# 2019 NECO Mathematics Answer & Question

__MATHS OBJ:__

1-10: DBBCBDEDCC

11-20: DCDAECDEDC

21-30: CBAAECDCAB

31-40: CDDCEBDDCC

41-50: CDDDCABDBA

51-60: BBBCBCDADD

*COMPLETED… *

__MATHS THEORY:__

PART I(Answers All)

(1a)

At the end of year 1

Using; A = P(1 + R/100)N

A = #110,000(1+5/100)

A = #110,000(1.05)

Amount or savings = #115,500.00

At the beginning of year 2,

Principal, p = 115,500 + #50,000 = #165,500.00

At the end of year 2

A = #165,500(1+5/100)�

A = #165,500 � 1.05

A = #173,775.00

At the beginning of year 3,

Principal, p = #173,775 + #50,000 = #223,775.00

At the end of year 3,

A = #223,775(1+5/100)

A = #223,775 � 1.05

A = 234,963.75

Total savings after 3 years = #234,963.75 + #50,000 = #284,963.75

(1b)

By end of third year

Savings is lesser than #300,000.00 by;

#300,000.00 – 284,963.75

=#15,036.25

= #15,036.25

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(2a)

3^2x – y = 1 ———— Given

16^x/4 = 8^3x – y —- Given

Now,

3^2x – y = 3^0

2^4x – 2 = 2^3(3x – y)

Therefore

2x – y = 0 ——-(1)

4x – 2 = 3(3x – y) —–(2)

From (1) y = 2x ——(3)

Put eqn (3) into (2)

4x – 2 = 3(3x – 2x)

4x – 2 = 3x

4x – 3x = 2

x = 2

Put x = 2 into eqn (3)

y = 2(2)

y = 4

Therefore x = 2 and y = 4

(2b)

x� – 4/3 + x + 3/2

=2(x� – 4) + 3(x + 3)/6

= 2x� – 8 + 3x + 9/6

= 2x� + 3x + 1/6

= (2x + 1)(x + 1)/6

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(3)

Draw the diagram

Using SOHCAHTOA

|TM|/|MD| = Tan 28�

298.5+1.5|MD| = 0.5317

|MD| = 300/0.5317 = 564.2m

Similarly,

|TM|/|MC| = Tan 34�

300/|MC| = 0.6745

|MC| = 300/0.6745 = 444.8m

Distance between both, CD

= 564.2 – 444.8

=119.4m

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(4)

In a tabular form

Mark(x): 1-5, 6-10, 11-15, 16-20, 21-25, 26-30

Mid-mark(x): 3, 8, 13, 18, 23, 28

Frequency (f): 6, 4, 5, 5, 6, 4

Ef = 30

fx: 18, 32, 65, 90, 138, 112

Efx = 455

d = x – x̄: -12.167, -7.167, -2.167, 2.833, 7.833, 12.833

(x – x̄)�: 148.028, 51.361, 4.694, 8.028, 61.361, 164.694

f(x-x̄)�: 888.167, 205.444, 23.472, 40.139, 368.167, 658.778

Ef(x-x̄)� = 2184.167

Mean, x̄ = Efx/Ef = 455/30

x̄ = 15.167

Variance = Ef(x-x̄)�/Ef

2184.167/30

=72.82

Standard deviation = √2184.167/30

=√72.822

=8.53

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(5a)

x� – 5x – 24 = 0

x� – 5x = 24

x� – 5x + 25/4 = 24 + 25/4

(x – 5/2)� = 121/4

x- 5/2 = �√121/4

x – 5/2 = � 11/2

x = 5/2 � 11/2

x = 5/2 + 11/2 OR 5/2 – 11/2

x = 16/2 OR -6/2

x = 8 or -3

(5b)

S� (3x� – 4x + 2)dx

0

= 3x^2+1/2+1 – 4x^1+1/1+1 + 2x^0+1/0+1]�

0

=3x�/3 – 4x�/2 + 2x/1]�

0

= x� – 2x� + 2x ]�

0

=[(2� – 2(2)� + 2(2)] – [0]

= 8 – 8 – 4

= 4

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PART II (Answer Any FIVE)

(7a)

4x� – 9y� = 19

2x� x� – 3� y�=19

(2x-3y)(2x+3y)=19

Substitute for 2x+3y=1

2x-3y=19…………(1)

2x+3y=1…………..(2)

Subtract equ (2) from (1)

2x-3y-(2x+3y)=19-1

3x-3y-2x-3y=18

-6y/-6=18/-6

y = -3

Substitute for y in equ (1)

2x-3(-3)=19

2x+9=19

2x/2=10/2

x=5

(7b)

√4.033/0.611 � 0.356

Put No and Log In a tabular form

No | log

4.033 | 0.6056 -> | 0.6056

0.611 | 1.7860+ –

0.356 | 1.5514

0.611�0.356|1.3374->|1.3374

4.033/ | ——–> | 1.2682

0.611�0.356 �2

√4.033/0.611 | ——-> | 0.6341

�0.356

Antilog = 4.306

Ans = 4.306

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(8ai)

Total surface area

= Total surface of cylinder + curved surface of hemisphere

=(πr� + 2πrh) + (2πr�)

=π(r� + 2rh) + π(2r�)

= π[(r� + 2rh) + 2r�]

= π[(7� + 2(7)(10) + 2(7�)]

= π[(49 + 140) + 98]

= π(287)

= 287πcm�

Using π = 22/7

Total surface area = 287 � 22/7 = 41 � 22

= 902cm�

(8aii)

Volume = volume of cylinder + volume of hemisphere

= πr�h + 2/3πr�

=π[r�h + 2/3r�]

= π[(7�)(10) + 2/3(7)�]

= π[490 + 686/3]

= π[2156/3]

= 22/7 � 2156/3

= 22 � 308/3 = 6776/3cm�

OR 2258.67cm�

(8b)

Draw the diagram

Perimeter of Arc = Φ/360 � 2πr

= 120/360 � 2 � 22/7 � 7

= 1/3 � 44

= 44/3cm OR 14.67cm

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(9ai)

X + 1⅔x ≤2⅓x – 1�

X + 5x/3 ≤7x/3 – 5/4

Multiply through with (12)

12x + 20x ≤ 28x – 15

32x ≤ 28x – 15

32x – 28x ≤-15

4x <_ -15

X ≤ -15/4

X ≤ -3�

(9aii)

4x – 1/3 – 1+2x/5 ≤ 8 + 2x

Multiply through with 15

5(4x – 1)-3(1+2x)≤15(8+2x)

20x – 5 – 3 – 6x ≤ 120 + 30x

14x – 8 ≤ 120 + 30x

14x – 30x ≤ 120 + 8

-16x ≤ 128

X ≥ 128/-16

X ≥ -8

(9b)

Gradient, m = y2 – y1/x2 – x1

m = 9 – 7/6 – 3 = 2/3

Acute angle Φ= Tan-�(2/3)

Φ = Tan-�(0.6667)

= 33.69�

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(10ai)

Given S = t� – 3t� + 9t + 1

Velocity is zero when;

ds/dt = 0

ie 3t� – 6t + 9 = 0

t� – 2t + 3 = 0

Using formula

t = -(-2) �√(-2)� – 4(1)(3)/2(1)

t = 2�√4 – 12/2

t = 2�√-8/2

t = 2�2√-2/2 = 1�√-2 secs

t is complex

(10ii)

Acceleration is zero;

when ds/dt = 0;

d/dt(3t� – 6t + 9) = 0

6t – 6 = 0

6t = 6

t = 6/6 = 1 secs

(10b)

V = 3t� – 6t + 9

at t= 2secs

Velocity v = 3(2)� – 6(2)+9

=12 – 12 + 9

= 9m/s

acceleration a when t = 2secs

a = 6t – 6

a = 6(2) – 6

a = 12 – 6 = 6m/s�

Acceleration, a after 6 secs

a = 6t – 6

a = 6(6) – 6

a = 36 – 6

a = 30m/s�

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(11)

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(12a)

In a tabular form

Class interval: 21-30, 31-40, 41-50, 51-60, 61-70, 71-80

Tally: II, IIII IIII, IIII IIII II, IIII IIII IIII, IIII III, III

Frequency: 2, 10, 12, 15, 8, 3

If = 50

(12b)

DRAW A TABLE:

Mid-mark(X): 25.5, 35.5, 45.5, 55.5, 65.5, 75.5

f: 2, 10, 12, 15, 8, 3

Ef = 50

d = x – x̄o : -20, -10, 0, 10, 20, 30

fd: -40, -100, 0, 150, 160, 90

Efd= 260

Where x̄0 is assumed mean = 45.5

Mean, x̄ = x̄0 + Efd/Ef

x̄ = 45.5 + 260/50

Mean x̄ = 45.5 + 5.2

Mean x̄ = 50.7

(12c)

Semi interquartile range

= Q3 – Q1/2

= 38th score – 13th score/2

= 60-42/2

= 18/2

= 9

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*COMPLETED… *