NECO Mathematics Obj And Essay/Theory Solution Questions and Answer – Oct/Nov 2020 Expo Runz.

WELCOME TO AYOSTUFFS BEST EXAM EVER
=====================================
KEEP REFRESH THIS PAGE IN EVERY 5MIN
===================================
MATHS OBJ
1-10: ACBAECABAE
11-20: CCABBCAAEC
21-30: BBEBBBACBD
31-40: CCADBAEABD
41-50: DEBBBCCBBD
51-60: DACEADAEEA
      
          MATHEMATICAL THEORY
=============================
(1a)
Equation:
y - y1/x - x1=y2 - y1/x2 - x1
y - 5/x - 6 = 7 - 5/-2 - 6
y - 5/x - 6 = 2/-8 = 1/-4
y - 5 = 1/4(x - 6)
y - 5 = -1/4x + 6/4
y = -1/4x + 6/4 + 5
y = -1/4x + 3/2 + 5
y = -1/4x + 13/2
OR 4y = -x+26
X + 4y = 26

(1b)
 2
 S(2x + 9)dx
-1
= 2X¹+¹/1+1 + 9x]2
                              -1
= x² + 9x]2, -1
=[2²+9(2)] - [(-1)² + 9(-1)]
=[4 + 18] - [1 - 9]
= 22 + 8
= 30
=============================
(2a) 
No that traveled by bus
= u+18+5+12 = 52
=u+35 = 52
u = 52 - 35
u = 17

No that traveled by train 
= 6+5+12+v = 35
V + 23 = 35
V = 35 - 23
V = 12

Total no of tourist
=u+v+w+18+12+6+5 = 100
17+12+w+18+12+6+5 = 100
W + 70 = 100
W = 100 - 70
W = 30

(2b) 
No who travelled by at least two means of transportation
= 18 + 6 + 12 + 5
= 41
=============================
(3ai) 
EF = 50
Median = 25th + 26th/2 
= 6+6/2 = 6

(3aii) 
Range = 10 - 3 = 7
Median + Range = 6 + 7 = 13

(3b) 
Prob(Olu passes) = 2/5
Prob(Tony passes) = 3/4
Prob(Olu fails) = 3/5
Prob(Tony fails) = 1/4
Prob(both passes) = 
2/5 × 3/4 = 3/10
=============================
(4a) 
X² + 3x - 28 = 0
X² +7x - 4x - 28 = 0
X(x+7)-4(x+7) = 0
(X - 4)(X + 7) = 0
X - 4 = 0 OR x + 7 = 0
X = 4 OR x = -7

(4b) 
8x/9 - 3x/2 = 5/6 - x
Multiply by 18
18(8x/9) -18(3x/2) = 18(5/6) -18x
16x - 27x = 15 - 18x
18x + 16x - 27x = 15
7x = 15
X = 15/7
X = 2 whole 1/7
=============================
(5a) 
d/dx(4x³ - 2x + 4)⁵
= 5(4x³-2x+4)⁴ × (12x² - 2)
= 10(6x² - 1)(4x³ - 2x + 4)⁴

(5b) 
5/3(2-x) - (1-x)/(2-x) = 2/3
Multiply through with 3(2-x)
3(2-x)[5/3(2-x)] -3(2-x)[1-x/2-x] = 3(2-x)(2/3)
5 - 3(1-x) = (2 - x)(2)
5 - 3 + 3x = 4 - 2x
3x + 2x = 4 + 3 - 5
5x = 2
X = 2/5
=============================
(6ai)
Volume of sphere = 9 ⅓ ×(its surface area)
4/3πr³ = 28/3×4πr²
r = 28 units 

Surface area = 4πr²
= 4x22/7×28×28
= 9856 units squared. 

(6aii)
Volume = 9 ⅓ × 9856
= 28/3 × 9856
= 91989.33
=91989 cubic units

(6b)
log10(3x - 5)² - log10(4x -3)² = log10 25
Log10(3x - 5/4x - 3)² = log10 25
(3x - 5/4x - 3)² = 25
Square root of both sides gives 
3x-5/4x-3 = ±5
3x-5 = 5(4x-3) OR 3x-5 = 5(4x-3)
3x-5 = 20x-15 OR 3x-5 = 20x+15
3x-20x = 5-15 OR 3x+20x = 15+5
-17x = -10 OR 23x = 20
 X = 10/17 OR x = 20/23
=============================
(7a)
Given: 3x + 5y = 10
5y = -3x + 10 OR y = -3/5x+ 2
Gradient of the straight line = -1 ÷ (-2/5)
= 5/3
Equation of line:y-2/x-3 = 5/3
3y-6 = 5x - 15
3y = 5x - 15 + 6
3y = 5x - 9
y = 5/3x - 3
Intercept of line = -3

(7b)
Amount = P(1+R/100)³
=8000(1+5/100)³
=8000(1.05)³ OR 8000(1.05/100)³
=8000×1.157625
=#9261

Compound interest = Amount - principal
= 9261 - 8000
= #1261
=============================
(8a)
Ta; a+8d=50 --------(1)
T12; a+11d=65---------(2)
Subtract equation (1) from (2) 
a+11d-(a+8d)=65-50
a+11d-a-8d=15
11d-8d=15
3d/3=15/3
d=5

Substitute for d=5 in each equation (1)
a+8d=50
a+8(5)=50
a+40=50
a=50-40=10
Sn=n/2(2a+(n-1)d)
S70=70/2(2*10+(70-1)5)
=35(20+69*5)
=35*365=12,775

(8b)
V=t²-3t+2
d³/dt=v
d³/dt=t²-3t+2
v=6²-3(6)+2
=36-18+2
v=20mls

Recall; ds/dt=v
ds=vdt
∫ds=∫vdt
S=∫vdt
S=∫20dt
S=20t+c
Where c is constant.
=============================
(10a)
Let the woman's age be W
Let the daughter's age be d
Given: w = 4d-----(1)
Given:(w+5)²=(d+5)²+120--(2)
Put eqn(1)into(2)
(4d+5)² = (d+5)² + 120
(4d+5)² - (d+5)² = 120
[4d+5+d+5][4d+5-d-5] = 120
[5d+10][3d] = 120
5(d+2)(3d) = 120
15(d+2)(d) = 120
d(d+2) = 8
d² + 2d - 8 = 0
d² + 4d - 2d - 8 = 0
d(d+4) -2(d+4) = 0
(d - 2)(d + 4) = 0
d - 2 = 0 (only)
d = 2
Daughter is 2 years old

(10b) 
t = w + wy²/PZ
Multiply through by PZ
Pat = PWZ + wy²
Wy² = ptz - pwz
y² = Pz(t - w)/w
y = ±√PZ(t - w)/w
If P = 5, Z = 10, t = 9, w=3
y = ±√(5)(10) (9-3)/3
y = ±√(5)(10)(6)/3
y = ±√100
y = ±10
=============================
(11)
|Score|5-9|10-14|15-19|20-24|25-29|30-34|

|Frequency|3|10-T|3T|8|2T+2|T+2|

(a)Ʃ+=50
3+10-T+3T+8+2T+2+T+2=50

3+10+8+2+2-T+3T+2T+T=50

25+5T=50
5T=50-20
5T/5=25/5
T=5

(11b)
|Score|5-9|10-14|15-19|20-24|25-29|30-34|

|Frequency|3|5|15|8|12|7|

The frequency of the modal class is 15.

(11c)
Tabulate.
|Score|5-9|10-14|15-19|20-24|25-29|30-34|

|F|3|5|1|5|8|1|2|7|50

|X|7|12|17|22|27|32|

|FX|21|60|255|176|324|224|1060

|X-x̅|-14.2|-9.2|-4.2|0.8|5.8|10.8|

|(X-x̅)²|201.64|84.64|17.64|0.64|33.64|116.64|

F|(X-x̅)|604.92|423.2|264.6|5.12|403.68|816.48|2518

Mean (x̅) =Ʃfx/Ʃf=1060/50=21.2

Variance =Ʃf(x-x̅)²/Ʃf
=2518/50
=50.36
=========================
      COMPLETED

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

Ayostuffs.blogspot.Com send all exam expo answer earlier than

others

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




Comments

Post a Comment

Popular posts from this blog

WAEC GCE 2019 CRS/IRS Obj And Essay Answer – August/September Expo

Image
WAEC GCE CRS/IRS Obj And Essay/Theory Solution Questions and Answer –  AUGUST/SEPTEMBER 2019 Expo Runz. If you want to Get the Answers Direct To Your Phone As Sms, Kindly Send MTN Recharge Card worth of N800 Please always use MTN Number to subscribe for Direct Mobile for fast delivery of the answer Or If you want to Get the Answers Online Via Our Password/Link Kindly Send MTN Recharge Card worth of N500 How to Subscribe for the answer: Send The card pin, Subject name(CRS/IRS obj and essay/theory) and your phone number via SMS to 07060807249 Subscription Will End 1 hours Before Each Exam Starts NOTE :- Direct Mobile Candidates Gets Answers on their Phone number as SMS before Exam start. NOTE :- Please Do Not Call Our Number, Just Send Message, All SMS Sent To The Above Number Are Attended To, Our Phone Number Might Be Diverted To Avoid Distraction. Always Send Us SMS Of Your Complaint we will reply to your message immediately. WE DON’T TALK MUCH PLEASE Note -Each subject cost N800 f...
Read more

JAMB Releases 2019 UTME Results, Withholds 34,120 Results

Image
JAMB Releases 2019 UTME Results, Withholds 34,120 Results The Joint Admission Matriculation Board (JAMB) has released the 2019 UTME results. The Joint Admission Matriculation Board (JAMB) has released the 2019 UTME results.About 1,792,719 candidates sat for the exams, while 34,120 candidates had their results withheld for examination malpractice.The Registrar of the board, Professor Ishaq Oloyede, who made the announcement in Abuja on Saturday also stated that 15,145 results were withheld for further clarification. About 1,792,719 candidates sat for the exams, while 34,120 candidates had their results withheld for examination malpractice. The Registrar of the board, Professor Ishaq Oloyede, who made the announcement in Abuja on Saturday also stated that 15,145 results were withheld for further clarification. The Joint Admission Matriculation Board (JAMB) has released the 2019 UTME results.About 1,792,719 candidates sat for the exams, while 34,120 candidates had their results withheld f...
Read more

NECO 2020 Biology Obj And Essay Answer – Oct/Nov Expo

Image
WELCOME TO AYOSTUFFS BEST EXAM EVER ===================================== KEEP REFRESH THIS PAGE IN EVERY 5MIN =================================== NECO BIOLOGY ANSWERS BIOLOGY OBJ 1-10: ECDADACEAD 11-20: CAACCDDEDC 21-30: CCCAAEEDED 31-40: BBCEDACEEA 41-50: DDDCDCADAE 51-60: DCCEAADADB ==================================== You Are To AnswerThree(3) Questions  Only. We Answered All 1,2,3,4,5. Pls Pick Three(3) ==================================== (1a) DIAGRAM: (Make Sure Your Drawing is 8-10cm) [https://i.imgur.com/hm8wAm5.jpg] (1b) (i) Predation: This is a relationship between two organism in which one feeds on the other organisms. The one feeding on the other is called the predator while the one they are feeding on is known as the prey. Eg: owls hunting mice. (1bii) (ii) Mutualism: This is the relationship between two organisms in which both benefits from the feeding eg  association. Eg: Sea anemones and Clownfish (1ci) (i) Epigeal germination (ii) Hypogea germination (1cii) (...
Read more