Mathematics Nov/Dec 2009

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The part (a) of the question required the use of BODMAS rule, finding the LCM and simplifying.  Most candidates were reported to have followed this procedure and were able to manipulate the fractions well.  However, some candidates misinterpreted ” ÷” to mean “ + ”.   Others subtracted first before dividing.       31/3  ÷  11/4  –  4/9   =  10/3  x  4/5  –  4/9   =   8/3  – 4/9   =   20/9   =  22/9
            
            In part (b), many candidates reportedly found the expansion of (√6 -1)2
quite challenging.  Others could not reduce √96 to its lowest form.  The expansion of (√6 -1)2 = (√6 -1) (√6 -1) = 6 -√6 -√6 + 1 = 7 – 2√6.        √96 = √6 x 16 = 4√6
Therefore 2 + √96 – 4(√6 -1)2 = 2 + 4√6 – 4(7-2√6) = 2 + 4√6 – 28 + 8√6 which after a little computation resulted in -26 + 12√6.

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They were able to find the LCM of the denominators and cross multiply both sides of the equality sign by the LCM which was 12. Simplifying gave 4(x-1) – 6(3x – 1) -3(5-2x) which further gave x = 31/4.  In part (b), it was reported that the question was attempted by majority of the candidates who were able to form the simultaneous equation 2x + 3y = 31; 3x+2y = 29, solve these equation simultaneously to obtain x = 5 and y = 7.  However, some of the candidates did not add these values to obtain the required price which was GH¢12.00.

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It was reported that the performance of candidates in this question was not commendable.  They found it difficult; hence they scored very low marks.  They were expected to show that if P(M) = probability of the malaria patient surviving = 0.27, probability of not surviving, P(M¢) = 1- 0.27 = 0.73.  Similarly, if P(T) = probability of the thypoid patient surviving = 0.85, then probability of not surviving = P(T1)  =  1-P(T) = 1- 0.85 = 0.15.  Probability of either surviving = (0.27 x 0.15) + (0.85 x 0.73) = 0.66. Probability of neither surviving =

(0.73 x 0.15) = 0.11.  Probability of at least one surviving = 1- 0.11 = 0.89.

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 A few other candidates were not able to draw the diagram correctly.
 

This diagram would have aided them to establish that the length of the arc of the circle = circumference of the base of the cone and that the radius of the sector = slant height of the cone. i.e. 135/360 x 2 x p x 40/1 = 2 x p x r where r = base radius of cone. Simplifying gave r = 15cm.

Slant height of cone, = l  =  40cm, r = 15cm. Using pythagoras theorem,
h2 + r2 = l 2  where h = perpendicular height of cone.
\ h = √402 – 152 = 37.081 cm.  Therefore volume of Cone = 1/3 x p x r2 x h

1 x 22 x 15 x 15 x 37.081   =  8741cm3 to the nearest cm3.

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The report stated that majority of the candidates could not answer part (a) of this question correctly which showed a poor knowledge of circle theorems.  They were expected to join X and Y to O.  This made <XOY and <XAY opposite angles of a cyclic quadrilateral and are supplementary i.e. a° + <XOY = 180°.  Also, <XOY = 2b° i.e. angle at the centre of a circle is twice angle at the circumference.  This implied that 2b° = 180 – a° or a° + 2b°  = 180°. 
It was reported that candidates performed better in part (b) than in part (a).  A good number of them were able to find the tangent of the angle at R and hence the angle at R as 36.87°.  Thus, they were able to obtain /TR/ = 8 cos R i.e. 8 cos 36.87° = 6.4cm.

ANSWER

 (a) of the question and performed well in it, it was observed that majority of them converted first to base ten, solved it and converted back to base 2.  Very few candidates were reported to have worked in base 2.  1112 = 112 = 101012. 
110002 – 101012 = 112. 

In part (b), majority of the candidates were reported to have found the question quite challenging.  They were expected to show that if p is the cost price of a television set, then the total sales for 18 sets = 115 x 18p = 20.7p.  Similarly, total 100 sales for 2 sets = 95/100 x 2p = 1.9p.  Therefore total sales on the 20 sets x 20.7p + 1.9p = 22.6p.  Total cost price  on the 20 sets = 20p.  Hence, percentage profit
=  22.6p – 20p    x  100  =  13%.
        20p                   1

ANSWER

The grouped frequency table is given by
   Class        f        Cf       Class
  Interval                         Boundary
 

    21-30     2        2        20.5 – 30.5  
                31-40    10      12       30.5 – 40.5
                41-50    12      24       40.5 – 50.5
                51-60    15      39       50.5 – 60.5
                61-70      8      47       60.5 – 70.5
                71-80      3      50       70.5 – 80.5
 

               The graph is given below

The modal mark derived from the graph = 53 while the probability = no. of candidates who scored marks greater than 63 ÷ total number of students             i.e. 9/50 = 0.18.

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to derive the two equations i.e. if x = length of field and y = breadth, then
xy = 7200; x + y = 180.  They were able to solve these equations and also obtained the dimensions as 60m and 120m.  However, attempt on the parts (b) and (c) was poorly handed by some of the candidates.  Many took the margin to be all round the field but the question specified along the longer sides.  Thus cost of clearing the field = N6.50 x 56 x 120 = N43,680.00.  The percentage of the part not cleared
             = 7200 – (120 x 56) x 100  =  62/3 or 6.67%
                          7200
Candidates, who sketched the diagram, were reported to have performed well in the question.

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a) and (b).  Many of them missed the (c) part because they did not quite understand what would happen as the ladder was moved by 2m further away from the wall. If the foot of ladder is moved a further 2m away from the wall, LP becomes (6 + 2)m = 8m, tanx = 4.5/8 and from the tables, x = 29°, to the nearest degree

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