Adding past ten with the tens partner

You have 9 mar­bles. Your friend gives you 6 more. How many mar­bles do you have now? You could count on from 9, one fin­ger at a time: 10, 11, 12, 13, 14, 15. That works, but it is slow and it is easy to lose count.

There is a quicker way. You make a ten first. Ten is the eas­i­est num­ber to add to, because $10 + 5$ is $15$ with­out any think­ing at all. Once you know this trick, sums like $9 + 6$ and $7 + 5$ take only a moment. At the end of the les­son you will also learn to find a miss­ing num­ber in a sum.

The idea: make a ten first#

To get to ten you use the tens part­ners from the first les­son:

So every sum is writ­ten in the same shape: ten plus what is left over.

Exam­ple 1#

Work out $9 + 6$.

Nine needs one to make ten. Take that one from the six. Five are left over.

$$9 + 6 = 10 + 5 = 15$$

Noth­ing was lost and noth­ing was added. The six was only split into $1 + 5$: the 1 went to the nine, and the 5 waits after the ten.

The pic­ture shows the same thing for $9 + 5$. One bead moves across to fill the ten frame. Then you have a full ten and four more, so $9 + 5 = 10 + 4 = 14$.

The three steps#

1. Look at the big­ger num­ber. Ask: what does it need to make ten?
2. Take that much from the other num­ber. See what is left.
3. Write ten and what is left. Ten and 4 is 14.

Adding to 9#

Nine needs one, so one comes off the other num­ber.

Notice the pat­tern. In every answer the units digit is one less than the num­ber you added. That hap­pens because nine bor­rowed one.

Adding to 8#

Eight needs two, so two come off the other num­ber.

Exam­ple 2#

Work out $8 + 7$.

Eight needs two. Take two from the seven. Five are left.

$$8 + 7 = 10 + 5 = 15$$

Think of eight apples in a bas­ket that holds ten. Two more apples fill the bas­ket, and the rest wait out­side.

Adding to 7#

Seven needs three.

Exam­ple 3#

Work out $7 + 6$.

Seven needs three. Take three from the six. Three are left.

$$7 + 6 = 10 + 3 = 13$$

Look at the first row. Seven and three are part­ners, so there is noth­ing left over: $7 + 3 = 10$.

Adding to 6 and to 5#

Six needs four. Five needs five.

Exam­ple 4#

Work out $6 + 8$.

Here the sec­ond num­ber is big­ger. You can still make ten with six: six needs four, and eight gives four, leav­ing four. So $6 + 8 = 10 + 4 = 14$. Or turn it round: eight needs two, six gives two, leav­ing four. The answer is the same.

When the small num­ber comes first#

For $3 + 9$, turn the sum round and start from the big­ger num­ber. Nine needs one. Three gives one and keeps two.

$$3 + 9 = 10 + 2 = 12$$

In the same way, $2 + 9 = 10 + 1 = 11$.

Remem­ber. You may always start from the big­ger num­ber. Turn­ing a sum round never changes its answer.

The dou­bles past ten#

These fin­ish the set of dou­bles you began in the first les­son.

Find­ing the miss­ing num­ber#

Some­times a sum has a box where one num­ber should be. In $6 + \square = 9$ the box hides a num­ber. You must find the num­ber that 6 needs to reach 9.

Stand on 6 and count up to 9: 7, 8, 9. That is three steps. So the box holds 3, and $6 + 3 = 9$.

Exam­ple 5#

Find the miss­ing num­ber: $4 + \square = 9$.

Count up from 4 to 9: 5, 6, 7, 8, 9. Five steps. So $4 + 5 = 9$, and the miss­ing num­ber is 5.

Exam­ple 6#

Find the miss­ing num­ber: $9 + \square = 9$.

Nine is already nine, so noth­ing needs to be added. The miss­ing num­ber is 0, because $9 + 0 = 9$.

Exam­ple 7#

Find the miss­ing num­ber: $8 + \square = 10$.

This is a tens part­ner. Eight needs two to make ten, so the miss­ing num­ber is 2.

You can also find a miss­ing num­ber by tak­ing away. For $1 + \square = 7$, take 1 from 7: $7 - 1 = 6$. Check by adding: $1 + 6 = 7$.

Com­mon mis­takes#

• For­get­ting to take the part­ner away from the other num­ber. For $9 + 5$ the five gives one and keeps four. It does not keep five.
• Mix­ing up the part­ners. Seven needs three, not four. Say the pairs aloud: nine and one, eight and two, seven and three, six and four, five and five.
• Writ­ing the ten wrong. Ten and four is four­teen, writ­ten $14$, not $104$.
• Writ­ing the total in the box. In $4 + \square = 9$ the box holds 5, not 9. The 9 is already there.

Key terms#

• Tens part­ner (tens com­ple­ment): the num­ber that makes ten with another num­ber. The tens part­ner of 7 is 3.
• Sum: the answer you get when you add.
• Dou­ble: a num­ber added to itself, such as $8 + 8$.
• Miss­ing num­ber: the num­ber hid­den in a box in a sum.

Prac­tice#

1. $9 + 4$
2. $8 + 6$
3. $7 + 5$
4. $6 + 7$
5. $5 + 8$
6. $2 + 9$
7. $8 + 9$
8. $7 + 7$
9. $2 + \square = 8$
10. $3 + \square = 9$
11. $5 + \square = 7$
12. Sita has 7 red beads and 6 blue beads. How many beads does she have alto­gether?

Answers#