Question

\(\frac{n}{4} + \frac{r}{8} = \frac{s}{8} + \frac{t}{6}\), where n, r, s, and t are positive integers.
Column A: \(2n + r\)
Column B: \(2s + t\)

• A. The quantity in Column A is greater.
• B. The quantity in Column B is greater.
• C. The two quantities are equal.
• D. The relationship cannot be determined from the information given.

Hint:

When comparing complex algebraic expressions, try to manipulate the given equation to express one column in terms of the other column's variables. If the resulting comparison still depends on the values of the variables, the answer is (D).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We are given an equation relating four positive integer variables. We need to compare two expressions involving these variables. The goal is to see if the given equation fixes the relationship between the two expressions.
Step 2: Detailed Explanation:
Let's manipulate the given equation to isolate the expression in Column A.
\[ \frac{n}{4} + \frac{r}{8} = \frac{s}{8} + \frac{t}{6} \]
To combine the terms on the left, find a common denominator (8).
\[ \frac{2n}{8} + \frac{r}{8} = \frac{s}{8} + \frac{t}{6} \]
\[ \frac{2n+r}{8} = \frac{s}{8} + \frac{t}{6} \]
Now, multiply both sides by 8 to solve for \(2n+r\).
\[ 2n+r = 8 \left(\frac{s}{8} + \frac{t}{6}\right) \]
\[ 2n+r = s + \frac{8t}{6} = s + \frac{4t}{3} \]
So, Column A is equal to \(s + \frac{4t}{3}\). We now need to compare this to Column B, which is \(2s+t\).
Comparison: \(s + \frac{4t}{3}\) vs \(2s+t\).
Subtract \(s\) and \(t\) from both sides to simplify the comparison.
We compare: \(\frac{4t}{3} - t\) vs \(2s - s\).
\[ \frac{t}{3} \quad \text{vs.} \quad s \]
Since \(s\) and \(t\) are independent positive integers, the relationship between \(\frac{t}{3}\) and \(s\) is not fixed.
Case 1: Let \(s=1\). For \(2n+r\) to be an integer, \(s + \frac{4t}{3}\) must be an integer, which means \(t\) must be a multiple of 3. Let \(t=3\). Then \(\frac{t}{3} = \frac{3}{3} = 1 = s\). In this case, the two sides of the comparison are equal, so Column A = Column B.
Case 2: Let \(s=1\) and \(t=6\). Then \(\frac{t}{3} = \frac{6}{3} = 2>s\). In this case, Column A>Column B.
Case 3: Let \(s=2\) and \(t=3\). Then \(\frac{t}{3} = \frac{3}{3} = 1<s\). In this case, Column A<Column B.
Since we found cases where A>B, A<B, and A = B, the relationship cannot be determined.
Step 3: Final Answer:
The relationship cannot be determined from the information given.