AMC 10 Daily Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 3 Easy

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

  • A.

    y+x^2

  • B.

    y+xz

  • C.

    y+y^2

  • D.

    y+2y^2

  • E.

    xy+z^2

Answer:E

Notice that xy+z^2 must be positive because \left| z^2\right|\gt 1\gt \left| xy\right|. Therefore the answer is \rm E.

The other choices:

\rm (A) As x grows closer to 0, x^2 decreases and thus becomes less than y.

\rm (B) x can be as small as possible (x>0), so xz grows close to 0 as x approaches 0.

\rm (C)  For all -1<y<0, y>y^2, and thus it is always negative.

\rm (D) The same logic as above, but when -\frac 12<y<0 this time.

Related Problems
Problem 1 AMC 10 Daily Practice Round 2
Problem 2 AMC 10 Daily Practice Round 2
Problem 4 AMC 10 Daily Practice Round 2
Problem 5 AMC 10 Daily Practice Round 2
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