The Chief Examiner reported that majority of the candidates who attempted this question did not illustrate the given information in a diagram correctly.  This also affected their being able to solve the question correctly.  Candidates were expected to represent the information in a diagram as shown:

         Q     140o
                          
                                 8 km                         
     20km                                                     R                                 
                                           r
              o
         P
        
From the diagram, ∠PQR = 180o – 140o = 40o.  Using the cosine rule,
|PR|2 = 202 + 82 – 2(20)(8) cos 40o simplifying this equation gave
|PR| = 14.79411km.  Using the sine rule  =  where  is the bearing of R from P.  Solving this equation gave  = 20.34o i.e. bearing of R from P is 020o or N20oE.  Let r be the distance of R north of P.  Then r = |PR| cos  = 13 km, correct to 2 significant figures.