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NABTEB LIVE QUESTIONS AND ANSWERS 2017/2018

Discussion in 'Waec, Weac, Neco, Waec GCE And Neco GCE' started by Yoi.shina, May 4, 2017.

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    BIOLOGY PRACTICAL ANS

    1 i )
    i ) carbon( iv ) oxide
    ii ) chlorophyll
    iii ) sunlight
    iv ) water and mineral salts
    1 ii )
    i ) protection : skin protects the body against entrance by microtes etc
    ii ) excretion : skin excretes excess water mineral salts and some nitrogenous wastes through the sweat glands as sweat
    1 iii )
    i ) transport of oxygen
    ii ) temperature regulation
    1 iv )
    i ) presence of chlorophyll for photosynthesis
    ii ) presense of cellular cell wall
    1 v )
    i ) it promote measures for the control of the world ' s major disease ' s through vaccuiation programmes and use of aritibioties
    ii ) it helps to set up international quararrayive regulation
    1 vi )
    i ) sieria - leone
    ii ) nigeria
    1 vii )
    - monocotyledonous -
    i ) they store food in the endo sperm
    ii ) posses parallel veriation
    - dicotyledonous -
    i ) they store food in the cotyledon
    ii ) posses not veriation
    1 viii )
    i ) protein 15 %
    ii ) fats and oil 15 %
    iii ) vitamins , minerals and water 10 %
    iv ) carbonhydrate 60 %
    1 ix )
    i ) canning
    ii ) suning
    iii ) smoking
    iv ) dehydration
    1 x )
    i ) soil
    ii ) forestry
    iii ) wild life
    iv ) water
    1 xi )
    i ) water ferns
    1 xii )
    i ) they manufacture food for plant
    ii ) they help in interchange of gases in plant
    1 xiii )
    i ) lungs
    ii ) gills
    iii ) tracheal
    iv ) skin
    1 xiv )
    i ) to help attract pollinators to a plant and in fertilization so that the plant creates seed
    ii ) they mediate the joining of the sperm contain within pollen to the ovules contained in the ovary
    1 xv )
    i ) retina

    2 ai )
    A - tilapia fish
    B - adult toad
    C - cockroach
    2 aii )
    A - pisces
    B - Amphibian
    C - insecta
    2 b )
    A - source of food
    B - source of protein
    2 c )
    Make a drawing of the side view of tilapia fish and lebel fully ( your drawing should be between 6 - 8 cm long

    3 ai )
    D - grasshopper
    E - groundnut seed
    3 aii )
    D - it destroys crops
    E - it is a source of protein , fat and oil
    3 bi )
    A - aquatic habitat
    D - terrestrial habibat
    3 bii )
    i ) presence of lings for flight
    ii ) presence of a pair of Artemis for locomotion
    iii ) presence of a pair of legs for movement
    3 c )
    sp cockroach - C
    i ) Dark or brown in colour
    ii ) it is bigger in size
    ii ) it arterma is longer
    sp grasshopper- D
    i ) green in colour
    ii ) it is smaller in size
    iii ) it anterma is shorter

    4 ai )
    E - Groundnut seed
    F - cowpea seed
    4 aii )
    i ) melon
    ii ) mango
    4 bi )
    i ) epigeal germination
    ii ) the specimen are leguminous plants ,they contribute to the soil fertility by nitrogen fixation mechanism
    4 ci )
    i ) weavil
    4 cii )
    i ) fat and oil
    ii ) protein
    4 ciii )
    i ) by sunning
    ii ) by canning of bitting
    iii ) by dehydration
    iv ) by application pasticides

    5 ai )
    i ) body division is three head thorax and abdomen
    ii ) presence of a pair antenna
    iii ) presence of 3 pair of walking legs
    iv ) presence of lungs
    5 aii )
    i ) exoskeleton
    5 b )
    dorsal view of specimen D - Grasshopper.
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    Physics and Chem practical answers are now available
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    NABTEB PHYSICS PRACTICAL ANSWERS BY YOISHINA☆☆☆ ==================================== Thursday 11th May 2017 Physics (Practical)


    1a) No of oscillations = 20 Height (h)= 1.0m
    Table of observation
    S/N| Y(cm)| t(s)| T= t/20(s)| T^2
    1| 10.00| 38.00| 1.90| 3.62
    2| 20.00| 35.80| 1.79| 3.22
    3| 30.00| 33.60| 1.68| 2.82
    4| 40.00| 31.20| 1.56| 2.42
    5| 50.00| 28.40| 1.42| 2.01
    ++++++++++++++++++++
    1aiv)
    Slope = ΔY/ΔT^2
    =62-10/1.5-3.6
    =52/-2.1
    = -24.76cm/s^2
    Intercept I, on the vertical axis=88cm or 0.88cm
    +++++++++++++++++++++++
    1av) Precautions
    i)I avoided error due to parallax when reading the stopwatch and metre rule
    ii)I ensured that the Bob was displaced through the small angle to avoid conical oscillation
    +++++++++++++++++++++++
    1bi)
    Period:is the time taken or required to complete one oscillation or to and fro movement
    Frequency:is the number of cycle or oscillation complete in one seconds
    +++++++++++++++++++++++
    1bii)
    i)The shape of the earth (spherical)
    ii)Distance of the object from its centre
    +++++++++++++++++++++++
    1biii)
    Diameter=2.4m, radius=1.2m, V=8m/s
    V=wr
    Making W subject of the formula
    W=v/r
    =8/1.2
    =6.67rads^-1
    ====================================
    3a)
    Standard resistor=1.0~(ohm)
    Table of observation
    S/N| I(A)| V(v)| I^-1(A^-1)| V^-1(v^-1)
    1| 0.4| 0.40| 2.50| 2.50
    2| 0.6| 0.60| 1.67| 1.67
    3| 0.8| 0.80| 1.25| 1.25
    4| 1.0| 1.00| 1.00| 1.00
    5| 1.20| 1.20| 0.83| 0.83
    ++++++++++++++++++++
    3av)
    Slope, s =ΔI^-1/ΔV^-1
    =3.0-0.5/3.0-0.5
    =2.5/2.5
    =1.0V/A
    +++++++++++++++++++++
    3avi)
    From,
    S=ΔI^-1/ΔV^-1
    1.0=V/0.75
    V=0.75v
    +++++++++++++++++++
    (3avii)
    Precautions
    i) I ensured tight connection of connecting wires
    ii)I ensured that the key was opened, to avoid the running down of the battery
    ++++++++++++++++++++++
    (3bi)
    Internal resistance(r) :This is the opposition of the current flow through the electrolytic cell
    +++++++++++++++++++++++
    (3bii)
    i) heating effect
    ii) chemical effect
    Expalin one:
    i) Heating effect is used in electric cooker for cooking.
    (3biii)
    r=1Ω,
    R=6Ω
    V=9v
    E=?
    I=E/R+r
    V/R=E/R+r
    9/6=E/6 + 1
    Cross multiply
    6E=9x7
    6E=63
    Making E subject of formula
    E=63/6
    =10.5v
    So therefore,E.M.F of the cell=10.5v.
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    2a)
    TEST : C + water, Mixed thoroughly and filtrate
    OBSERVATION : partially soluble in water with colourless filtrate and white residue
    INFERENCE : C is a mixture of soluble and insoluble salt
    ===========================
    (2bi)
    TEST :Filtrate +NaCl(aq)
    OBSERVATION :white precipitate is formed
    The precipitate soluble in excess NaOH(aq)
    INFERENCE :pb2+, Zn2+, or Al2+ is present
    ==========================
    (2bii)
    TEST: Filtrate + NH3(aq) in drop, then in excess
    OBSERVATION :white precipitate is formed
    The precipitate is insoluble in excess NH3(aq)
    INFERENCE : Pb2+ or Al2+ present
    ==========================
    (2biii)
    TEST :Filtrate + KI(aq) + warm and allow to cool
    OBSERVATION :Yellow precipitate is formed
    The precipitate reappears after cooling
    INFERENCE :pb2+ confirmed
    ===========================
    (2c)
    TEST :Residue + Hcl(aq)
    OBSERVATION :The residue dissolves, librating a colourless gas which turn lime water milky
    INFERENCE :Gass is CO2 from CO3^2+ is present
    =========================
    (2d)
    TEST :Residue + NH3(aq) in drops, then in excess
    OBSERVATION :white gelatinous precipitate is formed
    The precipitate is soluble in excess NH3(aq)
    INFERENCE :Zn^2+ Confirmed
    ==========================
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    CHEMISTRY PRACTICAL SOLUTIONS
    (No1a)
    Volume of Acid used is 25cm3
    Indicator used =Methyl orange
    Table of observation
    ========================
    Burette reading: Rough, 1st, 2nd, 3rd
    ========================
    Final reading(cm3):24.50, 23.50, 23.40, 23.60
    ========================
    Initial reading(cm3):0.00, 0.00,0.00, 0.00
    =========================
    Volume of Acid used:24.50, 23.50, 23.40, 23.60
    ==========================
    Volume of Acid used
    = 1st +2nd +3rd
    ––––––––––
    3
    = 23.50 +23.40+23.60
    –––––––––––––
    3
    = 70.3
    –––– =23.50cm3
    3
    +++++++++++++++++++++++++
    (1bi)
    Concentration of B in mold – 3
    CAVA NA
    –––– = –––
    CBVB NB
    CA= 0.03moldm-3
    VA=23.50moldm-3
    NA=2
    VB=25cm3
    NB=1
    CB=?
    CB = CA VA NB
    ––––––
    VB NA
    CB = 0.03 × 23.50 × 1
    ––––––––––
    25×2
    = 0.075
    –––– = 0.0141
    50
    Concentration of solution B in moldm-3 = 0.0141moldm-3
    ++±++++++++++++++++++++++
    (1bii)
    Milan mass of 2Fmol
    Amount of NaCo3 in 1dm3
    = 0.0141 × 1
    = 0.0141mol
    From the equation
    NaCo3 + 2HCL –––> 2NaCl +H2O + CO2
    1mole of Na2CO3 produces
    1mole of Nacl
    0.0141mole of Na2CO3 liberate
    2 × 0.0141mole of NaCl
    =1.6497g
    =1.65g
    +++++++++++++++++++++++++
    (1biii)
    From the equation
    1mole of Na2Co3 produce 1mole of Co2 at S.T.P
    1mole of NaCo3 produce 22.4dm3 of CO2 at S. T. P
    Therefore 0.0141mol of Na2Co3 will produce
    22.4 × 0.0141
    Volume of CO2 = 0.316dm3
    (1biv)
    Molar mass of H2X = (2×1) + X
    = (2+X) g/mol.
    +++++++++++++++++++++++
    (No2)
    (2a)
    TEST : C + water, Mixed thoroughly and filtrate
    OBSERVATION : partially soluble in water with colourless filtrate and white residue
    INFERENCE : C is a mixture of soluble and insoluble salt
    ===========================
    (2bi)
    TEST :Filtrate +NaCl(aq)
    OBSERVATION :white precipitate is formed
    The precipitate soluble in excess NaOH(aq)
    INFERENCE :pb2+, Zn2+, or Al2+ is present
    ==========================
    (2bii)
    TEST: Filtrate + NH3(aq) in drop, then in excess
    OBSERVATION :white precipitate is formed
    The precipitate is insoluble in excess NH3(aq)
    INFERENCE : Pb2+ or Al2+ present
    ==========================
    (2biii)
    TEST :Filtrate + KI(aq) + warm and allow to cool
    OBSERVATION :Yellow precipitate is formed
    The precipitate reappears after cooling
    INFERENCE :pb2+ confirmed
    ===========================
    (2c)
    TEST :Residue + Hcl(aq)
    OBSERVATION :The residue dissolves, librating a colourless gas which turn lime water milky
    INFERENCE :Gass is CO2 from CO3^2+ is present
    =========================
    (2d)
    TEST :Residue + NH3(aq) in drops, then in excess
    OBSERVATION :white gelatinous precipitate is formed
    The precipitate is soluble in excess NH3(aq)
    INFERENCE :Zn^2+ Confirmed
    ==========================
    (3ai)
    NaCl4 and NaNo3 dissolves readily in cold water.
    +++++++++++++++++++++++
    (3aii)
    CuCo3 leave a black residue on heating
    +++++++++++++++++++++++
    (3bi)
    E = CuO
    +++++++++++++++++++++++
    (3bii)
    Blue precipitate is formed, the precipitate dissolves in aqueous ammonia to give a blue deep solution
    +++++++++++++++++++++++
    (3biii)
    G is water (H2O)
    +++++++++++++++++++++++
    (3biv)
    A dark brown precipitate is formed at the button of the test tube, which shows that copper has been displaced by zinc for its solution
    +++++++++++++++++++++++
    COMPLETED
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    Office-Obj
    1caadccdcdb
    11cadbacaaac
    21bddcbdabaa
    31acabddaccb
    41cdccabadab
    Office-Practices-Theory-Answers
    Answer (5) Question Only Pls
    ================================
    2a)
    committee is a group of people appointed for a specific function by a larger group and typically consisting of members of that group.
    2b)
    i)standing
    ii)select or special
    iii)joint.
    i)standing committees consider bills and issues and recommend measures for consideration by their respective chambers.
    ii)select or special committees conduct investigations and studies and, on other occasions, also to consider measures.
    iii)joint committees conduct studies or perform housekeeping tasks rather than
    consider measures.
    ================================
    3a)
    Office furniture is an important element
    of office environment. It is needed to
    conduct office work efficiently and comfortably. Every office needs furniture
    as a basic facility.
    3b)
    i)General furniture
    ii)Special purpose furniture
    iii)Executive furniture
    3c)
    Tabulate
    -DUPLICATING-
    i)it is cheap form of printing
    ii)any paper can be used
    iii)it can be easily corrected
    iv)it does not fade
    v)exact copies may not be obtained
    -PHOTOCOPYING MACHINE-
    i)it is expensive
    ii)any paper cannot be used
    iii)correction cannot be easily done
    iv)it tends to fade
    v)exact copy is obtained
    ================================
    4a)
    An open cheque is a cheque that is not crossed on the left corner and payable at the counter of the drawee bank on presentation of the cheque.
    4bi)
    standing order- is an instruction a bank account holder "the payer" gives to his or her bank to pay a set amount at regular intervals to another's "the payee's" account.
    4bii)
    credit transfer is a system whereby successfully completed units of study contributing towards a degree or diploma can be transferred from one course to another.
    4biii)
    postal order is an order for payment of a specified sum to a named payee, issued by the Post Office.
    4biv)
    money order is a printed order for payment of a specified sum, issued by a bank or Post Office.
    4bv)
    certified cheque is a form of cheque for which the bank verifies that sufficient funds exist in the account to cover the cheque, and so certifies, at the time the cheque is written.
    4bvi)
    I.O.U. Is a Non-negotiable debt instrument addressed to a creditor, dated, and signed by the borrower.
    ================================
    5ai)
    In Camera- this is a meeting which is not open to the public,this simply meens in secret,maybe for private reasons
    5aii)
    Amendement- if an amendment to motion is proposed a vote is taken or the motion should the amendement be carried. It becomes the resolution and there will be no further discussion on the original motion, other wise the motion is then voted upon
    5aiii)
    Proxy- this is a method of voting where or member who are unable to attend the meeting may vote by post or appoint another member to vote or his behalf
    5aiv)
    Teller- This is the title given to the person appointed to count the votes at a meeting
    5av)
    Casting- this is a second vote, usually allowed to the chairman except in the case of a company meeting. A casting vote is only used for or against a motion
    ================================
    6ai)
    Envelope sealing machine- this is use to seal all types and sizes of envelops
    6aii)
    jogger- it is use to dry ink after printing in a digital duplicator
    6aiii)
    perforator- this is use to perforate paper or create hole on paper
    6aiv)
    stapler- it is use to staple paper together i.e it is use to join pagies of paper or similar material by driving a thin metal staple pin through the sheet and folding the end
    6av)
    franking machine- this machine makes printed impression of stamps on envelops. The machine can be franked on a strip
    ================================
    7a)
    Recorded delivery it still gets mixed in with the normal post but has to be signed for at the end address. While
    Registered post is kept separate all
    the way through the system. It is designed for higher value items and carries insurance built into the cost.
    7b)
    i)POSTAL SERVICES
    ii)PAYMENT SERVICES
    iii)SATELLITE SERVICES
    iv)COURIER SERVICES
    v)COMPUTER SERVICES
    ================================
    8ai)
    -ACCOUNT DEPARTMENT-
    i)cash control
    ii)to keep proper books of account
    iii)to control expenditure
    8aii)
    -PERSONNEL DEPARTMENT-
    i)human resources planning
    ii)recruitment, interviewing and selection
    iii)industrial training and development
    8aiii)
    -MARKETING DEPARTMENT-
    i)to make market research
    ii)to advertise
    iii)distribution of goods
    8aiv)
    -TRANSPORT DEPARTMENT-
    i)inspection of vehicles at check posts
    ii)revenue collection for the government
    iii)registration of vehicle
    8av)
    -ADMINISTRATIVE DEPARTMENT-
    i)planning
    ii)directing
    iii)organizing
    ================================
    9)
    i)Usage
    ii)End user
    iii)Price
    iv)Availability of parts and consumable
    v)Technical support Ideally, technical
    9ai)
    Usage- The first thing to consider while making a purchase decision is the estimated usage of the equipment.
    9aii)
    End user- The skill level of the employee who is going to operate the equipment should be considered.
    9aiii)
    Price- When it comes to purchasing new or used office equipment, price is the dominating factor.
    9aiv)
    Availability of parts and consumable- This factor is one of the most important factors to think about.
    9av)
    Technical support- Ideally, technical support and software updates should be included with the initial purchase price of the equipment.
    ================================
    10a)
    Incoming mail- This is defined as those mail process that are received by any company, and in addition to the postal address, contain company-specific addressee information such as name, title department, subdivision and other specific details
    10b)
    i)Receiving the mail
    ii)Sorting the mail
    iii)Opening the mail
    iv)Scrutiny of the contents
    i)Receiving the mail- Generally mails
    are received once or twice a day delivered by the postman.
    ii)Sorting the mail- Sorting of letters means grouping of letters on definite
    order.
    iii)Opening the mail- In small organization letters are opened by the officer or head clerk.
    iv)Scrutiny of the contents- After the letters are opened, the contents are
    removed from the envelopes and are
    scrutinised.
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    *NABTEB ANSWER 2017*

    =====================
    VERIFIED NABTEB MATHS
    OBJ:
    1-10=DBDCDABBCD
    11-20=DCDBDCDAAB
    21-30=BCDAADCACD
    31-40=DABABCBDCD
    41-50=ABBBAABDAD

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

    VERIFIED-MATHS-THEORY ANSWERS
    =======================
    INSTRUCTION:ANSWER questions 1 to 5 and any other four questions
    MATHS THEORY
    10.30am-1:00p.m
    ======================
    SECTION A(ANSWER ALL QUESTION)
    QUESTION 1 -5)
    ======================
    1a)
    1 4/5 × 2 1/3 / 3 3/4 – 4/5 × 2/3
    =9/5 × 7/3 / 18/5 – 4/5 × 2/3
    =21/5 / 18 – 4/5 × 2/3
    =(21/5 ÷ 14/5) × 2/3
    =21/5 × 5/14 × 2/3
    =1


    1b)
    x^2 +5x – 6= 0
    (x^2 + 6x) – ( x– 6) = 0
    x(x+6) –1 ( x+6)= 0
    (x–1) (x +6)= 0
    x – 1 =0 or x+6 = 0
    x=1 or x =– 6

    =====================
    2a)

    U ={2,3 ,4 ,5, 6, 7,8 ,9}
    A ={2,3 ,5 , 7}
    B ={3,6,9 }

    i)AUB ={2,3,5,6,7,9}

    ii)
    A'nB'
    A' ={4,6, 8,9,}
    B' ={2,4,5,7,8,}

    A'nB' {4,8}

    2b)

    161n = 32less down 5
    1×n^2+6×n^1+1×n° = 3× 5^1+2×5°
    n^2 + 6n + 1 = 15 + 2
    n^2 +6n + 1 =17
    n^2 + 6n – 16 = 0
    (n^2 +8n) – (2n – 16) = 0
    n(n + 8) –2 (n + 8) = 0
    (n – 2)(n + 8) = 0
    n – 2 = 0 or n + 8 =0
    n = 2 or n = –8

    hence,
    n = 2

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

    3a)

    1/2logy^8 = 2
    logy^√81 =2
    logy^9 =2
    y^2 = 9
    y = √9=3
    y = 3

    3b)
    0.016 × 0.048 / 0.64
    =16 ×10^-3 / 64 × 10^-2
    =16 × 48 × 10^-6 / 64 × 10^-2
    =768 / 64 × 10^-6 × 10^2
    =12 × 10^-4

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

    4)
    h/8 = h + 20/12
    12h = 8h + 160
    4h =160
    h = 160/4
    h =40cm
    Hence ,H =h +20=40 +20 =60cm
    H=60cm

    Volume of bucket =
    1/3πR^2H – 1/3πr^2h
    =1/3π(R^2H – r^2h)
    =1/3(3.142)( (12)^2(60) – (8)^2(40))
    =1/3(3.142)(8640 – 2560)
    =1/3 × 3.142 × 6080
    volume is =6367.7cm^3

    since ,
    1litre = 1000cm^3
    Capacity =6367.7 / 1000
    =6.3677litre
    =6.4litre

    =====================
    5a)
    2/3(x – 2) – x –1/x–2
    =2–3(x –1)/3(x–2)
    =2 – 3x + 3/5x – 6
    =5–3x/3x–6


    5b)
    x^2 + 3x + 2/ x^2 – 4
    =(x^2 + x) + (2x + 2)/(x+2)+(x –2)
    =x(x+1) +2(x+1)/(x+2) +(x – 2)
    = (x+2) (x – 2)/(x+2)(x –2)
    = x+1/x –2

    ====================
    SECTION B
    ANSWER ONLY FOUR QUESIONS
    ===================
    8a)
    if (x – 6), 2x and (8x – 20) are consecutive terms of G.P

    the common ratio is
    r = 2x / x– 2 --------(1)
    r = 8x + 20 / 2x --------(2)

    Equating (1) and (2)
    2x/x – 6 = 8x + 20/2x
    4x^2 = (x – 6) (8x + 20)
    4x^2 = 8x^2 + 20x – 48x – 120
    4x^2 – 28x – 120 = 0
    x^2 – 7x – 30 = 0
    Solving quadratically,
    (x^2 + 3x) – (10x – 30) = 0
    x(x + 3) –10 (x + 3) = 0
    (x - 20) (x + 3) = 0
    x – 10 =0 or x + 3 = 0
    x = 10 or x = -3

    8b)
    √72 × 3√18 × 14√6 / 2√24 ×√12 √36×2 × √9×3 × 24√6 /2√4×6 × √4×3
    6√2 x 9√2 × 14√16 / 4√6 × 2√3
    6 × 9 × 2 × 14 √6 / 4√6 × 2√3
    3 × 9 × 9 / √3
    by rationalizing the denomenator
    = 189/√3 × √3/√3
    =189√3 / 3
    =63√3

    ===================
    9a)
    x =30° <base angle of issoceless triangle is equal >
    y =θ = 180 – 60
    θ=120° , r = 5cm
    the length of the chord AC
    L =2rsinθ/2
    =2 × 5 × sin120/2
    =10sin60
    =10 × 0.866

    Length of chord is =8.66cm

    9aii)
    Area of shaded segment = area of sector – Area ot triangel
    θ/360 × πr^2 – 1/2 (5)^2sinθ
    =120/360 × 3.142 × (5)^2sin120
    =120 × 3.142 × 25 /360 – 25 × sin120/2
    =26.18 – 10.825
    =15.358cm^2

    9b)

    Speed = Distance / time
    Distance = 500 × 2
    =1000km

    x^2 =(1000)^2 + (450)^2 -2(1000)(450)cos120
    =1000000 + 202500 +450,000
    x^2 =1652500
    x =√1652500
    x = 1285.5km

    The bearing the airport (θ)
    450/sinθ = 1285.5/sin120
    sinθ = 450 × sin120 / 1285.5
    sinθ =0.3032
    θ = sin(0.3032)= 17.6°
    =18°

    10a)
    blue marble = 3
    white marble = 2
    Red marble = 4/9

    i)
    Pr (both of them will be red)
    first drawn = 4/9
    second drawn = 3/8
    Pr (both red) =4/9 × 3/8
    =1/8

    ii)
    Pr (the two are of the same color)
    = RR or BB or WW
    =(4/9 × 3/8)+ (3/9 × 2/8)+ (2/9×1/8)
    =12/72 + 6/72 + 2/72
    =12+6+2 / 72
    =20/72
    =5/18


    10b)
    sine(2θ – 30°) and (3θ – 45)are supplementary that there sum is 180°
    2θ – 30 + 3θ–45=180
    5θ –75 =180
    5θ=180 + 75
    5θ =225
    θ=51°

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

    11a)

    hence, volume of hemispherical portion is half of the volume of the cone

    Volume of hemisphere=2/3πr^3
    Volume of cone =1/3πr^3
    2/3πr^3 = 1/2(1/3πr^2 h)
    2/3πr^3 = 1/6 πr2 h
    h = 2 × 6 × πr^3 / 3πr^2 = 4r
    h = 4r = 4 × 4 = 16cm
    hence the vertical angle is
    tanθ = r/h
    tanθ = 4/6 = 1/4
    θ =tan(0.25) = 14.036°

    θ =14°(correct to nearest degree)

    11b)
    Total volume of solid

    =1/3πr^2h (h + 2r)
    =1/3 × 22/7 × (4)^2 (16 + 8)
    =1/3 × 22/7 × 16 × 24
    = 22 × 16 × 24 / 21
    = 402.285cm^3

    =====================
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    ESSAY WRITING

    Answer one question from this section.
    All questions carry equal marks. Your answer should not be less than 400 words except where otherwise stated. You are advised to spend about 50 minutes on this section.

    1) Write a letter to your uncle abroad asking for financial assistance to enable you set up a business related to your course of study on the completion of your training. Describe the steps you intend to take to ensure the success of the business.

    2) You have just been elected the president of the Students' Union in your school. Write a speech indicating at least three areas you wish to effect changes.

    3) Write an article for publication in a national newspaper on the topic: 'The Qualities of a Responsible Citizen'.

    4) Your team organised a practical session recently in your course area. Write a comprehensive report.
    ____________________________

    5a)
    Kaman's greatest disappointment was the way the woman he met at the river honia give him a cold and Stern look on his way back to the village. He thought he would be given a hero's welcome, because he had fought for their land.

    5b)
    The first familiar thing he saw was river honia, where so often he had taken both in the past.

    5c)
    The women's attitudes towards him was very unfriendly, he was given a cold and hard looked.

    5d)
    He had been force to leave home earlier because he was arrested by security officers for fighting authority and was in favour of his village.

    5e)
    i)Adjectival clause
    ii)it give more information about the noun "Wanjiku"

    5f)
    i)Whined----------sounded
    ii)Apparent--------- Visble/obvious
    iii)Scarcely-----------Barely/hardly
    iv)Worn out--------Exhausted/fatigue
    v)Deliberately--------Intentionally
    ____________________________

    6a)
    The following are the six(6)steps adapted by the Auctioneer during the auction process :-

    i)Advertisement to the general public regarding the description of the goods to be sold.

    ii)communication the venue and the time of the auction through the advertisement.

    iii)Preparation of the auction ground where the Auctioneer will meet and address the people who come to bid for the gods to be sold.

    iv)Accept the bid from the members of the crowd

    v)Selling the goods to the highest bidder

    vi)Inviting the crowd to make an offer

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    NABTEB CHEMISTRY THEORY
    1ai)
    An ion is any atom or group of atoms which possesses an electric charge. While
    An Atom is the smallest particle of an element which can take part in a chemical reaction.

    1aii)
    A molecule is the smallest particle of a substance that can normally exist alone and still retain the chemical properties of that subsstance.

    1bi)


    1bii)

    1ci)
    Chemical bonding: this is the process of combining atoms of elements together, either by sharing or transfering the electrons

    TYPES OF CHEMICAL BONDING ARE:
    -Electrovalent bonding.
    -Covalent bonding
    -Coordinate coovalent bond.

    1ci)
    i)the metallic bond
    ii)the van der vaals
    iii)hydrogen bond

    1cii)
    TABULATE
    -Attractive Forces-
    i)they are very waek
    ii)they have very strong affinity for electrons

    -Bonds-
    -i)they are very strong
    ii)they do not have a strong affinity fir electrons
    ======================

    2ai)
    i)Radioisotope or radioactive isotope is an isotope that is made artificially by bombarding neutrons or protons or deutrons at elements.

    ii)Polymer: this is the giant molecule formed during the process of polymerization.

    2aii)
    CLICK HERE FOR THE IMAGE

    Alpha (α) Ray
    -it is symbolically represented by helium particle 4;2He
    -it is relatively massive
    -it is strongly deflected in an electrostatic field.
    -it is positively charged
    -it has little penetrating power
    -highest ionization energy

    Beta (β) Ray
    -it is symbolically represented by
    -it is slightly deflected in an electrostatic field.
    -they have relatively small mass.
    -they have more penetrating power than x-rays
    -low ionization energy

    Gamma Rays
    -it has no deflection in an electrostatic field
    -it is electrically neutral
    -it has heighest penetrating power
    -it has least ionization energy

    2aii)
    Half-Life: this is the time taken or required for radioactive elements to decay by half of its original mass.

    2bi)
    -it is used in the treatment of cancer
    -it is used to sterilize surgical equipment

    2bii)
    N = ?, No = 2.56kg, T1/2 = 6months.
    N/No = (1/2)^n
    n = 2×12/6 = 4
    N/2.56 = (1/4)^4
    N/2.56 = 1/16
    N = 2.56/16
    = 0.16
    ==========================

    3ai)
    Experiment to explain the law of conservation of mass method/procedure: put some sodium chloride solution in a conical flask. full a small test-tube with silver trioxonitrate(v) solution and by a means of two pieces string, suspend it in a conical flask. insert a stopper and weigh the whole apparatus on a balance and the mass is recorded. mix thye two liquid by pulling the string attached to the bottom end of the small test-tube. Weigh the whole apparatus again.

    Result: the weight of flask and contents before reaction is equal to the weight of flask and contents after reaction.

    Conclusion: since there is no overall change in mass when the products are formed. we can say that, matter can neither be created nor destroyed during the chemical reaction.

    3aii)
    i)Molecular foemula: this is the formula that gives the exact number of moles of atoms of the component elements in one mole of the compound.
    ii)An isomer: this is the compound that exhibit ispomerism.
    iii)A chemical equation: this is a symbolical representation of a chemical reaction in the form of symbols and formula.

    3bii)
    CLICK HERE FOR THE IMAGE

    3biii)
    i)Hydrogen sulphide (H2S)
    ii)carbon(iv) oxide (CO2)

    3bi)
    TABULATE
    —————- |Samplei| Sampleii| Sampleiii|
    Mass of boat alone | 26.7g | 27.2g | 26.3g |
    Mass of boat + Al2O3 | 77.8g | 72.9g | 82.7g |
    Mass of boat + Al | 53.8g | 51.4g | 52.2g |
    Mass of aluminium | 27.1g | 24.2g | 29.9g |
    Mass Of Oxygen | 24.0g | 21.5g | 26.5g |

    Percentage of Aluminium | 27.1/51.1×100 = 53.03% | 24.2/45.7×100 = 52.95% | 29.9/56.4×100 = 53.01% |
    The percentage of aluminium in all the three samples are approximately 53%
    =========================

    4ai)
    i)Avogadro’s Hypothesis: it states that equal volume of all gases at the same temperature and pressure contain the same number of molecule.
    ii)Molar Volume: this is the volume occupied by one mole of that gas at S.T.P, and is equal numerically to 22.4dm^3.
    iii)Electrolyte: this is a compound which conducts electricity and is decomposed in the process

    4aii)
    Amount = mass/volume = 250/32
    = 7.8mol

    4bi)
    The electrolysis of aqeous copper II tetraoxosulphate VI using platinum electrodes yield copper deposits at the cathode and oxygen of anode.
    At the cathode: both Cu^2+ and H^+ migrate to the cathode where Cu^2+ are discharged prefrentially. Cu^2+ + 2e^- —-> CU.
    At the anode(platinum): both SO4^2- and OH^- migrate to the anode where OH^- are prefrentially discharged as oxygen gas.
    OH^- —-> OH + e^-
    OH + OH —> H2O(l) + O
    O + O —–> O2(g)

    4bii)
    Faraday second law of electrolysis: this states that, when the same quantity of electricity is passed through different electrolytes, the relative number of moles of the elements discharged are inversely proportional to the charge on the ions of the elements.

    4biii)
    -it is used in purification of metals.
    -it is used in extraction of metals
    -electroplatino of one metal by another

    4ci)
    CLICK HERE FOR THE IMAGE

    Hydrogen can be prepared laboratorily by the action of dilute hydrochloric acid on zinc with libertaion of hydrogen gas.
    Zn + HCl —> ZnCl2 + H2

    4cii)
    Moisten a piece of filter paper with lead II trioxonitrate (V) solution and drop it into a gas jar of the unknown gas. if the gas is hydrogen sulphide the paper turns black.
    ===========================

    5ai)
    i)A period of elements: this is the horizontal rows of elements which are numbered from 1 to 7.
    ii)periodic table: this is a table involving the arrangements of atoms in the order of increasing atomic number.

    5aii)
    i)sodium belongs to group 1
    ii)magnesium belongs to group 2
    iii)aluminium belongs to group 3

    5aiii)
    Nitrogen = 1s^2 2s^2 2p^3
    Neon = 1s^2 2s^2 2p^6

    5bi)
    The iron ore is first roasted in air so that iron III oxide is produced.
    The iron III oxide is then mixed with coke and limestone and heated to a very high temperature in a blast furnance. The ingredients are loaded into the furnance. the temperature inside the furnance varies from about 2000°C near the bottom to about 200°C at the top.
    After a series of exothermic reaction, the molten iron formed sinks to the bottom of the furnance and is tapped off. it is run into moulds where it sets as pig iron.

    5bii)
    -maganese
    -carbon
    -steel

    5ci)
    -it is used for making electric wires
    -it is used for plumbing and roofing.

    5cii)
    -air(oxygen)
    -water.

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