Three quantities A, B, C are such that AB = kC, where k is a constant. When A is kept constant, B varies directly as C; When B is kept constant, A varies directly as C and when C is kept constant, A varies inversely as B. Initially, A was at 10 and A : B : C was 1 : 3 : 5. Find the value of A when B equals 18 at constant C.
Explanation:
Initial values are 10, 30 and 50.
Thus, we have 10 × 30 = k × 50.
Hence, k = 6
Thus, the equation is AB = 6C.
For the problem, keep C constant at 50. Then, A × 18 = 6 × 50.
i.e. A = 300/18 = 16.66
Hence, option (b).
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