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The pancreas ( the organ labeled 4) contains exocrine glands that produce enzymes important to digestion. These enzymes include trypsin and chymotrypsin to digest proteins; amylase for the digestion of carbohydrates; and lipase to break down fats.

(b) The mechanism of hearing in mammals begins with sound waves first reaching the pinna and passing along the auditory meatus to the ear drum. This causes vibration which passes along the ossicles. The ossides amplify the vibrations which are transmitted to the inner ear via the oval window. This sets the endolymph in the inner ear (cochlea) to vibrate. Impulses are then set up by the hair cells of the cochlea and transmitted via the auditory nerve to the brain for interpretation. (c) Care for the ear: (i) Clean ear regularly with cotton wool or swab to remove dust and wax. Also avoid puncturing the ear drum by avoiding sharp objects and loud noise. (ii) Always see the doctor in case there is any ear trouble.

(a)(i) Cell theory states that: (1) the cell is the smallest unit of life. (2) The cell is the unit of function of life. (3) New cells are produced from existing cells. (4) All living things have cells and cell products. (ii)(1) Robert Hook (2) M. J. Schleiden (c) Differences between a plant and animal cell are: Plant cell Animal cell (1) Possesses large vacuole (2) Possesses chloroplast (3) Possesses cellulose cell wall - Has small vacuole - Has no chloroplast - Has no cellulose cell wall. (d) Difference between Hydra and mammal: (i) Level of organization: Hydra is organised in the tissue level while mammal is organised in the system level (ii) Symmetry: Hydra is radially symmetrical while mammal possesses bilateral symmetry (iii) Hydra possesses two body layers, e.g. ectoderm and endoderm while mammal possesses three layers.

(a)(i) Components of an ecosystem are: (1) Producers. (2) Consumers (3) Decomposers. (ii) The producers or green plants use carbon (iv) oxide, water, chlorophyll and sunlight as a source of energy to produce carbohydrate. The consumers or animals feed on the producers or their products. Consumers release carbon (iv) oxide to the plants to take in. Decomposers or micro-organisms break down dead plants and other organisms to release nutrients which are absorbed by plants (b) Steps for studying plants and animals in a community: (i) Mapping out or selection or demarcating of a particular area (ii) Measurement of ecological factors. (iii) Collection of specimens. (c)(i) competition: Inadequate supply of nutrients, light, water and food leads to competition which results in survival of the fittest. This means that those that cannot stand the test of competition die. This eventually leads to reduction in population. Lack of these necessities also leads to migration which reduces population size. (ii) pathogens: These are disease causing agents or organisms which kill other organisms in a population thereby reducing the population. Attack by pathogen can result in the development of resistant strains which eventually lead to a growth of disease free population. (iii) Water: All living organisms need water for survival. (1) Water is needed for metabolism. (2) Lack of water leads to food scarcity and death, thereby reducing population. (3) Adequate supply of water leads to increase in food production and supply which eventually leads to increase in population. (4) Excess water supply may result in flooding which brings about displacement and reduction in population.

(a)(i) Excretory organs found in living organisms: Lung, kidney, skin, liver and nephridia (ii) Lungs - man, Kidney - man, Skin - man, Liver - man, Nephridia - earth worm, Contractile vacuole - paramecium, Stomata / Lenticel - plants, Flame cells - tape worm, Malpigian tubules - insects. (b)(1) Two ways in which the excretory organ of earth worm is similar to that of mammals are: (1) Both possess structure for filtering excretory products. (2) Each is well supplied with blood. (ii) Differences between mammals and earth worm: Mammals Earth worm -Not ciliated. -Has no special tissue for storing waste products. -Has only one outlet for discharging. - Ciliated. - Has special tissues for storing some excretory products. - Has an outlet per segment hence possesses many outlets. (c} Mechanism of opening and closing of the stomata: Each stoma is enclosed by guard cells. Guard cell has chloroplast and is also elastic. During photosynthesis, sugar is formed and this increases the concentration of cell. sap. Guard cell receives water by means of osmosis and becomes turgid. This makes the stoma to open. During the night, sugar in the guard cell is converted to starch. This makes the osmotic concentration of guard cell to reduce. Water is lost lay the guard cells through osmosis to the surrounding. This makes the guard cell to be flaccid, hence stoma closes.

first work out the expression and then correct the answer to 4 s.f = 241.34..............(A) (3 x 10-\(^3\))\(^2\)............(B) = 3\(^2\)x\(^2\) = \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\) (Note that x\(^2\) = \(\frac{1}{x^3}\)) = 24.34 x 3\(^2\) x \(\frac{1}{10^6}\) = \(\frac{2172.06}{10^6}\) = 0.00217206 = 0.002172(4 s.f)

\(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\) = \(\sqrt{9 \times 3}\) + \(\frac{3 \times {\sqrt{3}}}{{\sqrt{3}} \times {\sqrt{3}}}\) = 3\(\sqrt{3}\) + \(\sqrt{3}\) = 4\(\sqrt{3}\)

\(\frac{4\frac{3}{4} - 6\frac{1}{4}}{4\frac{1}{5} \text{ of } 1\frac{1}{4}}\) \(\frac{19}{4}\) - \(\frac{25}{4}\)............(A) \(\frac{21}{5}\) x \(\frac{5}{4}\).............(B) Now work out the value of A and the value of B and then find the value \(\frac{A}{B}\) A = \(\frac{19}{4}\) - \(\frac{25}{4}\) = \(\frac{-6}{4}\) B = \(\frac{21}{5}\) x \(\frac{5}{4}\) = \(\frac{105}{20}\) = \(\frac{21}{4}\) But then \(\frac{A}{B}\) = \(\frac{-6}{4}\) \(\div\) \(\frac{21}{4}\) = \(\frac{-6}{4}\) x \(\frac{4}{21}\) = \(\frac{-24}{84}\) = \(\frac{-2}{7}\)

\(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\) Removing the decimals = \(\frac{275 \times 64}{2500 \times 800}\) = \(\frac{88}{10^4}\) \(88 x 10^{-4} = 88 x 10^{1} x 10^{-4} = 8.8 x 10^{-3}\)

use "T" to represent the total profit. The first receives \(\frac{1}{3}\) T remaining, 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)T The seconds receives the remaining, which is \(\frac{2}{3}\) also \(\frac{2}{3}\) x \(\frac{2}{3}\) = \(\frac{4}{9}\) The third receives the left over, which is \(\frac{2}{3}\)T - \(\frac{4}{9}\)T = (\(\frac{6 - 4}{9}\))T = \(\frac{2}{9}\)T The third receives \(\frac{2}{9}\)T which is equivalent to N12000 If \(\frac{2}{9}\)T = N12, 000 T = \(\frac{12 000}{\frac{2}{9}}\) Total share[T] = N54, 000 The first receives \(\frac{1}{3}\) of T → \(\frac{1}{3}\) * N54, 000 = N18,000 The second receives \(\frac{4}{9}\) of T → \(\frac{4}{9}\) * N54, 000 = N24,000 The third receives \(\frac{2}{9}\)T which is equivalent to N12000. Adding the three shares give total profit of N54,000

You can use any whole numbers (eg. 1. 2. 3) to represent all the mangoes in the basket. If the first child takes \(\frac{1}{4}\) it will remain 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) Next, the second child takes \(\frac{3}{4}\) of the remainder which is \(\frac{3}{4}\) i.e. find \(\frac{3}{4}\) of \(\frac{3}{4}\) = \(\frac{3}{4}\) x \(\frac{3}{4}\) = \(\frac{9}{16}\) the fraction remaining now = \(\frac{3}{4}\) - \(\frac{9}{16}\) = \(\frac{12 - 9}{16}\) = \(\frac{3}{16}\)

Interest I = \(\frac{PRT}{100}\) ∴ R = \(\frac{100 \times 1}{100 \times 5}\) = \(\frac{100 \times 7.50}{500 \times 5}\) = \(\frac{750}{500}\) = \(\frac{3}{2}\) = 1.5%

The diagonal of a rhombus is a line segment that joins any two non-adjacent vertices. A rhombus has two diagonals that bisect each other at right angles. i.e this splits 6cm into 3cm each AND 8cm to 4cm Using Hyp\(^2\) = adj\(^2\) + opp\(^2\) Hyp\(^2\) = 3\(^2\) + 4\(^2\) Hyp\(^2\) = 25 Hyp = 5 ∴ Length (L) is 5cm because a rhombus is a quadrilateral with 4 equal lengths

Length of Arc AB = \(\frac{\theta}{360}\) 2\(\pi\)r = \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\) = \(\frac{4 \times 22 \times \times 3}{30}\) \(\frac{88}{10}\) = 8.8cm Perimeter = 8.8 + 2r = 8.8 + 2(10.5) = 8.8 + 21 = 29.8cm

Volume of a cylinder = \pir\(^2\)h First convert 3m to cm by multiplying by 100 Volume of External cylinder = \pi \times 13\(^2\) \times 300 Volume of Internal cylinder = \pi \times 10\(^2\) \times 300 Hence; Volume of External cylinder - Volume of Internal cylinder Total volume (v) = \pi (169 - 100) \times 300 V = \pi \times 69 \times 300 V = 20700\picm\(^3\)