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This is written as the remainder (rem.) .

Begin with work in the number range of the tables (to tenth multiple). You will need much concrete work and lots of repetition, because it is very important that the learners understand what they are doing before you go on to larger numbers.

The learners must do research in books and pamphlets about the different traffic signs and discuss them before they complete the signs.

Many pictures and different objects with these shapes are required to ensure that the learners recognise all the shapes.

Make the learners aware of the fact that there is no easy way of folding or dividing for obtaining fifths of 2-D shapes. This must be determined by measuring .

It may be necessary to help the learners to determine the location of the first square that must be coloured in. Do not offer help if they are able to find it independently.

Encourage learners to tell where they live and how they would explain the route to their home to someone else. Help them to explain an easy route to find a certain room in the school.

Leaner section

Content

Activity: divide with a rem [lo 1.7, lo 1.8, lo 3.1]

  • Bonny and Tommy discovered that some numbers couldn’t be divided equally. They are the numbers that are not multiples of the divisors. Sometimes the remainder can be divided into smaller parts, but this is not always possible.
  • Take a good look: Bonny has to divide 13 apples between herself and Tommy. How many will each one get?

13 ÷ 2 = 6½

  • Now look at this:

Tommy wants to divide 13 marbles equally between himself and Jaco. How many marbles will each one get and how many will be left over?

Each one gets 6 and 1 is left over. (Tommy cannot halve the marble.)

  • He thought about it like this:

The nearest multiple of 2 that is less than 13, is 12. He worked with 12 ÷ 2 and knew that 1 would be left over. (Regroup: 12 + 1) The 1 that is left over is known as the remainder . 13 ÷ 2 ¬ 6 rem. 1

Number sentence Nearest multiple Remainder Complete number sentence
1 3 ÷ 27 ÷ 21 1 ÷ 21 5 ÷ 21 9 ÷ 2 1 2 ÷ 2 = 6 1 1 3 ÷ 2 ¬ 6 rem 1
  • Do the same with the multiples of 3, 4, 5 and 10.
Number sentence Nearest multiple Remainder Complete number sentence
13 ÷ 317 ÷ 422 ÷ 526 ÷ 336 ÷ 1038 ÷ 523 ÷ 37 ÷ 49 ÷ 524 ÷ 10
  • Think!

  • Bonny has 67 one-cent pieces and she wants to divide them equally between herself and Tommy . How many will each one get and how many will be left over?

Each one will get 33 one-cent pieces and 1 cent left over.

  • Do the calculations. Use the method that you prefer:
46 ÷ 4 ¬ 68 ÷ 3 ¬
85 ÷ 2 ¬ 59 ÷ 5 ¬
  • Use your own method to solve the problems. Show how you do it.

Your educator has bought 57 pencils. How many learners will each get 5 pencils and how many pencils will be left over?

  • Granny has collected 95 eggs and she wants to provide 3 shops with an equal number of eggs. How many eggs will go to each shop and how many will be left over?
  • Dad has R87 and wants to buy pens at R4 each. How many pens can he buy and how much money will be left over?
  • Fill the answers in as quickly as possible:

  • Complete the work on the wheels of the bus:
  • Follow the number route to find out how far the bus has travelled. Begin at the triangle and end at the rectangle.
  • How many road signs do you see on the way to school? There are signs that: give warnings , give commands and provide information. Find out what these different road signs look like.
  • Use these circles, rectangles and triangles to draw your own road signs.

Assessment

Learning Outcome 1: The learner will be able to recognise, describe and represent numbers and their relationships, and to count, estimate, calculate and check with competence and confidence in solving problems.

Assessment Standard 1.7: We know this when the learner solves and explains solutions to practical problems that involve equal sharing and grouping and that lead to solutions that also include unitary and nonunitary fractions (e.g. ¼, ¾);

Assessment Standard 1.8: We know this when the learner can perform calculations, using appropriate symbols, to solve problems;

Learning Outcome 3: The learner will be able to describe and represent characteristics and relationships between two-dimensional shapes and three-dimensional objects in a variety of orientations and positions.

Assessment Standard 3.1: We know this when the learner recognises, identifies and names two-dimensional shapes and three-dimensional objects in the environment and in pictures.

Questions & Answers

a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
how did I we'll learn this
Noor Reply
f(x)= 2|x+5| find f(-6)
Prince Reply
f(n)= 2n + 1
Samantha Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Mathematics grade 3. OpenStax CNX. Oct 14, 2009 Download for free at http://cnx.org/content/col11128/1.1
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