NECO GCE – MATHEMATICS

Link/Password: N500 MTN CARD

Direct mobile: N600 MTN CARD

Send Subject name, Card Pins, Phone number via SMS to 0815 962 4211

(YOU CAN SUBSCRIBE ON WHATSAPP OR SEND VIA SMS!!!)

After sending ur card, relax and wait for ur password in the midnight or very early in the morning of the exam day!

This page will be passworded 4 hours before exam commence!

# NOTE: If We Don’t Pick Up Your Call, SMS us, we reply SMS faster!

# NECO GCE 2019 – MATHS ANSWERS

MATHS OBJ

01-10: BCBEDDCCDC

11-20: DEBCBBDDEC

21-30: BCBCCBAEED

31-40: EBABEDCDBB

41-50: DDCECDCBAB

# 51-60: DDBCEACBBE

THEORY

++++++++++++++++++++++++

(1)

âˆš((0.0024)Ã—35000)/0.0105

No | Log

0.024 |_3.3802

35000 |4.5441

- |=1.9243

0.0105|_2.0212 –

|3.9031 Ã·2

|1.9516

âˆš((0.0024)Ã—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

6xÂ² + 9x + 4x + 6 = 0

6xÂ² + 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]

= [â…”(3) â…”(4) â…”(0)]

[â…”(2) â…”(0) â…”(3)]

[â…”(1) â…”(4) â…”(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(2Ã—0-3Ã—2)-4(2Ã—2-3Ã—1)

+0(2Ã—2-0Ã—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= âˆš(X1X2)^2 (y1y2)^2

âˆš(-2-3)^2 + (-6-5)^2

âˆš(-5)^2 + (-11)^2

âˆš25 + 121

âˆš146

= 12.08

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

(4a)

Given: curve; y xÂ² – 3x

gradient ; dy/dx = 2x – 3

At (2-2), gradient = 2(2) – 3

4-3 = 1

(4b)

Given; y= 1+xÂ²/1-xÂ²

dy/dx = (1-xÂ²)(2x) – (1+xÂ²)(-2x)/(1-xÂ²)Â²

= (1-xÂ²)(2x) – (1+xÂ²)(2x)/(1-xÂ²)Â²

= 2x(1-xÂ² + 1+xÂ²)/(1-xÂ²)Â²

= 2x(2)/(1-xÂ²)Â²

= 4x/(1-xÂ²)Â²

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

(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 Ã— 5/15) + (10/16 Ã— 9/15)

=30/240 + 90/240

=120/240 = 1/2

(ii)

Prob (2 balls of different colours)

= BR or RB

= (6/16 Ã— 10/15) or (10/16 Ã— 6/15)

= 60/240 + 60/240 = 120/240

=1/2

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

(6a)

27^(2x+1) Ã— 3^-x = 81^(x-2)/9^(x+2)

= 3^3(2x+1) Ã— 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 = (1Ã—2^2)+(1Ã—2^0) = 4+1 = 5base10

11base2 = (1Ã—2Â¹)+(1Ã—2raise to power 0) = 2+1 = 3base10

1000base2 = 1Ã—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(3Ã—5Ã—5) base 10

=log(3Ã—5Â²) 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

=N545

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

(9)

Draw the diagram

Q(34.5Â°N, 22.3Â°W)

P(34.5Â°N, 38.7Â°E)

R(35.4Â°S, 38.7Â°E)

Distance between P and Q along the parallel of latitude

D = Î¦/360 Ã— 2Ï€r

Where r = RcosÎ¦

Î¦ = 38.7Â° + 22.3Â° = 61Â°

Q = 34.6Â°

R = 6400km

D = 61/360 Ã— 22/7 Ã—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)

(9c)

Circumference of the circle of latitude through R = 2Ï€r where r = RcosÎ¦

= 2Ï€RcosÎ¦

= 2 Ã— 22/7 Ã— 6400cos35.4

=281600/7(0.8151)

= 229,532.16/7

= 32,790.3

= 32,800km (3 s.f)

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

## Leave a Reply