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fmaths
1BDBCAADCAC
11DCDABBBBCA
21CDCBABDAAB
31DABABCBACA
more coming (7)
tabulate
class interval: 1.0-1.1, 1.2-1.3, 1.4-1.5, 1.6-1.7, 1.8-1.9
class boundary: 0.75-1.15, 1.16-1.35, 1.35-1.55, 1.55-1.75, 1.75-1.95
Median(x):1.05, 1.25, 1.45, 1.65, 1.85
f: 2,3,8,5,2
d=x-A: -0.4, -0.2, 0, 0.2, 0.4
fd: -0.8,0.6, 0, 1.0, 0.8
total:
under f put (Ef= 50), under fd, put Efd= o.4
Mean(x)= A Efd/Ef
=1.45 (0.4/20)
=1.45 0.02
=1.47 metres
10a
5 sqroot2/3-sqroot2 - 5-sqroot2/3 sqroot2 = (5 sqroot2)(3 sqroot2)(5-sqroot2)(3-sqroot2)/(3-sqroot2)(3 sroot2)
=5(3 sqroot2) sqroot2(3 sqroot2)-[5(3-sqroot2)-sqroot2(3-sqroot2]/3(3 sqroot2)-sqroot2(3 sqroot2)
15 5sqroot2 3sqroot2 2-[15-5sqroot2-3sqroot2 2/9 3sqroot2-3sqroot2-2
=15 2-15-2 5sqroot2 3sqroot2 5sqroot2 3sqroot/9-2
=(5 3 5 3)sqroot2/7
=16sqroot2/7=0 16/7sqroot2
10b.
3x-y-z = -2-------------equ (1)
x 5y 2z=5------------equ(ii)
2x 3y z = 0----------equ(iii)
these equ can be written in matrix form as
13a.
P(keji pass) = 9/10, P(keji fail)=1-9/10 =1/10
p(kwasi pass) = 4/5, p(kwasi fail) = 1 - 4/5 = 1/5
p(ania pass)=x, p(ania fail) = 1-x
prob(only one of them) = p(keji pass other fail) p(kwasi pass other fail) p(ania pass other fail)
= 9/10 *1/5*(1-x) 4/5 * 1/10(1-x) x*1/10*1/5=9/50
=9/50(1-x) 4/50(1-x) x/50 = 9/50
=9(1-x) 4(1-x) x=9
9-9x 4-4x x=9
13-12x=9
-12x=9-13
-12x=-4
x=-4/-12=1/3
13b.
prob(of at least one)
=p(only one) p(two of them) p(all of them)
=9/50 [(9/10)(4/5)(1-x) 9/10(x)(1/5) 4/5(x)1/10] (9/10)(4/5)(x)
=9/50 (36/50(2/3) 9/50(1/3) 4/50(1/3) 36/50(1/3)
=9/50 24/50 3/50 4/150 12/50=27/150 72/150 9/150 12/150 36/150
=27 72 9 12 36/150=156/150
(15b)
initial velocityu=0ms^-1, final velocity =20ms^-1
distance s=8m
(i) from V^2=U^2 2as
20^2=0^2 2as
400=0 16a
a=400/16
a=400/16
therefore a= 25ms^-2
(ii)time to covered 40m i.e S=40m
from S=((V U)/2)t
40=((20 0)/2)t
40=20t/2
40=10t
t=40/10=4s
hence t=4 seconds(1)
(Log(base 2)m)^2-log(base 2)m^5=10
(log(base 2)m)(log(base tita)m)-3log(base 2)m=10
(log(base 2)m)(log(base 2)m-3)=10
(log(base 2)m)9log(base 2)m-3)=10
log(base 2)m-3=10
log(base 2)m-3=10/log(base 2)m
log(base 2)m-log(base 2)m^3=10log(base 2)m^-1
log(base 2)m-3log(base 2)m=-10log(base 2)m
log(base 2)m/(3^-1)=log(base 2)m^-10
m/3^-1=-10
m=-10*3^-1
m=-10*1/3
m=-10/3
(2a)
given m*n =((m^2)-(n^2))/2mn
-3*2=(-3)^2-(2)^2
=(9-4)/-12
=5/-12
=-5/2
(2b)
to show that whether * is associative
=(m*n)*p=m*(n p)
=(m n)*p=((m^2)-((n^2))/2mn)*p
=((m^2)-((n^2)^2-p^2)/(((m^2)-((n^2))/2mn)p
=(((m^2)-((n^2)^2)/4(m^2)(n^2))-p^2)/2p((m^2)-(n^2))/2mn)
=(m^2)-((n^2)^2)/4(m^2)(n^2)(p^2)/4(m^2)(n^2))* 2mn/(2p(m^2)-(n^2))
=((m^2)-(n^2))^2-(2mnp)^2)/(2mn(2p)((m^2)-(n^2))
=((m^2)-(n^2))^2-(2mnp)^2))/4pmn(m^2)-(n^2))
also, m*(n p)=m*((n^2)-(p^2))/2np)
=m^2-((n^2)/2np)^2)/2m((n^2)-(p^2))^2/2m((n^2)-(p^2)/2np))
=(m^2)-((n^2)-(p^2))^2)/4np))/(m)((n^2)-(p^2))/np
=(4(m^20np-((n^2)-(p^2))^2)/4np)/m((n^2)-(p^2))/np
=(4(m^20np-((n^2)-(p^2))^2)/4np)/m((n^2)-(p^2))/np
=(4(m^20np-((n^2)-(p^2))^2)/4np)*(np)/m((n^2)-(p^2))
hence * is not associative
(3.)
3x^2 5x^2 1=0
Solving a=3, b=5, c=1
x= -b( _)sqr root (b2-4ac/2a)
=( -5 _sqr root 5 ^2 - 4(3)(1))/(2x3)
= -5 _ sqr root 25 - 12/6
= -5 _ sqroot 13/6
= -5 sqroot 13/6
= -5 sqroot 13/6 or -5 - sqroot 13/6
27 (alpha sqr root3 betasqr root3)=27[(alpha beta (alpha beta)]
alpha beta = -5/6 sqroot13/6 -5/6 sqroot13/6
= -10/6
alphabeta= (-5/6 sqroot 13/6)(-5/6 - sqroot 13/6)
= 25/36 - 13/36 = 25 - 13/36 = 12/36 = 1/3
then, 27 (alphasqroot3 betasqroot3) =27 [(alphasqroot3)= 27[(-10/6)sqroot3(1/3)(-10/6)
= 27 (-1000/216 10/6)
=27 (-1000 360/216)
=27 (-640/216) = -17280/216
=-80. (6)
No of man =5,
No of woman =3,no of committee =3 total no of people =5 3=8.
A no of ways of foring the committee=8C base3,
but nC base r=n!/(n-r)!r!,
8C base3=8!/(8-3)!3!,
=8!/5!3!= 8*7*6*5/5!*3*2=56ways.
(9a)
y=(x-3)(x^2 5),
let u =x-3,
du/dx=1,
for product rule,
dy/dx=Udy/dx Vdu/dx,
dy/dx=(x-3)(2x) (x^2 5)(1),
=2x^2-6x x^2 5=3x^2-6x 5. (9b)
IF (X 1)^2 is a factor of f(x)=x^3 ax^2 bx 3
then (x 1)^2 = 0 is a solution of f(x)
ie x 1=0
or x = -1 (twice)
also (x 1)^2 = x^2=2x 1
when x=-1
f(x)=f(-1)=(-1)^3 a(-1)^2 b(-1) 3
= -1 a-b 3
=a-b 2=0
=a-b=-2
9b continues
also, f(0)= 0^3 a(0^2) b(0) 3 (not equal) to 0
F(1)= 1^3 a(1^2) b(1) 3
=1 a b 3=0
=a b=-4
solving (1) &2(2)
a-b=-2
a b=-4
2a=6
a=-6/2=-3
from(1)
-3-b=-2
-b=-2 3
-b-1
b=-1
a=-3,b=-1
(ii)
therefore f(x)= (x^3)-(3x^2)-x 3
zero of f(x) are
x=-1(twice) and x=1 (11a)
f(x)=x-3/2x-1,
g(x)=x-1/x-1,
g of =g(f(x),
g of =x-3/2x-1-1/x 3/2x-1 1,
g of =x-3-(2x-1)/2x-1/x-3 (2x-1)/2x-1,
x-3-2x 1/x-3 2x-1,
-x-2/3x-4
=x 2/4-3x.
(11b) y=9x-x^3,
y:x(9-x^2),
y:x(3^2-x^2),
y :x(3-x)(3 x),
for the roots of y =0,
y:x(3-x)(3 x)=0,
:x=0,3 or -3,
y=9x-x^2,
dy/dx=9-3x^2,
at turnin point,
dy/dx=0,
9-3x^2=9,
x^2=3,
x=root3,
x=1.732,
For maximum and minimum values,
when x=1.732,
y=9(1.732)-(1.732)^3,
=15.588-5.196,
=10.392 minimum
call08135784898

Neco Exam Time Table
Neco Time Table 2014 June/July Neco Exam Time Table
Below is the Neco Time Table 2014;
NATIONAL EXAMINATIONS COUNCIL (NECO)
KM 8 BIDA ROAD PMB 159, MINNA, NIGER STATE
SENIOR SCHOOL CERTIFICATE EXAMINATION: JUNE/JULY 2014
EXAMINATION TIME-TABLE
22ND MAY TO 10TH JULY, 2014
Thursday 22nd May To Wednesday 28th June (Actual date and time will be fixed by the Council)
4114 Paper IV: Oral Arabic 1hr 20mins
4104 Paper IV: Oral French 1hr 30mins
6181 Paper I: Practical – Clothing and Textiles 3hrs
3091 Paper I: Practical – Foods and Nutrition 3hrs
3081 Paper I: Practical – Home Management 3hrs
4044 Paper IV: Aural: Music 40 mins
4041 Paper I: Performance Test, Music Technology/Alternative 1hr 30mins
===================
Friday 30th May
2011 Paper I: Practical – Biology 2hrs 10.00 am – 12.00 pm
===================
Wednesday 4th June
2021 Paper I: Practical – Chemistry 2hrs 10.00 am – 12.00 pm
===================
Thursday 5th June
2031 Paper I: Practical – Physics 2hrs 45mins 10.00 am – 12.45 pm
===================
Friday 6th June
2061 Paper I: Practical – Agricultural Science 1hr 30mins 10.00 am – 11.30 am
===================
Tuesday 10th June
4063/4061 Paper III&I: Objective & Practical – Geography 2hrs 30mins 10.00 am – 12.30 pm
===================
Thursday 12th June
4073/4072 Paper III&II: Objective & Essay – Government 2hrs 40mins 10.00 am – 12.40 pm
===================
Friday 13th June
4062 Paper II: Essay – Geography 2hrs 10.00 am – 12.00 pm
===================
Monday 16th June
2023/2022 Paper III&II: Objective & Essay – Chemistry 3hrs 10.00 am – 1.00 pm
===================
Tuesday 17th June
1012/1013 Paper II & III: Essay & Objective - English Language 2hr 45mins 10.00 am – 12.45 am
1014 Paper IV: Test of Orals – English Language 45mins 12.45 pm – 1.30 pm
===================
Wednesday 18th June
5033/5032 Paper III & II: Objective & Essay – Commerce 2hrs 40mins 10.00 am – 12.40 am
===================
Thursday 19th June
2033/2032 Paper III & II: Objective & Essay – Physics 3hrs 10.00 am – 1.00 pm
===================
Friday 20th June
2063/2062 Paper III & II: Objective & Essay – Agricultural Science 2hrs 30mins 10.00 am – 12.30 pm
===================
Monday 23rd June
1023 Paper III: Objective – General Mathematics 1hr 45mins 10.00 am – 11.45 am
1022 Paper II: Essay – General Mathematics 2hrs 30mins 12.00 pm – 2.30 pm
===================
Wednesday 25th June
2043 Paper III: Objective – Further Mathematics 2hrs 10.00 am – 12.00 pm
2042 Paper II: Essay – Further Mathematics 2hrs 30mins 12.00 pm – 2.30 pm
4093/4094 Paper III & IV: Objective & Prose – Literature-in-English 2hrs 15mins 3.00 pm – 5.15 pm
===================
Thursday 26th June
2013/2012 Paper III & II: Objective & Essay – Biology 2hrs 30mins 10.00 am – 12.30 pm
4113/4112 Paper III & II: Objective, Essay & Literature - Arabic 3hrs 30mins 2 pm – 5.30 pm
===================
Friday 27th June
4013/4012 Paper III & II: Objective & Essay – Christian Religious Studies 2hrs 30mins 10.00 am – 12.30 pm
4023/4022 Paper III & II: Objective & Essay – Islamic Studies 2hrs 10.00 am – 12.30 pm
===================
Monday 30th June
4083/4082 Paper III & II: Objective & Essay – Economics 3hrs 10.00 am – 1.00 pm
===================
Tuesday 1sh July
4123/4122 Paper III & II: Objective & Essay – Hausa 3hrs 10.00 am – 1.00 pm
4133/4132 Paper III & II: Objective & Essay – Igbo 3hrs 10.00 am – 1.00 pm
4143/4142 Paper III & II: Objective & Essay – Yoruba 3hrs 10.00 am – 1.00 pm
4153/4152 Paper III & II: Objective & Essay – Edo 3hrs 10.00 am – 1.00 pm
4163/4162 Paper III & II: Objective & Essay – Efik 3hrs 10.00 am – 1.00 pm
4173/4172 Paper III & II: Objective & Essay – Ibibio 3hrs 10.00 am – 1.00 pm
===================
Thursday 3rd July
3011 Paper I: Practical – Technical Drawing 3hrs 10.00 am – 1.00 pm
4092 Paper II: Drama & Poetry – Literature In English 1hr 40mins 2.30pm – 4.10pm
===================
Monday 4th July
3013/3014 Paper III & IV: Objective & Drawing – Technical Drawing 2hrs 30mins 10.00 am – 12.30 pm
===================
Schools and Candidates should note the following:
i. For Physical Education and Auto Mechanics Practical Paper I, French, Music, and Arabic Paper IV, the specific venues for the examinations should be confirmed from NECO Office in the respective state.
ii. Practical in the Sciences, Stenography, Computer Studies, and Data Processing should be in sets where the number of candidates is large and equipment cannot go round.
iii. VISUAL ART PAPER IV [CREATIVE DESIGN] SHOULD BE GIVEN TO CANDIDATES AT LEAST ONE
WEEK, WHILE WELDING AND FABRICATION, ENGINEERING CRAFT PRACTICE THREE WEEKS TO EXAMINATION DAY.
iv. The Nigerian Languages Hausa, Igbo and Yoruba include Literature aspect.
v. BRINGING GSM INTO THE EXAMINATION HALL IS PROHIBITED.
vi. Use of scientific calculator is not allowed.
vii. The Albinos and blind candidates are to be given 15minutes extra time across all subjects.call08135784898.

(1biii)
number of hydroxide ions, 1+1.00dm(^3) of solution B was used.
from the equation of the reaction
2KOH====>2K(+) + 2OH(-)
1 mole of B contaoins 6.00*10(^23) ions of OH(-)
therefore 2 mole of B will contain 2*6.00*10^23 ions of OH(-)
therefore No of OH(-) = 2* 6.00*10(^23)
1.20*10^(23)
(1bii) concentration of Bin mol dm(^-3)
= CAVA/CBVB= mole ratio
CAVA/CBVB= NA/NB
=1/2
where CA= concentration of A in mol/(dm^3)= 0.0531 moldm(^-3)
VA=volume of acid =27.00
CB= concentration of Bin mol/dm(^3)
VB= volume of base=25cm(^3)
nA=1
nB=2
CAVA/CBVB=1/2.
CB= 2*CAVB/(1*VB)
=((2*0.0531*27.00)/(1*25)) moldm*^-3)
=2.878/25 moldm^(-3)
=0.115 mol dm(^-3)
(1bi)
concentration of A in moldm(^-3)
250cm(^3) of H2SO4 will contain (1000/280)*1.30g/dm(^3)
=4*1.30g/dm(^3)
=5.20g/dm(^3)
therefore mass concentration of H2SO4 is 5.20g/dm(^3).
molar mass of H2SO4= (2*1)+(1*32)+(4*16)
=(2+32+64)gmol^(-1)=98g/dm(3)
therefore molar concentration of solution A in mol dm(^3) is given by
=(mass concentration)/(molar mass)= (5.20g/dm(^3))/98 gmol(^-1)
=0.05306 mol/dm(^-3)
=0,0531 moldm(^-3)
(2a)
Test:
(i) C+ water+ litmus paper
(ii)1st portion of solution + Ba(NO3)2 solution+dilute HNO3 till in excess
(iii)2nd portion + NaOH(aq) +heat gently
Observation:
(i) dissolved to give a colourless solution. turn blue litmus paper red.
(ii)white precipitate formed insoluble in excess dilute HNO3.
(iii) A colourless gas which has an irritating smell evolved.
it turns moist red litmus paper blue. it form a dense white fumes with conc. Hcl stopper.
Inference:
(i) C is a suluble salt.
(ii) SO4(2-), SO3(2-) present. SO4(2-) confirmed.
(iii) NH3 gas from NH4(+)
(2b)
Test:
(i) D+ distilled water + heat
(ii) 1st portion of mixture+ iodine solution
(iii)2nd portion of mixture + fehlings solutions + heat
(iv) 3rd portion of mixture + dil HCl+ heat+fehlings solution
Observation:
(i) insoluble with a curdy white paste
(ii) blue black colouration
(iii)No precipitate or any visible reaction
(iv) Brick-red precipitate.
Inference:
(i)
(ii)starch is present
(iii) reducing sugar is absent
(iv)starch has been hydrolysed to reducing sugar
(3a)
(i)sand+iodine----sublimation
(ii)water+methanol---distillation
(boiling point 100 degree Celsius) (boiling point 65 degree celsius)
(3b)
PbCO3====> PbO(s)+CO(g)
(i)
there is evolution of colourless and odourless gas which turns lime water milky.
The gas is Carbon(iv)oxide with yellow deposit of Lead(ii) oxide.
(ii)
when a thin strip of filter paper soaked in acidified K2Cr2O7 solution is introduced into a gasjar containing
sulpur(iV)oxide gas, the colour of acidified K2Cr2O7 is changed from orange to green reducing it to chromium(iii) tetra oxo sulphate(vi), (Cr2(SO4)3).
CALL08135784898.

Verified Agric obj
1-10. ADBDDAAAAC
11-20.ADDBDABDBC.
21-30.BAAADDDCCC.
31-40.ACCDCCBDCA.
41-50.BDCBBCBDBD.
(3a)
i)soil requirement;-well drained rich friable loamy soil
ii)spacing;-90cm by150cm in small cutting but in long cutting 1m by 1m
iii)cassava mosaic
iv)sheep
b)natural pasture is a piece of land of field containing grass for sheep, goats and cattle to it as it grows while artificial pasture are grasses and legumes grown primarily for feeding farm animals and livestock
3c)disease is a departure from normal state of health,presenting marked symptoms or outward visible signs
d)mendel's law of independent assortment states that when genes segregate,they recombine independently or independent of one another
e)this is the process by which trees make provision for their lost parts.it could also be defined as the regrowth of trees following forest exploitation
(3f)i)it encourages soil erosion
ii)reduces water percolation rate
iii)results in loss of soil nutrient
iv)reduces humus content of the soil
v)reduces wild life population in the area concerned
2a.
i. Farmstead can be defined as a farm house and all its production and processing structures.
ii. Secondary tillage -: this involves the filling of the top soil at minimal level so as not to destroy the top soil.
2b.
i. Hitch :- it helps to remove soil or mud that cling to the disc.
ii. Furrow wheel :- it provides a balancing effect for the plough as it supports both vertical and side thrusts.
iii. Disc coulter ;- it makes vertical cuts/furrows.
iv. Plough share -: This makes horizontal cuts and uproots weeds.
v. Mouldboard ;- it carries the soil.
2c.
Photosynthesis.
2d.
i. Organic manuring.
ii. crop rotation
iii. Bush fallowing.
2e.
i. Weak slender plants.
ii. Causes stunting of the root system.
2f.
i. Millipede.
ii. Nematode
iii. Rat.
iv. Termite6)a)
i)it helps west african farmers to increase their plant and livestocks
ii)it acts as a connecting link between the researchers and the farming communities of west african thereby increasing their farm inputs
iii)it helps west african farmers to attain certain skills and specialization in the production of certain crops and animals are also acquired by the farmers through the agent
iv)agricultural extension development program helps to improve the outlook of west african farmers towards their problems or difficulties
v)it helps in the introduction of land management to west african farmers and the management techniques which increase land fertility and productivity
b)
i)low level of illiteracy among farmers which makes them find it difficult to follow instructions on new farming techniques
ii)the farm inputs being grossly inadequate and usually getting to the farmers very late
iii)poor facilities for extension communication which limits the scope and efficiency of coverage
iv)many agents being ignorant of the traditions and customs of their target communities often run foul of them,thereby making it difficult for them to recieve audience
c)
i)soil requirement;-rich sandy loam
ii)method of propagation;-by seeds
iii)-store in jute bags
-clean rhombust silos
d)i)field pests;-stem borer,grasshopper
ii)storage pest;-maize weevil,rats1a.
Commercial farming: this is the production of cash crops animals in large quantity. These products are produced majorly for sale.
1b(I)
Canning industry:
(II)
Beverage industry: coffee and orange
(III)
Feed mill: palm kernel and
1d.
Surface irrigation: this is the system by which water is made to flow on the surface of the farm land by means furrow, channels and flooding.
1ei.
Horizon A: this is the tops soil in the horizon. It is preoccupied with partly decomposed organic matter. After this horizon is horizon B which is subsoil
1eii
Horizon D: this is the last horizon in the soil profile. It is unweathered soil.
4a
Animal nutrition can be defined as the science of feeding.
4b
-it aids ovulation and heat period
-it helps to develop secondary sexual characters
-it brings about puberty in female animal
-it helps to develop mammary gland
4c
I selection-this is the process of picking from a group of animals which schow desirable qualities.
Ii breeding-this is the process of producing offsprings with desirable qualities
Iii introduction-this is the bringing into the country or farm,new breeds of good qualities
4d
Ornamental plants are plants majorly flowers planted for decoration of houses and offices
4e
-budding
-grafting>