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First page of ZIMSEC O-Level Mathematics Paper 2 2021 Marking Scheme

ZIMSEC O-Level Mathematics Paper 2 2021 Marking Scheme

Use this ZIMSEC O-Level Mathematics Paper 2 2021 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.

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Exam board
ZIMSEC
Subject
Mathematics
Level
O-Level
Document
Marking Scheme
Paper
Paper 2
Year
2021
File
PDF, 1.5 MB

About this marking scheme

This is the ZIMSEC O-Level Mathematics marking scheme for Paper 2 from 2021. It shows how marks are awarded, so you can mark your own answers.

More Mathematics O-Level 2021 papers

Other years: Mathematics O-Level

Text from the first page

1

 \multirow[b]{2}{*}{(b)(i)}  |  \multirow{2}{*}{
$(\hat{A D E}=)(\hat{A C E}=\prime \hat{A B E})$
B|B|
}  |  & Mark Additional guidance
  |  &  |  (2) ACE and ABE condome micsing BI and Bobun capú $\mathrm{B}_{2} \mathrm{~B}_{4}$ сесь еряя
  |  $\widehat{B E C}=37^{\circ}$  |  withent $\Phi$  |  $(B E C=) \begin{array}{l}90-53 \text { or equiv } M /  180-(90+53)\end{array}$
 (ii)  |  &  |
BH
Anw; no penalty $37^{\prime}$ Al

  |  $(B \hat{A} E=) 127^{\circ}$  |  [H  |  equiv (GAE) $=127 \quad \mathrm{BI} 127^{\circ}$ BI
 (iii)  |  &  |  ** Anw: no penalty
  |  $(D \hat{O} C=) 84^{\circ}$  |  [2]  |  2 (their $\left(90^{\circ}-48^{\circ}\right)$ ) or MI equiv M1 $84^{\circ}$ AI

Anw: No penally

2
(a)(i)

Mark Addmional guidance
(a)(ii)
\[
(V=) 165 \quad \begin{array}{l}
\text { Marking Guidance }
7
{ }_{165}^{3} \times \frac{22}{7} \times 9 \times \frac{3^{2}}{2} / 2 \frac{5}{2}
\text { Anw: } 1
\end{array}
\]

M1
Al
A1
\[
r=\sqrt{\frac{3 V}{7 h \pi}}
\]
\[
\begin{array}{l}
r^{2}=\frac{3 V}{7 h \pi} \text { or equiv }
r=\sqrt{\frac{3 V}{7 h \pi}} \quad \text { Al }
\end{array}
\]

MI

M1
Al

Anw: no penalty
(b)(i)
\[
(A B)=)\binom{7}{-6}
\]
[2] B 2 B 2
-1 each error missing brackets a use of
(b)(ii)

B3 -1 each error -1 each error
SC : After ${ }^{\beta} 0$ award SCB1 for $\frac{1}{\text { their } \operatorname{det}}\left(\begin{array}{ll}3  |  2  6  |  5\end{array}\right)$ or $\begin{array}{c}\text { shds }  \text { met } \bar{e}  \text { correct }\end{array}$ in - not be
\[
\left(A^{-1}\right)=\frac{1}{3}\left(\begin{array}{ll}
3  |  2
6  |  5
\end{array}\right) \text { B3 } \frac{1}{3}\left(\begin{array}{ll}
a  |  b
c  |  d
\end{array}\right) \text { where } a, b, c \text { and } d \text { are }
\]
elements of $A$ even with all signs changed.
Anw: no penalty

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