(a)
| Limes | Apples | |
| Good | 10 | 8 |
| Bad | 6 | 6 |
The table shows the number of limes and apples of the same size in a bag. If two of the fruits are picked at random, one at a time, without replacement, find the probability that : (i) both are good limes ; (ii) both are bad fruits ; (iii) one is a good apple and the other a bad lime.
(b) Solve the equation \(\log_{3} (4x + 1) – \log_{3} (3x – 5) = 2\).
Explanation
(a) Total number of fruits = 16 limes + 14 apples = 30
(i) P(both are good limes) = \(\frac{10}{30} \times \frac{9}{29} = \frac{3}{29}\)
(ii) P(both are bad fruits) = \(\frac{12}{30} \times \frac{11}{29} = \frac{22}{145}\)
(iii) P(one is a good apple and the other a bad lime) = \((\frac{8}{30} \times \frac{6}{29}) + (\frac{6}{30} \times \frac{8}{29})\)
= \(\frac{8}{145} + \frac{8}{145}\)
= \(\frac{16}{145}\)
Note : The arrangement is not specific. Hence, it can be a good apple first and then a bad lime or a bad lime first and then a good apple.
(b) \(\log_{3} (4x + 1) - \log_{3} (3x - 5) = 2\)
\(\log_{3} (\frac{4x + 1}{3x - 5}) = 2\)
\((\frac{4x + 1}{3x - 5}) = 3^{2}\)
\(\frac{4x + 1}{3x - 5} = 9 \implies 4x + 1 = 9(3x - 5)\)
\(4x + 1 = 27x - 45 \implies 23x = 46\)
\(x = 2\)