ZIMSEC Examination: Ordinary Level: Mathematics: Paper 2

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ZIMBABWE SCHOOL EXAMINATIONS COUNCIL
General Certificate of Education Ordinary Level
Mathematics Paper 2

4004/2
Time: 2 hours 30 minutes

INSTRUCTIONS TO CANDIDATES.
Write your Name, Centre number and Candidate number in the species provided on the answer booklet.
Answer all questions in Section A and any four questions from Section B.
If you use more than one booklet, fasten them together.
All working must be clearly shown on the same sheet as the rest of the answer.
Omission of essential working will result in loss of marks.
If the degree of accuracy is not specified in the question and if the answer is not exact, the answer should be given correct to three significant figures.
Answers in degrees should be given correct to one decimal place.
Mathematical tables and Non programmable electronic calculators may be used to evaluate explicit numerical expressions.

INSTRUCTIONS FOR CANDIDATES
The number of marks is given in brackets [ ] at the end of each question or part question.

(1.) Set Theory: (a.) The Universal Set, $\mathcal{E}$ has subsets P and Q such that:

$ \mathcal{E} = \{1; 2; 3; 4; 5; ...; 10\} \\[3ex] P = \{2; 3; 5; 7\} \\[3ex] $ Q is a set of factors of 42.

(i.) List all elements of $P \cap Q$
(ii.) Find $n(P \cup Q)'$.

(b.) Number 1

The Venn diagram shows the Universal Set $\mathcal{E}$ and three subsets A, B, and C.
The number of elements in each region of the sets is shown.
Given that $n(A) = 22$.
find,
(i.) the value of $x$
(ii.) $n(A \cap C)$.

$ (a.) \\[3ex] \mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \\[3ex] P = \{2, 3, 5, 7\} \\[3ex] Q = \{1, 2, 3, 6, 7\} \\[5ex] (i.) \\[3ex] P \cap Q = \{2, 3, 7\} \\[5ex] (ii.) \\[3ex] P \cup Q = \{1, 2, 3, 5, 6, 7\} \\[3ex] (P \cup Q)' = \{4, 8, 9, 10\} \\[3ex] n(P \cup Q)' = 4 \\[5ex] (b.) \\[3ex] n(A) = 22 \\[5ex] (i.) \\[3ex] x + 2 + 4 + 6 = 22 \\[3ex] x + 12 = 22 \\[3ex] x = 22 - 12 \\[3ex] x = 10 \\[5ex] (ii.) \\[3ex] n(A \cap C) = 4 + 6 = 10 $

(2.) Expressions and Equations: (a.) Expand $(3 - 2y)^2$
(b.) Factorise $2x^2 - 3x + 1$
(c.) Solve the equation $3(2x - 1) - (2 - x) = 7$
(d.) Solve the simultaneous equations:

$ 3m + n = -5 \\[3ex] m = 1 - 3n \\[3ex] $

$ (a.) \\[3ex] (3 - 2y)^2 \\[3ex] = (3 - 2y)(3 - 2y) \\[3ex] = 9 - 6y - 6y + 4y^2 \\[3ex] = 4y^2 - 12y + 9 \\[5ex] (b.) \\[3ex] 2x^2 - 3x + 1 \\[3ex] .......................................... \\[3ex] 2x^2 \cdot 1 = 2x^2 \\[3ex] \text{Factors are: } -2x \text{ and } -x \\[3ex] .......................................... \\[3ex] 2x^2 - 2x - x + 1 \\[3ex] 2x(x - 1) - 1(x - 1) \\[3ex] (x - 1)(2x - 1) \\[5ex] (c.) \\[3ex] 3(2x - 1) - (2 - x) = 7 \\[3ex] 6x - 3 - 2 + x = 7 \\[3ex] 7x - 5 = 7 \\[3ex] 7x = 7 + 5 \\[3ex] x = \dfrac{12}{7} \\[5ex] $ Check

$x = \dfrac{12}{7}$
LHS	RHS
$ 3(2x - 1) - (2 - x) \\[3ex] ........................................ \\[3ex] 2x - 1 \\[3ex] = 2\left(\dfrac{12}{7}\right) - 1 \\[5ex] = \dfrac{24}{7} - \dfrac{7}{7} \\[5ex] = \dfrac{17}{7} \\[5ex] 3\left(\dfrac{17}{7}\right) \\[5ex] = \dfrac{51}{7} \\[5ex] 2 - x \\[3ex] = 2 - \dfrac{12}{7} \\[5ex] = \dfrac{14}{7} - \dfrac{12}{7} \\[5ex] = \dfrac{2}{7} \\[5ex] ........................................ \\[3ex] = \dfrac{51}{7} - \dfrac{2}{7} \\[5ex] = \dfrac{49}{7} \\[5ex] = 7 $	$7$

$ (d.) \\[3ex] 3m + n = -5 \quad\dots eqn.(1) \\[3ex] m = 1 - 3n \quad\dots eqn.(2) \\[5ex] \underline{\text{Substitution Method}} \\[3ex] \text{Substitute for } m \text{ in } eqn. (1) \\[3ex] 3(1 - 3n) + n = -5 \\[3ex] 3 - 9n + n = -5 \\[3ex] 3 + 5 = 9n - n \\[3ex] 8n = 8 \\[3ex] n = \dfrac{8}{8} \\[5ex] n = 1 \\[5ex] \text{Substitute for } n \text{ in } eqn. (2) \\[3ex] m = 1 - 3(1) \\[3ex] = 1 - 3 \\[3ex] = -2 \\[3ex] $ Check

$m = -2$, $n = 1$
LHS	RHS
$ 3m + n \\[3ex] 3(-2) + 1 \\[3ex] -6 + 1 \\[3ex] -5 $	$-5$
$ m \\[3ex] -2 $	$ 1 - 3n \\[3ex] 1 - 3(1) \\[3ex] -2 $

(3.) Percent Applications: (a.) A vendor bought 120 cabbages at $7 200 for sale.
The vendor sold all the cabbages at $ 45 each because the market was flooded.
Calculate the percentage loss made by the vendor.

Measurements and Units: (b.) The exchange rate between the Zimbabwean dollar and the United States of American dollar on a certain day was ZWL$ 85 to USA$1.
Express USA $170 in Zimbabwean dollars on that day.

Mathematics of Finance: (c.) Calculate the Simple Interest if $ 6000 is invested at 7.5% per annum for a period of 5 years.

(d.) A bank offers loans to farmers at a compound interest rate of 4.5% per annum.
Calculate the total amount to be paid by a farmer who borrows $ 45 000 for a period of 2 years.

$ \underline{\text{Cabbages}} \\[3ex] \text{Number of cabbages} = 120 \\[3ex] \text{cost price} = \$7200 \\[3ex] \text{selling price} = 120 \cdot \$45 = \$5400 \\[5ex] \text{Loss} = 7200 - 5400 = \$1800 \\[3ex] \text{Percentage Loss} = \dfrac{1800}{7200} \cdot 100 = 25\% \\[5ex] (b.) \\[3ex] \text{Unity Fraction Method} \\[3ex] \text{Set up units: } USA\$... \cdot \dfrac{ZWL\$...}{USA\$...} \\[5ex] \text{Set up measurements and units: } USA\$170 \cdot \dfrac{ZWL\$85}{USA\$1} \\[5ex] = ZWL\$14450 \\[5ex] (c.) \\[3ex] $

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ZIMSEC Examination: Ordinary Level: Mathematics: Paper 2

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ZIMBABWE SCHOOL EXAMINATIONS COUNCIL General Certificate of Education Ordinary Level Mathematics Paper 2

ZIMBABWE SCHOOL EXAMINATIONS COUNCIL
General Certificate of Education Ordinary Level
Mathematics Paper 2