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\( \definecolor{colordef}{RGB}{249,49,84} \definecolor{colorprop}{RGB}{18,102,241} \)
International Baccalaureate · PYP 1

Adding and Subtracting to 100

⌚ ~98 min ▢ 10 blocks ✓ 54 exercises ➣ Prerequisites : Numbers to 100, Addition Within 20, Subtraction Within 20, Number Bonds to 10 and Doubles
I Adding and Subtracting 10 or Whole Tens

Léa has \(47\) stickers. She wins \(10\) more, then \(30\) more. How many does she have? Tom says: “No need to count one by one: only the tens change.”

\(47\) is \(4\) tens and \(7\) ones. Adding \(10\) adds one ten: \(5\) tens and \(7\) ones, \(57\). Adding \(30\) adds three tens: \(8\) tens and \(7\) ones, \(87\). The ones never moved.


Method — Adding or Subtracting Whole Tens
  • \(+ 10\) or \(- 10\): one ten more or one ten less; the ones digit does not change. \(37 - 10\): one ten taken from the three tens, \(27\).
  • \(+ 20\), \(30\), … \(90\) or \(- 20\), … \(90\): count in tens. \(76 - 30\): \(7\) tens minus \(3\) tens make \(4\) tens, so \(46\). \(25 + 60\): \(2\) tens plus \(6\) tens make \(8\) tens, so \(85\).
We say the tens, not the numbers: “seven tens minus three tens”.
Example
Compute: (a) \(63 + 10\), (b) \(91 - 10\), (c) \(34 + 50\), (d) \(98 - 70\).

(a) One ten more: \(73\). (b) One ten less: \(81\). (c) \(3\) tens \(+ 5\) tens \(= 8\) tens: \(84\). (d) \(9\) tens \(- 7\) tens \(= 2\) tens: \(28\).

About How Much Is a Sum?
Before computing a sum, we can say about how much it is: each number is replaced by a whole ten close to it, and the tens are added.
Figure
  • \(\textcolor{colordef}{28}\) is close to \(30\) and \(\textcolor{colorprop}{31}\) is close to \(30\). So \(\textcolor{colordef}{28}+\textcolor{colorprop}{31}\) is about \(30 + 30 = 60\). The exact sum, \(59\), is close to \(60\).
  • Is \(38 + 21\) more or less than \(50\)? The tens alone already make \(30 + 20 = 50\), and the ones \(8\) and \(1\) add more. So \(38 + 21\) is more than \(50\): it is about \(40 + 20 = 60\) (the exact sum is \(59\)).
The whole tens give a number close to the sum: to know the sum exactly, we still compute.
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Skills to practice
  • Adding or Subtracting 10
  • Adding or Subtracting Whole Tens
  • Estimating a Sum
II The Complement to the Next Ten

Nour has \(74\) marbles. The box can hold \(80\). How many marbles are missing to fill it? Nour says: “I know that \(4 + 6 = 10\).”

\(74\) is \(7\) tens and \(4\) ones. To reach the next ten, \(80\), we need to complete the \(4\) ones into a full ten: the complement of \(4\) to \(10\) is \(6\). So \(6\) marbles are missing: \(74 + 6 = 80\).


Method — The Complement to the Next Ten
To find what is missing from a number to the next ten:
  1. Look at the ones digit: in \(74\), it is \(4\).
  2. Take its complement to 10, which we know by heart: \(4 + 6 = 10\).
  3. That complement is the answer: \(74 + 6 = 80\).
The complement to \(10\) we already know now works for every number: it takes \(6\) to go from \(4\) to \(10\), from \(34\) to \(40\), from \(74\) to \(80\).
Example
How much to reach the next ten? (a) from \(58\), (b) from \(21\), (c) from \(95\).

(a) The ones digit is \(8\), \(8 + 2 = 10\): it takes \(2\), \(58 + 2 = 60\). (b) \(1 + 9 = 10\): it takes \(9\), \(21 + 9 = 30\). (c) \(5 + 5 = 10\): it takes \(5\), \(95 + 5 = 100\).

Skills to practice
  • Reaching the Next Ten
III Adding a Small Number Through the Ten

Tom has \(47\) marbles and wins \(8\). He says: “\(47\) and \(8\)… I go to \(50\) first.” Why \(50\)?

\(50\) is the next ten after \(47\). It takes \(3\) to get there (\(7 + 3 = 10\)). Tom splits the \(8\) into \(3 + 5\): \(47 + 3 = 50\), then \(50 + 5 = 55\). Going through the ten is the same trick as going through \(10\) with small numbers (\(7 + 8 = 7 + 3 + 5\)), now on bigger numbers.
Figure


Method — Adding a Number Smaller than 10
  • Without crossing a ten: when the ones do not go past \(9\), only the ones digit changes. \(32 + 4 = 36\), because \(2 + 4 = 6\).
  • Crossing a ten: when the ones would go past \(9\), go through the next ten: first the complement to the ten, then what is left. \(47 + 8\): \(3\) to reach \(50\), then \(5\) more: \(55\).
How to tell which case: \(2 + 4\) stays under \(10\), but \(7 + 8\) goes past \(10\).
Example
Compute: (a) \(63 + 5\), (b) \(63 + 9\), (c) \(86 + 7\).

(a) \(3 + 5 = 8\), no ten crossed: \(68\). (b) \(3 + 9\) goes past \(10\): \(63 + 7 = 70\), then \(+ 2\): \(72\). (c) \(86 + 4 = 90\), then \(+ 3\): \(93\).

Skills to practice
  • Adding Without Crossing a Ten
  • Adding by Crossing a Ten
IV Taking a Small Number from a Whole Ten

Léa has \(\dollar 50\) in five \(\dollar 10\) bills. She spends \(\dollar 6\). How much is left? She says: “I break one bill.”

She keeps four bills (\(\dollar 40\)) and breaks the fifth into ten \(\dollar 1\) coins. She pays \(6\) coins: \(10 - 6 = 4\) coins are left. In all, \(\dollar 40\) and \(\dollar 4\) make \(\dollar 44\). So \(50 - 6 = 44\).


Method — Subtracting from a Whole Ten
To take a number from \(1\) to \(9\) away from a whole ten (\(30\), \(50\), \(80\)…):
  1. Break one ten into ten ones: \(50 = 40 + 10\).
  2. Take away from the ten: \(10 - 6 = 4\) — the remainder is the complement of \(6\) to \(10\).
  3. Put back the other tens: \(40 + 4 = 44\).
So the result always ends with the complement to \(10\): \(80 - 3 = 77\), \(30 - 9 = 21\).
Example
Compute: (a) \(70 - 2\), (b) \(40 - 7\), (c) \(100 - 5\).

(a) \(10 - 2 = 8\), then \(60 + 8 = 68\). (b) \(10 - 7 = 3\), then \(30 + 3 = 33\). (c) \(10 - 5 = 5\), then \(90 + 5 = 95\).

Skills to practice
  • Subtracting Without Crossing a Ten
  • Breaking a Ten
V Two-Step Problems
Method — Solving a Problem in Two Steps
Some problems cannot be solved with one calculation: the question hides another number that you must find first.
  1. Find the hidden number: Read the story and find the number the question needs but does not give. Calculate it first.
  2. Answer the question: Use the hidden number in a second calculation to answer the question.
  3. Check: Make sure the answer makes sense in the story.
Example
Emma makes a necklace with 8 red beads and 7 blue beads. Then Hugo gives her 10 golden beads for her bead box. How many beads does Emma have in all?

  • Step 1 — the hidden number: The question needs the number of beads on the necklace. We know its two parts (8 and 7): we add.
    Figure
    \(\textcolor{colordef}{8}+\textcolor{colorprop}{7}=\textcolor{olive}{15}\). The necklace has \(15\) beads.
  • Step 2 — the answer: Now we know the two parts (15 and 10): we add again.
    Figure
    \(\textcolor{colordef}{15}+\textcolor{colorprop}{10}=\textcolor{olive}{25}\).
  • Check: We recompose: \(15+10=25\). And \(25\) is more than \(15\): it makes sense because she received more beads.

Example
Louis has 24 marbles. He gives 9 marbles to Hugo and 6 marbles to Emma. How many marbles does Louis have left?

  • Step 1 — the hidden number: The question needs the number of marbles given away. We know its two parts (9 and 6): we add.
    Figure
    \(\textcolor{colordef}{9}+\textcolor{colorprop}{6}=\textcolor{olive}{15}\). Louis gives away \(15\) marbles.
  • Step 2 — the answer: We know the whole (24) and one part (15): we subtract to find the other part.
    Figure
    \(\textcolor{olive}{24}-\textcolor{colordef}{15}=\textcolor{colorprop}{9}\) (from \(15\), \(5\) more make \(20\), \(4\) more make \(24\): \(9\)).
  • Check: We recompose: \(15+9=24\), the marbles at the start. And \(9\) is less than \(24\): it makes sense because he gave marbles away.

Skills to practice
  • Adding Twice
  • Adding Then Subtracting
VI Review & Quizz
Skills to practice
  • Quiz