2019 WAEC GCE Mathematics Questions and Answers

West Africa Examination Council, WAEC GCE Mathematics Questions and Answers for 2019 June/July Examination are now available, always subscribe to get the already available questions and answers by Examrelief an hour before each paper.

WAEC GCE Mathematics Questions and Answers

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👑 2019 WAEC GCE MATHEMATICS ANSWERS👑

 

100% VERIFIED MATHEMATICS OBJ ANSWERS

1-10: BBDCCBCBBA
11-20: BCBDCDBDDB
21-30: BCBDDBAACA
31-40: AABBDABCCB
41-50: BBBDBCBCDB

100% VERIFIED MATHEMATICS THEORY ANSWERS

(1a)
Table for multiplication (x) in mod 8
Using (2, 3, 5, 7)

In a tabular form.
Under (x) 2, 3, 5, 7
Under 2 4, 6, 2, 6
Under 3 6, 1, 7, 5
Under 5 2, 7, 1, 3
Under 7 6, 5, 3, 1

(1bi)
3(x)n = 5
3 × n = 5 + cont. Of mod 8 to be divisible by 3;
3n = 5 + 8 + 8
3n = 21
n = 21/3
n = 7

(1bii)
n(x)n = 1
n * n = 1 + 8
n² = 9
n = √9
n = 3
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(2a)
Given; slant height l = 18.7cm
Diameter, d = 24cm
π = 22/7
But, Curved surface area of cone = πrl
=π(d/2)l
= 22/7 × 24/2 × 18.7
=9873.6/14 = 705cm²

(2b)
128^x × 2/16^(1-x) = 8⅔x
= 2^7x × 2/2^4(1-x) = 2³(⅔x)
= 2^7x + 1/2^4(1-x) = 2^2x
Cross multiply
2^4(1 – x) + 2x = 2^7x + 1
4(1 – x) + 2x = 7x + 1
4 – 4x + 2x = 7x + 1
4 – 2x = 7x + 1
9x = 3
X = 3/9 = 1/3
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(3a)
3√3/2 – 4√2/3 – √24
=3√3/2 – 4√2/√3 -√6*4
=(3√3/√2 × √2/√2) – (4√2/√3 × √3/√3) – 2√6
=3√6/2 – 4√6/3 – 2√6
=(3/2 – 4/3 – 2)√6
=(9 – 8 – 12)√6
= -11/6√6

(3bi) prob(only one passes) = m passes and N fails or M fails and N passes
Prob (M passes) = 2/3; prob(M fails) = 1 – 2/3 = 1/3
Prob(N passes) = 4/5; prob(N fails) = 1 – 4/5 = 1/5
Prob(only one passes) = (2/3 × 1/5) + (1/3 × 4/5)
= 2/15 + 4/15
= 6/15
= 2/15

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(3bii)
prob(at least one passes) = prob(only one passes) + prob(both parties)
= 2/5 + (2/3 × 4/5)
= 2/5 + 8/15
= 6/15 + 8/15
= 14/15
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(4a)
17²=X²+15²
17²=X²+225
289-225=X²
64=X²
C=√64=8
Tanθ=18/15
Tanθ/1+2tanθ=18/15/1+2(8/15)=8/15/1+16/15
=8/15÷31/15=8/15*15/31=8/31

(4b)
Log10 y/log10 64 = 1/2
Cross multiply
2log y = log10 64
Log10 y² = log10 64
y = √64
y = ±8
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(5a)
Obtuse Reflex
Obtuse Obtuse = 164°

OMP + OPM + Obtuse MOP = 180°(angles in a triangle)
But OMP = OPM = x(radius of the circle and hence base angles are the same)

2x + 164 = 180
2x = 180 – 164
2x = 16°
X = 16/2 = 8°

Also MNP = 1/2×obtuseMOP
= 1/2 × 164°
= 82°

Now; MNP + NMP + NPM = 180° (angles in a triangle)
82 + (52+8) + (m+8) = 180°
m + 150 = 180°
M = 180 – 150
M = 30°

(5b)
Portion of land for cassava = 2/5
Portion of land for plantain = 1/3 × (1 – 2/5)
= 1/3 × 3/5
= 1/5

Portion of land for yam = 1 -(2/5 + 1/5)
1 – (3/5)
=2/5
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(5b)
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(6)
a. CLICK HERE FOR IMAGE
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(8ai)
a + 4d = 11 …..i
a + 7d = 20 …. ii

a + 7d = 20
– a + 4d = 11
3d = 9

d = 9/3
d = 3
Using eqn i
a + 4(3) = 11
a + 12 = 11
a = 11 – 12
a = -1

T12 = a + 11d
= -1 + 11(3)
= -1 + 33
= 32

(8aii)
Sn = n/2[20 +(n – 1)d]

S12 = 12/2[2(-1)+(12-1)3]
=6[-2 + (11)3 ]
= 6[-2 + 33]
= 6[31]
= 1806

(8b)
Let the S.p = M
15% of m = 15m/150
= 3m/20 (Discount)

Therefore; collected = M -3m/20
So M/1 – 3M/20 =36000/1
20m – 3m = 720000
17m = 720000
m = 720000/17
M = 42352.94%

Discount = 42352.94 × 3/20
Discount = #6,352.94
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(9ai)
Draw the Venn diagram

(9aii)
No selected for biology but neither physics nor mathematics = 36

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(9aiii)
No not selected for any of three subjects = 11

(9b)
Mean = 22+18+(2x+1)+10+20/5 = 15
2x + 71 = 75
2x = 75 – 71
2x = 4
X = 4/2 = 2

The numbers are: 22, 18, 5, 10 and 20
Arranging in ascending order, we have 5, 10, (18), 20, 22

The median is 18

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(10a)
Loan borrowed = 80/100 × 350,000
=280,000
Amount paid to the bank after 8 years = P + I
= P + PRT/100
=280000 + 280000×7×8/100
=280,000 + 156,800
Amount = #436,800

Total cost of house to the man = #350,000 + Interest
= #350,000 + 156,800
= #506,800

(10b)
Percentage increase in cost of house = 506,800 – 350,000 × 100%
= 156,800/350,000 × 100%
= 44.8%

(c) Percentage loss = loss/cost price × 100%
=(506,800 + 10,000) – 460,000/(506,800 + 10,000) × 100%
= 516,800 – 460,000/516,800 × 100%
56,800/516,800 × 100%
= 0.1099 × 100%
= 10.99%
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(12)
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(13a)
Draw the diagram

(i) Extend DO to touch AB at M
AOD + AOM = 180°( 130 + AOM = 180°
AOM = 180 – 130 = 50°

Also;
BMO = BAO + AOM(Ext < = sum of two opposite interior BMO = 26 + 50 = 76°
Also; ABD = 1/2AOD(angle at centre = twice angle at circum)
ABD = 1/2 × 130
=65°
Hence ODB + ABD + BMO = 180°(sum of ODB + 65° + 76° = 180°
ODB = 180 – 141
= 39°

(ii) BOD + ODB + DBO = 180°(sum of But ODB = DBO(base angles of an isosceles triangle)
BOD + 39° + 39° = 180°
BOD + 78° = 180°
BOD = 180 – 78 = 102°

(13b)

|1 2 6

3|1,3 2,3 (6,3)
4|1,4 2,4 (6,4)
5|1,5 2,5 (6,5)

prob (greater than 7) = 3/9 = 1/3
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