Use this ZIMSEC O-Level Mathematics Paper 2 November 2023 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
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Questions: ZIMSEC O-Level Mathematics Paper 2 November 2023 Question Paper
This is the ZIMSEC O-Level Mathematics marking scheme for Paper 2 from November 2023. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.
o(b)
\[
\begin{aligned} | \frac{(150-1,20)}{1,20} \times 100
= | \frac{0,30}{1,20} \times 100
= | \xrightarrow{25 \%}
\end{aligned}
\]
(a) (in)
\[
\text { 娘 } \begin{aligned} | \frac{100}{120} \times 264 \mathrm{~g}
= | 5 / 6 \times 260 \mathrm{~g}
= | 220 \mathrm{~g}
\end{aligned}
\]
1(a)(ii)
\[
\text { i) } \begin{aligned} | \left(\frac{100-7}{7}\right) \times 91
= | \frac{93}{7} \times 91
= | $ 1209
\text { or } | \left(\frac{100}{7} \times 91\right)-91
= | 1300-91
= | $ 1209
\end{aligned}
\]
1(b) A triangle $A B C$ is formed which is a natt angled △ since it lies within a Semi-Circle $\therefore A B$ becomes hypotunese
\[
\begin{aligned}
\therefore | A B=A C^{2}+B C^{2} | A B \text { From } z^{2}=x^{2}+y^{2}
\therefore | A B=\sqrt{7 x^{2}+5^{2}}
\end{aligned}
\]
\[
\begin{array}{l}
A B=(\sqrt{81,28}) \mathrm{cm}
A B=9,013078189 \mathrm{~cm}
A B=9,02 \mathrm{~cm} 2 \mathrm{~d} . \mathrm{pX}
A B=9,013 \mathrm{sg} .
\end{array}
\]
2) docus Sketch diagram
(b) Lous of Points $x \mathrm{~cm}$ from and above line, $A B$
- Actual diagram on Page 2 .
(2)
\[
q_{0}^{k}=\{30,31,32,33,34,35,36,37,38,39,9,0\}
\]
$130 \leq x \leq 40$ means $30 \alpha 40$ are
Inclusive
NB
\[
\begin{array}{l}
P=\{31,37\}
Q=\{32,36,40\}
\therefore 3(0 i) n(P)=2 .
\end{array}
\]
(ii),
In the above case its
Qn $p^{\prime}$ is ◯ $\xi=\{32,36,4$
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