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Class Intervals Class Boundaries Frequency Cumulative Frequency 0 - 9 0.5 - 9.5 1 1 10 - 19 9.5 - 19.5 3 4 20 - 29 19.5 - 29.5 3 7 30 - 39 29.5 - 39.5 6 13 40 - 4921 39.5 - 49.5 8 21 50 - 59 49.5 - 59.5 12 33 60 - 69 59.5 - 69.5 6 39 70 - 79 69.5 - 79.5 3 42 80 - 89 79.5 - 89.5 4 46 90 - 99 89.5 - 99.5 4 50 \(\sum f = 50\) (b) (c) (I) Mean mark = 49.9 (ii) Percentage of at least 66 mark = \(\frac{13}{50} \times 100\)% = 26%

\(\frac{1}{3}x - \frac{1}{4}(x + 2) \geq 3x -1\frac{1}{3}\) \(\frac{1}{3}x - \frac{(x + 2)}{12} \geq 3x - \frac{4}{3}\) \(\frac{4x - 3x - 6}{12} \geq 3x - \frac{4}{3}\) \(\frac{x - 6}{12} \geq \frac{3x}{1} - \frac{4}{3}\) \(\frac{x - 6}{12} \geq \frac{9x - 4}{3}\) 3(x - 6) \(\geq\) 12(9x - 4) 3x - 18 \(\geq\) 108x - 48 3x - 108x \(\geq\) -48 + 18 \(\frac{-105x}{-105} \geq \frac{-30}{-105}\) x \(\geq \frac{10}{35}\) x \(\geq\) \(\frac{2}{7}\) (b)(i) C\(B\)A = 108\(^o\) - (90\(^o\) + 35\(^o\)) = 180\(^o\) - 125\(^o\) = 55\(^o\) sin35\(^o\) = \(\frac{|AB|}{66}\) |AB| = 66 x sin30\(^o\) |AB| = 37.856m cos 35\(^o\) = \(\frac{|AC|}{66}\) |AC| = 66 x cos35\(^o\) |AC| = 55.06m Tan52\(^o\) = \(\frac{|AD|}{37.856}\) |AD| = 37.856 x Tan52\(^o\) = 48.45m the height of the tower to 2d.p (ii) Tan A\(C\)D = \(\frac{48.453}{54.06}\) =0.8962 A\(C\)D = tan\(^{-1}\)0.8962 = 41.867\(^o\) Angle of elevation of the tower from C = 35\(^o\) + 41.867\(^o\) = 76.87\(^o\) to 2d.p

(a) When x = 60\(^o\) y = 2cos 60\(^o\) + 3sin 60\(^o\) y = 3.6 When x = 120\(^o\) y = 2 cos 120\(^o\) + 3 sin 180\(^o\) y = 1.6 When x = 180\(^o\) y = 2 cos 180\(^o\) + 3 sin 180\(^o\) y = 2 When x = 300\(^o\) y = 2 cos 300\(^o\) + 3 sin 300\(^o\) y = 1.6 when x = 360\(^o\) y = 2 cos 360\(^o\) + 3 sin 360\(^o\) y = 2 (b) (c)(i) 1555.5\(^o\), 318\(^o\) (ii) When x = 3.42\(^o\), y = 0.9\(^o\)

(a) Weight of full tomatoes = 130kg Weight of half tomatoes = \(\frac{130}{2}\) = 65kg Cost price of half tomatoes = \(\frac{N52,000}{2}\) = N26,000 Profit = \(\frac{S.P - C.P}{C.P}\) x 100 30 = \(\frac{(S.P - 26,000)}{25,000}\) x 100 S.P = N7800 + N26,000 S.P = N7,800 + N26,000 S.P of half tomatoes = N33,800 Selling price per kg = \(\frac{33,800}{65}\) = N520 per kg Reduction in selling price = 12% = \(\frac{12}{100}\) x 520 = N62.40k per kg New selling price per kg = N520 - N62.40k = N457.60k (b) 5kg is thrown away remaining 60kg The selling price of entire 60kg = 60 x N457.60k = N27,456 The entire selling price N33,800 + N27,456 = N61,256 % profit = \(\frac{(61.256 - 50,000)}{52,000}\) x 100 = 17.8%

(a) (i) ar\(^2\) = \(\frac{1}{4}\) ........(i) ar\(^5\) = \(\frac{1}{32}\) ........(ii) \(\frac{ar^5}{ar^2} = \frac{1}{32} \div \frac{1}{4}\) r\(^3 = \frac{1}{32} \times \frac{4}{1}\) r\(^3 = \frac{1}{32} \times \frac{4}{1}\) r\(^3 = 2^{-3} = (\frac{1}{2})^3\) r = \(\frac{1}{2}\) Since ar\(^2\) = \(\frac{1}{4}\) a(\(\frac{1}{2}\))\(^2\) = \(\frac{1}{4}\) a = 1 The first term is 1 ad the common ratio is \(\frac{1}{2}\) (ii) T\(_7\) = ar\(^6\) = 1 x (\(\frac{1}{2}\))\(^6\) The seventh term = \(\frac{1}{64}\) (b) x = 2 and x = -3 a(2)\(^2\) + b(2) + c = 0 4a + 2b + c = 0 4a + 2b + c = 0.....(i) a(-3)\(^2\) + b(-3) + c = 0 9a - 3b + c = 0.......(ii) Solving eq.(i) and (ii) 4a + 2b + c = 0 -(9a - 3b + c) = 0 -5a + 5b = 0 (x - 2)(x + 3) = 0 x\(^2\) + 3x - 2x - 6 = 0 x\(^2\) + x - 6 = 0 a = 1, b = 1, c = -6

x\(^2\) = 172\(^2\) - 8\(^2\) x\(^2\) = 289 - 64 x\(^2\) = 225 x = \(\sqrt{225}\) x = 15 tan y \(\frac{8}{15}\) 1 + 2 tan y = 1 + 2(\(\frac{8}{15}\)) = \(\frac{1}{1} + \frac{6}{15}\) = \(\frac{15 + 16}{15} = \frac{31}{15}\) \(\frac{\tan y}{1 + 2 \tan y}\) = \(\frac{8}{15} \div \frac{31}{15}\) = \(\frac{8}{15} \times \frac{15}{31} = \frac{8}{31}\) (b) Otobo shares = N60,000 Remainder = N300,000 - N60,000 = N240,000 Ada's share = \(\frac{5}{12}\) x N240,0000 = N100,000 Adeola's share =; = 300,000 - N160,000 = N140,000 Ratio of the money is Otobo : Ada : Adeola 60,000 : 100,000 : 140,000 60 : 100 : 140 6 : 10 : 14 3 : 5 : 7

T\(O\)S = 68\(^o\) x 2 = 136\(^o\) (Angle at centre is twice angle at circumference) T\(R\) = 180\(^o\) - (90 + 68\(^o\)) = 180\(^o\) - 158\(^o\) = 22\(^o\) x = 2 x T\(R\)O = 2 x 22\(^o\) = 44\(^o\) (b) Let the capacity of Tank B = x litres Capacity of Tank A = x + 600 litres If 100 litres is pumoed out of each tank then, (x + 600) - 100 = 3(x - 100) x + 500 = 3x - 300 500 + 300 = 3x - x \(\frac{800}{2} = \frac{2x}{x}\) x = 400 litres Capacity of Tank A = 400 + 600 = 1,000 litres Capacity of Tank B = 400 litres

SECTION C (a(i) Sex of specimen P (cockroach): Male. (ii) Reason: Presence of anal style (b)(i) Habitat of Specimen P (Cockroach) and Specimen Q (Grasshopper): Specimen P (Cockroach): Pit toilet/dark boxes/soak away/wardrobe/kitchen/dark cupboard/book shelves/crevices cracks. Specimen Q (Grasshopper): Garden/farmland/vegetation/grassland. (b(i) Adaptations of Specimens P and Q to their Habitats:, Specimen P (Cockroach): - flat/dorsoventral flattened body; to move between small cracks/spaces/crevices. Jointed legs; for fast movement (to escape from predators); -tiny legs to crawl (membraneous) wings for flight, dark brown colour of body; serves as camouflage (from predators); ;- well developed mandibles/mouth parts; - for feeding; - compound eyes; for vision/sight/wide view hardened forewings/chitinous exoskeleton; for protection; spiracles for gaseous exchange; -antenna for detection Specimen Q (Grasshopper):-(Green) colour of body/for camouflage (from predators), (A pair of) antenna to detect vibrations and smell/sensitivity;-(numerous abdominal) spiracles; for gaseous exchange/breathing ;- well developed mandibles/mouth parts; - for chewing / feeding / tearing Jointed legs for fast movement (to escape from predators); -(membraneous) wings; for flight (from predators); eyes; for vision/sight. ci) Differences between specimens P and Q: Specimen P (Cockroach) Specimen Q (Grasshopper) y -It has brown coloured body Green coloured bod Dorsoventral flattened body Cylindrical body - Short hind leg/limb -Elongated hind leg/limb -Has two pairs of wings One pair of wings - Slender hind leg (s) Large/muscular/stout hind legs - Long antennae Short antennae Spines on every side of hind leg Row of spines on the inner Side of the hind leg (ii) Similarities between Specimens P and Q both have: - membraneous Wings; segmented body; - three body division/head, thorax and abdomen - three pairs of walking legs: have jointed appendages; presence of antennae: -biting chewing mouthparts/mandible;- exoskeleton of chitin: -compound eyes.. (d) (i) Feeding mechanism of Specimen Q (Grasshopper): Biting and chewing (mouthparts). (ii) Food items Specimen Q (Grasshopper):, Feeds on foliage leaves, young/tender stem/green stem/ grass.

a. (i) Challenges that make the conservation of resources difficult: - Limited land area for development leads to the rapid destruction of the natural forest; - Over grazing/over fishing; - Poaching/ killing of endangered species; - Bush burning/poor farming methods; - Over cultivation of land; - High demand for timber that has resulted in the felling of trees/indiscriminate cutting of trees for firewood; - Damming of rivers for hydro-electricity and irrigation has resulted in the alteration of the ecosystem; - Lack of awareness of the populace and policy makers of the need for conservation/lack of education; - Improper enforcement of existing laws/poor leadership; - Forest fire; - Inadequate funding in conservation; - Lack of international/local cooperation on conservation; - Illegal/unlawful exploitation of minerals. (ii) Ways by which Government ensures the conservation of resources: - Creation of parks/gardens/wildlife/zoo/reserves; - Putting in place a chemical and toxic waste dump programmes to avoid water pollution; - Enactment of conservation laws; - Enforcement of conservation laws; - Creating awareness/education through the media; - Establishment of conservation fund; - Training of conservation personnel/employing experts to manage natural resources; - Establishing local and international cooperation; - Aforestation/planting of trees. b. Endangered species - Species whose population has become so small; that it can completely be wiped out/become extinct; if not properly managed. c. (i) Genotype of child: OO (ii) Genotype of Father: OO; AO; BO; d. Importance of genetic variation to plants - It enables plants to spread to colonies/new areas; - It may enable plants to survive harsh/adverse conditions/environments; - It may enable plants to be resistant to diseases; - It gives rise to varieties of desirable plant species; - It makes plants produce better fruits/yield; - It makes plants produce early yield;

a. (i) Type of genetic cross: Dihybrid (ii) Reason for answer: Two characters are involved in the cross/involves two characters separable in inheritance. b. (i) Phenotype of F1: Tall and Red. (ii) Genetic Diagram Let T represent the gene for Tallness; t represent the gene for Dwarfism; R represent the gene for Red; r represent the gene for White. OR Parental genotype TtRr X TtRr X TR Tr tR tr TR TTRR Tall Red TTRr Tall Red TtRR Tall Red TtRr Tall Red Tr TTRr Tall Red TTrr Tall White TtRr Tall Red Ttrr Tall White tR TtRR Tall Red TtRr Tall Red ttRR Dwarf Red ttRr Dwarf Red tr TtRr Tall Red Ttrr Tall White ttRR Dwarf Red ttrr Dwarf White Crosses of parental genotypes Tall and Red - 9 Tall and White - 3 Dwarf and Red - 3 Dwarf and White - 1 Parental genotype Parental gametes (TR, Tr, tR, tr) shown Correct Crosses of parental gamete Phenotypic ratio 9:3:3:1 (iii) Calculation of Tall and White flowers in F2 (3/16) x (160/1) = 30

a. (i) Table Organism Respiratory Tadpole (2 days old) Gill (External) Cockroach Trachea/tracheoles Domestic fowl Lung/lungs Talinum Stoma/stomata Earthworm (moist) Skin surface/body surface Amoeba Cell membrane/plasmalemma/body surface (ii) Characteristic Features of respiratory surfaces - Large surface area; - Thin-walled/thin body surface; - Moist surface; - Well-ventilated surface; - Permeable surface; - Highly vascularised/network of capillaries present. b. Diagram of respiratory organ/gill of Tilapia

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a. Vertebrae with corresponding location Vertebrae Location Cervical Neck Thoracic Chest/Thorax Lumbar Abdomen/Upper abdomen/loin Sacral Hip/lower abdomen Caudal Tail/coccyx b. Function of: (i) Palisade - Manufactures food/carries out photosynthesis/site of photosynthesis; - Contains chloroplasts for photosynthesis; (ii) Vascular bundle - Contains xylem for transport of water/mineral salts/phloem for translocation of manufactured organic substances/hormones; - Provides mechanical support; (iii) Epidermis - It keeps the leaf’s shape; - Provides mechanical support; c. Function of (i) Sieve tube - Translocates/transports manufactured foods to other parts of the plant (ii) Cortex - The cells of cortex store starch; - Outer cortex cells of green stems contain chloroplasts and make food by photosynthesis/chlorophyllus cortex cells manufacture food; (iii) Epidermis - Cells of pith act as packing/food storage tissues; - Supports the stem; d. Structural differences (i) External structural differences between stem and root of a maize seedling Stem of a maize seedling Root of a maize seedling Fibrous root absent fibrous root present Chlorophyllus/greenish non chlorophyllus/non-greenish Leaves are present leaves are absent Thicker in size thinner in size Coleoptiles/shoot sheath present coleoptiles/shoot sheath absent Coleorhizae/radicle sheath absent coleorhizae/radicle sheath present. (ii) Internal structural differences between stem and root of a maize seedling Stem of a maize seedling Root of a maize seedling Vascular bundles are scattered vascular bundles are arranged/in a ring form More vascular bundles less vascular bundles Cortex is narrower cortex is wider Pith is absent pith is present Endodermis absent endodermis present Phloem and xylem do not alternate phloem alternates with xylem Piliferous layer absent piliferous layer present

a. Features of the small intestine that increase the rate of absorption of digested food - Ridges/furrows on inner linings; - Small finger-like projections/villi; - Minute projections/microvilli on the villi; - Thin epithelial lining of the villi; - Highly vascularised/well supplied with blood; - Intestine is long/it has a large surface area. b. Fate of glucose formed in a plant during photosynthesis - Glucose combines with Nitrogen/Sulphur/Phosphorus to form proteins for growth; - Glucose is stored as starch; - Glucose is oxidised to form carbon dioxide/water/energy in respiration; - Glucose is also converted into lipids/cellulose which is stored. c. Mode of feeding of (i) Euglena - Exhibits both autotrophic and heterotrophic modes of nutrition; - In autotrophic mode of feeding, the chlorophyll in its stellate chloroplast absorbs sunlight energy; - And the dissolved carbon dioxide in the pond water diffuses into the cell of Euglena; - Inside the chloroplast, the absorbed solar energy/sunlight is used in the synthesis of glucose from water and carbon dioxide; - Glucose is converted into starch and stored in the paramylum granules; - In heterotrophic mode of feeding, it takes in organic matter/diatoms e.t.c.; - Through the gullet; - And digests it in the reservoir. (i)Spirogyra - Manufactures its own food through photosynthesis; - Its spiral chloroplast/chlorophyll absorbs the sunlight energy; - It absorbs carbon dioxide from water/its environment to manufacture food; - The manufactured food is used; - while the rest/excess is stored in its pyrenoid in the form of starch; d. Characteristics of an Estuarine Habitat - It is shallow; - It has a high level of nutrients; - Salinity/pH/O2 level fluctuates; - Turbidity is high during rainy season; - It is affected by tides; - Species diversity is low; - Muddy/soft soil/substratum d. Differences between an aquatic and terrestrial habitat Aquatic habitat Terrestrial habitat Temperature is lower temperature is higher Water is the medium for growth/reproduction soil is a medium for growth Presence of dissolved oxygen presence of atmospheric oxygen Organisms are not affected by humidity organisms are affected by humidity Organisms are affected by tide/waves/water/current organisms are not affected by tide/waves/water current Organisms are not affected by wind action organisms are affected by wind action Light affects the distribution of organisms more light affects the distribution of organisms less Water is abundant water is not abundant

a. (i) Meat type that may be recommended for an obese patient A, D (ii) Reason for meat types Meat types A and D have lower fat content compared to the other meat types. b. (i) Meat type that would provide most energy B, C (ii) Reason They have the highest fat composition/content; when oxidised would release much energy. c. (i) Meat type that may be recommended for a child suffering from kwashiorkor A, C, D and E (ii) Reason A child suffering from kwashiorkor is deficient in/lacks adequate amount of protein; needs more protein; and meat types A, C, D and E have high protein contents. d. (i) Meat type most suitable for an active teenager A, B, C, D, E, F/all meat types. (ii) Meat type most suitable for a 70-year old human A, D e. Other class of food that provides energy Carbohydrate f. Uses of fat in the human body - provides insulation; - generates heat; - provides energy; - protects vital organs; - forms part of the cell membrane/formation of phospholipids; - smoothens the skin; - forms solvent for some fat soluble vitamins/vitamin A,D,E,K. g. Procedure of testing for fat in a meat sample using white paper - The meat sample is rubbed against piece of (white) paper; - and allowed to dry; - a translucent spot on the paper; - confirms the presence of fat in the meat sample.