Use this ZIMSEC O-Level Mathematics Paper 1 November 2023 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
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Questions: ZIMSEC O-Level Mathematics Paper 1 November 2023 Question Paper
This is the ZIMSEC O-Level Mathematics marking scheme for Paper 1 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.
1(a) 4,695 to $2 d_{i} p=4,70$.
1(b) 794,3 to $5 f=800$.
1(c) 0,00231 In standard form
\[
=2,31 \times 10^{-3}
\]
2(b) 5710 to base 6
\[
\begin{array}{l}
=6 \mid 57
6 \mid 9 r 3
6 \mid 1 r 3
0 r 1
\end{array}
\]
2 (d) $7 \times 8^{3}+846$ as number
\[
\begin{array}{l}
\text { In base } 8
=\quad 8^{3} \quad 8^{2} \quad 8^{1} \quad 8^{0}
=\quad 70168
\end{array}
\]
2(c)
\[
\begin{array}{r}
\begin{array}{r}
12
+\quad 12
=1000_{2}
\end{array}
\end{array}
\]
3(a) 0217 as a time in 12how notation $=2.17 \mathrm{am}$
$3(b)=0$ Paralielogram number
$3(b)=2$ Partlines of symmers of
4) $f(x)=3 x-1$ ratainal format
(a) $f(-2)=3(-2)-1=-7$
\[
\left.\begin{array}{l}
5=3 x-1
5+1=3 x
\frac{3 x}{3}=\frac{6}{3}
\end{array}\right\} \begin{array}{r}
3 x / 3=6 / 3
x=2
\end{array}
\]
\[
\begin{array}{l}
\text { 5(a) Simpliny }(-3 m)^{4}
=(-3 m)(-3 m)(-3 m)(-3 m)
=81 m^{4}
5(b)(c) 4 a^{3 / 2}=(\sqrt[2]{4 a})^{3}=(7)^{3}
=7 \times 7 \times 7=\frac{343}{1 / 2^{3}}=1 / 2 \times 2 \times 2
5(b)(i) 2^{-3}=1
\end{array}
\]
6) Solve the Simulterious equation
\[
\begin{array}{l}
3 x+2 y=8
5 x+3 y=4,5
\end{array}
\]
\[
\begin{array}{r}
\times 5) 15 x+10 y=40
\times 3) 15 x+9 y=13,5
-\frac{15 x+10 y}{15 x+0 y}=60
y=26,5
\end{array}
\]
\[
\begin{array}{l}
3 x+2(26,5)=8
3 x+53=8
3 x=8-53
\frac{3 x}{3}=-\frac{45}{3}
x=-15
\therefore x=-15 ; y=26,5
7 \text { (a) AOT }=90^{\circ}-(74 / 2)
\text { AOT }=90^{\circ}-37^{\circ}
\text { AOT }=53^{\circ}
\end{array}
\]
7 b) $\frac{B A T}{B A T}=\left(\frac{180-74}{2}\right)=\frac{106}{2}$
\[
\hat{B A T}=53^{\circ}
\]
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