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2017 AMC 10B Problem 3

Problem 3 of 25EasierAlgebra

Real numbers x,x, y,y, and zz satisfy the inequalities 0<x<1,0 < x < 1, −1<y<0,-1 < y < 0, and 1<z<2.1 < z < 2. Which of the following numbers is necessarily positive?

Answer choices

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Solution

Since −1<y-1 < y and 1<z,1 < z, we can add the inequalities to see that 0<y+z.0 < y+z. This naturally proves choice E correct. Furthermore, we can eliminate every other choice with the following values: x=0.1,x=0.1,y=−0.25,y=-0.25,z=1.25.z=1.25. Thus, the correct answer is E .

More practice

Concepts: inequality · counterexample

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.