__Waec gce 2019 Maths Obj And Essay Answers – Aug/Sep Expo__

__MATHS OBJ:__

1-10: BBDCCBCBBA

11-20: BCBDCDBDDB

21-30: BCBDDBAACA

31-40: AABBDABCCB

41-50: BBBDBCBCDB

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

(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

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

(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)

Log10 y/log10 64 = 1/2

Cross multiply

2log y = log10 64

Log10 y² = log10 64

y = √64

y = ±8

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

(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

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

(6)

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

(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

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

(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

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

(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)

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

(13a)

Draw the diagram

(i) Extend DO to touch AB at M

AOD + AOM = 180°(angles on a straight line)

130 + AOM = 180°

AOM = 180 – 130 = 50°

Also;

BMO = BAO + AOM(Ext angle = sum of two opposite interior angles of a triangle)

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 angles in a triangle)

ODB + 65° + 76° = 180°

ODB = 180 – 141

= 39°

(ii) BOD + ODB + DBO = 180°(sum of angles in a triangle)

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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