Adding past ten with the tens partner

Where you will use this

You have 9 mar­bles. Your friend gives you 5 more. How many mar­bles do you have now?

You could count on from 9, one fin­ger at a time: 10, 11, 12, 13, 14. That works. But it is slow, and it is easy to lose count.

There is a quicker way. You make a ten first. Tens are easy to add to. Once you know the trick, sums like $9 + 5$ and $7 + 6$ take only a moment. You will use it every day: count­ing money, adding scores in a game, or find­ing how many pen­cils are in two boxes.

So far every answer has stopped at ten. Now the answers go past it, and count­ing on your fin­gers starts to run out. This les­son gives you the trick the whole acad­emy course is built on.

The idea

Ten is an easy num­ber to add to. $10 + 4$ is $14$ with­out any think­ing at all — the four sim­ply sits in the units place. So the plan is always the same: get to ten first, then add on what is left over.

To get to ten you use the pairs from the first les­son, the tens com­ple­ments:

Try it on 9 + 5

$$\begin{aligned}9 + 5 &= (9 + 1) + 4 \\ &= 10 + 4 \\ &= 14\end{aligned}$$

Nine needed one, so one was bor­rowed from the five. That left four, and ten and four make four­teen. Noth­ing was lost and noth­ing was added — the five was only split into $1 + 4$.

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. Put the ten and what is left together. Ten and 4 is 14.

Try it with beads or but­tons on the table. Put 9 in a row and 5 in another row. Slide one but­ton across. Now you have a row of ten and four more. Count them to check.

Adding 9

Nine needs one. Take one from the other num­ber and the rest goes after the ten.

9 + 6

$$\begin{aligned}9 + 6 &= (9 + 1) + 5 \\ &= 10 + 5 \\ &= 15\end{aligned}$$

There is a short­cut worth notic­ing here. Every answer is one less than the num­ber you added with a ten in front of it: $9 + 7$ gives $16$, and 6 is one less than 7. That hap­pens because nine bor­rowed one.

Adding 8

Eight needs two.

8 + 5

$$\begin{aligned}8 + 5 &= (8 + 2) + 3 \\ &= 10 + 3 \\ &= 13\end{aligned}$$

One more: 8 + 6

Eight needs two. Take two from the six. Four are left.

$$\begin{aligned}8 + 6 &= (8 + 2) + 4 \\ &= 10 + 4 \\ &= 14\end{aligned}$$

Think of eight apples in a bas­ket that holds ten. Two more apples fill the bas­ket. Four apples wait out­side. A full bas­ket and four is four­teen.

Adding 7

Seven needs three.

7 + 6

$$\begin{aligned}7 + 6 &= (7 + 3) + 3 \\ &= 10 + 3 \\ &= 13\end{aligned}$$

One more: 7 + 5

Seven needs three. Take three from the five. Two are left. So $7 + 5 = 10 + 2 = 12$.

Adding 6

Six needs four.

6 + 7

$$\begin{aligned}6 + 7 &= (6 + 4) + 3 \\ &= 10 + 3 \\ &= 13\end{aligned}$$

One more: 6 + 9

Here the other num­ber is big­ger. You can still make ten with six: six needs four, and nine gives four, leav­ing five. So $6 + 9 = 10 + 5 = 15$. Or you can turn it round: nine needs one, six gives one, leav­ing five. The answer is the same.

Adding 5

Five needs five.

5 + 8

$$\begin{aligned}5 + 8 &= (5 + 5) + 3 \\ &= 10 + 3 \\ &= 13\end{aligned}$$

Adding 4, 3 and 2

The same trick still works, but by now the other num­ber is the big­ger one, and it is eas­ier to turn the sum round and start from that instead. $3 + 9$ is the same as $9 + 3$, which you already know is $12$.

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

The dou­bles past ten

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

A dou­ble is a num­ber added to itself. Here is one more dou­ble, worked with the tens part­ner.

$$\begin{aligned}9 + 9 &= (9 + 1) + 8 \\ &= 10 + 8 \\ &= 18\end{aligned}$$

Nine needs one. The sec­ond nine gives one and keeps eight. Ten and eight make eigh­teen.

Worked answers to check your­self against

These are the sums the sheet works through. Cover the answers, do them your­self, then look.

Your turn

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 $104$.
• Count­ing on from the smaller num­ber. Start from the big­ger one. It is quicker and there is less to count.
• Think­ing that turn­ing a sum round changes it. $3 + 9$ and $9 + 3$ both make $12$.

Key terms

Add

To put two amounts together and find how many there are in all.

Sum

The answer you get when you add.

Tens part­ner (tens com­ple­ment)

The num­ber that makes ten with another num­ber. The tens part­ner of 7 is 3.

Dou­ble

A num­ber added to itself, such as $8 + 8$.

Units place

The right-hand digit of a num­ber. In 14, the units digit is 4.

Ten frame

A box of ten squares in two rows of five, used to see when a ten is full.

Answers

Here are the answers to the thirty sums in Your turn. Check each one you got wrong with beads or with the tens part­ner.

Some of these stay below ten, like $5 + 2$ and $4 + 4$. You do not need the tens part­ner for those. Just count on. Sums 1, 3, 6, 7, 9, 16, 18, 20, 21, 25 and 28 are like that.

Sums 12, 24 and 30 land exactly on ten. The two num­bers are tens part­ners. Sum 1 adds nought, so the answer is the num­ber you started with.