2021 Nabteb Gce Advanced Mathematics Answers

2021 NABTEB GCE ADVANCED MATHEMATICS ANSWERS

 

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NABTEB O-LEVEL MATHEMATICS
01-10 DCCCABBAAD
11-20 CDBDCCCBCA
21-30 CDABDACBCC
31-40 BCCCABABBC
41-50 DAACCACBDB

 

(1a)
√75 − √3 (2√3−3)−5√27
= √(5×5×3) − (2×3)+3(√3) −5√(9×3)
= 5(√3) −6 +3(√3) −(5×3)√3
= 5√(3) + 3(√3) −15(√3) −6
= −7(√3) −6

(1b)
10(x−4)=4(2x−1)+5
10x−40=8x−4+5
10x−8x=1+40
2x=41
x=41/2
x= 20.5
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(2ai)
8C3
= (8!)/(3!×(8−3)!)
= (8×7×6×5!)/(3!×(8−3)!)
=(8×7×6×5!)/(3!×5!)
= (8×7×6)/(3×2×1)
= (336)/6
= 56

(2aii)
7P4
=7!/(7−4)!
= (7×6×5×4×3!)/(3!)
= 7×6×5×4
= 840

(2b)
25=(1/5)(125^x)
5^2 =5^(−1) *5^(3x)
2= −1 + 3x
3x = 2+1
3x = 3
x = 1
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(3i)
C = a + bN
80,000 = a + 4b……..(i)
110,000 = a + 6b…….. (ii)
Subtracting equ(i) from (ii)
30,000= 3b
b = 30000/3
b = 10,000
Substituting b in equ(i)
80,000 = a + 4b
80,000 = a + 40,000
a = 80,000 – 40,000
a = 40,000
The relationship between C and N is given by;
C = a + bN
C = 40,000 + 10,000N

(3ii)
When N = 9
C = 40,000 + 10,000(9)
C = 40,000 + 90,000
C = ₦130,000
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(4i)
Mean = Σx/n
x̄ = (12+19+20+21+22+22+16+8)/8
x̄ = 140/8
x̄ = 17.5

(4ii)
8, 12, 16, 19, 20, 21, 22, 22
Median = 19+20/2
= 39/2
= 19.5

(4iii)
Mode = 22

(4iv)
TABULATE

x | 12 | 19 | 20 | 21 | 22 | 22| 16 | 8

(x – x̄) | – 5.5 | 1.5 | 2.5 | 3.5 | 4.5 | 4.5 | -1.5 | -9.5

(x – x̄)² | 30.25 | 2.25 | 6.25 | 12.25 | 20.25 | 20.25 | 2.25 | 90.25

Σ(x – x̄)² = 184
= √(Σ(x – x̄)²/N
= √184/8
= √23
= 4.796
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(5a)
= (8x²+5x-3)/x³
= ∫(8x²+5x-3)/x³ dx
= ∫(8x²/x³) dx + ∫(5x/x³) dx – ∫3/x³ dx
= ∫(8/x)dx + ∫(5/x²)dx – ∫(3/x³)dx
= 8 ∫(1/x)dx + 5 ∫(1/x²)dx – 3 ∫(1/x³)dx
= 8 ∫ (1/x)dx + 5∫(x-²)dx – 3∫(x-³)dx
= 8Inx + 5(x-²+¹/-2+1) – 3(x-³+¹/-3+1)+c
= 8Inx + 5(x-¹/-1) – 3(x-²/-2)+c
= 8Inx – 5(1/x) – 3(1/x²)(-1/2) + c
= 8Inx – 5/x + (3/2x²) + c

(5b)
x² + y² – 6x – 2y – 15 =0
Grouping;
x² – 6x + y² – 2y = 15
Adding half the square of the coefficient of x and y to both sides
(x² – 6x + 3²) + (y² – 2y +1²) = 15 + (3)² + (1)²
(x – 3)² + (y – 1)² = 15+9+1
(x – 3)² + (y – 1)² = 25
h = -3, k = -1
Center = C (-3, – 1)
R² = 25
R = √25
R = 5
Center = C (-3, – 1), Radius = 5

(5c)
Check the image
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(6bi)
f(x) = 2x – 8
f(-2) = 2(-2) – 8
= – 4 – 8
= – 12
(6bii)
f(6) = 2(6) – 6
= 12 – 6
= 6
(6biii)
f(9) = 2(9) – 8
= 18 – 8
= 10
(6biv)
f(0) = 2(0) – 8
= 0 – 8
= – 8
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(7c)
x³ + tx² – x – 8 is divisible by x-1
x – 1 =0
x = 1
Substituting x= 1
(1)³ + t(1)² – 1 – 8 =0
1 + t – 1 = 8
t = 8
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(8a)
(6x – 10)/(x² – 2x – 3)
= (6x – 10)/(x+1)(x-3)
A/(x+1) + B/(x-3) = 6x-10
A(x-3) +B(x+1) = 6x-10
Substituting x= -1
A(-1 – 3) + B(-1 +1) =6(-1) – 10
A(-4) = – 6 – 10
-4A = -16
A = 4
Substituting x= 3
A(3-3) +B(3+1) = 6(3) – 10
B(4) = 18 – 10
4B = 8
B = 2
= A/(x+1) + B/(x-3)
= 4/(x+1) + 2/(x-3)

(8b)
(5^2y) – (5^1+y) + 6 =0
(5^2y) – (5^1)(5^y) + 6 =0
Let 5^y = x
x² – 5x + 6 =0
x² – 3x – 2x + 6 =0
(x² – 3x) (-2x + 6) =0
x(x – 3) – 2(x – 3) =0
(x – 2) (x – 3) =0
x – 2 =0
x = 2
x – 3 =0
x = 3
Recall:
5^y = x
Substituting
5^y = 2
y = Log₅(2)
y₁= 0.4307
5^y = 3
y = Log₅(3)
y₂ = 0.6826
y₁= 0.4307, y₂ = 0.6826

(8ci)
Check the image
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(12)
TABULATE
X | 7 | 3 | 5 | 5 | 7 | 4 | 6 | 4 | 6 | 7

Y | 5 | 7 | 6 | 8 | 5 | 6 | 5 | 7 | 4 | 7

xy | 35 | 21 | 30 | 40 | 35 | 24 | 30 | 28 | 24 | 49

x² | 49 | 9 | 25 | 25 | 49 | 16 | 36 | 16 | 36 | 49

y² | 25 | 49 | 36 | 64 | 25 | 36 | 25 | 49 | 16 | 49

Σx² = 310, Σx = 54, Σxy = 316
Σy² = 374, Σy = 60, n = 10

(12ai)
x̄ = Σx/n
x̄ = 54/10
x̄ = 5.4

(12aii)
ȳ = Σy/n
ȳ = 60/10
ȳ = 6

(12b)
Check the image

(12c)
The negative coefficient -12.06 suggests that as the independent variable increases, the dependent variable tends to decrease.
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(13a)
Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data.

(13b)
(i) Statistics helps in providing a better understanding and accurate description of nature’s phenomena.
(ii) Statistics helps in the proper and efficient planning of a statistical inquiry in any field of study.
(iii) Statistics helps in collecting appropriate quantitative data.
(iv) Statistics helps in presenting complex data in a suitable tabular, diagrammatic and graphic form for an easy and clear comprehension of the data.

(13c)
-Measure of Central Tendency-
(i) Mean
(ii) Median

-Measure of Dispersion-
(i) Standard deviation
(ii) Variance

(13d)
The mean is more important and the most used measure of central tendency because it uses all values in the data set to give you an average. For data from skewed distributions, the median is better than the mean because it isn’t influenced by extremely large values.
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