Use this ZIMSEC O-Level Mathematics Paper 2 November 2022 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
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Questions: ZIMSEC O-Level Mathematics Paper 2 November 2022 Question Paper
This is the ZIMSEC O-Level Mathematics marking scheme for Paper 2 from November 2022. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.
1.
a) $A \hat{C} B=10^{\circ}$
b) $B \widehat{D} C=80^{\circ}$
c) to get $C \hat{P} D$ first obtain $A \hat{C} B$ :
$A \hat{C} T=90^{\circ}$ therefore $A \hat{C} B=10^{\circ}$
\[
\begin{aligned}
C \hat{P} D | =32^{\circ}+10^{\circ} | =42^{\circ}
\end{aligned}
\]
d) $A \hat{O} D=2\left(90^{\circ}-32^{\circ}\right)$
\[
=116^{\circ}
\]
MARKS AWARDED a)1 b)1 c)2 d)3 : total marks 7
2.
a)
\[
\begin{aligned}
(m-n)(4 m+2 n)-(m-n)^{2} | =(m-n)[4 m+2 n-(m-n)]
= | (m-n)[4 m+2 n-m+n]
= | (m-n)(3 m+3 n)
= | 3(m-n)(m+n)
\end{aligned}
\]
b) The mass of the removed player is $891 \mathrm{~kg}-801 \mathrm{~kg}=90 \mathrm{~kg}$
The total mass of 8 players is $11 \times 81 \mathrm{~kg}=891 \mathrm{~kg}$
The total mass of 10 players after one player is removed is $80.1 \mathrm{~kg} \times 10=801 \mathrm{~kg}$
c) Maria has $100 \%$ before spending.
She spends $40 \%$ and she remained with the balance of 60\%
She again spends $30 \%$ of $60 \%=0.3 \times 0.6=18 \%$
She is left with $100 \%-40 \%-18 \%=42 \%$
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