In the diagram,  is parallel to. If ∠AEF = 3xo, ∠ABC = 120o and ∠CHG = 7xo, find the value of ∠GHB.
The Chief Examiner reported that this question was attempted by majority of the candidates and they performed better in part (a) than in part (b).  However, it was also reported that while some of the candidates were able to simplify the fraction in part (a), they did not express the solution in standard form as required. 
In part (a), majority of the candidates were reported to have multiplied both the numerator and the denominator of the fraction by 106 to obtain:
 x   =    =     =   5 x 101

In part (b), majority of the candidates were reported not to have gotten the question right.  Candidates were expected to construct a line parallel to  and passing through B or produce line AB to meet GH.  This would have led them to obtain the equation
3xo + 180o – 7xo =1200. Solving this equation gave xo=15o.  Therefore angle GHB would be 180o – 7(15o) =750.