WAEC/NECO MATHS GCE PAST QUESTIONS AND ANSWERS AS REQUESTED 2

WAEC/NECO MATHS GCE PAST QUESTIONS AND ANSWERS AS REQUESTED 2

CONTINUED FROM LAST POST

QUESTION 7

ANSWER 7

QUESTION 8

Question 8

(a) Using ruler and a pair of compasses only, construct a parallelogram PQRS with diagonals /PR/ = 8 cm and /QS/ = 10 cm.  The diagonals intersect at O and
ÐPOS = 75o .

(b)  Construct perpendiculars from R to  at C.
(c)   Measure:
(i)        │PQ│;
(ii)       │PS│;
(iii)       ÐQRS.

 

 

 

 

 

 

 

 

ANSWER 8

The Chief Examiner reported that this question was not attempted by majority of the candidates and majority of those who attempted it performed poorly in it.
 
Candidates were expected to recall that the diagonals of a parallelogram bisect each other.  Candidates were also expected to draw  = 10 cm and then locate O, the mid-point of   .  At O, they would then construct angle 75o and thereafter, locate points P and S such that   = 8 cm.  Candidates were then expected to join the points to form the parallelogram.  Using R as the centre, candidates would construct an arc to cut   produced, at 2 points X and Y, bisect   and locate C, the point of bisection of  .  Using a ruler, candidates would then measure and read off │PQ│ and │PS│ which were 7.3 cm and 5.6 cm respectively while  angle  QRS was measured using the protractor as 102 degrees.

QUESTION 9

  • Copy and complete the following table of values for the relation y =  + 2x – 2.
x -4 -3 -2 -1 0 1 2 3
y -3 -2
  • Using a scale of 2 cm to 1 unit on both axes, draw the graph of

y =  + 2x – 2 for -4 ≤ x ≤ 2.

  • Use your graph to:
  • find the roots of the equation  + 2x – 2;
  • indicate the region where  + 2x < 0.5.

ANSWER 9

This question was reported to be quite popular with the candidates who were able to complete the table of values and draw the required graph.  However, reading and answering questions from the graph were poorly handled by majority of the candidates.  The expected table of values was:

x

-4

-3

-2

-1

0

1

2

3

y

6

1

-2

-3

-2

1

6

13

These points would then be plotted on a graph sheet using the given scale.  The roots of the equation  + 2x = 2 were the points of intersection of the graph and the x – axis which were at x = -2.7 ± 0.1 and 0.75  ±  0.01.  The required region was the region where y  <  -1.5.  They were obtained from the graph as -2.2 ± 0.1 < x < 0.2 ± 0.1.  Candidates could either state this interval or shade the area on their graph.

QUESTION 10

If 3x + 3x+ 1= 36, find x.
The bearing of P from X 10 km away is 025o. Another point Q is 6 km from X and on a bearing of 162o. Calculate the:
(a) distance PQ;
(b) bearing of P from Q.

ANSWER 10

This question was reportedly, attempted by majority of the candidates and their performance was described as fair. It was also observed that candidates who attempted the question on bearing did not draw the correct diagram. Some did not draw the diagram at all.
In part (a), candidates were expected to factorize 3x out i.e. 3x + 3x+1 = 3x (1+3) = 3x(4). Thus, 3x + 3x+1 = 36 was the same as 3x(4) = 36 or 3x = 9 = 32. This implied that x = 2.

QUESTION 11

A sector of a circle with radius 20 cm has an area of 396 cm2. Calculate, correct to 1 decimal place, the

(a)sectoral angle

(b)perimeter of the sector

(c)volume of the cone formed when the sector is bent such that its straight edges coincide.

QUESTION 12

The ages (in years) of men selected from a community are as follows:

44 28 58 50 93 35 34 52 57 61
40 63 90 67 64 56 51 82 73 43
73  73 44 71 95 52 71 25 35 79
28 40 72 88 82 63 53 48 98 65
63 44 73 70 68 46 54 62 41 70

(a) Prepare a group frequency distribution table with the intervals 20 – 29, 30 – 39, …
(b) Calculate the mean deviation of the distribution.

ANSWER 12

The Chief Examiner reported that this question was very popular among the candidates and their performance was commended.  According to the report, their responses showed that they understood this aspect of the syllabus.

Candidates were reported to have obtained the following table correctly.

Class
Interval

Class
Mark

Tally

Frequency
(f)

fx

|x-|

f| x-

20 – 29

24.5

׀׀׀

3

73.5

36

108

30 – 39

34.5

׀׀׀

3

103.5

26

78

40 – 49

44.5

׀׀׀׀  ׀׀׀׀

9

400.5

16

144

50 – 59

54.5

׀׀׀׀  ׀׀׀׀

9

490.5

6

54

60 – 69

64.5

׀׀׀׀  ׀׀׀׀

9

580.5

4

36

70 – 79

74.5

׀׀׀׀  ׀׀׀׀

10

745.0

14

140

80 – 89

84.5

׀׀׀

3

253.5

24

72

90 – 99

94.5

׀׀׀׀

4

378.0

34

136

     

Σf = 50

Σfx = 3025

 

Σf |x-|=
768

 

QUESTION 13

ANSWER 13

ALL QUESTIONS AND COMMENTS PROVIDED BY WAEC

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