Defective Balls in Box

Let $c, t, r$ be the number of defective cricket, tennis, and rubber balls. Total balls = 15. Given: $\frac{t}{5} < \frac{c}{6} < \frac{r}{4}$. Also: $\frac{c+t+r}{15} = 2 \left(\frac{t}{5}\right) = \frac{6t}{15} \implies c + t + r = 6t \implies c + r = 5t$. Since $t$ must be an integer and $t < 5$, try $t = 1 \implies c + r = 5$. Condition: $\frac{1}{5} < \frac{c}{6} < \frac{r}{4} \implies 0.2 < \frac{c}{6} < \frac{r}{4}$. Testing values for $(c, r)$ where $c + r = 5$:

• If $c = 2, r = 3$: $\frac{c}{6} = \frac{2}{6} \approx 0.333$, $\frac{r}{4} = \frac{3}{4} = 0.75$. Check: $0.2 < 0.333 < 0.75$ holds!
• If $c = 1, r = 4$: $\frac{c}{6} = 0.167 < 0.2$ (fails). Thus $z = r = 3$ uniquely.