# NECO Mathematics Questions and Answers – 2019 June/July Expo

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# 👑 2019 NECO MATHEMATICS ANSWERS👑

1-10: DBBCBDEDCC
11-20: DCDAECDEDC
21-30: CBAAECDCAB
31-40: CDDCEBDDCC
41-50: CDDDCABDBA

(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)¹
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)
x² – 4/3 + x + 3/2
=2(x² – 4) + 3(x + 3)/6
= 2x² – 8 + 3x + 9/6
= 2x² + 3x + 1/6
= (2x + 1)(x + 1)/6

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(3)

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

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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)²: 148.028, 51.361, 4.694, 8.028, 61.361, 164.694

f(x-x̄)²: 888.167, 205.444, 23.472, 40.139, 368.167, 658.778
Ef(x-x̄)² = 2184.167

Mean, x̄ = Efx/Ef = 455/30
x̄ = 15.167

Variance = Ef(x-x̄)²/Ef
2184.167/30
=72.82

Standard deviation = √2184.167/30
=√72.822
=8.53

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(5a)
x² – 5x – 24 = 0
x² – 5x = 24
x² – 5x + 25/4 = 24 + 25/4
(x – 5/2)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)
S² (3x² – 4x + 2)dx

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

=3x³/3 – 4x²/2 + 2x/1]²

= x³ – 2x² + 2x ]²

=[(2³ – 2(2)² + 2(2)] – [0]
= 8 – 8 – 4
= 4

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

(7a)
4x² – 9y² = 19
2x² x² – 3² y²=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

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

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

(8ai)
Total surface area
= Total surface of cylinder + curved surface of hemisphere
=(πr² + 2πrh) + (2πr²)
=π(r² + 2rh) + π(2r²)
= π[(r² + 2rh) + 2r²]
= π[(7² + 2(7)(10) + 2(7²)]
= π[(49 + 140) + 98]
= π(287)
= 287πcm²
Using π = 22/7
Total surface area = 287 × 22/7 = 41 × 22
= 902cm²

(8aii)
Volume = volume of cylinder + volume of hemisphere
= πr²h + 2/3πr³
=π[r²h + 2/3r³]
= π[(7²)(10) + 2/3(7)³]
= π[490 + 686/3]
= π[2156/3]
= 22/7 × 2156/3
= 22 × 308/3 = 6776/3cm³
OR 2258.67cm³

(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 – 1¼
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 ≤ -3¾

(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

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(9b)
Gradient, m = y2 – y1/x2 – x1
m = 9 – 7/6 – 3 = 2/3

Acute angle Φ= Tan-¹(2/3)
Φ = Tan-¹(0.6667)
= 33.69°

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(10ai)
Given S = t³ – 3t² + 9t + 1
Velocity is zero when;
ds/dt = 0
ie 3t² – 6t + 9 = 0
t² – 2t + 3 = 0
Using formula
t = -(-2) ±√(-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(3t² – 6t + 9) = 0
6t – 6 = 0
6t = 6
t = 6/6 = 1 secs

(10b)
V = 3t² – 6t + 9
at t= 2secs
Velocity v = 3(2)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/s²

Acceleration, a after 6 secs
a = 6t – 6
a = 6(6) – 6
a = 36 – 6
a = 30m/s²

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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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