If \(\frac{a}{b} = \frac{3}{2}\), which of the following must be true?
I. \(\frac{b}{a} = \frac{2}{3}\)
II. \(\frac{a-b}{a} = \frac{1}{3}\)
III. \(a+b = 5\)
I. \(\frac{b}{a} = \frac{2}{3}\)
II. \(\frac{a-b}{a} = \frac{1}{3}\)
III. \(a+b = 5\)
Show Hint
- I only
- II only
- III only
- I and II
- II and III
The Correct Option is D
Solution and Explanation
This question tests the ability to manipulate an algebraic proportion and determine which of the resulting statements are necessarily true.
Step 2: Detailed Explanation:
We are given the proportion \(\frac{a}{b} = \frac{3}{2}\). This means that \(a\) and \(b\) are in the ratio 3:2. We can express this as \(a = 3k\) and \(b = 2k\) for some non-zero number \(k\).
Statement I: \(\frac{b}{a} = \frac{2}{3}\)
This is the reciprocal of the given equation. If we take the reciprocal of both sides of \(\frac{a}{b} = \frac{3}{2}\), we get \(\frac{b}{a} = \frac{2}{3}\). This is always true. Alternatively, using substitution: \(\frac{2k}{3k} = \frac{2}{3}\). Statement I must be true.
Statement II: \(\frac{a-b}{a} = \frac{1}{3}\)
Let's substitute \(a = 3k\) and \(b = 2k\) into the expression: \[ \frac{3k - 2k}{3k} = \frac{k}{3k} = \frac{1}{3} \] This is always true as long as \(k \neq 0\), which is required for the original fraction to be defined. Statement II must be true.
Statement III: \(a+b = 5\)
Let's use our substitution. \(a+b = 3k + 2k = 5k\). The statement says \(a+b = 5\), which means \(5k = 5\), or \(k=1\). This is true only in the specific case where \(a=3\) and \(b=2\). However, other values are possible. For example, if \(k=2\), then \(a=6\) and \(b=4\). In this case, \(\frac{a}{b} = \frac{6}{4} = \frac{3}{2}\), but \(a+b = 10\). Since the statement is not true for all possible values of a and b, it is not a statement that must be true. Statement III is not necessarily true.
Step 3: Final Answer:
Statements I and II must be true.
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