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# 👑 2019 NABTEB MATHEMATICS ANSWERS👑

__100% VERIFIED MATHEMATICS OBJ ANSWERS__

*1-10: ACCBABCCCA
11-20: BCDCCAABCB
21-30: DAAACCBDCA
31-40: CCCBABAADD
41-50: CAADDACCDD*

__100% VERIFIED MATHEMATICS THEORY ANSWERS__

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^ MEANS “Raise To Power”

* MEANS multiply(x)

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(1a)

2^2x * 8^x-2 * 4^3(1-x) = 1/64

2^2x * 2^3(x-2) * 2^6(1-x) = 1/2^6

2^2x * 2^3x-6 * 2^6-6x = 2^-6

2^2x+3x-6+6-6x = 2^-6

2x +3x- 6+6-6x = -6

5x-6x = -6

-x = -6

x = 6

(1b)

Log 125/81 + 2log 1^2/3

Log125/81 + 2Log5/3

Log 125/81 + Log(5/3)^2

Log 125/81 + Log 25/9

= Log ( 125/81 × 25/9 )

= Log ( 5^3 × 5^2/3^4 × 3^2 )

= Log 5^5/3^6

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(2a)

A = {1,3,5}

(i)

B = { x: x^2 – 7x + 12}

= { x: (x^2 – 3x) – (4x + 12)}

= { x: x(x-3) – 4(x-3)}

= { x: (x-4) (x-3) = 0}

= { x: x-4=0 or x-3=0}

B = {3,4}

(ii)

AnB = {1,3,5}n{3,4}

= {3}

(2bi)

Sum of interior angles = (2n-4)90

(2n-4)90 = 1080

2n-4 = 1080/90

2n-4 = 12

2n = 12+4

2n = 16

n = 16/2 = 8

(2bii)

Interior angles = 1080/80 =135degree

Exterior + Interior = 180

Exterior + 135 = 180

Exterior = 180-135 = 45degree

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(3a)

120x + 10(base 4) = 110(base 4)

(1*x^2) + (2*x^1) + (1*x^0) + (1*4^1) + (0*4^1)= (1*4^0) + (1*4^1) + (0*4^0)

x^2 + 2x + 5 = 20

x^2 +2x + 5 -20 =0

x^2 + 2x -15 = 0

(x^2 + 5x)-(3x-15) = 0

x(x+5)-3(x+5) =0

(x-3)(x+5) =0

x -3=0 or x+5=0

x =3 or x= -5

Therefore , x=3

(3b)

Length of chord = 2rsinΦ/2

= 2×6sin80/2

= 12sin40

= 12 × 0.6428

L = 7.7cm

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(4a)

Proportion = 1^1/2 : 2 : 3

= 3/2 : 2 : 3

Sum = 3/2 + 2 + 3 = 3/2 + 5/1

= 3+10/2 – 13/2

Let the total length be L

Longest piece = 20.7cm

= 3/13/2 × L = 20.7

= 3×2/13 × L = 20.7

6L = 13×20.7

6L = 269.1

L = 269.1/6

L = 44.85cm

Hence, the shortest piece

L = 3/2/13/2 × 44.85

= (3/2 × 2/13) × 44.85

= 3/13 × 44.85

= 10.35cm

(4b)

PRT= ΔPQR { Alternative segment is equal }

ΔPQR =58°

From ΔPQR

ΔPQR + ΔQRP + ΔPRQ =180°

58+65+ΔPRQ = 180°

PRQ =180-123 =57°

ΔPRQ = 57°

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(5a)

Let’s Ada’s age = x

Chika = x+10

In five years time

Ada = x+5

Chika = (x+10)+5 = x+15

Chika = 2(Ada)

x+15 = 2(x+5)

x+15 = 2x+10

2x-x = 15-10

x = 5years

Ada = 5years

Chika = x+10

= 5+10 =15years

Chika = 15years

(5b)

21m^2 + 5m =6

21m^2 + 5m -6=0

(21m^2 + 14m)-(9m-6)=0

7m(3m+2)-3(3m+2)=0

(7m-3)(3m+2)=0

7m-3=0 or 3m+2=0

7m=3 or 3m= -2

m = 3/7 or m = -2/3

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(6a)

PQ = 420km

Latitude of P =30°N

Latitude of Q =?

PQ = Φ/360 × 5πR

420=Φ/360 ×2×22/7×6400

420=281600Φ/2520

281600Φ=420×2520

281600Φ= 1058400

Φ=1058400/281600

Φ=3.76

Φ=4°{ nearest degree }

Q°N – 30 =4

Q = 4+30

Q=34°N

(6b)

(1- √3)/√5 – √2) + (1+ √3/√5 + √2)

= (1 – √3)(√5 + √2) + (1+√3)(√5 – √2)/ (√5 – √2)(√5 + √2)

= (√5 + √2 – √15 – √6 + √5 – √2 + √15 – √6)/ ( 5 + √10 – √10 – 2)

= 2√5 – 2√6/5-2 = 2(√5 – √6)/3

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(7a)

Given that;

A = 3.48, B = 21.75 and C = 0.265

√BC/A² = √21.75*0.265/(3.48)²

Tabulate

Under; No

21.75

0.265

(21.75*0/265)½

(3.48)²

0.1983

Under; Log

1. 3375

+

T. 4232/0.7607

=0.7607÷2

=0.3804. Numerator

0.5416*2

0.3804. Denumerator

0.3804

1.0832/T. 2972

√21.75*0.265/(3.48)²

Total answer. = 0.1983

(7b)

s+ 2 = 2 +t/2-t

(s+2)(2-t)= 2+t

2s – st + 4 -2t=2+t

2s+4-2=t+st+2t

2s+2=3t+st

2s+2=t(3+s)

t =2s+2/3+s

t =2(s+1)/s+3

Given s=10.5

t =2(10.5+1)/10.5+3

= 2×11.5/13.5 =1.7

t=1.7

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(8a)

Interior angle = 140

Exterior angle = 180-140=40

Number of sides =360/40=9

Therefore,

2p+1=9

2p=9-1

2p=8

p =8/2=4

p =4

(8bi)

U=90

n(FnT)=5

n(TnA)=8

n(FnA)=15

n(F)=30

n(T)=25

n(FnTnA)=4

n(FnTnA)^c=x

Football (F) only = 30-(1+4+11)

=30-16=14

Tennis (T) only = 25-(1+4+4)

= 25-9=16

Athletic (A) only = 18

(8bii)

14+16+18+11+4+1+4+x=90

= 68+x=90

x=90-68

x=22

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

(9a)

H = height of the tree

From ΔTCA

tan45 = H/x+3

H=(x+3)tan45……(i)

From ΔTCB

tan54 = H/x

H=xtan54……(ii)

Equating (i) and (ii)

(x+3)tan45 – xtan54

x+3= 1.3764x

1.3764x – x=3

0.3764x=3

x=3/0.3764

x=7.97m

Hence ,

Height of the tree

H=xtan54

= 7.97tan54

= 7.97 × tan54

= 7.97 × 1.3764

H=10.97m

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(11)

CLICK HERE & HERE

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(12a)

Tabulate

Age(x) (in months) – 202,168,158,151,135,116

Frequency (f) – 1,1,1,1,1,1

Fx – 202,168,158,151,135,116

Ef = 6

Efx = 930months

(i)

Mean = Efx/Ef = 930/6 =155months

= 12years 11months

(ii)

Median = 158+151/2 = 309/2=154.5

= 154months 15days

(12b)

Price tag of the car = N2,400,000.00

Selling price of the old car = N350,000.00

Balance of the new car = N2,400,000 – N350,000

= N2,050,000

10% of 2,050,000 = 10/100 × 2,050,000

= N205,000

Total balance of the new car price = N2,050,000 – N205,000

= N2,225,000

Monthly installment payment for 54months = N2,225,000/54

= N41,759.26

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

(13a)

P=N500,000.00 , R=5% , T=4years

That’s = PRT/100 = 500,000 × 5 × 4/100 = 5,000 × 20

= N100,000

Amount at the end of 4yrs

A= P+I = 500,000 + 100,000 N600,000

(13b)

Total cost of the motor tyre = 15,000 × 100 = 1,500,000.00

Selling price of 45 = N18,000 × 45 = N810,000.00

Selling price of 40 = N20,000 × 40 = N800,000.00

Selling price of the 85 tyres = N810,000 + N800,000

= N1,610,000

Profit = N1,610,000 – 1,500,000 + 285,000

= N110,000 + N285,000 = N395,000

% Profit = 395,000/1,500,000 × 100

= 26.33%

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**SNAPPED ANSWERS**

OBJ. CLICK HERE

1 & 2. CLICK HERE

3 & 4. CLICK HERE

5 & 6. CLICK HERE

6con’t & 7. CLICK HERE

7con’t & 8. CLICK HERE

9. CLICK HERE

11a. CLICK HERE

11b. CLICK HERE

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