Exam (elaborations)
MIP1501 Assignment 2 2023
- Course
- MIP1501 (MIP1501)
- Institution
- University Of South Africa
MIP1501 Assignment 2 2023 ()
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MIP1501Assignment 2
2023
(689656)
[School]
[Course title]
,MIP1501 Assignment 2 2023
Question 1
1.1 Which Property is illustrated by 187 + 376 = 376 + 187? Name and explain
this property in your own words. (4)
The property illustrated by 187 + 376 = 376 + 187 is the commutative property of
addition. This property states that when adding two numbers, the order of the
numbers does not affect the result. In other words, you can add the numbers in any
order and still get the same sum. For example, if you have 2 + 3, it will give you
the same result as 3 + 2. This property holds true for any numbers, including larger
ones like 187 and 376.
1.2 Explain with examples how you can teach the Intermediate Phase
mathematics learners, the difference between place value and value of a
digit. (6)
To teach the difference between place value and value of a digit to Intermediate
Phase mathematics learners, you can use the following examples:
Example 1:
Consider the number 346. The place value of the digit '4' in this number is in the
tens place because it is the second digit from the right. The value of the digit '4' is
40 because it represents 4 tens.
Example 2:
Take the number 527. The place value of the digit '7' is in the ones place because it
is the rightmost digit. The value of the digit '7' is 7 because it represents 7 ones.
In these examples, emphasize that the place value depends on the position of the
digit within the number, while the value of the digit represents how much it
contributes to the overall value of the number.
, 1.3 For her homework, Thandeka calculated and got as the answer.
1.3.1 Identify and write the misconception Thandeka has. (3)
The misconception Thandeka has is that when adding fractions with unlike
denominators, she simply added the numerators together and kept the same
denominator.
1.3.2 Assist Thandeka by explaining what to do in order to add fractions with
unlike denominators. (3)
To assist Thandeka in adding fractions with unlike denominators, explain the
following steps:
Step 1: Find the least common multiple (LCM) of the denominators.
Step 2: Rewrite each fraction so that they have the same denominator, which is the
LCM found in step 1.
Step 3: Add the numerators of the fractions together.
Step 4: Keep the common denominator found in step 1.
Step 5: Simplify the resulting fraction, if possible.
1.3.3 Show how she should have calculated, show all the steps. (4)
Let's say Thandeka's fractions were 1/3 and 2/5. Here are the correct steps to add
them:
Step 1: The LCM of 3 and 5 is 15.
Step 2: Rewrite 1/3 as 5/15 (multiplying the numerator and denominator by 5) and
2/5 as 6/15 (multiplying the numerator and denominator by 3).
Step 3: Add the numerators: 5/15 + 6/15 = 11/15.
Step 4: Keep the common denominator of 15.
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