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CAT Previous Year Paper | CAT Progressions and Series Questions | Question 29

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Question 29 : The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u = \\frac{(x+y)}{2}\\) and v = \\frac{(y+z)}{2}\\). If x ≥ z, then the minimum possible value of x is [TITA]


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Explanatory Answer

Method of solving this CAT Question from Progressions and Series

Given that AM (x, y, z) = 80
So, x + y + z = 80 × 3 = 240 ----- (1)
Also, AM (x, y, z, u, v) = 75
So, x + y + z + u + v = 75 × 5 = 375 ----- (2)
Equation (2) – (1), we get
u + v = 375 - 240 = 135
Its also given that, u = \\frac{x + y}{2}) and v = \\frac{y + z}{2})
So, u + v = \\frac{x + y + z}{2}) + \\frac{y}{2})
\\frac{240}{2}) + \\frac{y}{2}) = 135 => y = 30
Substituting this value in (1), we get x + z = 210
It’s given that x ≥ z
x would take the minimum value when x = z
=> 2x = 210
=> x (min) = 105

The question is "The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u = \\frac{(x+y)}{2}\\) and v = \\frac{(y+z)}{2}\\). If x ≥ z, then the minimum possible value of x is [TITA]"

Hence, the answer is 105

 

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