Question

The smallest prime number satisfying the inequality $\frac{2n-3}{3}\ge\frac{n-1}{6}+1$ is

• A. 2
• B. 3
• C. 5
• D. 7
• E. 11

Hint:

Inequality Tip: Always simplify the inequality entirely before testing conditions like "prime" or "integer". Testing values too early can lead to unnecessary arithmetic mistakes.

Answer 1

The Correct Option is C

Concept:
To solve a linear inequality involving fractions, find a common denominator to eliminate the fractions, isolate the variable, and then evaluate the condition (in this case, finding the smallest prime number) within the resulting solution set.

Step 1: Set up the inequality.
Write down the given algebraic inequality: $$\frac{2n-3}{3} \ge \frac{n-1}{6} + 1$$

Step 2: Eliminate the denominators.
Multiply the entire inequality by the least common multiple (LCM) of 3 and 6, which is 6: $$6 \left( \frac{2n-3}{3} \right) \ge 6 \left( \frac{n-1}{6} \right) + 6(1)$$ $$2(2n - 3) \ge (n - 1) + 6$$

Step 3: Expand and simplify.
Distribute the 2 on the left side and combine the constant terms on the right side: $$4n - 6 \ge n + 5$$

Step 4: Isolate the variable n.
Subtract $n$ from both sides, and add 6 to both sides: $$3n \ge 11$$ $$n \ge \frac{11}{3}$$ $$n \ge 3.66...$$

Step 5: Find the smallest prime number satisfying the condition.
The value of $n$ must be greater than or equal to roughly 3.66. The sequence of prime numbers is 2, 3, 5, 7, 11, etc. The absolute smallest prime number that is strictly greater than 3.66 is 5. Hence the correct answer is (C) 5.