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MIP1501 Assignment 3 (70%) $6.02   Add to cart
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MIP1501 Assignment 3 (70%)
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  • Course
  • Bed Intermediate Phase (MIP1501)
  • Institution
  • University Of South Africa (Unisa)

I achieved 70% for this module. This assignment will be well worth your money! Well compiled and excellently executed. Do yourself a favour, save your time and spare yourself the stress.

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Document information

  • April 30, 2021
  • 10
  • 2020/2021
  • Other
  • Unknown

Subjects

  • mip
  • mip
  • mip
  • mip1501
  • mip1501
  • mip1501
  • 1501

Written for

  • University of South Africa (Unisa)
  • Bed Intermediate Phase (MIP1501)
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ChristinaAvgoustis

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Table of Contents
Question 1..................................................................................................................................2
Question 2..................................................................................................................................5
Question 3..................................................................................................................................6
Bibliography..............................................................................................................................10


Assignment: 03
Unique number: 790368
Due Date: 06 July 2020
Student number: 65111133
Name: Christina
Avgoustis

, MIP1501 Assignment 03


Question 1
1.1.
1.1.1. [Grade 4] Write down the next four numbers in the following sequence: 90; 93;96…

Solution: 93-90=3 (common difference)
∴96+3=99
99+3=102
102+3=105
105+3=108
∴90; 03; 96;99;102;105;108

1.1.2. [Grade 4] Tom bought 7 soccer balls at R40 each. How much did he spend in total on
soccer balls?

Solution: 7 soccer balls
R40 each
∴ 7x40= 280
∴Tom spent R280 in total on the soccer balls.

2
1.1.3. [Grade 6] There were 20 chocolates in the cupboard. Jack and his sister ate
5
of them. How many chocolates were left?

2
Solution: of 20
5
20÷ 5=4
1
∴ of 20 is 4
5
2
of 20 is (4x2)
5
2
∴ of 20 is 8
5
20-8=12
∴They ate 8 chocolates and so they have 12 chocolates left.

1.1.4. [Grade 5] Jane divided a certain number by 17 and found that the answer was 246
with a remainder of 8. What is the number?

α
Solution: =246 r 8
17
α =( 246 × 17 ) +8
a=4182+8

BED 90103 Page 2 of 10

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