In part (a), candidates were expected to square both sides of the equality sign to obtain
t2 =   - r2q, multiply through by r to have rt2 = pq – r3q and factorize pq – r3q to have q(p – r3).        Hence, q =  .
This part of the question was very popular among the candidates and majority of them performed very well in it. However, some of them did not square both sides of the equality sign correctly. Many candidates were also reported not to factorize correctly.
In part (b), candidates were expected to recognize the question as involving the solution of simultaneous equations. Candidates were to first express the indicial equation in the same base and compare their indices i.e. 9(1 – x) = 27y becomes 32(1 – x) = 33y. Hence 2(1 – x) = 3y or 3y + 2x = 2. Candidates were then expected to solve 3y + 2x = 2 and x – y =   simultaneously to obtain x =  , y = 1. Adding x and y gave x + y =  .
Majority of the candidates attempted this question satisfactorily though there were a few others who were not able to obtain the second equation due to their inability to handle the indices involved.