$$x\cdot\frac{7}{15}=\frac{7}{15}x\ and\ x\cdot\frac{41}{12}=\frac{41}{12}x$$
$$What\ is\ the\ smallest\ possible\ value\ of\ x?$$
$$going\ through\ all\ available\ options\ one\ after\ the\ other$$
$$if\ x=\frac{7}{60}\ \ for\ \ \frac{7}{15}x$$
$$\frac{7}{15}\cdot\frac{7}{60}=\frac{14}{900}$$
$$this\ is\ not\ an\ integer\ so\ x\ cannot\ be\ \frac{7}{60}$$
$$if\ x=\frac{60}{7}\ \ for\ \ \frac{7}{15}x$$
$$\frac{7}{15}\cdot\frac{60}{7}=4$$
$$if\ x=\frac{60}{7\ }\ \ for\ \ \frac{41}{12}x$$
$$\frac{41}{12}\cdot\frac{60}{7}=\frac{205}{7}$$
$$this\ is\ not\ an\ integer\ so\ x\ cannot\ be\ \frac{60}{7}$$
$$if\ x=\frac{7}{30}\ \ for\ \ \frac{7}{15}x$$
$$\frac{7}{15}\cdot\frac{7}{30}=\frac{14}{450}$$
$$this\ is\ not\ an\ integer\ so\ x\ cannot\ be\ \frac{7}{30}$$
$$if\ x=\frac{30}{7}\ \ for\ \ \frac{7}{15}x$$
$$\frac{7}{15}\cdot\frac{30}{7}=2$$
$$if\ x=\frac{30}{7\ }\ \ for\ \ \frac{41}{12}x$$
$$\frac{41}{12}\cdot\frac{30}{7}=\frac{205}{14}$$
$$this\ is\ not\ an\ integer\ so\ x\ cannot\ be\ \frac{30}{7}$$
$$if\ x=\frac{15}{7}\ \ for\ \ \frac{41}{12}x$$
$$\frac{41}{12}\cdot\frac{15}{7}=\frac{615}{84}$$
$$this\ is\ not\ an\ integer\ so\ x\ cannot\ be\ \frac{15}{7}$$
Note: There is no correct answer in the available options. Hence, we cannot ascertain a definite answer