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Botswana General Certificate of Secondary Education: Mathematics: Paper 2

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These are the solutions to the BGCSE Mathematics Paper 2 questions.
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Botswana Examinations Council
Botswana General Certificate of Secondary Education
Mathematics
Paper 2

Candidates answer on the Question Paper.
No Additional Materials are required.
2 hours

Instructions
(1.) Write your name, Centre number and candidate number in the spaces provided at the top of this page.
(2.) Answer all questions.
(3.) Write your answers in the spaces provided on the question paper.
(4.) If working is needed for any question, it must be shown below that question. Omission of essential working will result in the loss of marks.
(5.) Do not use staples, paper clips, highlighters, glue or correction fluid.
(6.) If the degree of accuracy is not specified in the question and if the answer is not exact, the answer should be given to three significant figures.
Answers in degrees should be given to one decimal place.
(7.) In any question where the value of π is required, use the value from your calculator or take π as 3.142.

Information
(1.) The number of marks for each question or part question is shown in brackets [ ].
(2.) The total mark for this paper is 75.
(3.) Electronic calculators may be used.

Because electronic calculators are permitted, the TI-84 Plus CE calcuator will be used.

Formula Sheet: Information and Formulae
(1.) In an election, three candidates, Janice, Kris and Loma received a total 225 630 votes.
Janice received 110 000 votes.
(a.) Percent Applications: Express the number of votes Janice received as a percentage of the total votes.

(b.) Data Presentation: Pie Chart: The proportion of votes each candidate received is represented in a pie chart.
The sector angle representing Kris's votes in 16°.
Calculate the number of votes Kris received.
(c.) How many votes did Loma receive?


$ \text{Total votes} = 225630\text{ votes} \\[3ex] \text{Votes received by Janice} = 110000\text{ votes} \\[5ex] (a.) \\[3ex] \underline{\text{Janice}} \\[3ex] \text{Percent of Janice votes} = \dfrac{110000}{225630} \times 100\% \\[5ex] = 48.7523822\% \\[3ex] \approx 48.8\% \quad\text{(to three significant figures)} \\[5ex] (b.) \\[3ex] \underline{\text{Kris}} \\[3ex] \text{sector angle} = \dfrac{\text{number of votes}}{\text{Total votes}} \times 360^\circ \\[5ex] 16^\circ = \dfrac{\text{number of votes}}{225630} \times 360^\circ \\[5ex] \text{number of votes} = \dfrac{16 \times 225630}{360} \\[5ex] = 10028\text{ votes} \\[5ex] (c.) \\[3ex] \underline{\text{Loma}} \\[3ex] \text{number of votes} + 10028 + 110000 = 225630 \\[3ex] \text{number of votes} = 225630 - 10028 - 110000 \\[3ex] = 105602\text{ votes} $
(2.) Measurements and Units: A copper pipe, 1m long, has a mass of 3.9kg.
What is the mass of a 0.25m length of this pipe?


$ \underline{\text{Unity Fraction Method}} \\[3ex] \text{Set up units:} \\[3ex] 0.25m \times \dfrac{...kg}{...m} \\[5ex] \text{Set up measurements and units:} \\[3ex] 0.25m \times \dfrac{3.9kg}{1m} \\[5ex] = 0.975kg $
(3.) Mathematics of Finance: Mmapula pays for the connection to the national electricity grid.
She pays the same amount each month for 18 months with interest.

(a.) Each month, Mmapula pays P274.56 plus 4.78% interest.
Work out how much Mmapula pays each month.

(b.) Calculate the total amount Mmapula pays for 18 months.


$ (a.) \\[3ex] \underline{\text{Each Month}} \\[3ex] \text{Basic fee} = P274.56 \\[3ex] \text{interest} = 4.78\% \times 274.56 \\[3ex] = \dfrac{4.78}{100} \times 274.56 \\[5ex] = P13.123968 \\[5ex] \text{Monthly payments} = 274.56 + 13.123968 \\[3ex] = P287.683968 \\[3ex] \approx P288 \quad\text{(to three significant figures)} \\[5ex] (b.) \\[3ex] \underline{\text{18 Months}} \\[3ex] \text{Total payments} = 287.683968 \times 18 \\[3ex] = 5178.311424 \\[3ex] \approx P5180 \quad\text{(to three significant figures)} $
(4.) Mensuration: A rectangular sports field measuring 80m by 120m is to be covered with artificial turf.
(a.) Calculate the area of the field.

(b.) The artificial turf sold as rectangular pieces, each measuring 1.6m by 2m.
How many pieces of artificial turf are required to cover the field completely?

(c.) Each piece of artificial turf costs P329.50.
Calculate the total cost of the turf.


$ (a.) \\[3ex] \underline\text{Rectangular Sports Field} \\[3ex] \text{Area of the field} = 80 \times 120 \\[3ex] = 9600m^2 \\[5ex] (b.) \\[3ex] \underline{\text{Rectangular Pieces of Artificial Turf}} \\[3ex] \text{Area of each artificial turf} = 1.6 \times 2 \\[3ex] = 3.2m^2 \\[3ex] \text{Number of artificial turf required to completely cover the field} \\[3ex] = \dfrac{\text{Area of the field}}{\text{Area of each artificial turf}} = \dfrac{9600}{3.2} \\[5ex] = 3000\text{ pieces} \\[5ex] (c.) \\[3ex] \text{Total cost of the surf} \\[3ex] = \text{Cost of 3000 pieces of turf} @ \;\;P329.50 \text{ per turf} \\[3ex] = 3000 \times 329.50 \\[3ex] = P988500 $
(5.) Kinematics: A car travels from town A to town B, a distance of 56km.
The car leaves town A and travels at a constant speed covering 40km in 30 minutes.
It then had a breakdown and stops for 15 minutes.
After the breakdown, the car continues at a constant speed for another 10 minutes to reach town B.

(a.) Draw a distance-time graph to represent the journey from town A to town B.

Number 5

(b.) What happens to the speed of the car after the breakdown?


(time, distance)

Before Breakdown
From town A: (0, 0)
distance = 40km
time = 30 minutes
(30, 40)

During Breakdown
Stops for 15 minutes; no distance covered.
distance = 40km
time = 30 + 15 = 45 minutes
(45, 40)

After Breakdown
The car continues at a constant speed for another 10 minutes to reach town B.
Reaches town B, a distance of 56km from town A
total distance = 56km
time = 45 + 10 = 55 minutes
(55, 56)

The distance-time graph to represent the journey from town A to town B is drawn as shown:

Number 5

$ s.......t.......d \\[3ex] \text{speed, }s \times \text{ time, }t = \text{ distance, }d \\[3ex] s = \dfrac{d}{t} \\[5ex] (b.) \\[3ex] \text{Before Breakdown: } (30, 40) \\[3ex] d = 40km \\[3ex] t = 30min \\[3ex] = 30min \times \dfrac{1hr}{60min} = 0.5min \\[5ex] s = \dfrac{40}{0.5} = 80km/hr \\[5ex] \text{During Breakdown: } (45, 40) \\[3ex] s = 0km/hr \quad\text{the car stopped} \\[5ex] \text{After Breakdown: } (55, 56) \\[3ex] d = 56 - 40 = 16km \\[3ex] t = 10min \\[3ex] = 10min \times \dfrac{1hr}{60min} = \dfrac{1}{6}hr \\[5ex] s = 16 \div \dfrac{1}{6} \\[5ex] = 16 \times 6 \\[3ex] = 96km/hr \\[3ex] $ After the breakdown, the speed of the car increased to 96km/hr
(6.) Trigonometry: Bearings and Distances: The diagram represents the positions of a tower, T, a waterhole W and campsite C.
The campsite is 5.2km from the tower on a bearing of 134°.
The waterhole is 6.8km from the campsite on a bearing of 044°.

Number 6

(a.) Show that the size of angle TCW is 90°
(b.) Calculate the bearing of the waterhole, W, from the tower, T.


Let us do some constructions as shown in the diagram

Number 6-1st

$ \underline{\text{Diagram}} \\[3ex] \beta = 90^\circ \quad(\text{Right Angle}) \\[3ex] \alpha + \beta = 134^\circ \\[3ex] \alpha + 90 = 134 \\[3ex] \alpha = 134 - 90 \\[3ex] \alpha = 44^\circ \\[5ex] \underline{\text{Angle Between Parallel Lines}} \\[3ex] \alpha = \alpha = 44^\circ \quad(\text{alternate angles are equal}) \\[5ex] \underline{\text{Diagram}} \\[3ex] \alpha + \tau = 90^\circ \quad(\text{Right Angle}) \\[3ex] \tau = 90 - \alpha \\[3ex] = 90 - 44 \\[3ex] = 46^\circ \\[5ex] (a.) \\[3ex] \angle TCW = \tau + 44^\circ \\[3ex] = 46 + 44 \\[3ex] = 90^\circ \\[5ex] (b.) \\[3ex] \triangle TCW \text{ is a right triangle} \\[3ex] \tan \angle WTC = \dfrac{opp}{adj} \quad(\text{SOHCAHTOA}) \\[5ex] \tan \angle WTC =\dfrac{6.8}{5.2} \\[5ex] \angle WTC = \tan^{-1} \left(\dfrac{6.8}{5.2}\right) \\[5ex] \angle WTC = 52.59464337^\circ \\[3ex] $ The bearing of the waterhole, W, from the tower, T is angle φ as shown in the diagram

Number 6-2nd

$ \underline{\text{Diagram}} \\[3ex] \phi + \angle WTC = 134^\circ \\[3ex] \phi + 52.59464337 = 134 \\[3ex] \phi = 134 - 52.59464337 \\[3ex] \phi = 81.40535663 \\[3ex] \phi \approx 81.4^\circ \quad(\text{to one decimal place}) $
(7.) Mensuration: The diagram represents a bar of chocolate in the form of a prism with a trapezium cross-section, ABCD.
The height of the chocolate bar is 1.2cm and its length is 18cm.
AB = 7.5cm and CD = 10cm.

Number 7

Calculate
(a.) the area of cross-section, ABCD
(b.) the volume of the chocolate bar,
(c.) the number of chocolate bars that can be made using 5 litres of chocolate.


(8.) Geometry Transformations: The diagram shows triangle L.
Draw the image of triangle L after a reflection in the y-axis.

Number 8


(9.) Fractions: Express $\dfrac{3}{2x + 1} + \dfrac{1}{x - 1}$ as a single fraction in its simplest form.


(10.) Mensuration: The diagram represents a solid triangular-based pyramid, JMKL.
Angle JML = angle KML = angle JMK = 90°.
$JM = KM = LM = 12cm$

Number 10

(a.) What is the geometrical relationship between triangle JMK, triangle JML and triangle KML?
(b.) Calculate
(i.) the length of JK,
(ii.) the total surface area of the pyramid.


(11.) Literal Equations: Re-arrange the formula $q = \dfrac{3p + r}{p}$ to make p the subject.


(12.) Linear Functions: Find the equation of the line parallel to $y = 2x + 5$, passing through the point $(-2, -3)$.


(13.) Quadratic Functions: The table shows some values of x and the corresponding values of y for $y = 3 - 2x - x^2$.

$x$ $-4$ $-3$ $-2$ $-1$ $0$ $1$ $2$
$y$ $-5$ $0$ $3$ $4$ $3$ $0$ $-5$

(a.) Draw the graph of $y = 3 - 2x - x^2$ for values of x from –4 to 2.

Number 13

(b.) Use the graph to solve the equation $3 - 2x - x^2 = 0$.


(14.) Statistics: Data Presentation: Histogram: The histogram represents the amounts of money that customers paid at a supermarket one morning.

Number 14

(a.) How many customers paid P20 or less?
(b.) How many customers in total, bought goods at the supermarket that morning?


(15.) Statistics: Measures of Center: Median: These are the times, in minutes, spent by 10 people in a queue at a store.

$ 0.5 \hspace{2em} 1.2 \hspace{2em} 15 \hspace{2em} 5.2 \hspace{2em} 3 \hspace{2em} 4.8 \hspace{2em} 4 \hspace{2em} 6 \hspace{2em} 2.5 \hspace{2em} 1 \\[3ex] $ Find the median time.


(16.) Kinematics: Theo walked for 23 minutes from home to the bus stop.
He arrived at the bus stop 9 minutes early for the 10:30 am bus.
At waht time did he leave home?


(17.) Mensuration: The diagram shows a kite, WXYZ.
O is the point of intersection of the diagonals of the kite.
$WZ = 10.2cm$, $OZ = 8.21cm$ and $OY = 8.75cm$.

Number 17

Calculate
(a.) YZ,
(b.) angle ZWX.


(18.) Factoring: Factorise completely $2ax - 6ay + bx - 3by$


(19.) Vectors: $\overrightarrow{OA} = \begin{bmatrix} 5 \\ 2 \end{bmatrix}$ and $\overrightarrow{OB} = \begin{bmatrix} 9 \\ 6 \end{bmatrix}$.

(a.) Find $\overrightarrow{OA} - \dfrac{2}{3}\overrightarrow{OB}$

(b.) Find $\overrightarrow{AB}$


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