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# WAEC GCE 2019 GENERAL MATHEMATICS OBJ AND THEORY VERIFIED QUESTIONS AND ANSWER WAEC GCE 2019/2020 MATHEMATICS OBJ AND ESSAY EXPO | MATHS QUESTIONS AND ANSWERS VERIFIED

(1a)

Tabulate

X|2|3|4|5|7|

2|4|6|2|6|

3|6|1|7|5|

5|2|7|1|3|

7|6|5|3|1|

(1bi)

n=7,i;e 3 (*)7=5

(1bii)

n(*)n=1, n=3

=====================================

(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

====================================================

(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

(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

=========================================

(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)

Log^10/Log^64=½

Log10^10=Log10^64^½

y=√64=±8

=====================================

(5a)

Obtuse Reflex

Obtuse

Obtuse

= 164°

OMP + OPM + Obtuse MOP = 180°(

OMP + OPM + Obtuse MOP = 180°(

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° (

82 + (52+8) + (m+8) = 180°

m + 150 = 180°

M = 180 – 150

M = 30°

=====================================

(6a)

2^m*(1/8)^n = 128

2^m *2^-n = 2^7

m-3^n=7 ———(eq1)

4^m÷2^-4n=1/6

2^2m÷2^-4n=2-4

2m+4n=-4

m+2n=2 —–(eq2)

Subtract from (2)

m+2n-(n-3n)=-2-7

m+2n-m+3n=-9

5n/5=-9/5

n=-9/5

Sub for n in eq2

m+2(-9/5)=-2

m-18/5=-2

m=-2+18/5=8/5

;(m-n)=5/5-(-9/5)=17/5

(6b)

(-2,½) and (1-⅔)

y2-y1/x1-x2 = y-y1/x-x1

-⅔-½/1-(-2) = y-½/x-(-2)

-7/6/3=y-½/x+2

-7/18*y-½/x+2

(y-½)18=-7(x+2)

18y-9=-7x-14

18y=-7x-14+9

18y/18=-7x/18-5/18

y=-7x/18-5/18

18y=-7x-5

=====================================

5th –> 11

8th –> 20

a + (n -1)d = 11

a + (5ˉ¹)d = 11

a + 4d = 11 –> (i)

a + 7d = 20 –> )ii)

a + 7d – a – 4d = 20 – 11

7d – 4d = 9

3d = 9

D = 9/3 = 3

D = 3

From eqn (i)

a + 4d = 11

a + 4 x 3 = 11

a + 12 = 11

a = 11 – 12

a = 1

(8ai)

Tn = a + (m-1)d

T₂ = -1 (12 – 1) 3

= -1 + 11 x 3

= -1 + 33

T12 = 32

(8aii)

Sn = n/2 [2a + (n – 1)d]

= 12/2 [ 2 x (1) + 11 x 3]

= 6 [ -2 + 33)

= 6 (31)

= 186

====================================

(9i)
DRAW THE DIAGRAM

n (U) = 110
n (P) = 25
n (B) = 45
n (M) = 48
n (Pnm) = 10
n (Bnm) = 8
n (pnBnm) = 5

(9ii)
for biology = 45 = 36

(9iii)
x + 14 + 1 + 5 + 5 + 36 + 3 + 35 = 110
x + 25 + 36 + 38 = 110
x = 110 – 99
x = 11

(9b)
15 = 22 + 18 + 2 x 1 + 10 +20/5
15 x 5 = 40 + 2x + 31
45 = 71 + 2x
45 – 71 + 2x
2x/2 = -26/2
X = 13
22, 18, 2x + 1, 10 , 20
For 2x (-B) +1 = -26 +1 = -25
22, 18, -25, 10, 20
-25, 10, 18, 20, 22
The media is 18

=====================================

(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%

=============================

(12)

Draw the diagram

x² = 20² + 8² – 2x 20 x 8 x Cos 50º

x² = 400 + 64 – 320 x 0.6427

x² = 464 – 205.7

√x² = √258.3

x = 16.1

x = 16km

(12ai)

TJ = 16km

(12aii)

sinØ/20 = sin50/16

sinØ = 20 x sin 50/16

sinØ 0.9576

Ø = sinˉ¹ 0.9576 = 73.3º

The bearing 90 -73.3 = 16.7º

(12b)

Draw the diagram

y = √12² + 5²

y = √144 + 25

y = √169 = 13

tan Ø = 5/12 = (0.04167)

Ø = tanˉ¹ (0.4167) = 22.6º

Sin Ø = 5/6 + x

Sin 22.6º = 5/6 + x

(0.3843) (6 + x) = 5

2.3058 + 0.3843x = 5

0.3843x = 5-2.3058

0.3843x/0.3843 = 2.6942/0.3943

X = 7.01

X ≈ 7

=====================================

(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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