Question

The numerator of a fraction is $6x+1,$ the denominator is $7-4x,$ and $x$ can have any value between $-2$ and $2,$ both included. The values of $x$ for which the numerator is greater than the denominator are:

$\dfrac35\lt x\le2$

$\dfrac35\le x\le2$

$0\lt x\le2$

$0\le x\le2$

$-2\le x\le2$

Answer: A

Concepts:inequality linear equation

Difficulty rating: 1110

Small Hint:

Compare the numerator and denominator directly by solving $6x+1\gt7-4x$

Big Hint:

Intersect the resulting inequality with the given interval $[-2,2]$

Solution:

The required comparison gives $6x+1\gt7-4x,$ so $10x\gt6$ and $x\gt\frac{3}{5}.$ Intersecting this with $-2\le x\le2$ gives $\frac35\lt x\le2.$

Thus, the correct answer is A.

1958 AMC 12 Problem 34

Answer: A

Difficulty rating: 1110

Small Hint:

Big Hint:

Solution:

Problem 34 in Other Years