Lesotho General Certificate of Secondary Education: Mathematics: Paper 4

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Examinations Council of Lesotho
Lesotho General Certificate of Secondary Education
Mathematics
Paper 4 (Extended)

2 hours 30 minutes
Marks: 130

Read These Instructions First
(1.) Write your name, centre number and candidate number in the relevant spaces provided at the top of this page.
(2.) Write in dark blue or black pen.
(3.) You may use an HB pencil for any diagrams or graphs.
(4.) Do not use staples, paper clips, glue or correction fluid.
(5.) Answer all questions.
(6.) Electronic calculators should be used.
(7.) If working is needed for any question, it must be shown below that question.
(8.) If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place.
For π, use either your calculator value or 3.142.
(9.) The number of marks is given in brackets [ ] at the end of each question or part question.

(1.) Mathematics of Finance: (a.) Teboho and Sebolelo share M72 000 in the ratio 3 : 5.
Calculate
(i.) Teboho's share,
(ii.) the percentage of Sebolelo's share.

(b.) Lebeha invested part of M10 000 at 2.5% per annum simple interest and the remaining amount at 1.9% per annum simple interest.
He obtained the total interest of M678 after three years.
Calculate the amount invested at 2.5%.

(c.) The value of the car depreciates at the rate of 5% every year.
The value of the car is now M83 800.48.
Calculate the value of the car 7 years ago.

$ (a.) \\[3ex] \text{Teboho : Sebolelo} = \text{Ratio of } 3 : 5 \\[3ex] \text{Sum of ratios} = 3 + 5 = 8 \\[3ex] \text{Amount shared} = M72 000 \\[5ex] (i.) \\[3ex] \text{Teboho's share} = \dfrac{3}{8} \times 72000 \\[5ex] = 3 \times 900 \\[3ex] = M2700 \\[5ex] (ii.) \\[3ex] \text{Percentage of Sebolelo's share} = \dfrac{5}{8} \times 100\% \\[5ex] = 62.5\% \\[5ex] (b.) \\[3ex] SI = P \times r \times t \\[3ex] \text{Where} \\[3ex] SI = \text{Simple Interest} \\[3ex] P = \text{Principal} \\[3ex] r = \text{rate} \\[3ex] t = \text{time} = 3\text{ years} \\[3ex] $ Let us represent the information in a table for easier computation.
Let the:
part of M10 000 invested at 2.5% per annum simple interest = $x$
part invested at 1.9% per annum simple interest = $10000 - x$

Account	Principal, P (M)	rate, r (%)	time, t (years)	Simple Interest, SI (M)
Part 1	$x$	$ 2.5\% = \dfrac{2.5}{100} = 0.025 $	$3$	$0.075x$
Part 2	$10000 - x$	$ 1.9\% = \dfrac{1.9}{100} = 0.019 $	$3$	$0.057(10000 - x)$

$ 0.075x + 0.057(10000 - x) = 678 \quad\text{(total interest of M678)} \\[3ex] 0.075x + 570 - 0.057x = 678 \\[3ex] 0.018x = 678 - 570 \\[3ex] 0.018x = 108 \\[3ex] x = \dfrac{108}{0.018} \\[5ex] x = M6000 \\[3ex] $ The amount invested at 2.5% is M6000

(c.) We shall use the Compound Decay Formula modeled as annual depreciation.

$ A = \text{Depreciation Amount} = M83 800.48 \\[3ex] P = \text{Principal} = ? \\[3ex] r = \text{rate} = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] t = \text{time} = 7\text{ years} \\[3ex] m = \text{Number of depreciating periods per year} = 1 \quad\text{depreciating annually} \\[5ex] P = \dfrac{A}{\left(1 - \dfrac{r}{m}\right)^{mt}} \\[7ex] = \dfrac{83800.48}{\left(1 - \dfrac{0.05}{1}\right)^{1 \times 7}} \\[7ex] = 120000.0064 \\[3ex] \approx M120000 \quad\text{(to 3 significant figures)} \\[3ex] $ The value of the car 7 years ago is approximately M120,000

Lesotho General Certificate of Secondary Education: Mathematics: Paper 4

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Examinations Council of Lesotho Lesotho General Certificate of Secondary Education Mathematics Paper 4 (Extended)

Examinations Council of Lesotho
Lesotho General Certificate of Secondary Education
Mathematics
Paper 4 (Extended)