﻿
Ceebookanswers - 100% NECO GCE 2019 MATHEMATICS SOLUTION AVAILABLE HERE

# Gurusfamily.Net

Welcome Gurusian & Ceebookers ! Whatsapp Group:  No 1. And Best Exam Portal Site In Nigeria. We Helped Thousands Of Nigerians Pass Their Exams And We Have Proofs.
Call/WhatsApp: 08061695859 Call/Text/WhatsApp:
08061695859
08064643382

100% NECO GCE 2019 MATHEMATICS SOLUTION AVAILABLE HERE
Author: Mr. Dan
Posted on: 00:04:21 Sat, 30 Nov 2019

2020 JAMB & WAEC UPDATE

Are You Writing Upcoming 2020 JAMB & WAEC? Then Submit Your Number Below For Updates Call: 08064643382 Now..

*MATHS OBJ*

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

(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=2x2+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

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

(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)

(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)

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

100% Neco Gce 2019 Real ( Mathematics) Answers Available Here..

Direct Mobile: N1500 MTN Card.

Forward your Name, MTN CARD PIN, Subject Name, Phone number to: 08064643382..

PLEASE NO CALLS. JUST SEND THE CARD

REMEMBER: Always Pay At least a Day Before Exam

Subscribtion ends 1hr to Exam.

NOTE:- All SMS sent to the above Phone Number are Attendable, Our Phone Number might be diverted to avoid distraction. Always Send us SMS of your Complain.

Make Sure you Subscribe if you Dont Want to be on Hot Seat.

Wishes you Best of Luck..

Stay Updated with all Nigeria University News, Polytechnic News, JAMB UTME, Post-UTME, WAEC, NECO, NABTEB, GCE and more; Join/Like Us now on Facebook and Twitter Gurusfamily Home > Forum > \$title
Users Online: 135
For Inquiry:
Call/Text/Whatsapp: 08061695859

Whatsapp Group: 