# NECO GCE Mathematics Questions and Answers – 2019 NECO GCE Expo

National Examination Council, NECO GCE Mathematics Questions and Answers for 2019 NECO GCE Examination is now available, subscribe now and get the questions and already solved answers by Examrelief.

# 👑 2019 NECO GCE MATHEMATICS ANSWERS👑

1-10: BCBEDDCCDC
11-20: DEBCBBDDEC
21-30: BCBCCBBEBD
31-40: EBABEDCDEB
41-50: DDAECACBAB
51-60: DDBCEACBBE

(1)
√((0.0024)×35000)/0.0105

No | Log
0.024 |_3.3802
35000 |4.5441
– |=1.9243
0.0105|_2.0212 –
|3.9031 ÷2
|1.9516

√((0.0024)×35000)/0.0105

Antilog of .9516 =89.45

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

(2a)
Given that the roots of the equation are
X = -2/3 and X = -3/2
3x = -2 and 2x = -3
3x + 2 = 0 and 2x + 3 =0
(3x+2)(2x+3) = 0
6x² + 9x + 4x + 6 = 0
6x² + 13x + 6 = 0

(2b)
R = [3 4 0 ]
[2 0 3 ]
[1 2 2]

(2ci)
2/3R

=2/3[3 4 0]
[2 0 3]
[1 2 2]

= [⅔(3) ⅔(4) ⅔(0)]
[⅔(2) ⅔(0) ⅔(3)]
[⅔(1) ⅔(4) ⅔(2)]

= [2 8/3 0 ]
[4/3 0 2 ]
[2/3 4/3 4/3]

(2cii)
|R|

= |3 4 0|
|2 0 3|
|1 2 2|

=3|0 3| -4|2 3| +0|2 0|
|2 2| |1 2| |1 2|

=3(2×0-3×2)-4(2×2-3×1)
+0(2×2-0×1)
=3(0 – 6)-4(4 – 3) +0(4 – 0)
=3(-6) -4(1) +0(4)
= -18 – 4 + 0 = -22

(2ciii)
The transpose of R
= [3 2 1]
[4 0 2]
[0 3 2]

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

(3a)
Given : R(3,5) and S(-2, -6)
equation to line through them is :
y-5/x-3 = -6-5/-2-3
y-5/x-3 = -11/-5
y-5/x-3 = 11/5
5(y-5)= 11(x-3)
5y-25 = 11x-33
5y-11x = 25-33
5y-11x = -8 or 11x-5y = 8

(3b)
RS= √(X1X2)^2 (y1y2)^2
√(-2-3)^2 + (-6-5)^2
√(-5)^2 + (-11)^2
√25 + 121
√146
= 12.08

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

(4a)
Given: curve; y x² – 3x
gradient ; dy/dx = 2x – 3
At (2-2), gradient = 2(2) – 3
4-3 = 1

(4b)
Given; y= 1+x²/1-x²
dy/dx = (1-x²)(2x) – (1+x²)(-2x)/(1-x²)²
= (1-x²)(2x) – (1+x²)(2x)/(1-x²)²
= 2x(1-x² + 1+x²)/(1-x²)²
= 2x(2)/(1-x²)²
= 4x/(1-x²)²

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

(5)
No of blue balls = 6
No of red balls = 10

(i)
Prob (2 balls of some colour)
= BB or RR
Total no of balls = 6+10 = 16
BB or RR
(6/16 × 5/15) + (10/16 × 9/15)
=30/240 + 90/240
=120/240 = 1/2

YOU MAY ALSO LIKE:  NECO 2017: Verified Biology Obj & Theory Answers/Expo/Runz/Choke Available Here

(ii)
Prob (2 balls of different colours)
= BR or RB
= (6/16 × 10/15) or (10/16 × 6/15)
= 60/240 + 60/240 = 120/240
=1/2

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

(6a)
27^(2x+1) × 3^-x = 81^(x-2)/9^(x+2)
= 3^3(2x+1) × 3^-x = 3^4(x-2)/3^2(x+2)
=3^6x+3-x = 3^4x-8-2x-4
5x+3 = 2x – 12
5x – 2x = -12-3
3x = -15
3x/3 = -15/3
X = -5

(6b)
X/x+101 = 11/1000
Since all the members are in binary, convert all to denary (base 10)
Xbase2 = Xbase10
101base2 = (1×2^2)+(1×2^0) = 4+1 = 5base10
11base2 = (1×2¹)+(1×2raise to power 0) = 2+1 = 3base10

1000base2 = 1×2^3 = 8base10
X/X+101 = 11/1000 –> X/X+5 = 3/8
3(x+5) = 8(x)
3x+15 = 8x
15 = 8x – 3x
15 = 5x
15/5 = 5/5
X = 3
Convert X=3 to base 10 to base 2
2|3
2|1R1
|0R1
.:. x = 11

(6c)
Given that log5 base 10 = 0.699

and log3 base 10 = 0.477
10
Log75 base 10 = log(3×5×5) base 10

=log(3×5²) base 10

=log3 base 10 + 2log5 base 10

=0.477 + 2(0.699)
= 0.477 + 1.398
= 1.875
log75 base 10 = 1.875

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

(7a)
Given that Y=2×2+7x-6
To find the gradient of the curve at the point x=3
dy/dx=4x+7 (at x=3)
dy/dx=4(3)+7
dy/dx=19

(7bi)
Ade bought 7kg of maize + 4kg of meat=N4240
Kemi bought 3kg of maize + 5kg meat =N4610
Let x = 1kg of maize and y=1kg of meat
Therefore 7x+4y=4240—–(eq1)
3x+5y=4610———(eq2)
Substituting simultaneously
Multiply eq1 by 3 and eq2 by 7
21x+12y=12720—eq3
21x+35y=32270—eq4
Substract eq3 from eq4
23y=19550
23y/23=19550/23=850
y=850
Substitute y=850 in1
7x+4y=4240
7x+4(850)=4240
7x=4240-3400
x=840/7
=120
Hence the total cost price per Kg of maize is N120.00 while the total cost per kg of meat is N850.00

(7bii)
Total cost of 10kg of maize and 5kg of meat
=10x+5y
=10(120)+5(850)
1200+4250
=N5450

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

(8)
=========================

(9)
Draw the diagram
Q(34.5°N, 22.3°W)
P(34.5°N, 38.7°E)
R(35.4°S, 38.7°E)
Distance between P and Q along the parallel of latitude
D = Φ/360 × 2πr
Where r = RcosΦ
Φ = 38.7° + 22.3° = 61°
Q = 34.6°
R = 6400km
D = 61/360 × 22/7 ×6400cos34.6
D =1717760(0.8231)/252
D = 5,610.66768
= 5,610km

(9b)
|PR|=(34.6+35.4)/360 * 2 *22/7*6400
=70/360 * 44/7 *6400
=19712000/2520
=7820km(3 sf)
Therefore the shortest distance between P and R = 7820km(3 sf)

YOU MAY ALSO LIKE:  NECO 2020: Verified NECO Animal Husbandry Specimens Available Here

(9c)
Circumference of the circle of latitude through R = 2πr where r = RcosΦ
= 2πRcosΦ
= 2 × 22/7 × 6400cos35.4
=281600/7(0.8151)
= 229,532.16/7
= 32,790.3
= 32,800km (3 s.f)

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

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

======COMPLETED=======

BEST OF LUCK

National Examination Council, NECO GCE Mathematics Questions and Answers for 2019 NECO GCE Examination is now available, subscribe now and get the questions and already solved answers by Examrelief.

## ATTENTION:- PLEASE AFTER SENDING YOUR CARD DO NOT THROW THEM AWAY UNTIL YOU RECEIVE A CONFIRMATION MESSAGE FROM US.

How to Subscribe:
Forward MTN Recharge Card Pin Along with your Phone Number And Your Subject Name to Examrelief On 08020886337

E.g; Text, “I sub For 2019 NOV/DEC NECO GCE expo, Mathematics, 1234567890123455 (N500), 1234567890123455 (N500), 08012345678, WhatsApp” to 08020886337

Subscription Price List;

Direct SMS (You get the complete OBJ and Theory Answers Direct to your phone Via SMS) ===> N1,000 MTN CARD

WhatsApp (You get the complete OBJ and Theory Questions and Answers Direct to your phone Via WhatsApp) ===> N1,000 MTN CARD

Online Answers Page: (Here we send you the password to our Answers Page where the answers are posted online.) ===> N600 MTN CARD

NOTE:- Subscription Ends 1Hrs to each paper.

## 2019 NECO GCE Mathematics Questions and Answers

WARNING: For more info/clarification, You can always reach us through Text Message (SMS) or WhatsApp chat.

WE DON’T TALK MUCH

Learn how to pay for what you need most. Be wise and try harder to pass

Always Subscribe To Avoid Being On Hot Seat.