Question

If $x\lt a\lt0$ means that $x$ and $a$ are numbers such that $x$ is less than $a$ and $a$ is less than zero, then:

$x^2\lt ax\lt0$

$x^2\gt ax\gt a^2$

$x^2\lt a^2\lt0$

$x^2\gt ax$ but $ax\lt0$

$x^2\gt a^2$ but $a^2\lt0$

Answer: B

Concepts:inequalities negative numbers multiplication

Difficulty rating: 1340

Small Hint:

Both numbers are negative, but $x$ has the larger absolute value

Big Hint:

Multiply $x\lt a$ once by $x\lt0$ and once by $a\lt0,$ reversing each inequality

Solution:

Since $x\lt a\lt0,$ multiplying $x\lt a$ by the negative number $x$ reverses the inequality and gives $x^2\gt ax.$ Multiplying the same inequality by the negative number $a$ gives $ax\gt a^2.$ Therefore $x^2\gt ax\gt a^2.$

Thus, the correct answer is B.

1956 AMC 12 Problem 44

Answer: B

Difficulty rating: 1340

Small Hint:

Big Hint:

Solution:

Problem 44 in Other Years