(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 ...
Question 1 Report
(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 \(\log_{3} (4x + 1) - \log_{3} (3x - 5) = 2\).
Answer Details
| Limes | Apples | Total | |
|---|---|---|---|
| Good | 10 | 8 | 18 |
| Bad | 6 | 6 | 12 |
| Total | 16 | 14 | 30 |
There are 30 fruits. Picking is without replacement, so the second denominator is 29.
(i) Both good limes (10 good limes):
\[ \frac{10}{30}\times\frac{9}{29}=\frac{90}{870}=\frac{3}{29}\approx 0.103 \](ii) Both bad fruits (12 bad fruits):
\[ \frac{12}{30}\times\frac{11}{29}=\frac{132}{870}=\frac{22}{145}\approx 0.152 \](iii) One good apple and one bad lime (either order; 8 good apples, 6 bad limes):
\[ \frac{8}{30}\times\frac{6}{29}+\frac{6}{30}\times\frac{8}{29}=\frac{48+48}{870}=\frac{96}{870}=\frac{16}{145}\approx 0.110 \](b) Solve \(\log_3(4x+1)-\log_3(3x-5)=2\). Combine the logs:
\[ \log_3\!\left(\frac{4x+1}{3x-5}\right)=2 \;\Rightarrow\; \frac{4x+1}{3x-5}=3^2=9 \] \[ 4x+1=9(3x-5)=27x-45 \;\Rightarrow\; 46=23x \;\Rightarrow\; x=2 \]Check: \(3x-5=1>0\) and \(4x+1=9>0\), so \(x=2\) is valid.
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