General Mathematics Paper 2, May/June. 2015

Question 1

(a)  Without using Mathematical tables or calculator, simplify
3¸ (5 - 2) + 5.

(b)  A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. 
Find the probability that the sum of the two numbers is greater than 3 and less
than 7.

Observation

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 = .