Wednesday, 20 April 2016

Waec 2016/2017 May/June Mathematics Paper 1 & 2 Essay,obj, theory Question and answer

MATHS OBJ: 1-10: DCBCABBADA 11-20: CCBDABABBA

21-30: CACCDDDACC

31-40: BCDCDCBADA

41-50: ABACBBDADD

(1a) (0.09*1.21)/(3.3*0.00025) =(9*10^-2*1.2*10^1)/(3.3*25*10^-5) =(1089^10^-4)/(82.5*10^-5) =1.32*10^(-4+6) =1.32*10^2 (1b) p=5600,T=3,A=1200+5600=6800 A=P(1+R/100)^3 6800=5600(1+R/100)^3 (1+R/100)^3=6800/5600 (1+R/100)^3=1.214 1+R/100=cuberoot(1.214) 1+R/100=1.0667 R/100=1.0667-1 R/100=0.0667 R=0.0667*100 R=66.7% (2a) 7(x+4)-2/3(x-6)<_2[x-3(x+5)] 7x+28-2x/3-4<_2[x-3x-15] 24+7x-2x/3<_2x-6x-30 7x-2x/3-2x+6x<_-30-24 11x-2x/3<_-54 33x-2x<_-162 31x<_-162 x<_-162/31 (2b) x = tricycles and y = taxi cabs 2x+4y=6b … (1) x+y=20 … (2) from equ..(2) y=20-x substitute for y in equ…(1) 2x+4(20-x) = 66 2x+80-4x=66 80-66=4x-2x 14/2 = 2x/2 x=7 number of tricycles = 7 3b) h=7cm T=462cm^2 T=pie sqr + 2 pie rH 462= pie( r ^2 +2*7*r) 462/3.142 =r sqr= =14r r sqr+ 14r=147 r sqr +14r- 147=0 r= 14+-( (sqr(14))-4*1*147)/2*1 r=-(14 +-(196-4))/2 r=(-14+_12)/2 r=(-14+12)/2 or r=2/2 =1cm ================================== (4a) Pr(3)=x/total total=25+30+x+28+40+32 =155+x 0.225=x/155+ x =34.875+0.225x=x x-0.225x=34.875 =0.775x=34.875 degree x=34.875/0.775 =45 ================================== (4b) Pr(even)=30+28+40+32 =90 Pr(even)=90/200=9/20 Pr(prime no)=25=30+45+40 =140 Pr(prime no)=140/200=14/20 Pr(even or prime)=9/20 + 14/20 =========================== (5a) Area PSR=1/2*PR*SM PR=8+6=14cm 36=1/2 * 14* 6 36=1/2 * 14* 6 *QR QR=8.6cm ================================== (5b) diagram sin45 degree= 10.65/h h=10.65/sin 45 degree =10.65/0.7071 h=15.06cm =============================== (6a) first term9a)=1.0=1 common diff(d)=3-1=2 last term(L)=101 L=a+(n-1)d 101=1+(n-1)*2 (n-1)*2=100 n-1=100/2 n-1=50 n=501+1 n=51 sum(s)=n/2(2a+(n-1)d) S=51/2(2*1+(51-1)*2) s=25.5(2+100) S=25.5*102=2601 ================================== (6b) UC=95 Bus & Train(B&T)=7 Train & Car( T & C)=3 B&T&C=8 Bus only(x)=B&C only Bus(b)=47 train(T)=30 (draw the venn diagram) Bus only= 47-(7+8+x) x=47-15+x 47=x+x+7+8 47=2x+15 2x=47-15 2x=32 x=32/2 =16 Train only= 30-(7+8+2) =30-18=12 ==================== (6bi)=x=16 ==================== (6bii) number who travelled by at leasttwo means =7+1+3+8 =19 ================================== (7a) given; (x-2)/4=(x+2)/2x 2x*(x-2)=4*(x+2) 2x^2 -4x= 4x+8 2x^2-4x-4x-8=0 2x^2-8x-8=0 x^2-4x-4=0 x^2-4x=4 x^2-(4* 1/2)=4+(2)^2 (x-2)^2=4+4 (x-2)^2=8 (x-2)= sqr root(8) x-2=+-2.83 x=+-2.83 x=+_2.83+2 x=-2.83+2 x=4.83 or x=-0.83 ========================== (7b) P+Q+S=180 degree(sum of angles in a triangle) 52 degree+ 90 degree+ S= 180 degree 142+ S=180 S=180-142 S=38 degree ============================= (i)52+38+SQT =180 degree SQT=180-90 SQT=90 degree (ii)PQT= 52 degree ============================== (8a) diagram z=110 degree(sum of angles on a straight line) z=180-110 z=70 dgree w degree= 70 degree x+r+w degree= 180 degree(sum of angle in a triangle) x+2x+70 degree=180 degree 3x=110 x=110/3=36.7 dgeree ================================== (8b) B=10, G=12 TB=600.00 TB=TG+600 mean of TB=100+ mean of TG where TB= total collection of boys TG=total collection by girls mean of TB= ET/B mean of TB= (TG+600)/10 100+ mean of TG= (TG+600)/100 mean of TG= mean of TG/12 100+ mean of G/12= (TG+600)/10 (100+TG)/12=(TG+600)/10 10*(1200+TG)=12*(TG+600) =12000+10TG=12TG+7200 =12TG-10TG=12000-7200 2TG=4800 TG=N2400 TB=TG+600 TB=600+2400 Tb=N3000 ============================= (11a) Given (3p+4q)/(3p-4q) = 2 3p+4q=2*(3p-4q) 3p+4q=6p-8q 3p-6p=-4q-8q =-3p=-12q p/q=-12/-3 p/q=4/1, P=4, q=1 p:q=4;1 ================================== (11b) (i) perimeter(p)=PQ+PU+QR PQ=UT+TS+SR 34=PQ+UR+4+4 PQ=UR 34-8=2PQ PQ=26/2=13cm ================================== (ii) 13=2+TS+2 TS=13-4=9cm radius(r)=9/2=4.5cm Area(A)= pie r sqr+22/7*(4.5)^2 =445.5/7 63.64cm^2

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WAEC 2016 May/June 100% Correct Agricultural Pratical Paper, Question And Answers

(1a) Experiment to compare the porosity and water holding capacity

(1b) Grind loam and clay soil into fine particles, after drying in the sun. Equal masses of the dry, dry clay and dry loam are place seperately into three funnels which are blocked at the neck with cotton wool. The three funnels containing different types of soil are then placed on top of the three measuring cylinders. Then quickly pour equal amount of water into the funnels, containing each soil type. Then leave for an hour

(1c) (i)Loamy (ii)Sandy (iii)Clay

(1d) (i) I will ensure the three funnels containing different types of soil are placed on top of the three measuring cylinders (ii) I will ensure equal amount of water is quickly poured into the funnels containing the soil type (1e) -Sandy soil: Because the porosity is high in sandy soil because of large pores spaces and larger particles sizes ===================================

(2a) Ranging pole is a surveying instrument consisting of a straight rod pointed in bands of alternate red and white each one foot wide,used for sighting by surveyors

(2b) A-For making positions in surveying -For sighting by surveyor B-For marking posititon on surveying land C-For measuring the lenght of land -For detecting area of land (2c) A-By cleaning after use -By keeping in cool dry place B-By keeping away from water -By not exposing it to direct sunlight for so long C-By cleaning after use -By keeping in a cool dry place ======================================= (3ai) G-Herbivore H-Herbivore (3aii) G-Leaves(Defilators) H-Fruits

(3aiii) H-It helps in pollination -They help in dispersal of fruit -They help in the dispersal of seeds

(3aiv) I-It can be controlled using scar scrow J-It can be controlled using traps.It can also be controlled using biological method eg use of cat ===================================== (4a) N: -Animal feed industry -ceramic Industry -pharmaceutical industry O: -paint industries -pharmaceutical industries -cosmetic industries

(4b) -it causes loss of weight -it causes loss of appetite -it causes anaemia -it damages some internal organs -it prevent respiration

(4c) -by dipping the animal -by drenching -by clearing bushes around -by giving the animal injection

Unijos Remedial Admission List 2015/2016 Available On The School Portal

This is to inform candidates who applied for the 2015/2016 remedial Programme of the University of Jos that they can now check their admission status on the school website.

To check your admission status, visit CLICK HERE-Enter your JAMB Reg Number, Full name in the spaces provided.

-Select the academic session -Click on “Find Details” to access your status

OAU Notice to Admitted Candidates on 2015/2016 Acceptance Fee Payment

This is to inform 2015/2016 admitted candidates of Obafemi Awolowo University, Ile-Ife that they are to;

1. Accept their various admissions by paying thier acceptance fee of #20,000 before interacting with their pages on the E-portal.

2.The facility to download RRR for acceptance fee payment on CLICK HEREwould be ready by Wednesday, 20th of April 2016.

3. Download the RRR on the E-portal only ( CLICK HERE)

4. Visit banks that accept payment into REMITAL to make your payment there after return to CLICK HERE click on “undergraduate login”. Enter your JAMB Registration number as the username and your surname as the password to continue with your registration

WAEC 2016 May/June 100% Correct Biology Paper ,Essay,Theory (Obj), Question And Answers

BIOLOGY OBJ:

1-10: DCDACACCDC

11-20: CCBDADBBCD

21-30: DCBDDDBACB

31-40: CDABBBBDDB

41-50: DBCAABADBC

1a) -Single/independent free living organism eg amoeba -As a colony eg volvox -As a filament eg spirogyra

1b) -Amoeba -Volvox -Spirogyra

1ci) -Plant cell are rectangular in shape while Animal cell are spherical in shape -Plant cell contain chloroplast while animal cell do not contain chloroplast -Plant cell contain cell wall while animal cell do not contain cell wall -Plant cell nucleus is at the side while the animal cell is at the middle

1cii) -They both have nucleus -They both have mitochondrion -They both have endoplasmic reticulum 3ai) A habitat comprises of an organism immediate environment or where it lives or resides 3aii) -They help in the decay of the organisms -They also help to return certain substances like gases eg Nitrogen to the atmosphere thereby regulating certain biological processes such as the Nitrogen cycle,carbon cycle -Some of the substances released are important in processes such as respiration,photosynthesis 3b) Energy flows basically from the producers in a fresh water habitat to the consumers.The producers usually are phytoplanktons which are taken up by the primary consumers usually small fishes.The primary consumer are taken up by the secondary consumers which are larger fishes such as sharks.This sharks may inturn be taken up by man which is the tertiary consumer SCHEMATIC DIAGRAM Phytoplanktons->Small fishes->Sharks->Man 3c) -They causes diseases such as rice smut,black pod of cocoa which may lead to waste -They also reduce the economic value of plants leading to financial loss 3cii) -They help in fermentation of certain substances like yughourt eg lactobacillus species -They help in decomposition and decay of substances eg sprophytic micro organism -They causes diseases eg smallpox and chicken pox hence have been useful tools in the development of antibiotics and vaccines -They are useful in certain biological processes eg photosynthesis -They are used in sewage treatment -Sources of enzymes

2a) Mouth,Oesophagus,Stomach,Duodenum,Small intestine,Caecum,Appendix,Large intestine,Rectum,Anus

2bi) TABULATE Alimentary canal of birds|Alimentary canal of man -There is presence of crop|Crop absent -There is presence of proventiculus|Proventiculus absent -There is presence of gizzard|Gizzard is absent

2bii) -Both have caecum -Both have Oesophagus -Both have duodenum

2c) Herbivore posses heterodant type of dentition in which they posses teeth of different shapes and sizes.Herbivores feed mainly on vegetable hence its teeth is 28 in numbers.Its incisors are flat with sharp cutting edges for cutting vegetables or grasses .canine is absent.the space creates due to absense of the canine is called diastema.Diastema allows manipulation of grasses in the mouth.It has a very large molar and premolar closely packed and have very large ridged surface area for cutting and grinding

2d) -They help in regulation of blood sugar by producing insuling and glucagon -They help in digestion

6a) -Tactic movement movement is found in some plant and animals while nastic movement is found in plants alone -Tactic movement rsponse to directional stimuli while nastic movement respond to non directional stimuli 6bi) -Euglena -Chlamydomonas 6bii) -Mimosa plant -Sunflower 6ci) -Regular cleaning -Exposure to fresh air -Eating balanced diet -Regular exercise 6cii) -Temperature -pressure -Pain -Cold 6d) -High temperature affect evaporation and reduces performance -High temperature may lead to seed dormancy -It leads to loss of soil nutrients through evaporation -It affects the ripening and maturity of plant -it is necessary for germination of seeds 6ei) Courtship behaviour refers to the instructive behaviour on response to certain external stimuli in animals.It is the pattern of behaviour that preceeds mating and reproduction in animals.It is performed by male and female organism 6eii) -pairing -Teritoriality -Display 6f) -Holozoic nutrition -Amoeba -Entamoeba histolytica -Humans 6g) Crime detection using finger print method: this applies to forensic.In this method,individuals having different finger print pattern can be identified using database of stored information of such individuals. These method correctly identifies such individuals aptly and is useful in crime detection

WAEC 2016 May/June 100% Correct Financial Account ,Essay,Theory (Obj), Question And Answers

(ACCOUNT OBJ: 1-10: CCBBCACCCD

11-20: CCBCBBDBBB

21-30: CAADADBCBA

31-40: ABDCDACBBB

41-50: BBDDADCCCC

(1a) General journal is the accounting version of our personal journals. It doesn’t record everything that happens to the business, of course, but it does record every financial transaction that takes place (sometimes alone, sometimes as a group of similar transactions). Like our personal journal entries, it notes the date, the accounts involved, and the amounts of money, as well as providing a brief description of what happened.

(1b) -Opening entries -Closing entries -Correction of errors -Transfer between accounts -Purchase of fixed assets on credit -Recording of disposal of fixed asset

2ai -Discount Allowed -Bills receivable -Bad debts -Return inwards

(aii) -Discount Received -Bills Payable -Cash to suppliers -Return outwards

(2b) -Error of original entry -Error of omission -Error of commission -Error of principle -Compensating errors -Complete reversal of entry

(4a) Depreciation is the measure of the wearing out, consumption or other loss of value of a fixed asset whether arising from use, effluxion of time or obsolescence through technology and market changes

(4b) (i) Physical deterioration (ii) Obsolescence (iii) The time factor (iv) Economic factor (v) Inadequacy

4c) (i) Straight line: This allows an equal amount to be charged as depreciation for each year of expected use of the asset. The basic formulae is Cost- Estimated residual value/ number of years of expected use. Advantages: (i)it is simple to calculate (ii)It is time oriented Disadvantages: (i)Assumption of equal or constant revenue per year is unrealistic (ii)Might lead to a misleading picture of the financial statement (4cii) Reducing balance: Under this method, the depreciation charged per annum is determined by applying a fixed rate of depreciation on the net book value of the asset at the beginning of each year.

Disadvantage of reducing balance: Difficulty in calculating the rate of depreciation

(4ciii) Revaluation: By this method, the asset is revalue each year, any difference will be charged as depreciation to the profit and loss account. The value of the asset at the beginning and end of the year must be known.

( 8 ) gross profit %= g.p/sales * 100= 96000/240000 * 100 = 40% net profit %= n.p/sales * 1000= 80000/240000 * 100=3.33% return on capital employed= 8000/142000 * 100= 5.6% current ratio= CA: CL =82000:40000 =2.1:1 Acid test ratio= CA-stock:CL 82000-36000:40000= 1.2:1 Rate of stock turn= cost of goods sold/average stock =144000/1/2(20000+36000) =5times working capital= CA-CL =52000-40000 =42000 shareholders fund=100000+10000 =110000 Liquid asset=82000-36000 =46000

Wishes you Best of Luck.

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