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Examen

BNU1501 EXAM PACK 2023

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Subido en
27 de mayo de 2023
Número de páginas
266
Escrito en
2022/2023
Tipo
Examen
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BNU1501 EXAM PACK 2023


UPDATED QUESTIONS
WITH SOLUTIONS

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UNIVERSITY EXAMINATIONS


mm
mm


MAY/JUNE 2020


BNU1501
BASIC NUMERACY
Duration: 2 Hours 30 Minutes 100 Marks
EXAMINERS
FIRST DR MT MASETSHABA
SECOND MS BS NCUBE
EXTERNAL DR HARRY WIGGINS (University of Pretoria)

This paper consists of 10 pages, which include a list of formulas on page 10.
Programmable calculator permissible.
Instructions:
The paper consists of 25 questions for a total of 100 marks. Answer all the questions.
Only one option indicated as [1], [2], [3] or [4] per question is correct.
Marks will not be deducted for incorrect answers.
The answers to the examination MCQ may only be submitted online.

Submit the answers on my Unisa by

1. On the landing page for Unisa, before login, go back to the link where you downloaded your examination
my
paper: Login and download my Exam Question Paper for May/June 2020.
2. Login using your student number and my Unisa password.
3. On the next screen, find the module code BNU1501. Click on the link to “submit MCQ”. This link will only
display if the examination session is still open for submissions.
4. A new screen will open that will guide you through the steps to upload your answers.
5. Enter the total number of questions (25) for the paper in the Number of Questions field.
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CONFIDENTIAL BNU1501
Page 2 of 10 May/June 2020

Question 1
The expres
35 ÷ 5 + 2 − ((3 × 4 − 2) − (18 − 3 × 4))
can be rewr

[1] 9 − 6 + 42 .
[2] 9 − 10 + 6 .
[3] 5 − 6 + 42 .
[4] 5 − 10 + 6 .


Question 2
Subtracting 2x + x3 and x+7 from −8 − 5x3 + 4x gives

[1] −15 − 6x3 + x .
[2] 3x + x3 + 7 .
[3] −x + 6x3 + 15 .
[4] 7x − 4x3 + 15 .


Question 3
The option
√ √3
2 25 − 54 + 9 + 42

is
√ √3
[1] 23 2 − 3 2 + 5 .

[2] 23 − 3 2 + 5 .
√ √
[3] 8 2 − 33 2 × 3 + 3 + 4 .

[4] 5 2+7 .


Question 4
The expres
a3 × b5
b9 × a 5
is
1
a × b2
[1] .

[2] a−1 b−2 .

[3] ab2 .
−1
[4] ab2 .


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Question 7
. x = −2 [4]
. [3]
8
x=
11
. [2]
8
x=−
1
. [1]
5
x=−
1
gives
2
−1(x − 3) + 2x = −3x +
5
in the equation
x Solving for t
Question 6
. [4]
2
(x + 3)
1
. [3]
2
x+3+ x
1
. [2]
2
x+3
1
. [1]
2
x+3×
1
is
a number increased by 3 and then halved
The correct
Question 5
May/June 2020 Page 3 of 10
BNU1501 CONFIDENTIAL
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