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Let the total number of marbles be x No of gold-plated marbles, = 3, No of diamond marbles, = 4 No of silver marble = x - 3 - 4 = (x - 7) Pr(both were silver) = \(\frac{(x - 7)}{x}\) x \(\frac{(x - 8)}{(x - 1)}\) ( since selection is without replacement will affect the second pick) = \(\frac{(x - 7)}{x}\) x \(\frac{(x - 8)}{(x - 1)}\) = \(\frac{1}{15}\) \(\frac{(x^2 - 15x + 56 )}{(x^2 - x) }\) = \(\frac{1}{15}\), cross multiply = 15x\(^2\) - 225x + 840 = x\(^2\) - x = 14x\(^2\) - 224x + 840 = 0 Divide through by 14 = x\(^2\) - 16x + 60 = 0 = x\(^2\) - 10x - 6x + 60 = 0 = x(x - 10) - 6(x - 10) = 0 = (x - 10)(x - 6) = 0 = x = 6 or 10 x = 10 ( total marble should not be less than the sum of both gold and diamond marbles) Recall, Silver marbles = x - 7 = 10 - 7 = 3 THEREFORE, the number of silver marbles = 3.

The term of the given G.P. are (x + 2), (4x + 3), and (7x + 24) (a) Common ratio = \(\frac{(4x + 3)}{(x + 2)}\) = \(\frac{(7x + 24)}{(4x + 3)}\) cross multiply (4x + 3)(4x + 3) = (x + 2)(7x + 24) 16x\(^2\) + 24x + 9 = 7x\(^2\) + 38x + 48 16x\(^2\) - 7x\(^2\) + 24x - 38x + 9 - 48 = 0 9x\(^2\) - 14x - 39 = 0 9x\(^2\) - 27x + 13x - 39 = 0 (from 9x\(^2\) x 39) 9x(x - 3) + 13(x -3) = 0 (9x + 13)(x - 3) = 0 (9x + 13) = 0 or (x - 3) = 0 x = \(\frac{-13}{9}\) or 3 (b) When x = \(\frac{-13}{9}\) Common ratio = \(\frac{(4x + 3)}{(x + 2)}\) = \(\frac{(4(\frac{-13}{9}) + 3)}{((\frac{-13}{9}) + 2)}\) = \(\frac{\frac{-25}{9}}{\frac{5}{9}}\) = \(\frac{- 25}{9}\) x \(\frac{9}{5}\) = - 5. Common ratio,r = - 5 When x = 3, Common ratio = \(\frac{(4(3) + 3)}{((3) + 2)}\) = \(\frac{15}{5}\) = 3. Common ratio, r = 3

(a) Age Frequency 5 2 6 2x - 1 7 y + 2 8 4 9 2 10 y - 1 \(\sum\)f = 20 2 + 2x - 1 + y + 2 + 4 +2 + y - 1 = 20 2x + 2y + 8 = 20 2x + 2y = 20 - 8 = 12 divide through by 2 x + y = 6 - - - - - - - - - - - - - -(i) Also, \(\frac{\text{x}}{\text{y}}\) = \(\frac{1}{2}\) y = 2x - - - - - - - - - - - - (ii) put y = 2x into eqn -- - - -(i) x + y = 6 = x + 2x = 6 3x = 6 x = 2 put x = 2 into eqn - - - - - (ii) y = 2x = 2(2) = 4 (b) AGE Frequency F Fx 5 2 2 10 6 2(2) - 1 = 3 3 18 7 4 + 2 = 6 6 42 8 4 4 32 9 2 2 18 10 4 - 1 = 3 3 30 \(\sum\) f = 20, \(\sum\)fx = 150 Mean,\(\overline{x}\) = \(\frac{\sum fx}{\sum f}\) = \(\frac{150}{20}\) = 7.5 THEREFORE, mean age = 7.5 years.

(a) The diagram above is the illustration of the given information. (b) Let |OB| = x, and |OC| = h From \(\triangle\)OBC Tan 48º = \(\frac{h}{x}\) h = x tan48 - - - - - - - - - - - - -(i) From \(\triangle\)AOC tan 40º = \(\frac{h}{46 + x}\) h = tan 40(46 + x) = 46(tan40) + x(tan40) - - - - - - - - - - -(ii) Equate -- - - -eqn (i) and (ii) x tan 48 = 40(tan46) + x(tan40) x tan 48 - x tan 40 = 40 tan 46 x(tan 48 - tan 40) = 40 tan 46 x(1.1106 - 0.8391) = 38.5986 x(0.2715) = 38.5986 x = \(\frac{38.5986}{0.2715}\) = 142.1680 Height of pole = from eqn (i) = 142.1680 x 1.1106 = 157.8918 ≈ 157.9 m ( to one decimal place)

Given diameter d = 26cm radius r = \(\frac{26}{2}\) = 13cm \(\overline{OQ}\) = radius NOTE: \(\overline{OR}\) bisects \(\overline{PQ}\) and \(\triangle\)POQ, \(\overline{PR}\) = \(\overline{RQ}\) Cos(Q\(\widehat{O}\)R) = \(\frac{5}{13}\) (Q\(\widehat{O}\)R) = cos\(^{-1}\)((\(\frac{5}{13}\)) = 67.38º P\(\widehat{O}\)R = 2 x Q\(\widehat{O}\)R = 2 x 67.38 = 134.76 ≈ 135( nearest whole number) (b) Area of minor segment = Area of minor sector P\(\widehat{O}\)Q - Area of triangle P\(\widehat{O}\)Q = [\(\frac{P\widehat{O}Q}{360}\) x \(\pi\) x |OQ|\(^2\)] - [\(\frac{1}{2}\) x |OQ|\(^2\) x sin(P\(\widehat{O}\)Q)] = [\(\frac{135}{360}\) x \(\frac{22}{7}\) x 13\(^2\)] - [\(\frac{1}{2}\) x 13\(^2\) x Sin(134.76)] = 198.745 - 60 = 138.745 = 139cm\(^2\) (to the nearest whole number)

let the total Acres = x Wife's Share = \(\frac{2}{5} \times x = \frac{2x}{5}\) Remaining part = x - \(\frac{2x}{5}\) = \(\frac{3x}{5}\) Part for family = \(\frac{1}{3}\) x \(\frac{3x}{5}\) = \(\frac{x}{5}\) Remaining part = x - (\(\frac{2x}{5}\) + \(\frac{x}{5}\)) = \(\frac{2x}{5}\) Total ratio = 2 + 5 + 2 = 10 Child with least share = \(\frac{2}{10}\) x \(\frac{2x}{5}\) = 8 x = \(\frac{8 \times 25}{2}\) = 100 acres. (ii) Acres wife received = \(\frac{2}{5}\) x 100 = 40 acres. (b) If $400 is deposited, the remaining amount to be paid will be = $1600 - $400 = $1200. so, P = $1200, T = 12months = 1 year, R = 25% Therefore, S.I = \(\frac{1200 \times 25 \times 1}{100}\) = $300 But, Total cost = initial deposit + principal + interest = $400 + $1,200 + $300 = $1,900

(a) Solving the eqns of the two lines simultaneously: x + 2y = 7 - - -- - - - - - -(i) x - y = 4 - - - - - - - - - - - -(ii) subtract eqn (ii) from (i) x - x + 2y - (- y) = 7 - 4 3y = 3 y = 1 put y = 1 into eqn (ii) x - 1 = 4 x = 4 + 1 = 5 The point of intersection of lines is (5, 1) ⇒ If the line passes through Origin (0, 0) and (5, 1) Using: \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{y - y_1}{x - x_1}\) \(\frac{1 - 0}{5 - 0} = \frac{y - 0}{x - 0}\) \(\frac{1}{5} = \frac{\text{y}}{\text{x}}\) Therefore, the equation of the line = x - 5y = 0 (b) Given that the ratio of the interior angle to the exterior angle of a regular polygon = 4: 1 Thus, I = 4E But, the sum of the interior angles and the exterior angles of a regular polygon = 180 i.e. I + E = 180 4E + E = 180 5E = 180 E = \(\frac{180}{5}\) = 36º Since, I = 4E = 4 x 36 = 144º (ii) value of the exterior angle = 36º (i) Number of sides of the polygon = \(\frac{360}{\text{n}}\) = E, n = \(\frac{360}{36}\) = 10 (since E = 36º) (iii) Sum of the interior angles of the polygon. = (n - 2)180 = (10 - 2)180 = 8 x 180 = 1,440.

(i) 2 x T\(\widehat{S}\)Q = P\(\widehat{O}\)Q (2 times angle at circumference = angle at centre of circle) T\(\widehat{S}\)Q = \(\frac{P\widehat{O}Q}{2}\) = \(\frac{56}{28}\) = 28º T\(\widehat{Q}\)P = T\(\widehat{S}\)Q (angles in alternate segment} (ii) S\(\widehat{T}\)O + T\(\widehat{S}\)Q + S\(\widehat{V}\)P = 180º (sum of angles in a \(\triangle\)) S\(\widehat{T}\)O + 28º + 109º = 180º S\(\widehat{T}\)O = 180º - 28º - 109º = 43º Now, O\(\widehat{T}\)Q = O\(\widehat{Q}\)T (base angles of an isosceles triangle) O\(\widehat{T}\)Q + O\(\widehat{Q}\)T + P\(\widehat{O}\)Q = 180º 2(O\(\widehat{T}\)Q) = 180 - 56 = 124º O\(\widehat{T}\)Q = \(\frac{124}{2}\) = 62º Q\(\widehat{T}\)S = S\(\widehat{T}\)O + O\(\widehat{T}\)Q = 42 + 62 = 105º (b) \(\frac{2n^2 - 3n - 2 }{2n^2 + 3n + 1} \times \frac{n^2 - 1}{n^2 - 4}\) = \(\frac{(2n + 1)(n - 2)}{(2n + 1)(n + 1)} \times \frac{(n + 1)(n - 1)}{(n + 2)(n - 2)}\) After canceling common factors = \(\frac{n - 1}{n + 2}\)

(a) \(\frac{B\widehat{T}Q}{2}\) = B\(\widehat{T}\)C = C\(\widehat{T}\)Q (\(\overline{TC}\) bisects \(\angle\)BTQ) B\(\widehat{T}\)Q = B\(\widehat{A}\)T = 44º(angles in alternate segments) B\(\widehat{T}\)C = \(\frac{∠BTQ}{2}\) = \(\frac{44}{2}\) = 22º A\(\widehat{B}\)T = P\(\widehat{T}\)A = 60º (angles in alternate segments) Now, A\(\widehat{C}\)T + B\(\widehat{T}\)C = A\(\widehat{B}\)T (sum of opp. interior angles is equal to the exterior angle of a triangle) A\(\widehat{C}\)T + 22º = 60º A\(\widehat{C}\)T = 60 - 22 = 38º (b) (i) Given: Circumference of base, 2\(\pi\)r = 11 -- - - - - - - - - - -(i) Also, height, h = 6r + 3 - - - - - - - - - -(ii) From, eqn (i), 2\(\pi\)r = 11, r = \(\frac{11}{2\pi}\) = 1.75 m (ii) From equation (ii); height, h = 6r + 3 = 6(1.75) + 3 = 13.5 m (iii) Volume of cylinder, V = \(\pi r^2\)h = \(\frac{22}{7} \times 1.75^2\) x 13.5 = 130 m\(^3\)

Specimen L - Thoracic vertebra Specimen M - Humerus bone Specimen P - Femur bone Specimen Q - Lumbar Vertebra (a) Region of the human body where each of specimens L and Q are located. L: chest or thorax region Q: abdominal region or between thoracic and sacral vertebrae b(i)Four observable differences between specimens L and Q. L(Thoracic vertebra) Q(Lumbar vertebra) presence of small centrum presence of large centrum Neural spine is long Neural spine is short Transverse process is long Transverse process is short Absence of anapophysis Anapophysis is present Demifacets are present Demifacets are absent (ii) Observable similarities between specimens L and Q. - Both thoracic and lumbar vertebrae have facet joints that articulate with adjacent vertebrae. - Both thoracic and lumbar vertebrae have a body. - Both thoracic and lumbar vertebrae have a spinous process that extends backward. - Both thoracic and lumbar vertebrae have transverse processes that extend laterally. - Both of them possess neural arch. c(i) Bones each that articulate with specimens M(humerus bone) and P(femur bone) M: scapula, ulna, cervical P: pelvic girdle, tibia (ii) One vertebra each that is located and after each of specimens L and Q in the vertebral column. Before After L: ...cervical vertebra.. ...Lumbar vertebra... Q: ....thoracic vertebra.. ....sacral vertebra.... (d) Specimen Features Neural canal Rounded head Neural spine Long shaft L(thoracic) ✔ X ✔ M(humerus bone) X ✔ X ✔ P(femur bone) ✔ X ✔ Q(Lumbar vertebra) ✔ X ✔

Spirogyra cells join together to form filaments. Spirogyra is a green algae that forms long, unbranched filaments. The filaments are made up of many cylindrical cells that are connected end-to-end. Spirogyra are unicellular organisms that connect to form multicellular filaments. To become long filaments, single spirogyra cells link together, end-to-end, forming strands up to several centimeters in length.

The cell membrane, also known as the plasma membrane, controls the movement of substances in and out of a cell. "cytoplasmic membrane" is the same as "cell membrane"; they are both terms used to describe the outer boundary of a cell that separates the cell's interior from the external environment, and it's also sometimes called the "plasma membrane". The cell membrane is a semipermeable lipid bilayer that surrounds the cell. It separates the inside of the cell from the outside environment. The cell membrane is selectively permeable, meaning it allows some substances to pass through but not others. The cell membrane regulates the transport of materials entering and exiting the cell.

The bursting of a red blood cell placed in a hypotonic solution is called haemolysis. A hypotonic solution has a lower solute concentration than the cell's contents. When a red blood cell is placed in a hypotonic solution, water moves into the cell.This causes the cell to swell and eventually burst.The process of a cell swelling and bursting is called haemolysis. Isotonic: An environment where the flow of water in and out of a cell is equal. Osmosis: The movement of water across a semipermeable membrane. Osmosis is important in maintaining homeostasis, which is the tendency of systems toward a relatively stable dynamic equilibrium. Crenation: The process of a red blood cell shrinking in size when placed in a hypertonic solution.