Question

A box contains 6 cricket balls, 5 tennis balls and 4 rubber balls. Of these, some balls are defective. The proportion of defective cricket balls is more than the proportion of defective tennis balls but less than the proportion of defective rubber balls. Moreover, the overall proportion of defective balls is twice the proportion of defective tennis balls. What BEST can be said about the number of defective rubber balls in the box?

Accepted Answer

C. 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.

Defective Balls in Box

Explanation

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