The Chief Examiner reported that this question was attempted by majority of the candidates and their performance was commended. The report further stated that candidates’ performance in part (a) was better than what it was in part (b).
In part (a), it was reported that majority of the candidates applied the appropriate rules of logarithms correctly. They were able to show that log10 – 2log10 + log10 = log10( ×  × )  = log1010 = 1.
In part (b), it was reported that majority of the candidates showed poor understanding of the question. Some others were able to obtain the population in each year but did not multiply this result by GH¢ 15.00 which was the amount per head. They were expected to respond as follows:

Year	Details	Amount  (GH¢)
2003	3000 × 15	45,000.00
2004	(3000 + ( × 3000)) × 15 = 3600 × 15	54,000.00
2005	(3600 + ( × 3600)) × 15 = 4320 × 15	64,800.00
2006	(4320 + ( × 4320)) × 15 = 5182 × 15	77,760.00
2007	(5182 + ( × 5182)) × 15 = 1036.4 × 15 (i.e. ≈ 1037 × 15)	93,315.00 (93,312.00 was accepted)
Total		334,875.00 (334,872.00 was accepted)

                             
GH¢ 334,872.00 was accepted where the population for year 2007 was
not rounded up to 1037.

 

The Chief Examiner reported that this question was attempted by majority of the candidates and their performance was better in part (a) than in part (b).
In part (a), candidates’ performance was commended. Majority of them were reported to apply the rule of BODMAS correctly. Candidates were expected to simplify the bracket first to obtain 5 - 2 =  -  = = . Therefore, 3 ÷ (5 - 2) + 5 became  ÷  +  =  ×  +  =  +  = 7.
In part (b), it was reported that candidates’ performance was not as good as it was in part (a). Majority of them were not able to obtain the sample space of the outcomes which was tabulated as follows

	2	3	4
1	1,2	1,3	1,4
3	3,2	3,3	3,4
5	5,2	5,3	5,4

From the table, the total number of possible outcomes = 9. The favourable outcomes were {(1, 3), (1, 4), (3, 2) and (3, 3)}. Therefore, the number of favourable outcomes = 4. Hence, the required probability = .