(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)