Sat Inequality Practice Problems

\(Q1.\:\) Solve for \( x: 5x-1 \le 3 \)

\( a) \quad x \le \frac{5}{4} \)

\( b) \quad x \le \frac{4}{5}\)

\(c) \quad x \le \frac{2}{5} \)

\( d) \quad x \le \frac{5}{2} \)

\( \text{Answer/Explanation}\)

\(Q2.\: \) if \( \frac{1}{3}x \le 1\) then

\( a) \quad x \in (3, \infty) \)

\( b) \quad x \in (-3, \infty)\)

\(c) \quad x \in (-\infty,3) \)

\( d) \quad x \in (-\infty, -3) \)

\( \text{Answer/Explanation}\)

\(Q3.\: \)For \(-14-2x\leq 11-7x\) which of the following is NOT a solution?

\( a) \quad 3 \)

\( b) \quad 4\)

\(c) \quad 5 \)

\( d) \quad 6 \)

\( \text{Answer/Explanation}\)

\(Q4.\:\) \(3x+4 \le 7x-5\) is equivalent to

\( a) \quad -4x \le 9 \)

\( b) \qquad 4x \ge 9\)

\(c) \qquad 4x \le 9 \)

\( d) \qquad 3x \le 7x-1 \)

\( \text{Answer/Explanation}\)

\(Q5.\:\) \(11x+6 \le 13x-9\) is NOT equivalent to

\( a) \quad -2x \le -15 \)

\( b) \qquad 2x \ge 15\)

\(c) \qquad 6 \le 2x-9 \)

\( d) \qquad 2x \le 15 \)

\( \text{Answer/Explanation}\)

\(Q6.\:\) For \(\frac{x+9}{6}>2 \) which of the following is a solution?

\( a) \quad 1 \)

\( b) \quad 2\)

\(c) \quad 3 \)

\( d) \quad 4 \)

\( \text{Answer/Explanation}\)

\(Q7.\:\) If \(x<y\) then which of the following is True?

\( a) \qquad 2x > 2y \)

\( b) \quad -2x > -2y\)

\(c) \qquad x > y \)

\( d) \qquad \frac{1}{x} < \frac{1}{y} \)

\( \text{Answer/Explanation}\)

\(Q8.\:\) If \(a<b\) then which of the following is NOT true?

\( a) \qquad 5a < 5b \)

\( b) \quad -a > -b\)

\(c) \quad -3a < -3b \)

\( d) \qquad \frac{1}{a} > \frac{1}{b} \)

\( \text{Answer/Explanation}\)

\(Q9.\:\) If \(\frac{x-3}{2} \le \frac{x-2}{3}\), what is the greatest possible value of \(x\)?

\( a) \quad 5 \)

\( b) \quad 6\)

\(c) \quad 7 \)

\( d) \quad 8 \)

\( \text{Answer/Explanation}\)

\(Q10.\:\) If \(\frac{x-3}{3} \le \frac{x+2}{4}\) , what is the greatest possible value of \(\frac{7x+5}{3}\)?

\( a) \quad 5 \)

\( b) \quad 6\)

\(c) \quad 7 \)

\( d) \quad 8 \)

\( \text{Answer/Explanation}\)

Leave a comment