Use this ZIMSEC O-Level Mathematics Paper 1 November 2023 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
A free account is needed to download. Sign in or create one
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.
(i)(a) 4,70
(b) 800
(c) $2,31 \times 10^{-3}$
2(a)
\[
\begin{array}{l}
=3584+14
=3598
\end{array}
\]
8 | 3598
8 | $449 r 6$
8 | $56 r$ | 1
8 | 7 | $r$ | 0 | 0 | $r$ | 7
\[
=7016_{8}
\]
(b)
6 | 57
6 | $9 r 3$
6 | $1 r 3$
$=$ | $133_{6}$
(c) $\mathrm{lH}_{2}$
\[
\begin{array}{l}
\left.\frac{-1}{1000}\right|_{2}
=1000_{2}
\end{array}
\]
3(a) 02:17am
b(i) 2
(ii) 2 .
Ca By Sir Eddie 0787359604
Mathematics paper i Marking Guide N2023
$4 f(x)=3 x-1$
(a)
\[
\begin{aligned}
f(-2) | =3(-2)-1 . | =-6-1 | =-7
\end{aligned}
\]
(b) $x$ if $f(x)=5^{\circ}$
\[
\begin{array}{l}
3 x-1=5
\frac{3 x}{3}=\frac{8}{3}
\therefore x=2
\therefore x
\end{array}
\]
(a) $(-3 m)^{4}$
\[
=81 \mathrm{~m}^{4}
\]
$(b x)$
\[
\begin{aligned} | 49^{3 / 2}
= | \sqrt[2]{(49)^{3}}
= | 7^{3}
= | 343
\end{aligned}
\]
(ii) $2^{-3}=\frac{1}{8}$
(6)
6) $\binom{3 x+2 y=8}{5 x+3 y=4,5}^{3} 2$
$9 x+6 y=24$
$-10 x+6 y=9$
$-x=15$
\[
\begin{array}{l}
x=-15
5(-15)+3 y=4,5
-75+3 y=4,5
3 y=79,5
y=26,5
\end{array}
\]
\[
\therefore y=26,5 ; x=-15
\]
\[
\begin{array}{l}
\text { Proof }
3(-15)+2(26,5)=8
8=8
5(-15)+3(26,5)=9
4,5=4,5
\end{array}
\]