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# ðŸ‘‘ 2019 WAEC GCE MATHEMATICS ANSWERSðŸ‘‘

__100% VERIFIED MATHEMATICS OBJ ANSWERS__

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__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 Ã— 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

nÂ² = 9

n = âˆš9

n = 3

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(2a)

Given; slant height l = 18.7cm

Diameter, d = 24cm

Ï€ = 22/7

But, Curved surface area of cone = Ï€rl

=Ï€(d/2)l

= 22/7 Ã— 24/2 Ã— 18.7

=9873.6/14 = 705cmÂ²

(2b)

128^x Ã— 2/16^(1-x) = 8â…”x

= 2^7x Ã— 2/2^4(1-x) = 2Â³(â…”x)

= 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)

3âˆš3/2 – 4âˆš2/3 – âˆš24

=3âˆš3/2 – 4âˆš2/âˆš3 -âˆš6*4

=(3âˆš3/âˆš2 Ã— âˆš2/âˆš2) – (4âˆš2/âˆš3 Ã— âˆš3/âˆš3) – 2âˆš6

=3âˆš6/2 – 4âˆš6/3 – 2âˆš6

=(3/2 – 4/3 – 2)âˆš6

=(9 – 8 – 12)âˆš6

= -11/6âˆš6

(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 Ã— 1/5) + (1/3 Ã— 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 Ã— 4/5)

= 2/5 + 8/15

= 6/15 + 8/15

= 14/15

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(4a)

17Â²=XÂ²+15Â²

17Â²=XÂ²+225

289-225=XÂ²

64=XÂ²

C=âˆš64=8

TanÎ¸=18/15

TanÎ¸/1+2tanÎ¸=18/15/1+2(8/15)=8/15/1+16/15

=8/15Ã·31/15=8/15*15/31=8/31

(4b)

Log10 y/log10 64 = 1/2

Cross multiply

2log y = log10 64

Log10 yÂ² = log10 64

y = âˆš64

y = Â±8

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(5a)

Obtuse Reflex

Obtuse Obtuse = 164Â°

OMP + OPM + Obtuse MOP = 180Â°(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 = 16Â°

X = 16/2 = 8Â°

Also MNP = 1/2Ã—obtuseMOP

= 1/2 Ã— 164Â°

= 82Â°

Now; MNP + NMP + NPM = 180Â° (angles in a triangle)

82 + (52+8) + (m+8) = 180Â°

m + 150 = 180Â°

M = 180 – 150

M = 30Â°

(5b)

Portion of land for cassava = 2/5

Portion of land for plantain = 1/3 Ã— (1 – 2/5)

= 1/3 Ã— 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)

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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 Ã— 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

(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 Ã— 350,000

=280,000

Amount paid to the bank after 8 years = P + I

= P + PRT/100

=280000 + 280000Ã—7Ã—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 Ã— 100%

= 156,800/350,000 Ã— 100%

= 44.8%

(c) Percentage loss = loss/cost price Ã— 100%

=(506,800 + 10,000) – 460,000/(506,800 + 10,000) Ã— 100%

= 516,800 – 460,000/516,800 Ã— 100%

56,800/516,800 Ã— 100%

= 0.1099 Ã— 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 = 180Â°(~~ 130 + AOM = 180Â°~~

AOM = 180 – 130 = 50Â°

Also;

BMO = BAO + AOM(Ext < = sum of two opposite interior ~~ BMO = 26 + 50 = 76Â°~~

Also; ABD = 1/2AOD(angle at centre = twice angle at circum)

ABD = 1/2 Ã— 130

=65Â°

Hence ODB + ABD + BMO = 180Â°(sum of ~~ ODB + 65Â° + 76Â° = 180Â°~~

ODB = 180 – 141

= 39Â°

(ii) BOD + ODB + DBO = 180Â°(sum of ~~ But ODB = DBO(base angles of an isosceles triangle)~~

BOD + 39Â° + 39Â° = 180Â°

BOD + 78Â° = 180Â°

BOD = 180 – 78 = 102Â°

(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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