Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median...

Question

Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?

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3 kilograms We assign variables to the weights in sorted order to avoid ambiguity. x = weight of lightest box, y = weight of median box, z = weight of heaviest box We translate the average weight condition into an equation. x + y + z/3 = 7 We translate the median weight condition into an equation. y = 9 We use the two equations to find a relationship between x and z. x + y + z = 21 x + z = 21 - y x + z = 21 - 9 x + z = 12 Since y is the median, we know x y z. Using y=9 and x+z=12, we enforce z9 to find x. z = 12 - x z 9 12 - x 9 x 3

Explanation	Calculations	Help
$We assign variables to the weights in sorted order to avoid ambiguity.$	$x = weight of lightest box, y = weight of median box, z = weight of heaviest box$
$We translate the average weight condition into an equation.$	$\frac{x + y + z}{3} = 7$
$We translate the median weight condition into an equation.$	$y = 9$
$We use the two equations to find a relationship between x and z .$	$x + y + z = 21 x + z = 21 - y x + z = 21 - 9 x + z = 12$
$Since y is the median, we know x \leq y \leq z . Using y = 9 and x + z = 12, we enforce z \geq 9 to find x .$	$z = 12 - x z \geq 9 12 - x \geq 9 x \leq 3$

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