2018 NABTEB GCE Mathematics Questions & Answers Expi

2018 Nabteb Gce Mathematics Expo Answers

NABTEB GCE MATHEMATICS WELL ARRANGE VERSION FROM NO. 1 – 15

(1a)
6n+¹ x 9n x 4²n/18n x 2n x 12²n
= 6n+¹ x 3³n x 4²n/(6×3)n x 2n x (4×3)²n
= 6n+¹ x 3³n x 4²n/6n x 3n x 2n x 3²n x 4²n
= 6n+¹/6n x 6n = 6n+¹/6²n
= 6n+¹ – ²n
= 6¹-n

(1b)
Log243/log27
= log3⁵/log3³
= 5log⁵/5log³
= 5/3

(2a)
3t – 2p = 8 ——(i) x2
2t – 3p = 14 —- (ii) x3
6t – 4p = 16 ——–(iii)
6t – 9p = 42 ——–(iv)
Subtract equation (iv) from equation (iii)
– 4p + 9p = 16 – 42
5p = 26
P = -26/5
Therefore p = -5 1/5
From the equator (ii) above
3t – 2p = 8 substitute -26/5 for p
3t – 2 (-26/5) = 8
3t + 52/5 = 8
3t = 8 -52/5
3t = 40 – 53/5
3t = – 12/5
t = -12/5 ÷ 3/1
t = 12/5 x 1/3
t = -4/5
therefore t = -4/5, p = -5 1/5

(2b)
3√3(2/√3 – √12/6)
= 3√3 (2√3 – 2√3/6)
3√3 (2√3 – 3√3)
= 6/3 – 1
= 6 – 3/3
= 3/3 = 1


(3)
Given that;
Pr (Rabiru will pass the examination) = 2/3
Pr (Yahana will pass the examination) = 5/8
Pr (Ibrahim will pass the examination) = ¾
And the Pr (Kabra will fail) = 1 -2/3 = 1/3
Pr (Yahana will fail) = 1 -5/8 = 3/8
Pr (Ibrahim will fail) = 1 -3/4 = ¼
(a) Pr (three will pass) = 2/3 x 5/8 x ¾
(b) Pr (non of them will pass) = 1/3 x 3/8 x ¼ = 1/32
(c) pr (Kabiru and Yahana only will pass)
= Pr (K, Y, I)
= 2/3 x 5/8 x ¼
= 5/48

(4)
DRAW THE DIAGRAM
Let |GT| = |TB| = x
|GN| = |GT| = |TN| = x + x = 2x
|GN| = 2x
From ∆ FGN
Tan 68 = 8/|GN|
|GN| = 8/tan68 = 8/2.4751
|GN| = 3.23m
|GN| = 2 |GT|
|GT| = 3.23/2 = 1.616m

(4i) from the ∆ FGT
tanØ = 8/1.616
tanØ = 4.95
Ø = tan-¹ (4.95) = 78.58º
< FTG = 78.58 = 79º (correct to the nearest degree)

(4ii)
< TFN = 〆
〆 + β- = y
Y = 90 – 68 = 22
β + 0 = 90
β = 90 – 0 = 90 – 79 = 11º
hence, 〆 = y – β
〆 = 22 – 11
〆 = 11º
Therefore <TFN = 11º

(5a)
4 ½ – 3 (y – 2) = 2y +1/3
9/2 – 3 (y-2) = 2y+1/3
Multiply both sides by 3
27/2 – 9 (y-2) = 2y + 1
27 – 18 (y-2) = 2(2+1)
27 – 18y + 36 = 4y + 2
Collect the like terms
-18y – 4y = 2 – 36 – 27
– 22y = 61
Y = -61/-22
Y = 2 17/22 or 2.7727
(5bi)
7 – 2y ≥ 16

(5bii)
7 – 2y ≥ 16
-2y ≥ 16-7
-2y ≥ 9
Y ≤ -9/2
Therefore y ≤ -4 ½

(5biii)
The greatest possible value of y that satisfy the equation = -4 ½

(6)
Let the cost of water melon be x
And let the cost of mangoes be y
12x + 24y = 432 ——-(1) x2
24x + 12y = 360 ——-(ii) x1
24x + 42y = 864 ——–(iv)
Subtract equation (iv) from equation (iii)
36y = 504
Y = 504/36
Y = N14
From equation (i) above
12x + 24y = 432, substitute 14 for y
12x + 24(14) = 432
12x = 432 – 336
12x = 96
Therefore x = N8

(6ai)
The cost of water melon per kg is N8

(6aii) The cost of mangoes per kg is N14

(6aiii)
3x + 2y
3(8) + 2(14)
24 + 28 = 52
The cost of 3kg of water melon and 2kg of mangoes is N52

(6b)
123x = 83 base 10
123ºx = 83
1 x x² + 2x x+3 x xº = 83
x² + 2x + 3 = 83
x² + 2x – 80 = 0
x² + 10x – 8x – 80 = 0
x (x+10) – 8 (x+10) = 0
(x-8) (x+10) = 0
X = 8 or -10

(7a)
(TABULATE AND DRAW A TABLE) p = 3+5t – 2t²
t | -2 | -1 | 0 | 3 | 4 | 5 |
t² | 4 | 1 | 0 | 9 | 16 | 25 |
3 | 3 | 3 | 3 | 3 | 3 | 3 |
5t | -10 | -5 | 5 | 15 | 20| 25 |
-2t² | -8 | – 2 | -2 | -18 | -32 | -50 |
P | -15 | -4 | 6 | 0 | -9 | -22 |

(7b)
DRAW THE GRAPH

(7c)
The root of the equation are -1/2 and 3

(7d)
(TABULATE AND DRAW A TABLE) p = 3 – 2t
t | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
3 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 3|
-2t | 4 | 2 | 0 | -2| -4 | -6 | -8 | -10 |
P | 7 | 5 | 3 | 1 | -1 | -3 | -5 | -7 |

P = 3 – 2t
P = 3 + 5t – 2t²
3 + 5t – 2t² > 3 – 2t
5t – 2t² > -2t
7t – 2t² > 0
t (7 – 2t) > 0
t > 0 or 7 – 2t > 0
t > 0 or -2t > -9 or t <7/2
the range is 0 < t 7/2
0 < t < 3.5

(8a)
X (40ºS, 15 ºW)
Y (40ºS, 45ºW)
(DRAW THE CIRCLE)
Distance along the parallel of latitude = Ø/360 x 2πr
ˉXY = Ø/360 x 2πR cos 〆
= (45-15)/360 x 2 x 22/7 x 6400 x cos 40
= 30/360 x 44 x 6400 x 0.7660/7
= 30 x 44 x 6400 x 0.7660/2520
= 6471.168/2520 = 2567.9km
Distance ˉXY = 257km to the nearest 100km

(8bi)
Z (xºN, 45ºW)
Y (40ºS, 45ºW)
ˉYZ = 6700km
Speed = total distance/total time
850 = (2567.9 + 6700/total time
Total time = 9267.9/850 = 10.9hr
Total time = 10.9hrs approximate 11hrs

(8bii)
Latitude Z
Distance ˉ YZ
ˉYZ = Ø/360 x 2πR
6700 = (x + 40)/360 x 2 x 22/7 x 6400
6700 x 2520 = 281600 (x+40)
x + 40 = 16884000/281600
x + 40 = 59.96
x = 59.96 – 40
x = 19.96º
hence, the distance of Z to the nearest degree = 20ºN

(9)
DRAW THE TRIANGLE DAIGRAM
|YX| = x + 620
From ∆ YFP
Tan56 = R/620
R = 620 x tan56
= 620 x 1.4826
h = 919.19m
To find x
Tan 20 = h/x
X = h/tan20 = 919.19/0.3639
X = | XF | = 2525.45m
|YX| = |XF| + |FY|
|YX| = 2525.45 + 620
= 3145.5.45m
Hence the value of |YX| correct to the four significant figure is 3145m
|XY| = 3145m

(9b)
DRAW THE DAIGRAM
|YX|² = |OY|² + |OX|² – 2|OY| |OX| cosØ
(6)² = (5)² + (5)² – 2 (5) (5) cosØ
36 = 25 +25 – 50cosØ
36 = 50 – 50cos Ø
50cosØ = 50 – 36
50cos = 14
cosØ = 14/50 = 0.28
Ø = cos-¹ (0.028) = 73.34º
Therefore <XZY = ½ (73.74) (Angle at the center twice the angle at the circumference)
<XZY = ½ (73.74) = 36.87º
<XZY = 37º (nearest degree)

(9bii)
Let |XZ| = |YZ| = y
|XY|² = |YZ|² + |XZ|² – 2 |YZ| |XZ| Cos〆
(6)² = y² + y² – 2y² cos 36.87
36 = 0.4y²
y² = 36/0.4 = 60
y = √60 = 7.75m
hence, |XZ| = 7.8m (1dp)

(11a)
TABULATE
Scores x | frequency (f) | YF
2 | 2 | 4
3 | 4 | 12
4 | 5 | 20
5 | 3 | 15
6 | 4 | 24
7 | 2 | 14
– | 20 | 89

(11ai)
Mode is the most frequently occurring number from an observation
Therefore the modal score is 4

(11aii)
The median score = 4 + 4/2 = 8/2 = 4
Therefore the median score is 4

(11aiii)
Mean = Epx/Ef
= 3a/20
Therefore mean = 4.45

(11b)
A = P(1+r/100)n
= 53,000(1+7/100)⁵
= 53,000 (1.07) ⁵
= 53,000 (1.402551731)
A = 74,335.24
Compound interest = 74,335.24 – 53,000
Therefore compound interest = 21,335.24

(12a)
Monthly salary – 6,000.00
Commission = (5% x 8,400 x 52) – 21,840.00
Take home income – 27,840.00

(12b)
Gross pay – N125,018.35
Less:
Income tax (5% x 125,018.35) – N 6250.92
Union dues (2% x 125,018.35) – N 2,500.37
Housing fund (1% x 125,018.35) – N 1,250.18
Pension schemes (7.5% x 125,018.35) – N 3,125.46
Cooperative contribution (10% x 125,018.35) – N 12,5018.84 (23,754.15)
Net pay – N 101,264.20

(14a)
N 1 shares at N 1.10 divided 11%
Yield
Divided 11% x 1 – N 0.11
Less premium – (N 0.10)
Yield – N 0.01
N 1 shares at N 2.50 divided at N 20%
Yield
Divided 20% x 1 – N 0.20
Premium (2.50 – 1) – (N 1.50)
Loss (N 1.30)
N share at N 1.10 divided at 11% gives the highest yield.

(14b)
A = p[(1+r)n-1]/0.04
= 60,000 [(1.04)⁴-1]/0.04
= 60,000 [1.04)⁴-1]/0.04
= 60,000 (1.1698585 – 1)/0.04
A = N 254,787.84

(15a)
Annual tax free allowance:
Rent (15% x 365,500 – 54,825
Personal (12% x 365,500) – 43,680
= 98,685
Therefore annual tax free allowance = N98,685

(15b)
Calculation of annual income taxi
Gross salary – 365,500
Less tax free allowance – 98,685
Taxable income – 266,815

Tax payable:
Taxable income – 266,815
1st 65,000 (a) 0% – (64,000)
Balance after 1st scale – 201,815
Next 100,000 (a) 10% – (100,000)/101,815 10,000
Balance (a) 201% (1,815) 363
Total tax = 25,363
His annual income tax is N25,363

(15c)
Percentage of salary paid as tax
= 25,363/364.500 x 100/1
Therefore percentage of his salary paid as tax approximately 6.9%

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