SECTION 1[50 Marks]

1 Without using a calculator, evaluate \(\frac{1\frac{1}{5}-1\frac{1}{3}}{\frac{1}{8}-(-\frac{1}{2})^{2}}-\frac{7}{15}\) of 2.(4 marks)

2. Find the reciprocal of 0.216 correct to 3 decimal places, hence evaluate \(\frac{\sqrt[3]{0.512}}{0.216}\) (3 marks)

\(\frac{1}{0.216} = 4.630\)

\(\frac{\sqrt[3]{0.512}}{0.216} = 0.8 \times 4.630\)

= 3.704

3 Expand and simplify the expression \((2x^{2}-3y^{3})^{2}+12x^{2}y^{3}\)(2 marks)

\((2x^{2}-3y^{3})^{2}+12x^{2}y^{3}\)

\(=4x^{4}-12x^{2}y^{3}+9y^{6}+12x^{2}y^{3}\)

\(=4x^{4}+9y^{6}\)

4 In the parallelogram PQRS shown below, \(PQ=\) 8cm and angle SPQ = 30°.

If the area of the parallelogram is \(24~cm^{3},\) find its perimeter. (3 marks)

\(\frac{24}{2} = \frac{1}{2} \times 8 \times x \sin 30^{\circ}\)

\(x = \frac{12}{4 \sin 30} = 6cm\)

perimeter = 2(6+8) = 28

5 Given that \(9^{2y}\times2^{x}=72,\) find the values of x and y.(3 marks)

\(9^{2y} \times 2^{x} = 9 \times 8\)

\((3^{2})^{2y} \times 2^{x} = 3^{2} \times 2^{3}\)

\((3^{2})^{2y} = 3 \text{ and } 2^{x} = 2^{3}\)

\(4y = 2 \text{ and } x = 3\)

\(y = \frac{1}{2} \text{ and } x = 3\)

6 Three bells ring at intervals of 9 minutes, 15 minutes and 21 minutes. The bells will next ring together at 11.00 pm. Find the time the bells had last rang together.(3 marks)

LCM of 9, 15 and 21

3² x 5 x 7 = 315 minutes

Last time of ringing together

11:00

5:15

5:45 p.m.

7 Koech left home to a shopping centre 12km away, running at 8km/h. Fifteen minutes later, Mutua left the same home and cycled to the shopping centre at 20km/h. Calculate the distance to the shopping centre at which Mutua caught up with Koech.(3 marks)

\(\frac{x}{8} = \frac{x}{20} + \frac{1}{4}\)

\(\frac{x}{8} - \frac{x}{20} = \frac{1}{4}\)

\(\Rightarrow \frac{3x}{40} = \frac{1}{4}\)

\(x = 3\frac{1}{3}\)

Distance to shopping centre

\(12 - 3\frac{1}{3} = 8\frac{2}{3}\) km

8 Using a pair of compasses and ruler only, construct a quadrilateral ABCD in which \(AB=4cm. BC=6cm. AD=3cm,\) angle ABC = 135° and angle DAB = 60°. Measure the size of angle BCD.(4 marks)

Construction of \(135^{\circ}\) angle between lines \(AB=4\) cm and \(BC=6\) cm

Construction of \(60^{\circ}\) angle between lines \(AB=4\) cm and \(AD=3\) cm

Completion of quadrilateral ABCD

\(\angle BCD = 31^{\circ} \pm 1^{\circ}\)

9 Given that \(OA=2i+3j\) and \(OB=3i-2j\) Find the magnitude of AB to one decimal place.(3 marks)

\(\begin{pmatrix}3 \\ -2\end{pmatrix} - \begin{pmatrix}2 \\ 3\end{pmatrix}\)

\(= \begin{pmatrix}1 \\ -5\end{pmatrix}\)

magnitude \(=\sqrt{1^2 + (-5)^2}\)

\(=\sqrt{26} \simeq 5.1\)

10 Given that \(tan~x^{\circ}=\frac{3}{7}\) find cos \((90-x)^{\circ}\) giving the answer to 4 significant figures.(2 marks)

\(x = \tan^{-1} \frac{3}{7} = 23.20^{\circ}\)

Cos(90 - 23.2)° = 0.3939

11 Given that \(A=\begin{pmatrix}1&0\\ -2&3\end{pmatrix},\) \(B=\begin{pmatrix}3&0\\ 2&1\end{pmatrix}\) and \(C=2AB-A^{2}.\) Determine matrix C. (4 marks)

\(A^{2} = \begin{pmatrix} 1 & 0 \\ -2 & 3 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -2 & 3 \end{pmatrix} \)

\(= \begin{pmatrix} 1 & 0 \\ -8 & 9 \end{pmatrix}\)

\(2AB = 2 \begin{pmatrix} 1 & 0 \\ -2 & 3 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 2 & 1 \end{pmatrix} \)

\(= 2 \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \)

\(= \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}\)

\(C = 2AB - A^{2} \\ \) \(= \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ -8 & 9 \end{pmatrix}\)

\(= \begin{pmatrix} 5 & 0 \\ 8 & -3 \end{pmatrix}\)

12 Without using mathematical tables or a calculator, solve the equation \(2~log_{10}x-3~log_{10}2+log_{10}32=2.\)(3 marks)

\(\log_{10}(\frac{x^2}{2^3} \times 32) = 2\)

\(\frac{x^2}{2^3} \times 2^5 = 100\)

\(4x^2 = 100\)

\(x = \sqrt{25} = \pm 5\)

\(x = 5\)

13 A line L passes through point (3,1) and is perpendicular to the line \(2y=4x+5.\) Determine the equation of line L.(3 marks)

\(2y = 4x + 5 \Rightarrow y = 2x + \frac{5}{2}\)

gradient, \(M_1\) of line = 2

gradient, \(M_2\) of perpendicular is given by

\(2M_2 = -1 \Rightarrow M_2 = -\frac{1}{2}\)

equation of line L

\(\frac{y-1}{x-3} = -\frac{1}{2}\)

\(y = -\frac{1}{2}x + \frac{5}{2}\)

14 A Forex Bureau in Kenya buys and sells foreign currencies as shown below:

A businesswoman from China converted 195 250 Chinese Yuan into Kenya shillings.

(a) Calculate the amount of money, in Kenya shillings, that she received.

(1 mark)

(a) 195250 Chinese Yuan into Kenya Shillings

= 195250 × 12.34 = 2409385

(b) While in Kenya, the businesswoman spent Ksh 1 258 000 and then converted the balance into South African Rand. Calculate the amount of money, to the nearest Rand, that she received.(3 marks)

(b) Balance:

= 2409385 - 1258000

= 1151385

Balance in S.A. Rand

\(\frac{1151385}{11.37}\)

= 101265

15 The figure below represents a solid cone with a cylindrical hole drilled into it. The radius of the cone is 10.5 cm and its vertical height is 15 cm. The hole has a diameter of 7 cm and depth of 8 cm.

Calculate the volume of the solid.(3 marks)

Volume of solid

\(\frac{1}{3} \times \frac{22}{7} \times 10.5^2 \times 15 - \frac{22}{7} \times 3.5^2 \times 8\)

= 1732.5 - 308

= 1424.5 cm³

16 Bukra had two bags A and B, containing sugar. If he removed 2kg of sugar from bag A and added it to bag B, the mass of sugar in bag B would be four times the mass of the sugar in bag A. If he added 10kg of sugar to the original amount of sugar in each bag, the mass of sugar in bag B would be twice the mass of the sugar in bag A. Calculate the original mass of sugar in each bag.(3 marks)

\(4(A-2) = B+2\)

\(2(A+10) = B+10\)

\(4A - B = 10 .... (i)\)

\(\mp 2A \pm B = \pm 10 .... (ii)\)

\(2A = 20\)

\(\Rightarrow A = 10\)

Substitute \(A=10\) in (i)

\(4 \times 10 - B = 10\)

\(\Rightarrow B = 30\)

SECTION II (50 marks)

Answer only five questions in this section in the spaces provided.

17 The table below shows the height, measured to the nearest cm, of 101 pawpaw trees.

(a) State the modal class.(1 mark)

(a) modal class 40 - 44

(b) Calculate to 2 decimal places:

(i) the mean height;(4 marks)

(b) (i) mid points:

22, 27, 32, 37, 42, 47, 52, 57

\(\frac{22 \times 2 + 27 \times 15 + 32 \times 18 + 37 \times 25 + \\ 42 \times 30 + 47 \times 6 + 52 \times 3 + 57 \times 2}{101}\)

= 37.25

(ii) the difference between the median height and the mean height.

(5 marks)

(ii) Cumulative frequencies

2, 17, 35, 60, 90, 96, 99, 101

\(\frac{16}{25} \times 5\)

= 3.2

34.5 + 3.2

= 37.7

difference 37.7 - 37.25

= 0.45

18 The figure below represents a solid cuboid ABCDEFGH with a rectangular base. \(AC=13cm, BC=5cm\) and \(CH=15cm\).

(a) Determine the length of AB.(1 mark)

(a) \(|AB| = \sqrt{169-25} = 12\)

(b) Calculate the surface area of the cuboid.

(3 marks)

(b) \(2 \times 5 \times 12 + 2 \times 5 \times 15 + 2 \times 12 \times 15\)

\(= 630 cm^2\)

(c) Given that the density of the material used to make the cuboid is \(7.6g/cm^{3},\) calculate its mass in kilograms.

(4 marks)

(c) volume \(= 5 \times 12 \times 15 cm^3\)

mass \(= 7.6 \times 5 \times 12 \times 15\)

\(= 6840 gm\)

\(=\frac{6840}{1000}\)

\(= 6.84 kg\)

(d) Determine the number of such cuboids that can fit exactly in a container measuring 1.5m by 1.2m by 1m.

(2 marks)

(d) \(\frac{150 \times 120 \times 100 cm^3}{15 \times 12 \times 5 cm^3}\)

= 2000

19 Two alloys, A and B, are each made up of copper, zinc and tin. In alloy A, the ratio of copper to zinc is 3:2 and the ratio of zinc to tin is 3:5.

(a) Determine the ratio, copper: zinc: tin, in alloy A.

(2 marks)

(b) The mass of alloy A is 250kg. Alloy B has the same mass as alloy A but the amount of copper is 30% less than that of alloy A.

Calculate:

(i) the mass of tin in alloy A;

(2 marks)

(ii) the total mass of zinc and tin in alloy B;

(3 marks)

(c) Given that the ratio of zinc to tin in alloy B is 3:8, determine the amount of tin in alloy B than in alloy A.

(3 marks)

19. (a) Ratio: copper : zinc : tin

Copper : zinc : tin = \(9 : 6 : 10\)

(b) (i) mass of tin

\( = 250 \times \frac{10}{25} \) \( = 100 \text{ kg} \)

(ii) mass of zinc and tin in alloy B:

\( \text{mass of copper} = \frac{70}{100} \times 90 \) \( = 63 \) \( \therefore \text{mass of zinc and tin}: \) \( = 250 - 63 \) \( = 187 \)

(c) amount of tin in alloy A than B:

\( \text{mass of tin in alloy B} \) \( = \frac{8}{11} \times 187 \) \( = 136 \)

difference:

\( 136 - 100 \) \( = 36 \)

20 (a) Express \(\frac{1}{x-2}+\frac{2}{x+5}=\frac{3}{x+1}\) in the form \(ax^{2}+bx+c=0,\) where a, b and c are constants hence solve for x.(4 marks)

(b) Neema did y tests and scored a total of 120 marks. She did two more tests which she scored 14 and 13 marks. The mean score of the first y tests was 3 marks more than the mean score for all the tests she did. Find the total number of tests that she did.(6 marks)

20. (a) \(\frac{1}{x-2} - \frac{2}{x+5} = \frac{3}{x+1}\)

\(\frac{x+5-2(x-2)}{(x-2)(x+5)} = \frac{3}{x+1}\)

\(\frac{-x+9}{x^2+3x-10} = \frac{3}{x+1}\)

\(4x^2+x-39=0\)

\((4x+13)(x-3)=0\)

x = 3 or \(x = -3\frac{1}{4}\)

(b) mean for second set of tests

\(=\frac{147}{y+2}\)

\(\frac{120}{y} - \frac{147}{y+2} = 3\)

\(\frac{120y+240-147y}{y(y+2)} = 3\)

\(-27y+240 = 3y^2+6y\)

\(-9y+80 = y^2+2y\)

\(y^2+11y-80=0\)

\((y-5)(y+16)=0\)

\(y=5\) or -16

No. of tests: 5 + 2 = 7

21 The vertices of quadrilateral OPQR are \(O(0,0), P(2,0), Q(4,2)\) and \(R(0,3)\) The vertices of its image under a rotation are \(O'(1,-1), P'(1,-3), Q'(3,-5)\) and \(R'(4,-1).\)

(a) (i) On the grid provided, draw OPQR and its image \(O'P'Q'R'\)
(2 marks)

(ii) By construction, determine the centre and angle of rotation.
(3 marks)

(b) On the same grid as (a)(i) above, draw \(O"P"Q"R"\) the image of \(O'P'Q'R'\) under a reflection in the line \(y=x\).
(2 marks)

(c) From the quadrilaterals drawn, state the pairs that are:

(i) directly congruent;
(1 mark)

(ii) oppositely congruent.
(2 marks)

a) (i) \( \text{OPQR} \quad \checkmark \text{ drawn} \)

\( O'P'Q'R' \quad \checkmark \text{ drawn}\)

(ii)\( \text{Perpendicular bisectors} \quad \checkmark \text{ drawn (at least 2)}\)

\( \text{centre of rotation } (0, -1) \text{ shown}\)

\( \text{angle of rotation } -90^\circ\)

b)\( \text{line of reflection } x = y \text{ drawn}\)

\(\ \text{quadrilateral } O''P''Q''R'' \text{ drawn}\)

c) (i) \(\text{directly congruent quads:}\)

\(\text{OPQR and } O'P'Q'R'\)

(ii) \( \text{Oppositely congruent quads:}\)

\( \text{OPQR and } O''P''Q''R''\)

\( O'P'Q'R' \text{ and } O''P''Q''R''\)

22 The equation of a curve is \(y=2x^{3}+3x^{2}\).

(a) Find:

(i) the x-intercept of the curve; (2 marks)

(a) (i) \( \text{x-intercepts} \)

\( \text{when } y = 0 \)

\( x^2(2x + 3) = 0 \)

\( x = 0 \text{ and } x = -\frac{3}{2} \)

(ii) the y-intercept of the curve. (1 mark)

(ii) \( \text{y-intercept} \)

\( \text{when } x = 0, y = 0 \)

(b) (i) Determine the stationery points of the curve. (3 marks)

(b) (i) \( \text{stationary points of curve} \)

\( \frac{dy}{dx} = 6x^2 + 6x \)

\( \text{stationary points when } \frac{dy}{dx} = 0 \)

\( \text{i.e. } 6x^2 + 6x = 0 \)

\( 6x(x + 1) = 0 \)

\( x = 0 \text{ or } x = -1 \)

\( \therefore \text{ stationary points are:} \)

\( (0, 0) \text{ and } (-1, 1) \)

(ii) For each point in (b) (i) above, determine whether it is a maximum or a minimum. (2 marks)

\(\text{minimum point } (0, 0)\)

\(\text{maximum point } (-1, 1)\)

(c) Sketch the curve. (2 marks)

points plotted at \(\left(-1\tfrac{1}{2}, 0\right)\), \(\left(-1, 1\right)\) and \((0, 0)\)

smooth curve

23 Three pegs R, S and T are on the vertices of a triangular plain field. R is 300m from S on a bearing of 300° and T is 450m directly south of R.

(a) Using a scale of 1cm to represent 60m, draw a diagram to show the positions of the pegs. (3 marks)

\(\checkmark\) location of R \(\quad\) B1 \(\quad \text{length } 5 \text{ cm and bearing } 300^\circ\)

\(\checkmark\) location of T \(\quad B1 \quad \text{length } 7.5 \text{ cm; south of R}\)

\(\checkmark\) \(\textit{complete} \quad \triangle \quad B1\)

(b) Use the scale drawing to determine:

(i) the distance between T and S in metres; (2 marks)

(ii) the bearing of T from S. (1 mark)

(c) Find the area of the field, in hectares, correct to one decimal place. (4 marks)

(b) (i) Distance TS: \(\left(6.6 \left( \pm 0.1 \right)\right) \,\text{cm}\) (B1)

conversion \(\left(6.6 \times 60 = 396 \,\text{m}\right)\) (B1)

(ii) Bearing of T from S (B1)

\(\left(180 + 41^\circ \left( \pm 1^\circ \right) = 221^\circ\right)\)

(c) area of field (B1)

\(\angle TRS = 60^\circ\)

\(\text{area} = \left(\dfrac{1}{2} \times 300 \times 450 \sin \left(60^\circ\right)\right)\) (M1)

\(= \left(\dfrac{58456.71476}{10000}\right)\) (M1)

\(= \left(5.8 \,\text{ha}\right)\) (A1)

24 In the figure below, PQ is parallel to RS. The lines PS and RQ intersect at T. \(RQ = 10\)cm, \(RT:TQ = 3:2\), angle PQT = 40° and angle RTS = 80°.

(a) Find the length of RT.
(2 marks)

(b) Determine, correct to 2 significant figures:

(i) the perpendicular distance between PQ and RS;
(2 marks)

(ii) the length of TS.
(2 marks)

(c) Using the cosine rule, find the length of RS correct to 2 significant figures.
(2 marks)

(d) Calculate, correct to one decimal place, the area of triangle RST.
(2 marks)

24. (a) length of RT:

\( = \left( \dfrac{3}{5} \times 10 \right) \quad (M1) \)

\( = 6 \,\mathrm{cm} \quad (A1) \)

(b) (i) Perpendicular distance between PQ & RS:

\( = 10 \sin(40^\circ) \quad (M1) \)

\( = 6.4 \,\mathrm{cm} \quad (A1) \)

(ii)

\( \dfrac{TS}{\sin(40^\circ)} = \dfrac{6}{\sin(60^\circ)} \)

\( TS = \dfrac{6 \sin(40^\circ)}{\sin(60^\circ)} \quad (M1) \)

\( = 4.5 \,\mathrm{cm} \quad (A1) \)

(c) length RS using cosine rule:

\( RS^2 = 6^2 + 4.5^2 - 2 \times 4.5 \times 6 \cos(80^\circ) \quad (M1) \)

\( = 46.87299841 \)

\( RS = 6.8 \quad (A1) \)

(d) area of \(\triangle RST\):

\( = \dfrac{1}{2} \times 6 \times 4.5 \sin(80^\circ) \quad (M1) \)

\( = 13.3 \quad (A1) \)