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NABTEB EXPO 2020: Mathematics (OBJ And Theory) Questions And Answers Now Available

Posted by on September 26th, 2020


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How to Get 2020 NABTEB MATHEMATICS ESSAY AND OBJECTIVE Question Paper.

Follow the steps below to get your NABTEB 2020 MATHEMATICS questions and answers 5 hours before your exam time.

2020 Nabteb Mathematics Answers
=======================================

Obj
01-10: CBBCBDDACC
11-20: ABDBABDBCB
21-30: CABBCBAACD
31-40: BCABDABBDB
41-50: CBCBACCBAA

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

NUMBER 1⃣ 2⃣ 3⃣ 4⃣ 5⃣ 6⃣ 7⃣ 8⃣ 9⃣ 🔟 1⃣1⃣ 1⃣2⃣ 1⃣3⃣ 1⃣4⃣

Theory

(1a)
214base5 and 23base5

214base5 = 2×5² + 1×5¹ + 4×5⁰ = 2×25 + 5 + 4
= 59base10

23base5 = 2×5¹ + 3×5⁰ = 10 + 3 = 13base10

59base10 × 13base10 = 767
Convert to base Five

5| 767
5| 153 > 2
5| 30 > 3
5| 6 > 0
5| 1 > 1
5| 0 > 1

11032base5

(1b)
20+3 – x/3 = x+7
20+3-7= x+x/3 = 3x+x/3

16 = 4x/3
x =16×3/4
x=12

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

(2a)
Draw an equilateral triangle

A
/\
m+n / \ (3m-5n)
/ \
B ——– C
(m-2n+3)

Since the triangle is equilateral
m+n=3m-5n=m-2n+3
m+n=m-2n+3
Collect Like terms
m-m+n+2n=3
3n=3
n=3/3
n=1

Also,
m+n=3m-5n
Collect like terms
n+5n=3m-m
6n=2m
6(1)=2m
2m=6
m=6/2
m=3

Length=m+n
=3+1
=4

(2b)
Perimeter=3L
L=m+n=3m-5n=m-2n+3
L=4
P=3L
P=3×4
:. Perimeter=12 unit

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

(3a)
√0.81×10-⁵/√2.25×10⁷

= √81×10-²×10-⁵/√225×10⁷×10-²

= √81×10-⁵/225×10⁷

= √81/225 × √10-⁵-⁷
= √81/√225 × √10-¹²
= (9²)½/(15²)½ × (10-¹²)½
= 9/15 ×10-⁶

Multiply through by 3
9/15×10-⁶
3/5×10-⁶
0.6×10-¹×10-⁶
= 6×10-⁷

(3b)
3/√3(2/√3 – 12/√6)
= 3/√3(2/√3) – 3/√3(12/√6)
=6/√9 – 36/√18 = 6/3 – 36/√9×√2
= 6/3 – 36/3√2
= 2 – 12√2 = 2-12√2 × √2/√2
= 2 – 12√2/2
= 2-6√2

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

(4a)
U=80
n(P)=40
n(C)=45
n(B)=30
n(PΠC)=20
n(PΠB)=12
n(P’ΠC’ΠB’)=3
n(PΠCΠB)=x
n(BΠC)=y

Draw a Venn diagram
[img]https://i.imgur.com/MgLuFIb.jpg[/img]

Number of student offering physics only = 40–(20 – x + x + 12 – x)
=8 + x

Number of student offering Chemistry only = 45 –(20 – x + x + y – x)
= 25 + x – y

Number of student offering Biology only = 30 –(12 – x + x + y – x)
=18 + x – y

:. 8 + x + 25 + x – y + 18 + x – y + 20 – x + 12 – x + y – x + x + 3 = 80

x = y – 6

(ii)
Students offering one subject = 8 + y – 6 + 25 + y – 6 – y + 18 + y – 6 – y = 33 + y

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

(5a)
20pencils = ₦55
440pencils= X
X=55×440/20 = ₦1210

Unsaleable = 10/100 ×440 = 44pencils
Saleable = 440 – 44= 396pencils
Total expenses = 1210+150+200= ₦1560 = c.p

Sold
12pencils = ₦60
326pencils = X
X= 60×396/12= ₦1980 = s.p

Profit = s.p – c.p = 1980-1560 = ₦420

(5b)
% Profit = Profit/c.p ×100%
= 420/1560 ×100
= 26.92%

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

(6)
Area of triangle FBD =1/2 × base × height
= 1/2 × 8 × 3
=12cm²

Area of ABCD = L × B
=8×6
=48cm²

Area of CDIH = L×B
=12 × 6
=72cm²

Area of FDGH= L × B
=12 × 5
=60cm²

:. Total surface Area = 2(12) + 2(48) + 2(72) + 2(60)

=24 + 96 + 144 + 120

=384cm²

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

(7a)
y= 5x – 2x² , –2 ≤ x ≤ 4

TABULATE

x | -2 | -1 | 0 | 1 | 2 | 3 | 4 |

x² | 4 | 1 | 0 | 1 | 4 | 9 | 16 |

5x | -10 | -5 | 0 | 5 | 10 | 15 | 20 |

-2x² | -8 | -2 | 0 | -2 | -8 | -18 | -32 |

y | -18 | -7 | 0 | 3 | 2 | -3 | -12 |

(7b)
View Image ➡ [https://i.imgur.com/oG4h0WO.jpg]

(7ci)
The line of symmetry is at point (1,3).
:. The lube is x=1

(7cii)
The solution of the equation 5 + 5x – 2x² = 0, From the graph can be obtained as follow;

y=5x – 2x²
y=5 + 5x – 2x²
5x – 2x² = 5 + 5x – 2x²
y= -5.

y= -5 is traced from the graph and the values are -0.75 and 3.25

(7ciii)
The maximum turning point is y=3

(7civ)
Gradient at point x=1
Gradient=∆y/∆x
=4 – 2.2 ÷ 2.3 – 0
=1.8/2.3
=0.78

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

(8a)
TABULATE

Class interval | Class mark | Frequency | Tally | Cumulative frequency | Class Boundaries |

UNDER Class interval
21-30
31-40
41-50
51-60
61-70
71-80
81-90
91-100

UNDER Class mark
25.5
35.5
45.5
55.5
65.5
75.5
85.5
95.5

UNDER Frequency
2
5
7
9
11
8
5
3

UNDER Tally
//
~////~
~////~ //
~////~ ////
~////~ ~////~ /
~////~ ///
~////~
///

UNDER Cummulative Frequency
2
7
14
23
34
42
47
50

UNDER Class boundaries
20.5 – 30.5
30.5 – 40.5
40.5 – 50.5
50.5 – 60.5
60.5 – 70.5
70.5 – 80.5
80.5 – 90.5
90.5 – 100.5

(8bi)
Median = 1/2 of N
=N/2
=50/2
25th Cummulative Frequency

Median = 62.5

(8bii)
Semi interquantite Range = Q3 – Q1 ÷ 2

Q3= 3N/4
=30 × 50 ÷ 4
37.5th Cummulative Frequency

Q3= 74.5

Q1= N/4
=50/4
12.5th Cummulative Frequency
Q1=48.5

Semi Interquantite Range= 74.5 – 48.5 ÷ 2
=26/2
=13

(8biii)
View Graph ➡ [https://i.imgur.com/DfExmsk.jpg]

(8c)
The percentage of the students that passed the examination. If 45% is the passed mark = 7 + 9 + 11 + 8 + 5 + 3 × 100 ÷ 50
= 43/50 × 100
=86%

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

(9)
t∝v
t∝1/p
t∝v/p1= >t=kv/p
P=5,t=10minutes
V=20
10=20k/5
10*5=20k
50=20k
K=50/20=5/2

(9ai)
t=5/2v/p
t=5v/2p

(9aii)
V=50,t=?,p=2
t=5/2*50/2
=125/2 = 62.5minutes

(9aiii)
V=40, t=20minutes, p=?
20=5/2*40/p
20p=100
P=100/20=5

(9bi)
A=P(1+r/100)^n
(A/P)^1/n =(1+r/100)^n*1/n
(A=P)^1/n =1+r/100
(A=P)^1/n – 1 =r/100
100[(A/P)]^1/n – 1 ] =r

(9bii)
100[(506.19/450.0)^⅓-1]=r
r=100(1.040-1)
r=100(0.04)
=4

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

(10ai)
X²-10/X²+4x-5. =0
X²-10=0
X=±√10
X=+√10 or
X=-√10

(10aii)
X²+4X-5=0
X²-X+5X-5=0
X(X-1)+5(X-1)=0
X+5=0. or X-1=0
X=-5 or X=1

(10b)
(i)r=a+b
=7i + 2j-k
(ii)r=a+b+c
=2c+3j

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

(11ai)
Given
x² – 10 / x² + 4x – 5
The value of x for which the above fraction is zero is
x² – 10 = 0
x² = 10
x² = ±√10

(11aii)
The value of x for which the above fraction is undefined is
x² + 4x – 5 = 0
(x² – x) + (5x – 5) = 0
x(x–1) +5(x–1)=0
(x+5)(x–1)=0
x+5=0 or x–1=0
x=–5 or x=1

(11bi)
The resultant of the vectors
a=3i + j + 2k
And
b=4i + j – 3k

Resultant = | a + b |
a + b = (3i + j + 2k) + (4i + j – 3k)
=7i + 2j – k
:. The resultant
R=√(7)² + (2)² + (-1)²
R=√49+4+2
R=√54
R=7.35 units.

(11bii)
a=3i, b= -2i – j
c= i + 4j
Resultant = | a + b + c |
a + b + c = 3i + ( -2i – j) + i + 4j
=3i – i + 3j
=2i – 3j

Resultant = √(2)² + (3)²
=√4+9
=√13
=3.6 units

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

(12a)
Five montly moving average

0.3 +0.3+2.8 + 8.6+20.3/5= 32.8/5 = 6.46

0.3+2.8 +8.6+20.3+22.6/5 = 54.6/5 = 10.92

2.8+8.6+20.3+22.6+33.0/5 = 87.3/5 = 17.46

8.6+20.3+22.6+33.0 +29.2/5 = 113.7/5 = 22.74

(12b)
Deposite = ₦10,250

Instalment payment = ₦3600 per week

Total numbers of Week = 6*4 = 24 Weeks

Total instalment payment=24*3600 = ₦86,400

Total cost of motor bike = Intial deposit +
Total instalment = 10,250 +86,400= ₦96,650

(12c)
Worth of goods 7,000,000
custom duty = 25/100*7000,000 = 1750,000
profit made = 35/100*7000,000 = 2,450,000
selling price = Worth of goods & Custom dutyt
proft made 7,000,000 + 1,750,000 + 2,450,000= ₦11,200,000

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

(14)
Flat fee = ₦1000
Distance charge = ₦250 per km
Neight charge = ₦100 per gram
If weight =₦75g
Weight Charge = 75*100 = ₦7,500
If distance = 900KM
distance Charge = 900*250 = ₦225,000

(14ai)
Company charge = flat feet distance Changet
Weight charge =
1000 + 7500 +225000= ₦233,500

(14aii)
Customer’s bill = company charge + VAT
Stree VAT=5/100 * 233,500=₦11,675
Customer’s bill = 233,500 +11675
= ₦235,175

(14bi)
Personal allowance = ₦18,000

(14bii)
Sponse allowance = ₦5000

(14biii)
children allowance = ₦4,000 per child

(14biv)
Dependent relative = ₦6000 each

(14bv)
Given gross per annum= ₦1,020,000
INHIS insurance = 1/100 * 1,020,000 = ₦10,200
Union dues = 2/100*1,020,000 = ₦20,400
pension scheme = 7.5/100 * 1,020,000 =₦76,500
Tax paid =10/100*1020,000 = ₦102,000

(14ci)
monthly tax; 116700/12 ₦9,725.50

(14cii)
Total monthly pay =1167000/12= 97,250.00

Monthly net pay = 97,250 -9725=₦87,525.00

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

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