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**Check Mathematics Questions WAEC GCE:….**The West African Examinations Council (WAEC) is an examination board that conducts the West African Senior School Certificate Examination, for University and Jamb entry examination in West Africa countries. … In a year, over three million candidates registered for the exams coordinated by WAEC.

**Question 1**

A = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x: 3 < x < 9} are subsets of the universal-set

U = {2, 3, 4, 5, 6, 7, 8, 9}. Find

(a) A n(B’nC’);

(b) (AuB) n(BuC).

Mathematics waec gce 2017 ObservationPlease ensure to use curly brackets to enclose the elements of the sets.

Furthermore always list the elements of set C hence, this is for you to be able to find its complement.

Do it this way C i.e. C = {4, 5,6, 7, 8}, obtain the compliments of the sets Band C thus B’ = {4,5,6, 8}, C’ = {2, 3, 9,}.

Using these sets, the following procedures were to be followed:

(a)(B’ nC’) = { } Hence An (B’ n C’) = { }. Don’t be like Some candidates know to write like this { 0 } instead of { } or 0.

(b) (A u B) = { 2, 3,4,6, 7, 8, 9 }, (BuC) = {2, 3, 4,5,6, 7, 8, 9 } Therefore { Au B } n (BuC) = { 2, 3, 4, 6, 7, 8, 9 }.

**Question 2**

(a) The angle of depression of a boat from the mid-point of a vertical cliff is 35°. If the boat is 120 m from the foot of the cliff, calculate the height of the cliff.

(b) Towns P and Q are x km apart. Two motorists set out at the same time from P to Q at steady speeds of 60 km/h and 80 km/h. The faster motorist got to Q 30 minutes earlier than the other. Find the value of x.

WAEC GCE 2017 Observation.Furthermore, the report stated that in part (a), majority of the candidates could not draw the diagram correctly and this affected their performance significantly. A few others were unable to apply the trigonometric ratios correctly.

Candidates were expected to draw the diagram.

From the diagram, I FMI :: 120 tan 35° = 84.02m. Therefore the height of the cliff:: 2 x 84.02 = 168.04.

In part (b), the most observed weakness was their inability to convert from minutes to hours. They were expected to recall that time :: distance (i.e. t:: ~) and apply this to the problem.

speed v

Time taken by faster motorist « ~ while that of the other motorist = ~ where x is the distance

60 80

X x

from P to Q. – – – = Y2 (30 rnlnutes « Y2 hr). Simplifying this expression gave x = 120km.

60 80

Candidates’ responses to this question were reported to be generally below average.

**Question 3.**

(a) In the diagram, L PQR = 125°, LQRS = r, LRST = 800 and LSTU = 44°. Calculate the value of r

b) .In the diagram, TS is a tangent to the circle at A. ABI ICE, LAEC = sx”, LADB = 60° and LTAE = xo. Find the value of x”,

WAEC GCE Observation

Part (a) of this question required that candidates drew straight lines, one each passing through points Sand R and parallel to PQ and UT as shown in the diagram below – . LMQR = LQRN = 180 -125 = 55° ( alternate angles), LVST = 44° (alternate angle to LSTU). Therefore LNRS = LVSR = 80 – 44° = 360• Hence, r = 55 + 36 = 91°.

In part (b), candidates’ performance was reported to be worse than part (a). Candidates were reported to have exhibited poor understanding of circle theorems. Teachers were encouraged to do a lot of work in this area.

From the diagram, LBDA = LBAS = 600 (angles in the alternate segment). LBAE = 180° – 5x (adjacent angles on a transversal). Therefore LBAS + LBAE + LEAT = 60 + 180 – 5x + x = 180 (angles on a straight line). Solving this simple equation gave x = 150.

**Question 4**

The diagram shows a cone with slant height 10.5 cm. If the curved surface area of the cone is 115.5 cm²;, calculate, correct to 3 significant figures, the:

(a)base radius, r;

(b)height, h;

(c)volume of the cone. [Taken= 22]

Candidates are expected to show good understanding of mensuration of right circular cones.

The curved surface area,

iu! = 115.5cm², slant height,l= 10.scm, n = ll, hence r = 115.5

7nl = 11.5 x 7 = 3.S0cm. r2 + h2 = (10.5}2 => h = .J(10.5)2 – (3.5)2 = 9.90cm

22 x 10.5

Volume of cone = .1 x 22 x 3.5 x 9.899 == 127 cm²

3 7 1.

Question 5.

Two fair dice are thrown.

M is the event described by “the sum of the scores is 10” and

N is the event described by “the difference between the scores is 3”.

(a) Write out the elements of M and N.

(b) Find the probability of M or N.

(c) Are M and N mutually exclusive? Give reasons.

Observation.Always remember to list the elements of M and N as required.

Furthermore make sure you state correctly the condition for which events are mutually exclusive.

You are also to show that M = { (4,6) , (5,5), (6,4}. N = { (1,4), (2,5), (3,6), (4,1),

(5,2), (6,3) }. Probability of M = 3/36 = 1/12. Probability of N = 6/36 = 1/6.

Probability of Nor M:1/12 + 1/6 = 1/4. M and N are mutually exclusive because noelement is common to both sets. Thus the two events cannot happen at the same time i.e. n(M n N) = O.

**Question 6.**

(a) The scale of a map is 1:20,000. Calculate the area, in square centimetres, on the map of a forest reserve which covers 85 km²,

(b) A rectangular playing field is 18 m wide. It is surrounded by a path 6m wide such that its area is equal to the area of the path. Calculate the length of the field.

(c)

lkm = 100,000 ern, therefore 1km2 = 1km x 1km = 100,000 x 100,000 = 10,000 000,OOOcm2•

85km2 = 850,000,000,000 cm². 20,000 em on the ground = 1 cm on the map, hence 400,000,000 cm² on the ground = 1 cm2850,000,000,000 = 2125 cm2 on the map. 400,000,000.

Candidates should draw the diagram correctly.

From the diagram, area of path = 2(30 x 6) + 12a = 360 + 12a. Area of field = 18a. Since they are equal, 18a = 12a + 360. This gave a = 60 m.

see the reflex angle as 360 – x, hence, did not subtract their final answer from 360° when they had calculated the value of the reflex angle. Here, 360 – x x 22x Z x Z = 27.5 cm2

360 7 2 2

This meant that 360 – x = 360 x 2 x 27.5 from where we obtain x = 103° to the nearest degree

77.

**Question 8**

Using ruler and a pair of compasses only,

(a) construct

(i) a quadrilateral PQRS with IPSI :: 6 cm, LRSP:: 90°,

IRSI = 9 ern, IQRI = 8.4 cm and IPQI :: 5.4 cm;

(ii) the bisectors of LRSP and LSPQ to meet at X;

(iii) The perpendicular XTto meet PS at T.

(b) Measure IXT/’

Math waec gce 2017 Observation.Please construct the angle and not measure them.

construct a perpendicular from a given point to a given line segment.

**Question 9.**

In the diagram, /AB/ = 8 km, /BC/ = 13 km, the bearing of A from B is 310° and the bearing of B from C is 230°. Calculate, correct to 3 significant figures,

(a) the distance AC;

(b) the bearing of C from A;

(c) how far east of B, C is

Always remember to calculate LABC correctly.

Apply the cosine rule correctly.

LASC = 100°. Therefore /AC/ = 82 + 132 – 2 (8)(13)cos100 which gave / AC/ = 16.4 km.

sin (LCAS) = sin100. Hence, sin (LCAB) = 13Sin100

13 16.4 16.4

Simplifying gave LCAB = 51.32°. Bearing of C from A = 180 – (50 + 51.32) = 079°.

If the distance of C east of B = BD, then BO = BC cas 40° = 13 x cas 40° = 9.96 km.

Question 10.

(a) Copy and complete the table of values for the relation V = -X² + X + 2 for -3 ≤ x ≥ 3.

(b) Using scales of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the v-axis, draw a graph of the relation

y = -X² + X + 2.

(c) From the graph, find the:

(i)Minimum value of y;

(ii)Roots of the equation X² – x -2 = 0;

(iii)Gradient of the curve at x = -0.5.

Math gce , waec gce neco gce Observation.

x = -0.5.

Question 11.

In the diagram, L.PTQ = L.PSR = 900, /PQ/ = 10 ern, /PS/ = 14.4 cm and /TQ/ = 6 cm.

Calculate the area of quadrilateral QRST.

(b) Two opposite sides of a square are each decreased by 10% while the other two are each increased by 15% to form a rectangle. Find the ratio of the area of the rectangle to that of the square.

Math gce Observation.

We advise that student’s should always remember to apply the concept of similar triangles correctly.

Furthermore student should try and recognize the quadrilateral as a trapezium, this will help in finding its area.

Part (b) Candidates were expected to show that:

/PT/ = …./102 – 62 = 8 cm. m: =!J2L i.e ~ = 14.4.Hence, /SR/ = 10.8 cm.

/TO/ /SR/ 6 /SR/

Area of quadrilateral QRST = Yz (6 + 10.8) x 6.4 = 53.76 cm2. Don’t subtract the area of triangle PQT from triangle PRS.

This was also in order. In part (b) if the side of the square was y, then new breadth = 90 x y = 0.9y.

100

New length = 115 x Y = 1.15y. New area = 1.15y x 0.9y = 1.035/.

100

Hence, ratio = 1.035y² : y² = 1.035 : 1 or 207:200 .

Question 12.

The frequency distribution of the weight of 100 participants in a high jump competition is as

shown below:

Weight (kg) 20-29 30 – 39 40 – 49 SO – 59 60 – 69 70-79

Number of

participants 10 18 22 25 16 9

(a) Construct the cumulative frequency table.

(b) Draw the cumulative frequency curve.

(c) From the curve, estimate the:

(i) median;

(ii) semi-interquartile range;

(iii) probability that a participant chosen at random weighs at least 60 kg.

Make sure to draw the Ogive using class boundaries.

The median was 49.5 while the first quartile(Ql) was 37.8 and the third Quartile (~)59.5.

Hence, the semi-interquartile range Q1-Q1 = 59.5 – 32.8 = 10.85 ± 1.

2 2

Number of participants who weighed at least 60kg = 25. Therefore probability of choosing a

participant who weighed at least 60kg = 25= -.1 .

100 4.

Question 13.

(a)The third term of a Geometric Progression (G.P) is 24 and its seventh term is 4(20/27) .Find Its first term.

(b)Given that y varies directly as x and inversely as the square of z. If y = 4, when x = 3 and z = 1, find y when x = 3 and z = 2.

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