2019 Waec Gce Mathematics Obj And Theory Answers (Second Series)

2019 Waec Gce Mathematics  Answers Second Series
2019 Waec Gce Mathematics Obj And Theory Answers Second Series

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01-10: BBDCCBCBAA
11-20: BCBDCDBDDB
21-30: BCBDBBAACA
31-40: AABBDABCCB
41-50: AABBDABCCB

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(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

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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

(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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(5a)
Obtuse Reflex
Obtuse Obtuse = 164°

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°

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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°

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(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

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(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

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(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

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(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

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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°

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(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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