Abacus Math Program – Lesson 12 – The 5/10 Pair Combination Part 2

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Lesson 11 the 5/10 Pair Combination Part 1, we introduced the 5/10 pair combination for addition. Here in Lesson 12 Part 2 we will look at the 5/10 pair combination for subtraction. Just like in Lesson 11 Part 1 these examples will demonstrate how we use the 10 pair subtraction rule but instead of directly adding the 10 pair complement we will need to add the 10 pair in the form of a 5 pair.

Let’s work through the example problem 12 – 7. To subtract 7 from the 2 in 12 we use the 10 pair subtraction rule by first subtracting 1 from 1 from on the next rod and add the 3, the 10 pair complement of 7, to the 2 on the target rod. However, since there are only 2 available unused lower beads we must add the 3 in the form of a 5 pair. So we add 3 by adding 5 and then subtracting 2, the 5 pair complement of 3, from 2 for an answer of 5. In all of these 5/10 combination examples, the student should be thinking about them as nothing more than special cases of the typical 10 pair addition and subtraction bead movements. The only difference is when we add or subtract the 10 pair we do it in the form of a 5 pair.

In this first example we start by adding 14 to the abacus. Next we subtract 6 from 14 by subtracting 6 from 4 on rod B. The subtraction will require the 5/10 pair subtraction combination. First subtract 1 from 1 on rod A. Then add 4, the 10 pair complement of 6, to 4 on rod B as a 5 pair. Add 4 by adding 5 and subtracting 1, the 5 pair complement of 4, on rod B for an interim sum of 8. Next subtract 3 from 8 on rod B for the interim sum of 5. Lastly add 5 to 5 using the 10 pair addition rule. First add 1 to 0 on rod A and then subtract 5, the 10 pair complement of 5, from 5 on rod B for the final answer of 10.

In this next example we start by adding 18 to the abacus. Next subtract 5 from 18 by subtracting 5 from 8 on rod B for the interim sum of 13. Next subtract 7 from 13 by subtracting 7 from 3 on rod B. We will need to use the 5/10 pair subtraction combination. First subtract 1 from 1 on rod A. Next add 3, the 10 pair complement of 7, to 3 on rod B using a 5 pair. So add 5 and subtract 2, the 5 pair complement of 3. The interim sum is now 6. Lastly add 6 to 6 on rod B this time using the 5/10 pair addition combination. First add 1 to 0 on rod A then subtract 4, the 10 pair complement of 6, in the form of a 5 pair by adding 1, the 5 pair complement of 4, and subtracting 5. The final answer is 12.

In this last example we start by adding 14 to the abacus. Next subtract 9 from 14 by subtracting 9 from 4 on rod B. This will require the 5/10 pair subtraction combination. First subtract 1 from 1 on rod A. Next add 1, the 10 pair complement of 9, to 4 on rod B in the form of a 5 pair. So add 5 and subtract 4, the 5 pair complement of 1, on rod B for an interim sum of 5. Next add 4 to 5 on rod B for an interim sum of 9. Lastly add 9 to 9 on rod B with the 10 pair addition rule. First add 1 to 0 on rod A and then subtract 1, the 10 pair complement of 9, from 9 on rod B for the final answer of 18.

Next up Lesson 13.