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U n t er r i ch t spl a n
De c imal s - Ad d it io n
Altersgruppe: 4 t h Gr ade , 5 t h Gr ade
Virginia - Mathematics Standards of Learning (2009): 3 .9c , 4 .3 c ,
4 .5 d
Virginia - Mathematics Standards of Learning (2016): 4 .3 .d, 4 .6.a,
4 .6.b
Fairfax County Public Schools Program of Studies: 3 .9.c .1, 4 .3 .c .1,
4 .3 .c .2, 4 .5 .d.1, 4 .5 .d.2
Online-Ressourcen: W e i ghi ng M at t e r s
Opening
T eacher
present s
St udent s
pract ice
Class
discussion
10
10
12
13
min
min
min
min
M at h Obj e c t i v e s
E x pe r i e nc e a visual model of tenths and hundredths
P r ac t i c e using a balance scale
L e ar n to add decimals
De v e l o p a better understanding of place value
Ope ni ng | 10 min
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Display the following scale:
A sk: Let’s suppose we put a 1 kg weight on the left side of this
scale. How many 0.1 kg weights would we need on the right for the
scale to be balanced? How do you know?
We would need to place ten 0.1 kg weights on the right side of
the scale to balance the 1 kg weight on the left. We use a base 10
number system, which means we need 10 of any unit to make the
next larger unit: 10 tens equal 100, 10 ones equal 10, and 10
tenths equal 1 whole.
A sk: Instead, let’s place a 0.1 kg weight on the left side of the
scale. How many 0.01 kg weights would we need on the right for the
scale to be balanced? How do you know?
Again, we would need ten 0.01 kg weights on the right side to
balance the 0.1 kg weight on the left side.
A sk: Suppose we had two 1 kg weights on the left side of the
scale. How many 0.1 kg weights would we need on the right for the
scale to be balanced? How do you know?
If we need ten 0.1 kg weights to balance the1 kg, we would need
twenty 0.1 kg weights for the scale to balance two 1 kg weights.
A sk: Instead of using 0.1 kg weights, let’s use 0.01 kg weights. Now
how many would we need to balance the two 1 kg weights on the
left side of the scale?
We would need two hundred 0.01 kg weights for the scale to be
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balanced: ten 0.01 kg weights equal 0.1 kg weight and ten 0.1 kg
weights equal 1 kg weight, so one hundred 0.01 kg weights equal
1 kg weight and two hundred 0.01 kg weights equal 2 kg.
A sk: Suppose we placed twelve 0.1 kg weights on the left side of
the scale. How could we balance the scale using less than 12
weights?
Twelve 0.1 kg weights equal 1.2 kg. In order to balance the scale
we could place one 1 kg weight and two 0.1 kg weights on the
right side. This would only be three weights, and the scale would
be balanced.
S ay: Write the following numbers in e x pande d no t at i o n :
1. 2.3 =
2. 1.18 =
3. 4.01 =
Have the students work in their notebooks. When they have finished working
they should share their work with one another.
1. 2.3 = (2×1) + (3×0.1)
S ay: To make 2.3, we need 2 ones and 3 tenths.
2. 1.18 = (1×1) + (1×0.1) + (8×0.01)
S ay: To make 1.18, we need 1 one, 1 tenth and 8 hundredths.
3. 4.01 = (4×1) + (0×0.1 ) + (1×0.01)
S ay: To make 4.01, we need 4 ones, 0 tenths and 1 hundredth.
S ay: When we look at expanded notation, we are looking at place
value. Here, we are looking at the o ne s place, t e nt hs place and
the hundr e dt hs place. Remind the students that we use a base 10
number system. Place values increase 10 times as we move to the
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left. That’s why we need 10 hundredths to make 1 tenth (the tenths
place is to the left of the hundredths place), we need 10 tenths to
make 1 whole (the ones place is to the left of the tenths place), just
as we need 10 ones to make a ten and 10 tens to make 1 hundred.
A sk: When we place zeros at the right end of a number, after the
decimal point, do we change its value? Why or why not?
Zeros to the right of a decimal point that are at the end of the
number do not change its value. For example, 0.8 is the same as
0.80. The number 0.80 says that there are 0 hundredths: we haven’t
added any hundredths to the number 0.8, so they are equal. We can
compare this with 0.81, where we have added 1 hundredth to 0.8.
Zeros that come in the middle of a number after the decimal
point do change its value. The number 0.8 is not equal to 0.08.
The first, 0.8, is equal to 8 t e nt hs with no whole numbers. The
second, 0.08, is equal is equal to 8 hundr e dt hs with no whole
numbers.
T e ac he r pr e se nt s M at h game : W e i ghi ng M at t e r s - M o de l
w i t h De c i mal s | 10 min
Pesent Matific ’s episode W e ig h in g M a t t e r s - M o d e l w it h De c im a ls
to the class, using the projector.
This episode practices problem solving in the context of decimals. Find the
weight of a cubic cm of a certain material using a balance scale and weights
of 1gr, 0.1gr and 0.01gr.
E x a m p le :
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S ay: Please read the instructions at the bottom of the screen.
Students can read the instructions.
S ay: We are being asked how much a cubic cm of chalk weighs. In
order to answer this question we need to place the chalk on one
side of the scale and the weights on the other side until the scale is
balanced.
Place the chalk on one side of the scale.
A sk : What weight should we place first on the left side – the 1gr,
0.1gr or 0.01gr?
We should first place the 1gr weight in order to have an
estimation whether the chalk weighs more or less than 1gr. If we
place the 0.1gr or the 0.01 gr first, we might need to add several
more weights later. The fewer weights we use, the easier the
problem becomes.
Place the 1gr weight on the other side of the scale.
A sk : What can we tell about the weight of the chalk? How do we
know?
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We can see that the chalk weighs more than 1 gr since the scale
is still tilted to the side of the chalk.
Place another 1gr weight on the right side of the scale.
A sk : What can we tell about the weight of the chalk now? How do
we know?
We can see that the chalk weighs more than 2 gr since the scale
is still tilted to the side of the chalk.
Place another 1g weight on the right side of the scale.
A sk : What can we tell about the weight of the chalk now? How do
we know?
We can see that the chalk weighs less than 3 gr since the scale is
now tilted to the side of the weights. So we should replace one
of the 1 gr weights with a lighter weight.
Remove the 1gr weight and place 0.1 gr weight on the same side of the scale.
A sk : What can we tell about the weight of the chalk now? How do
we know?
We can see that the chalk weighs more than 2.1 gr since the scale
is tilted to the side of the chalk.
Repeat this until there are 2 weights of 1 gr and 3 weights of 0.1 gr, and the
scales are balanced.
A sk : What can we tell about the weight of the chalk now? How do
we know?
We can see that the chalk weighs exactly 2.3 gr since the scale is
balanced and there are 2 weights of 1 gr and 3 weights of 0.1gr on
the right side of the scale.
Explain the connection to the expanded notation: 2.3 = (2×1) + (3×0.1)
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E x a m p le :
S t ude nt s pr ac t i c e M at h game : W e i ghi ng M at t e r s - M o de l
w i t h De c i mal s | 12 min
Have students play W e ig h in g M a t t e r s - M o d e l w it h De c im a ls on
their personal devices.
Circulate, answering questions as necessary.
C l ass di sc ussi o n | 13 min
Discuss any problems the students faced while working individually.
Ask the class to tell how they dealt with any issues their classmates brought
up.
Display the following equations:
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1. (3×1)+(4×0.1)+(5×0.01)=
2. (0×1)+(6×0.1)+(0×0.01)=
3. (2×1)+(1×0.01)=
Have the students solve the problems in their notebooks. When they have
finished working, they should share their work with one another.
1. (3×1)+(4×0.1)+(5×0.01)= 3.45
S ay: 3 ones plus 4 tenths plus 5 hundredths equals 3.45 .
2. (0×1)+(6×0.1)+(0×0.01)= 0.6
S ay: 6 tenths is 0.6.
3. (2×1)+(1×0.01)=2.01
S ay: 2 ones plus 1 hundredths equals 2.01.
Display the following equation:
0.85 + 0.35 = 1.2
A sk: Why doesn't the answer have any digits in the hundredths
place?
When we add 0.85 and 0.35, we get 1.20. The digit in the
hundredths place is a zero. There are no hundredths, so we can
simply write 1.2. The numbers 1.2 and 1.20 are equal.
S ay: Using what we have learned, let’s try to find a method for
adding decimals. What steps are involved in adding decimals?
Some coaching may be needed. With prompting, the students should be able
to come up with these rules:
We add each unit to itself: the hundredths to the hundredths, the tenths to
the tenths, etc. We should start with the smallest unit, similar to addition
with whole numbers, because we might have regrouping of the smaller units
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into a larger unit. Similar to addition with whole numbers, we can use the
vertical method, as follows:
1. Write the numbers vertically, lining up the decimal points in each of the
addends. This way we know that the tenths are in line with the tenths, the
hundredths are in line with the hundredths, etc.
2. Place zeros at the end of each addend so that the addends all contain the
same number of decimal places. We know that adding zeros at the end
doesn’t change the value of the number.
3. Place the decimal point in the answer directly under the decimal points in
the problem.
4. Add normally.
Write the steps on the board as the students develop them.
S ay: Let’s look at a problem. What mistake can you find in this
solution?
Display the following:
4.27 + 0.56 =
A sk: What is the mistake?
The mistake is that the decimal points are not lined up, so that
we are adding the digit that represents tenths (2) to the digit that
represents hundredths (6), and the digit that represents ones (4)
to the digit that represents tenths (5). The right way is to add
ones to ones, tenths to tenths, hundredths to hundredths and so
on. If we estimate, we can see that the answer should be less than
5. Here, the answer is more than 9.
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Show the correct way to solve this problem:
Display the following problems. Have students complete them in their
notebooks.
1. 1.31 + 4.22 =
2. 4.77 + 2.1 =
3. 0.03 + 8.6 =
4. 1.4 + 7.8 =
5. 8.01 + 0.90 =
6. 0.01 + 0.1 + 1 =
When the students have finished working, review the solutions.
Discuss any questions they may have.
1. 1.31 + 4.22 = 5.53
2. 4.77 + 2.1 = 6.87
3. 0.03 + 8.6 = 8.63
4. 1.4 + 7.8 = 9.2
5. 8.01 + 0.90 = 8.91
6. 0.01 + 0.1 + 1 = 1.11
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