1. Without using mathematical tables or a calculator, evaluate
\(\frac{0.13 \times 0.3 - 0.003}{0.09}\) giving the answer in decimal form.(3 marks)
\(\frac{0.036 \times 1000}{0.09 \times 1000}\)
\(= \frac{36}{90}\)
\(= \frac{4}{10}\)
\(= 0.4\)
2. Simplify the expression:
\(\frac{4x^{2}-9}{2x^{2}+x-6}\)
Numerator:
\(4x^{2}-9\)
\(a^{2}-b^{2}=(a+b)(a-b)\)
\((2x+3)(2x-3)\)
Denominator:
\(2x^{2}+x-6\)
\(2x^{2}+4x-3x-6\)
\(2x(x+2)-3(x+2)\)
\((2x-3)(x+2)\)
Solution:
\(\frac{(2x+3)(2x-3)}{(2x-3)(x+2)}\)
\(\frac{2x+3}{x+2}\)
3. A straight line L1 whose equation is \(y = 2 - \frac{1}{3}x\) meets the y-axis at Q. Another straight line L2 is perpendicular to L1 at Q. Find the equation of L2 in the form \(y = mx + c\) where m and c are constants.(3 marks)
\(y = 2 - \frac{1}{3}(0)\)
\(y = 2\)
\(M_{1} \times M_{2} = -1\)
\(-\frac{1}{3} \times M_{2} = -1\)
\(M_{2} = 3\)
\(y = M_{2}x + c\)
\(y = 3x + c\)
\(2 = c\)
\(y = 3x + 2\)
4. A circle of radius 3 cm passes through all the vertices of a regular hexagon.
Determine the area of the circle that lies outside the hexagon.(3 marks)
Area of Circle - Area of Hexagon
\(\frac{22}{7} \times 3^{2} \) - \(\frac{1}{2} \times 3 \times 3 \times \sin 60 \times 6\)
\(= 28.29 - 23.38\)
\(= 4.91 cm^2\)
5. Solve for x in the equation \(25^{x} = 125^{\frac{2}{3}} \div 5^{-1}\)
\(25 = 5^{2}\)
\(5^{2x} = (5^{3})^{\frac{2}{3}} \div 5^{-1}\)
\(5^{2x} = 5^{2} \div 5^{-1}\)
\(5^{2x} = 5^{2-(-1)}\)
\(2x = 3\)
\(x = \frac{3}{2}\)
\(x = 1.5 \text{ or } 1\frac{1}{2}\)
6. A carpenter had two big pieces of wood of equal length. The carpenter cut the first piece into smaller pieces of length 15 cm each without remainder. The carpenter cut the second piece into smaller pieces of length 24 cm each without remainder. Determine the minimum length of each of the big pieces of wood.(2 marks)
L.C.M of 15 and 24
\(15 = 3 \times 5\)
\(24 = 2^3 \times 3\)
L.C.M of 15 and 24 = \(3 \times 5 \times 2^3 \\ = 120cm\)
7. A cylindrical container of internal radius 10.5 cm has a hemispherical base. The container has water up to a height of 30.5 cm. Calculate the surface area of the container that is in contact with water. Take \(π = \frac{22}{7}\) (4 marks)
Hemisphere = \(2πr^2\)
\(= 2 \times \frac{22}{7} \times 10.5^2\)
\(= 693 cm^2\)
Cylinder = \(20 cm - 10.5 cm = 20 cm\)
\(= πDh \text{ or } 2πrh\)
\(= 2 \times \frac{22}{7} \times 10.5 \times 20\)
\(= 1320 cm^2\)
Total = \(693 + 1320 = 2013 cm^2\)
8. In the following figure, points A and D are vertices of a trapezium ABCD. Line DC is parallel to line AQ. Line \(DC = 4 \text{ cm}\). Point B lies on the line AQ such that angle \(DCB = 90^{\circ}\).
Using a ruler and a pair of compasses only, complete the trapezium. Hence measure the length of line AB.(4 marks)
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9.The length of a minor arc AB of a circle centre O is 10 cm. The arc AB subtends an angle of 1.25 radians at O. Calculate the area of the minor sector AOB. (3 marks)
Arc length \(= \frac{\theta}{360} \times 2\pi r\)
\(\pi^{c} = 180^{\circ}\)
\(1.25^{c} = \frac{1.25 \times 180^{\circ}}{\pi}\)
\(10 = \frac{1.25 \times 180^{\circ}}{360 \pi} \times 2 \times \pi \times r\)
\(10 = 1.25 \times r\)
\(r = \frac{ 10}{1.25}\)
\(r= 8 cm\)
\(A = \frac{\theta}{360} \times \pi r^2\)
\(= \frac{1.25}{2 \pi} \times \pi \times 8^2\)
\(= 40 cm^2\)
11. A Kenyan bank bought and sold foreign currencies on two different dates as shown.
| Date | Currency | Buying (In Ksh) | Selling (In Ksh) |
|---|---|---|---|
| 12/1/2021 | 1 South African rand | 7.08 | 7.10 |
| 12/2/2021 | 100 Tanzanian shillings | 21.12 | 21.25 |
A South African tourist arrived in Kenya on 12/1/2021 with 15,000 South African rands. He changed the whole amount to Kenyan shillings. While in Kenya, he spent a total of Ksh 53,075 and changed the balance to Tanzanian shillings before leaving for Tanzania on 12/2/2021.
Determine, in Tanzanian shillings, the amount he received.(3 marks)
12. The coordinates of points O, A, B and C are (0, 0), (2, 1), (4, 4) and (-1, 7) respectively. The coordinates of point D is (x, y). Point M is the midpoint of line CD and it satisfies the vector equation \( \overrightarrow{OM} = \overrightarrow{OA} + \frac{5}{2}\overrightarrow{AB} \). Determine the coordinates of point D.(3 marks)
14. The following frequency distribution table shows the mass, in kilograms, of maize flour sold by 30 traders.
| Mass (in kg) | No. of Traders |
|---|---|
| 10-19 | 3 |
| 20-29 | 8 |
| 30-39 | 10 |
| 40-49 | 7 |
| 50-59 | 2 |
Calculate the median mass of maize flour sold.
(3 marks)
SECTION II (50 marks)
Answer only five questions in this section in the spaces provided.
17.On a market day, Abdul had 32 goats while Chebet had 56 goats. Chebet sold twice as many goats as Abdul. After the sale, the number of the goats that Abdul and Chebet remained with was in the ratio 3 : 5 respectively.
(a) Determine the total number of goats sold by Abdul and Chebet at the market.(4 marks)
(b) Abdul and Chebet raised a total of Ksh 97 600 from the sale of goats at the market. Abdul’s selling price per goat was 5% higher than that of Chebet. Determine the ratio of the earnings of Abdul to Chebet from the sale of goats.(4 marks)
(c) Abdul decreased the selling price per goat in the ratio 19:21 for the goats that remained. Determine the new selling price per goat.(2 marks)
18. The price of pens in a bookshop changed in the months of February, March, and April. The price of a pen in the month of February was Sh 2 less than the price of a pen in the month of March. In the month of April, the price of a pen was Sh 2 more than the price of a pen in the month of March. The bookshop sold pens worth Ksh 4,200 in February. In April, pens worth Ksh 4,500 were sold.
Take Ksh \(x\) to be the price of a pen in March.
(a) Write an expression in \(x\) for the number of pens sold by the bookshop in:
(i) February; (1 mark)
(ii) April. (1 mark)
(b) The bookshop sold 50 more pens in February than in April. Determine the number of pens sold in February. (6 marks)
(c) Determine the percentage change in the number of pens sold by the bookshop in April compared to the number sold in February. (2 marks)
19. Airport S is 1700 km on a bearing of 300° from airport R. Airport Q is 800 km on a bearing of 215° from R.
In the following figure, airport R is represented by point R.
(a) Using a scale of 1 cm to represent 200 km, show on the figure the relative positions of airports S and Q.(3 marks)
(b) Use the scale drawing to determine:
(i) the distance from airport S to airport Q in kilometres;(1 mark)
(ii) the bearing of S from Q.(2 marks)
(c) An aircraft flying at a speed of 400 km/h left S for Q at 8.00 am. Determine the time when the aircraft was exactly 1000 km from airport R on its way to Q.(4 marks)
19. Airport S is 1700 km on a bearing of 300° from airport R. Airport Q is 800 km on a bearing of 215° from R.
In the following figure, airport R is represented by point R.
(a) Using a scale of 1 cm to represent 200 km, show on the figure the relative positions of airports S and Q.(3 marks)
(b) Use the scale drawing to determine:
(i) the distance from airport S to airport Q in kilometres;(1 mark)
(ii) the bearing of S from Q.(2 marks)
(c) An aircraft flying at a speed of 400 km/h left S for Q at 8.00 am. Determine the time when the aircraft was exactly 1000 km from airport R on its way to Q.(4 marks)
The following figure represents a piece of land in the shape of a trapezium ABCD. Lines AB and DC are parallel. Point M is the midpoint of AB. The land is divided into three triangular plots. Line AB = 100 m, AD = 39 m, MD = 41 m and angle ABC = 30°.
(a) Calculate the area of the triangular plot AMD and hence the perpendicular distance between the two parallel sides.(4 marks)
(b) Calculate the length of:
(i) BC;(2 marks)
(ii) MC.(2 marks)
(c) Calculate the size of the obtuse angle BMC.(2 marks)
21. Two matrices P and Q are such that
\(P = \begin{pmatrix} 3 & 7 \\ h & 4 \end{pmatrix} \) and \(Q = \begin{pmatrix} 5 & -4 \\ 3 & -2 \end{pmatrix} \)
The determinant of \(PQ = 3\)
(a) (i) Determine the value of h. (3 marks)
\(PQ = \begin{pmatrix} 3 & 7 \\ h & 4 \end{pmatrix} \begin{pmatrix} 5 & -4 \\ 3 & -2 \end{pmatrix} \)
\(= \begin{pmatrix} 15 + 21 & -12 - 14 \\ 5h + 12 & -4h - 8 \end{pmatrix} \)
\(= \begin{pmatrix} 36 & -26 \\ 5h + 12 & -4h - 8 \end{pmatrix} \)
\(det(PQ) = 36(-4h - 8) + 26(5h + 12) \)
\(= -144h - 288 + 130h + 312 \)
\(3 = -14h + 24 \)
\(h = \frac{21}{14} = \frac{3}{2} \)
(a) (ii) Find, the inverse of matrix P. (2 marks)
\(P^{-1} = \frac{1}{det(P)} \begin{pmatrix} 4 & -7 \\ -1.5 & 3 \end{pmatrix} \)
\(det(P) = 12 - 10.5 \)
\(= 1.5 \)
\(P^{-1} = \frac{1}{1.5} \begin{pmatrix} 4 & -7 \\ -1.5 & 3 \end{pmatrix} \)
\(or \begin{pmatrix} \frac{8}{3} & -\frac{14}{3} \\ -1 & 2 \end{pmatrix} \)
(b) Patel and Lagat purchased watches at sh m per watch and phones at sh n per phone. Patel purchased 12 watches and 28 phones for Ksh 24 600. Lagat purchased 15 watches and 40 phones for Ksh 34 500.
i. Form two equations in m and n. (2 marks)
Patel: 12m + 28n = 24600
Lagat: 15m + 40n = 34500
=> 3m + 7n = 6150 - (i)
=> 1.5m + 4n = 3450 - (ii)
(ii) Use the matrix method to determine the price of a watch and that of a phone. (3 marks)
\( \begin{pmatrix} 3 & 7 \\ 1.5 & 4 \end{pmatrix} \begin{pmatrix} m \\ n \end{pmatrix} = \begin{pmatrix} 6150 \\ 3450 \end{pmatrix} \)
\( \frac{1}{1.5} \begin{pmatrix} 4 & -7 \\ -1.5 & 3 \end{pmatrix} \begin{pmatrix} 3 & 7 \\ 1.5 & 4 \end{pmatrix} \begin{pmatrix} m \\ n \end{pmatrix} = \frac{1}{1.5} \begin{pmatrix} 4 & -7 \\ -1.5 & 3 \end{pmatrix} \begin{pmatrix} 6150 \\ 3450 \end{pmatrix} \)
\( \frac{1}{1.5} \begin{pmatrix} 1.5 & 0 \\ 0 & 1.5 \end{pmatrix} \begin{pmatrix} m \\ n \end{pmatrix} = \frac{1}{1.5} \begin{pmatrix} 450 \\ 1125 \end{pmatrix} \)
\( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} m \\ n \end{pmatrix} = \begin{pmatrix} 300 \\ 750 \end{pmatrix} \)
\( m = 300 \)
\( n = 750 \)
22. Members of the Environmental Club planted trees. The number of trees planted by each member is recorded as follows.
24 7 11 6 12 27 8 9 33 23
12 32 19 33 16 21 14 16 17 10
21 18 16 9 17 8 28 13 8 19
9 17 24 18 29 15 12 31 22 14
Using a class width of 5 and starting with the class 6-10, make a frequency distribution table for the data. (2 marks)
| Class | Tally | \(f\) | \(mid-point(x)\) | \(fx\) |
|---|---|---|---|---|
| 6-10 | HH |||| | 9 | 8 | 72 |
| 11-15 | HH ||| / | 8 | 13 | 104 |
| 16-20 | HH |||| | 10 | 18 | 180 |
| 21-25 | HH || | 6 | 23 | 138 |
| 26-30 | ||| | 3 | 28 | 84 |
| 31-35 | |||| | 4 | 33 | 132 |
| Σf = 40 | Σfx = 710 | |||
(a) (ii) State the modal class. (1 mark)
16 - 20
(iii) Estimate the mean number of trees planted. (3 marks)
\( \bar{x} = \frac{\sum fx}{\sum f} = \frac{710}{40} \)
\(= 17.75 \)
| Class Limit | \(f\) |
|---|---|
| 5.5 - 10.5 | 9 |
| 10.5 - 15.5 | 8 |
| 15.5 - 20.5 | 10 |
| 20.5 - 25.5 | 6 |
| 25.5 - 30.5 | 3 |
| 30.5 - 35.5 | 4 |
23. On the cartesian plane below, triangle ABC has vertices A(-4, -4), B(-2.5, -7) and C(-2, 5.5) while triangle A"B"C" has vertices A'(1, 2), B'(4, 8) and C'(5, 5). The line joining A'(1, 2), B'(4, -4) is part of another triangle A'B'C'.
(a) Triangle A'B'C' is the image of triangle ABC under an enlargement centre (-3, -2) and scale factor (S.F) = -2. On the same grid, draw △A'B'C'.(3 marks)
(b) Triangle A"B"C" is the image of triangle A'B'C' under rotation centre O(0, 0).
(i) State the angle of rotation.(1 mark)
(ii) Complete triangle A"B"C'.(2 marks)
(c) Triangle A"B"C" is the image of triangle A'B'C' under a reflection.
(i) Draw the mirror line.(1 mark)
(ii) Determine the equation of the mirror line in the form \(y = mx + c\).(2 marks)
(d) Describe fully a single transformation that maps triangle A'B'C' onto triangle A"B"C".(1 mark)
24. The gradient of the curve \(y = x + 5x^2 + Px - 18\) at \(x = -1\) is -15.
(a) Find:
(i) the value of P; (3 marks)
\(\frac{dy}{dx}=3x^2+10x+P\)
At \(x = -1\),
\(\frac{dy}{dx}=3(-1)^2+10(-1)+P \\ =-15\)
3 - 10 + P = -15
P = -15 + 10 - 3 = -8
(ii) the equation of the normal to the tangent to the curve at \(x = -1\).(3 marks)
Gradient of normal \(=\frac{1}{15}\)
At \(x = -1\), \(y=(-1)^3+5(-1)^2+(-8)(-1)-18 \\ =-6\)
\(\frac{y+6}{x+1}=\frac{1}{15}\)
\(x+1=15y+90 \\ \Rightarrow x-15y-89=0\)
(b) Find the coordinates of the turning points of the curve.(4 marks)
\(y=x^{3}+5x^{2}-8x-18\)
\(\frac{dy}{dx}=3x^{2}+10x-8\)
At a turning point, \(3x^{2}+10x-8=0\)
\(3x^{2}+12-2x-8=0\)
\(3x(x+4)-2(x+4)=0\)
\(x+4=0\) or \(3x-2=0\)
\(x=-4\) or \(x=\frac{2}{3}\)