Grade 2 · 1.NBT.C.4 · 2.MD.B.6

Adding on an open number line

The model railroad club has 38 pieces of track laid and 45 more in the box. Counting on forty-five times is slow, and no ruler or marked line is at hand. How can a quick drawing find the total?

You need

Number lines with different scalesGrade 2 Ten more and ten lessGrade 1

Unlocks

Counting up to find a differenceGrade 3

Where you are 6 in Multi-digit addition

Number9

Arithmetic5

Addition and subtraction12

Multi-digit addition6

Adding two-digit numbers without regrouping

Adding on an open number line

Regrouping in addition Adding a one-digit number to a two-digit number by making a ten The partial sums method The standard addition algorithm

When you reach for it

A child who adds two two-digit numbers needs a way to keep track. Counting on by ones from 38 for forty-five counts is slow and a slip is easy. A ruler or a printed line would help, but it is rarely at hand, and a printed line from 0 to 100 is crowded.

An open number line answers this. It is a plain line with no marks drawn ahead of time. The child marks only the number they start from, then draws a jump for each piece they add, and marks only where each jump lands. Nothing is to scale: a long jump means “a lot” and a short jump means “a little”. The labels on the jumps carry the numbers, not the lengths.

The lesson Number lines with different scales read lines with evenly spaced marks. An open line keeps the idea that bigger numbers sit to the right, and drops the spacing. The child has also met it twice already: hops of ten in Adding two-digit numbers without regrouping, and a stop at the next ten in Adding a one-digit number to a two-digit number by making a ten. This lesson puts those moves together for any two numbers whose total is at most 100, including sums where the ones cross a ten, such as 38 + 45.

Reach for it when:

the child adds two two-digit numbers in their head and wants a record of each step;
the ones run past nine and a stop at a ten will help;
a number to add is close to a ten, such as 29 or 48, and a whole ten is easier to jump;
the question is “how many more?”, which is a jump up from one number to another.

The procedure

Four days build one picture. Day 1 jumps the tens and then the ones. Day 2 handles sums where the ones jump crosses a ten. Day 3 looks for fewer, friendlier jumps. Day 4 reads a drawing someone else made and uses the line to find a missing part.

Day 1: jump the tens, then the ones

Mark the start. Draw a line across the page. To find 24 + 53, mark 24 near the left end. There is no 0, no 10, nothing else; the line starts wherever the problem starts.

Jump the tens. The number being added, 53, is 5 tens and 3 ones (the picture from Tens and ones). Add the tens first. One jump of ten from 24 lands on 34, the move from Ten more and ten less: the tens digit goes up by one and the ones digit stays. Five jumps of ten land on 34, 44, 54, 64, 74. A child sure of the tens can make one long jump labeled +50 straight to 74, because 2 tens and 5 tens make 7 tens.

Jump the ones. Now add the 3 ones: a short jump from 74 to 77. The ones were 4 and became 7; the tens stayed at 7.

Figure 124 + 53. Jump the 5 tens, then the 3 ones.

Read the answer. The last mark is the total: 24 + 53 = 77. Every jump is written on the line, so the drawing is also a record:

24 + 53 = 24 + 50 + 3 = 74 + 3 = 77

Why it is allowed. Adding 53 is the same as adding 50 and then 3, because 53 is 50 + 3. Adding the parts one after another gives the same total as adding them at once (the associative property; see Properties of addition). The line just shows the parts in order.

Start from the bigger number. In 16 + 62 it is quicker to mark 62 and jump 16 than to mark 16 and jump 62: one ten and some ones instead of six tens and some ones. The order of adding does not change the total (the commutative property), so either start is correct. Example 2 does this.

Day 2: when the ones cross a ten

The ones run past nine. Take 38 + 45. Jumping the 4 tens from 38 gives 48, 58, 68, 78, or one jump of +40 to 78. Now 5 ones are left, and 78 + 5 goes past 80.

Stop at the ten. The ones jump is split so that it lands on the next ten first. From 78, the next ten is 80, and that is 2 away. The 5 is 2 and 3, so after landing on 80 there are 3 left: a last jump from 80 to 83.

38 \to 78 \to 80 \to 83

This is the stop at a ten from making a ten, done after the tens jump. The child already knows the pairs that make ten, so “how far from 78 to 80” is a fact they can say at once.

Why stop there? From a ten, adding a few ones needs no thought: 80 + 3 is 8 tens and 3 ones, 83. The hard part of 78 + 5 is the crossing, and the stop at 80 makes the crossing its own small jump. The tens digit goes up by one at that stop, because ten more ones have been gathered: 8 from the 78 and 2 from the jump.

The line shows the new ten. In blocks, the same sum needs a trade: 8 ones and 5 ones are 13 ones, and ten of them become a new ten. That trade is the work of Regrouping in addition. On the open line the trade is the stop at 80: the tens digit changes there and nowhere else in the ones jump.

Check the size. 38 is near 40, and 40 + 45 = 85, so the total should be near 85. The answer 83 is close. A total far from that guess means a jump was mislabeled or a ten was skipped.

Day 3: fewer jumps

The total does not depend on which jumps are made, only on what they add up to. A child who sees that can choose jumps that are easy to say.

Jump to a friendly ten first. For 56 + 27, the start 56 is 4 away from the ten 60. Take that 4 out of the 27 first: jump +4 to 60. What is left of 27 is 23. From a ten, adding 23 crosses nothing: 6 tens and 2 tens make 8 tens, and the 3 ones join no ones, so 60 + 23 = 83 in one jump. Two jumps instead of three. Example 5 compares the two ways.

Jump a whole ten and step back. For 47 + 29, notice that 29 is 1 less than 30. Jump +30 from 47 to 77. That added 1 too many, so step back 1 to 76. The jump back goes to the left and is labeled -1. Two jumps instead of +20, +3, +6.

47 + 29 = 47 + 30 - 1 = 77 - 1 = 76

This works best when the number added ends in 8 or 9: 38 is 40 less 2, and 59 is 60 less 1. When the number ends in 2 or 3, stepping back several ones is no easier than jumping them forward.

Any jumps, the same total. Tens then ones, a friendly ten first, or a whole ten and a step back: all three reach the same mark, because the jumps add to the same number. A child may draw whichever is clearest to them. The point of looking for fewer jumps is not speed; it is seeing a number such as 29 as “thirty, less one”.

Day 4: reading a drawing, and finding a missing part

Read someone else’s line. An open line made by someone else holds the whole sum. The first mark is the number they started with. The labels on the jumps, added together, are the number they added. The last mark is the total. So a drawing that starts at 29 and jumps +1, +40, +5 shows 29 + 46, because 1 + 40 + 5 = 46, and it ends on 75.

Find a missing part by jumping up. Some questions give the start and the total and ask for the jump: the club’s track needs 81 pieces and 34 are laid, so how many more? That is

34 + \square = 81

Mark 34 at the left and 81 at the right, then fill the gap with jumps. From 34, jump to the friendly ten 40 (+6). From 40, jump tens to 80 (+40). From 80, jump to 81 (+1). The jumps together are the missing part: 6 + 40 + 1 = 47.

Subtraction as a jump up. The same question is a subtraction: 81 - 34 asks how far apart the two numbers are, and the jump up from 34 to 81 measures exactly that. So 81 - 34 = 47. This is Subtraction as an unknown addend drawn on a line. Larger numbers, past 100, are counted up the same way in Counting up to find a difference.

Check by adding. A missing part can always be checked by putting it back in the sum: 34 + 47 is 34 + 40 = 74, then 74 + 6 = 80, then 80 + 1 = 81. The total matches.

Worked examples

A line drawn on scrap paper or a whiteboard is enough for each one. Say each jump aloud as it is drawn: “plus forty, seventy-eight; plus two, eighty; plus three, eighty-three.”

Example 1

Day 1: More track for the loop

The model railroad club is building a loop. Berhane has laid 24 pieces of track. Maile hands him 53 more, and he lays all of them. How many pieces of track are laid now?

Solution

The two amounts of track are put together, so we add: 24 + 53.

Mark 24 on an open line. The 53 is 5 tens and 3 ones. Jump the tens: five jumps of ten land on 34, 44, 54, 64, 74, or one jump of +50 lands on 74 (Figure 1). The ones digit is still 4 the whole time, because only tens were added.

Jump the ones: +3 from 74 lands on 77. The 4 ones and 3 ones make 7 ones, nine or fewer, so no ten is crossed.

24 + 53 = 77

There are 77 pieces of track laid. Check with tens and ones: 2 tens and 5 tens make 7 tens, 4 ones and 3 ones make 7 ones, so 77.

Example 2

Day 1: Two tide pools

On a beach walk, Udoka counts 16 animals in the first tide pool and 62 animals in the second. How many animals did she count in all?

Solution

Both counts go together: 16 + 62.

Jumping 62 from 16 would take six tens and two ones. Adding in the other order gives the same total (the commutative property), so mark 62 and add 16 instead: 1 ten and 6 ones.

Jump one ten: 62 \to 72. Jump the ones: 72 \to 78, since 2 ones and 6 ones make 8 ones.

16 + 62 = 62 + 16 = 78

Udoka counted 78 animals. Two jumps did the work that seven would have done from 16.

Example 3

Day 2: Straight pieces and curved pieces

The club’s second loop uses 38 straight pieces of track and 45 curved pieces. How many pieces does it use?

Solution

We add: 38 + 45.

Mark 38. Jump the 4 tens: +40 lands on 78. Five ones are left, and 78 + 5 goes past the next ten, 80. So split the 5 at 80. From 78, it takes 2 to reach 80; jump +2. The 5 is 2 and 3, so 3 are left: jump +3 from 80 to 83 (Figure 2).

38 \to 78 \to 80 \to 83

Figure 238 + 45. The 5 ones are split at 80, the next ten.

38 + 45 = 38 + 40 + 2 + 3 = 83

The loop uses 83 pieces. Check the size: 38 is near 40, and 40 + 45 = 85, so the total should be about 85, and 83 is close to that.

Example 4

Day 2: Snails on the rocks

Maile counts 67 periwinkle snails on the rocks and then 25 more in a tide pool. How many snails did she count?

Solution

We add: 67 + 25.

Mark 67. Jump the 2 tens: +20 lands on 87. Now 5 ones are left. The next ten after 87 is 90, which is 3 away, so jump +3 to 90. The 5 is 3 and 2, so jump +2 more, to 92 (Figure 3).

67 \to 87 \to 90 \to 92

Figure 367 + 25. The stop at 90 splits the 5 ones into 3 and 2.

Maile counted 92 snails. The stop moved from 80 in Example 3 to 90 here, and the split of the ones changed with it: 5 was 2 and 3 there and is 3 and 2 here. The split always depends on how far the last mark is from the next ten.

Example 5

Day 3: Two ways to lay the siding

Berhane has laid 56 pieces of track. The club adds a siding of 27 more pieces. Find the total two ways: tens then ones, and a friendly ten first. Which takes fewer jumps?

Solution

We add: 56 + 27.

Tens then ones. Jump +20 from 56 to 76. Seven ones are left, and the next ten is 80, 4 away. Jump +4 to 80, then the other 3 to 83. Three jumps: +20, +4, +3.

A friendly ten first. From 56 the next ten, 60, is 4 away. Take those 4 out of the 27 first: jump +4 to 60. What is left of 27 is 27 - 4 = 23. From a ten, adding 23 crosses nothing: 6 tens and 2 tens make 8 tens, and the 3 ones sit alone, so one jump of +23 lands on 83 (Figure 4). Two jumps: +4, +23.

Figure 456 + 27 in two jumps. Reach the friendly ten first, then add the rest at once.

56 + 27 = 56 + 4 + 23 = 60 + 23 = 83

Both ways give 83 pieces, because both sets of jumps add to 27: 20 + 4 + 3 = 27 and 4 + 23 = 27. The friendly ten first takes one fewer jump.

Example 6

Day 3: Mussels and limpets

On the beach walk, Berhane counts 47 mussels and 29 limpets on one rock. How many animals is that?

Solution

We add: 47 + 29.

The 29 is 1 less than 30, and 30 is three tens, an easy jump. So jump +30 from 47 to 77. That jump added 30, which is 1 more than 29, so step back 1: from 77 to 76. The step back is drawn as a jump to the left, labeled -1 (Figure 5).

Figure 547 + 29. Jump a whole 30, one more than needed, then step back 1.

47 + 29 = 47 + 30 - 1 = 77 - 1 = 76

Berhane counted 76 animals. Check with tens then ones: 47 + 20 = 67, then 3 to reach 70, then the other 6 of the 9 ones to 76. Same mark, but three jumps instead of two.

Example 7

Day 4: Reading Udoka’s drawing

Udoka drew this open line to add two numbers (Figure 6). What sum did she draw, and what is the total?

Figure 6Udoka’s drawing in Example 7. The start, the jumps and the stops tell the whole sum.

Solution

The first mark is the number she started from: 29. The jumps are +1, +40 and +5, so the number she added is

1 + 40 + 5 = 46

She drew 29 + 46. Follow the jumps: 29 + 1 = 30, a friendly ten. From 30, a jump of 40 is 3 tens and 4 tens, 7 tens: 70. From 70, +5 lands on 75.

29 + 46 = 75

The total is 75. Udoka made the friendly ten first, at 30, and then jumped the rest of the tens and ones. Her jumps add to 46, the same as a +40 and a +6 would, which is why her line and a tens-then-ones line end on the same mark.

Example 8

Day 4: How many more pieces?

The club’s long track needs 81 pieces to close the loop. 34 pieces are laid. How many more pieces are needed?

Solution

The laid pieces and the missing pieces together make 81:

34 + \square = 81

Mark 34 at the left of an open line and 81 at the right. Jump up from 34 to the friendly ten 40: that is +6. From 40, jump tens to 80: 4 tens to 8 tens is +40. From 80, jump +1 to 81 (Figure 7).

Figure 734 + \square = 81. Jump up from 34 to 81; the jumps together are the missing part, 47.

The jumps together are the missing part:

6 + 40 + 1 = 47

The club needs 47 more pieces. This is also the subtraction 81 - 34 = 47. Check by putting the part back: 34 + 47 is 34 + 40 = 74, then +6 to 80, then +1 to 81.

Boundaries

The total stays at 100 or below here. Every sum in this lesson ends at 100 or less. The line itself does not stop there: a jump from 96 by +4 lands on 100 and the next jump on 110. Totals past 100 are counted the same way in later lessons.
The labels carry the numbers, not the lengths. An open line is not to scale. A jump of 3 may be drawn as long as a jump of 10 and the answer is still right, because only the labels are added. To read a length off a line by measuring it, the marks must be evenly spaced, as in Number lines with different scales.
The jumps must add to the number being added. Any split works: +40, +2, +3 or +30, +15 or +50, -5 for 45. A split that adds to 44 or 46 gives a wrong total, however neat it looks.
A jump to the right adds; a jump to the left takes away. A step back is only correct as part of a pair, after a jump that went too far by exactly that much: +30 then -1 adds 29.
Whole numbers only. Every start, jump and landing here is a whole number (0, 1, 2, and so on), and every jump is a whole number of tens and ones.
A missing part needs the smaller number first. In 34 + \square = 81 the line runs from 34 up to 81. If the start were larger than the total, there would be no jump up to make.

Where this goes next

The open line is a way of adding by place value: tens with tens, ones with ones, a new ten made at a stop. The same sums done with blocks and a trade of ten ones for a ten are Regrouping in addition; written as separate tens and ones totals they are The partial sums method; done in columns they become The standard addition algorithm. The jump up of Day 4 grows into Counting up to find a difference, where the same landmarks (the next ten, then the next hundred) carry subtraction past 100. Later, number lines carry other numbers too: lengths in inches and centimeters are read along a ruler the same way, one jump at a time.

Exercises

Day 1: jump the tens, then the ones

32 + 45 = \square. Draw the jumps and name each number you land on.
51 + 36 = \square
13 + 75 = \square. Which number would you mark first, and why?
Berhane has laid 44 pieces of track. Maile brings 25 more and he lays them. How many pieces are laid now?
For 26 + 41, jump the tens one at a time, then the ones. Which numbers do you land on?

Day 2: when the ones cross a ten

48 + 35 = \square. Name the ten you stop at.
29 + 43 = \square
66 + 18 = \square
59 + 33 = \square
Udoka counts 75 snails on one rock and 18 on the next. How many snails is that?
64 + 36 = \square. Where does the last jump land?

Day 3: fewer jumps

Find 35 + 49 by jumping +50 and stepping back.
Find 26 + 38 by jumping a whole number of tens and stepping back.
Find 58 + 24 by jumping to a friendly ten first.
Find 77 + 19 in two jumps.
Maile has 39 curved pieces and 45 straight pieces of track. How many pieces does she have? Use a friendly ten first.

Day 4: reading a drawing, and finding a missing part

Figure 8Drawing A, for Exercise 17.

Drawing A shows a sum. What sum is it, and what is the total?

Figure 9Drawing B, for Exercise 18.

Drawing B shows a sum. What sum is it, and what is the total?
27 + \square = 63
45 + \square = 92
Berhane needs 70 pieces of track for a loop. He has laid 38. How many more does he need?
91 - 57 = \square. Find it by jumping up from 57.

Explain

In 38 + 45 the ones jump of 5 is split into 2 and 3. Why stop at 80, and not at some other number?
To add 29, you can jump +30 and step back 1. Explain why, using 52 + 29.
Find 46 + 38 two ways: tens then ones, and +40 then a step back. Why must both ways land on the same number?

Answers

77. From 32, four jumps of ten: 42, 52, 62, 72. Then +5: 77. So 32 + 45 = 77.
87. Jump +30 from 51 to 81, then +6 to 87: 51 + 36 = 87.
88. Mark 75 first and add 13: fewer jumps, and the order of adding does not change the total. 75 + 10 = 85, then +3: 13 + 75 = 88.
69 pieces. Jump +20 from 44 to 64, then +5 to 69: 44 + 25 = 69.
36, 46, 56, 66, 67. Four jumps of ten from 26, then one jump of 1: 26 + 41 = 67.
83, with a stop at 80. Jump +30 from 48 to 78. Then +2 to 80 and +3 to 83: 48 + 35 = 83.
72. Jump +40 from 29 to 69. Then +1 to 70 and +2 to 72: 29 + 43 = 72.
84. Jump +10 from 66 to 76. Then +4 to 80 and +4 to 84: 66 + 18 = 84.
92. Jump +30 from 59 to 89. Then +1 to 90 and +2 to 92: 59 + 33 = 92.
93 snails. Jump +10 from 75 to 85. Then +5 to 90 and +3 to 93: 75 + 18 = 93.
100. Jump +30 from 64 to 94. The 6 ones take 94 exactly to the next ten, which is 100, so the last jump lands on 100: 64 + 36 = 100.
84. Jump +50 from 35 to 85. That is 1 more than 49, so step back 1 to 84: 35 + 49 = 84.
64. 38 is 2 less than 40. Jump +40 from 26 to 66, then step back 2 to 64: 26 + 38 = 64.
82. From 58 the next ten is 60, 2 away: jump +2. What is left of 24 is 22, and 60 + 22 = 82 in one jump: 58 + 24 = 82.
96. Jump +20 from 77 to 97, then step back 1 to 96: 77 + 19 = 96.
84 pieces. From 39, jump +1 to 40. What is left of 45 is 44, and 40 + 44 = 84: 39 + 45 = 84.
45 + 37 = 82. The start is 45. The jumps are +30, +5, +2, which add to 37. They land on 75, 80, then 82.
58 + 36 = 94. The start is 58. The jumps are +2, +30, +4, which add to 36. They land on 60, 90, then 94.
36. Jump up from 27: +3 to 30, +30 to 60, +3 to 63. The jumps add to 3 + 30 + 3 = 36, and 27 + 36 = 63.
47. Jump up from 45: +5 to 50, +40 to 90, +2 to 92. The jumps add to 5 + 40 + 2 = 47, and 45 + 47 = 92.
32 more pieces. 38 + \square = 70. Jump +2 from 38 to 40, then +30 to 70. The jumps add to 32, and 38 + 32 = 70.
34. Jump up from 57: +3 to 60, +30 to 90, +1 to 91. The jumps add to 3 + 30 + 1 = 34, so 91 - 57 = 34. Check: 57 + 34 = 91.
After the tens jump, the line is at 78, and 80 is the next ten. Stopping at a ten splits the hard part, the crossing, into its own small jump: 78 needs 2 to make 80, a pair the child knows. From 80 the last 3 ones are read at once, 83, because a ten plus a few ones needs no counting. A stop at some other number, say 81, would leave a jump of 3 from 78 that still crosses 80 inside it.
29 is 30 - 1. Jumping +30 adds one more than 29, so stepping back 1 takes away the extra and the total jump is 29. For 52 + 29: 52 + 30 = 82, then 82 - 1 = 81. Check with tens then ones: 52 + 20 = 72, +8 to 80, +1 to 81. So 52 + 29 = 81.
Both give 84. Tens then ones: 46 + 30 = 76, +4 to 80, +4 to 84. A whole ten and a step back: 46 + 40 = 86, then back 2 to 84. Both must agree because the jumps in each way add to 38 (30 + 4 + 4 = 38 and 40 - 2 = 38), and adding a number in parts, in any order, gives the same total as adding it at once.