Question

x = a+y and a is constant while y $\alpha$ z, x = 30 when z = 6 and x = 24 when z = 2. Find x when z = 10.

Answer 1

$$\begin{gathered} \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}y\alpha z. \\ \Rightarrow y=kz\left(k\mathrm{is}a\mathrm{constant}.\right)...\left(\mathrm{i}\right) \\ \mathrm{Also}, \\ x=a+y \\ \therefore x=a+kz\left(\mathrm{From}\left(\mathrm{i}\right)\right) \\ \mathrm{When}x=30\mathrm{and}z=6: \\ 30=a+6k \\ \Rightarrow a=30-6k...\left(\mathrm{ii}\right) \\ \mathrm{When}x=24\mathrm{and}z=2: \\ 24=a+2k \\ \Rightarrow 24-2k=a...\left(\mathrm{iii}\right) \\ \mathrm{From}\left(\mathrm{ii}\right)\&\left(\mathrm{iii}\right),weget: \\ 30-6k=24-2k \\ \Rightarrow 30-24=-2k+6k \\ \Rightarrow 6=4k \\ \Rightarrow k=\frac{3}{2} \\ \mathrm{From}\left(\mathrm{ii}\right),weget: \\ a=30-6\times \frac{3}{2} \\ \Rightarrow a=21 \\ \mathrm{Equation}\mathrm{of}\mathrm{variation}: \\ x=21+\frac{3}{2}z \\ \mathrm{When}z=10: \\ x=21+\frac{3}{2}\times 10 \\ \Rightarrow x=21+15 \\ \Rightarrow x=36 \end{gathered}$$