logo



student ONLINE CANDIDATES : 34




PAYMENT FOR WAEC AND JAMB EXPO HAS STARTED, PAY NOW!!!



neco runs

WAEC / NECO / NABTEB /GCE SUBSCRIPTION PAYMENT EXPO


HOW TO PAY FOR WAEC/ NECO/ NABTEB /GCE ANSWER:

ALL SCIENCE ANSWERS + PRACTICAL COST: N5,500

ALL ART OR COMMERICAIL ANSWERS COST #4,500

 

WHATSAPP US AND SEND:- EXAM TYPE + MTN-CARD + PHONE NUMBER + SUBJECT TO 09055986588 (ONLY ON WHATSAPP)

UPDATE: Our Waec, Neco and Nabteb Exam Runs Payment is on, Earlyanswer is 100% Legit (Invite Your Classmates,Friends Here)

Answer Page

Answer Page

Confirmation page

Verify NECO / WAEC Payment

EARLYANSWER OFFICIAL WHATSAPP GROUP

JOIN OUR GENERAL WHATSAPP GROUP




« PREVIOUS POST:

| NEXT POST:

»

NABTEB 2019 MATHEMATICS OBJ AND ESSAY VERIFIED ANSWER

 

_MATHS OBJ_
1ACCBABCCCA
11BCDCCAABCB
21DAAACCBDCA
31CCCBABAADD
41CAADDACCDD
===========================

(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
===========================

(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
===========================

(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
===========================

(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°
===========================

(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
===========================

(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
===========================

(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
===========================

(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
===========================

(11a&11b)
CLICK HERE FOR THE IMAGE
===========================

(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%

Posted by on 27th May 2019.

Tags: ,

Categories: NABTEB

HOW TO WIN MTN CARD FROM EARLYANSWER
 
CLICK HERE TO SHARE THIS WEBSITE TO YOUR WHATSAPP FRIENDS / GROUPS AND WIN MTN CARD NOW

Chat Admin After You Share To 20 People On Whatsapp to Redeem The Price

Click here to submit your phone number for free updates about WAEC, NECO, NABTEB, GCE.

2023/2024 EXAM RUNS UPDATES👇

CLICK HERE TO GET WAEC 2023 EXAM RUNS


CLICK HERE TO GET NECO 2023 EXAM RUNS


CLICK HERE TO GET NABTEB 2023 EXAM RUNS


Our Answer Page for All Examination

0 Responses

Leave a Reply

« PREVIOUS POST:

| NEXT POST:

»