270 – 240 = 30\)
      \(x + z = 240-30 = 210\)
    Minimum possible value of x is when x = z, all other values of x will be greater than z
      \(2x = 210\)
      \(x = 105\)
Answer: Option B.
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Awesome response, there's just a small mistake at first when you subtracted the two equations you had "u+v= 135" but instead you wrote "u+y=135" kinda threw me off at first not gonna lie haha, thank you for your help !
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The average (arithmetic mean) of x, y and z is 80, and the average (ar 23 Nov 2024, 22:26
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