A function of two variable varies directly with x and inversely with y. Determine function when x = 5 and y = 3. Given that, for x = 0 and y = 1, f = 15 and for x = 1 and y = 15, f = 2.
10
Setup: Since \(f\) varies directly with \(x\) and inversely with \(y\), the relation is \(f(x,y) = \dfrac{kx}{y}\) for some constant \(k\).
Checking the given data:
Condition 1: \(f(0,1) = 15\) would force \(\dfrac{k\cdot 0}{1} = 15\), i.e. \(0 = 15\), which is impossible.
Condition 2: \(f(1,15) = 2\) gives \(k = 30\), so \(f(5,3) = \dfrac{30\cdot 5}{3} = 50\), which is not among the options.
Conclusion: The two given conditions are mutually inconsistent with a pure direct/inverse variation; the question contains a transcription error in the data. With the stated information, the answer 10 cannot be derived. This question should be treated as defective and corrected at source.
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