Inequality & Modulus: Level 2
1.
What is the only integer \(n\) such that \(2n + 5 > 15\) and \(3n < 20\)?
Solution (C):
1. Solve the first inequality: \(2n + 5 > 15 \implies 2n > 10 \implies n > 5\).
2. Solve the second inequality: \(3n < 20 \implies n < \frac{20}{3} \implies n < 6.66...\)
3. We need an integer \(n\) that is greater than 5 and less than 6.66...
4. The only integer in this range is 6.
2.
If \(|2x – 1| = 5\), what is the product of all possible values of \(x\)?
Solution (A):
Case 1: \(2x – 1 = 5 \implies 2x = 6 \implies x = 3\).
Case 2: \(2x – 1 = -5 \implies 2x = -4 \implies x = -2\).
The product of the values is \((3) \times (-2) = -6\).
3.
Which of the following is equivalent to the statement \( |x – 1| > 3 \)?
Solution (D): An absolute value inequality \(|a| > b\) unwraps to two separate inequalities:
Case 1: \(x – 1 > 3 \implies x > 4\).
Case 2: \(x – 1 < -3 \implies x < -2\).
The solution is \(x < -2\) or \(x > 4\).
4.
If \( \frac{x}{2} – 1 > \frac{x}{5} \), which of the following must be true?
Solution (C):
1. Multiply the entire inequality by the common denominator (10) to clear fractions:
\( 10 \times (\frac{x}{2} – 1) > 10 \times (\frac{x}{5}) \)
2. \( 5x – 10 > 2x \)
3. Subtract \(2x\) from both sides: \( 3x – 10 > 0 \)
4. Add 10 to both sides: \( 3x > 10 \)
5. \( x > \frac{10}{3} \).
5.
If \(-1 < x < 0\), which of the following must be true?
I. \(x^2 > x\)
II. \(x > 1/x\)
III. \(x^3 > x\)
I. \(x^2 > x\)
II. \(x > 1/x\)
III. \(x^3 > x\)
Solution (E): Let’s test \(x = -0.5\).
I. \(x^2 > x \implies (-0.5)^2 > -0.5 \implies 0.25 > -0.5\). This is **True**.
II. \(x > 1/x \implies -0.5 > 1/(-0.5) \implies -0.5 > -2\). This is **True**.
III. \(x^3 > x \implies (-0.5)^3 > -0.5 \implies -0.125 > -0.5\). This is **True**.
All three statements must be true.
6.
If \(|a| = \frac{1}{2}\) and \(|b| = \frac{1}{4}\), which of the following CANNOT be the result of \(a + b\)?
Solution (A):
\(a = \frac{1}{2}\) or \(a = -\frac{1}{2}\).
\(b = \frac{1}{4}\) or \(b = -\frac{1}{4}\).
Possible sums:
1. \(\frac{1}{2} + \frac{1}{4} = \frac{3}{4}\)
2. \(\frac{1}{2} – \frac{1}{4} = \frac{1}{4}\)
3. \(-\frac{1}{2} + \frac{1}{4} = -\frac{1}{4}\)
4. \(-\frac{1}{2} – \frac{1}{4} = -\frac{3}{4}\)
The set of possible values is \(\{ \frac{3}{4}, \frac{1}{4}, -\frac{1}{4}, -\frac{3}{4} \}\). The value 0 cannot be the result.
7.
If \(|ab| > ab\), which of the following must be true?
I. \(a < 0\)
II. \(b < 0\)
III. \(ab < 0\)
I. \(a < 0\)
II. \(b < 0\)
III. \(ab < 0\)
Solution (C): The absolute value of a number \(|k|\) is equal to \(k\) if \(k \ge 0\).
The absolute value \(|k|\) is greater than \(k\) only if \(k\) is negative.
In this problem, the number \(k\) is \(ab\).
The statement \(|ab| > ab\) means that \(ab\) must be a negative number.
Therefore, \(ab < 0\). This is statement III.
We don't know if \(a\) is negative or \(b\) is negative, just that they have opposite signs.
8.
If \(5 – 3x < 14\), then which of the following could NOT be the value of \(x\)?
Solution (B):
1. Solve the inequality: \(5 – 3x < 14\)
2. Subtract 5: \(-3x < 9\)
3. Divide by -3 and **flip the inequality sign**: \(x > -3\).
4. The value of \(x\) must be greater than -3.
5. We are looking for the value that *could not* be \(x\). This would be any number \(\le -3\).
6. Of the options, -4 is not greater than -3.
9.
If \((a-b)c < 0\) and \(c > 0\), which of the following must be true?
Solution (A):
1. We have a product of two terms, \((a-b)\) and \(c\), that is negative.
2. This means the two terms must have opposite signs.
3. We are given that \(c > 0\) (c is positive).
4. Therefore, the other term, \((a-b)\), must be negative.
5. \(a – b < 0 \implies a < b\).
10.
A parking meter holds a maximum of $2.00 (200 cents). It accepts only quarters (25 cents) and dimes (10 cents). If there are 5 quarters in the meter, what is the maximum number of dimes?
Solution (D):
1. Let \(q\) be quarters and \(d\) be dimes. The inequality is \(25q + 10d \le 200\).
2. We are given \(q = 5\). Substitute this value:
\(25(5) + 10d \le 200\)
3. \(125 + 10d \le 200\)
4. Subtract 125: \(10d \le 75\)
5. Divide by 10: \(d \le 7.5\)
6. Since the number of dimes (\(d\)) must be an integer, the maximum possible value for \(d\) is 7.
Score: 0 / 10