2017 AMC 10 B

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? (2017 AMC 10B Problem, Question#3)

  • A.

    y+x^2

  • B.

    y+xz

  • C.

    y+y^2

  • D.

    y+2y^2

  • E.

    y+z

Answer:E

Notice that y+z must be positive because \left| z\right|\gt\left| y\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 2017 AMC 10 B
Problem 2 2017 AMC 10 B
Problem 4 2017 AMC 10 B
Problem 5 2017 AMC 10 B
Think Academy Powered by ChatGPT