Here we will learn subtracting 2-digit numbers with borrowing. The subtractions with borrowing are solved step-by-step in four different ways.

When we need to subtract bigger number from a smaller number in ones place, we regroup tens into ones.

Let us understand this with the help of an example.

A shopkeeper sells bananas. He has 6 bunches of 10's and 5 single bananas.

Nairitee wants to buy 37 bananas from the shopkeeper. The shopkeeper gives her 3 bunches of 10's and breaks one bunch of 10's to give 7 bananas.

The shopkeeper is now left with how many bananas?

Column Method

Arrange the numbers in columns and subtract. In ones place, 5 is smaller than 7. We cannot take away 7 from 5. So, we break one ten into 10 ones. Now we have, 15 ones in ones place and 5 tens in tens place.

15 ones - 7 ones = 8 We write 8 in ones place. Subtract 3 tens from 5 tens. 5 tens - 3 tens = 2 tens Write 2 in tens place. |

So, 65 - 37 = 28

Worked-out examples on subtracting 2-digit numbers with borrowing:

**1.** Subtract 9 from 15.

**Solution:**

T O

1 5

__ - 9__

Since, 5 < 9, so 9 cannot be subtracted from 5. So, 1 ten, i.e., 10 ones is borrowed from the digit 1 of tens place. Now one ten, i.e., 10 ones are added to 5 ones to make it 15 ones. Now 15 ones – 9 ones = 6 ones.

Therefore, 15 – 9 = 6

**2.** Subtract 37
from 65

**Solution:**

The numbers are placed in column form, with the smaller number 37 written under the greater number 65.

T O

1 T → 10

6 5

__- 3
7__

2 8

(i) first ones are subtracted as 5 < 7 or 7 > 5. So, 7 cannot be subtracted from 5.

(ii) Now 1 ten is borrowed from 6 tens leaving 5 tens there.

(iii) 1 ten = 10 ones. So, 10 ones are added to 5 ones making the sum 15 ones

(iv) 7 ones are subtracted from 15 ones i.e., 15 ones – 7 ones = 8 ones. This 8 is written in one’s column.

(v) Now tens are subtracted. At ten’s place there are 5 tens left. So 5 tens – 3 tens = 2 tens. So, 2 is written in ten’s column.

(vi) Therefore, 65 – 37 = 28.

**3.** Subtract 28
from 83

**Solution:**

The smaller number 28 is written under greater number 83 in column form and ones are subtracted first, then the tens.

T O

1 T → 10

8 3

__- 2
8__

5 5

(i) 3 < 8, so 1 ten, i.e., 10 ones are borrowed from 8 tens with 7tens remaining there.

(ii) Now, 1 ten + 3 = 10 + 3 = 13 ones. So, 13 ones – 8 ones = 5 ones.

(iii) 7 tens – 2 tens = 5 tens.

Therefore, 83 – 28 = 55

**4.** Subtract 69
from 92

**Solution:**

The smaller number 69 is written under greater number 92 in column form and ones are subtracted first, then the tens.

T O

1 T → 10

9 2

__- 6
9__

2 3

(i) 10 + 2 = 12; 12 O – 9 O = 3 O

(ii) 8 T – 6 T = 2 T

Therefore, 92 – 69 = 23

Questions and Answers on Subtracting 2-Digit Numbers with Borrowing (Regrouping):

**1. Subtracting 1-digit number from 2-digit number with regrouping.**

(i) 38 - 9 = _____

(ii) 62 - 7 = _____

(iii) 44 - 6 = _____

(iv) 67 - 8 = _____

(v) 75 - 7 = _____

(vi) 94 - 8 = _____

(vii) 74 - 5 = _____

(viii) 51 - 2 = _____

(ix) 95 - 6 = _____

(x) 42 - 3 = _____

(xi) 53 - 4 = _____

(xii) 62 - 4 = _____

(xiii) 67 - 8 = _____

(xiv) 32 - 5 = _____

(xv) 22 - 9 = _____

(xvi) 93 - 3 = _____

**Answers:**

(i) 29

(ii) 55

(iii) 38

(iv) 59

(v) 68

(vi) 86

(vii) 69

(viii) 49

(ix) 89

(x) 39

(xi) 49

(xii) 58

(xiii) 59

(xiv) 27

(xv) 13

(xvi) 90

**2. Subtracting 2-digit number with regrouping.**

(i) 80 - 17 = _____

(ii) 38 - 29 = _____

(iii) 71 - 34 = _____

(iv) 47 - 19 = _____

(v) 86 - 27 = _____

(vi) 65 - 47 = _____

(vii) 51 - 25 = _____

(viii) 62 - 37 = _____

(ix) 73 - 57 = _____

(x) 94 - 46 = _____

(xi) 46 - 28 = _____

(xii) 56 - 27 = _____

(xiii) 57 - 19 = _____

(xiv) 31 - 24 = _____

(xv) 41 - 16 = _____

(xvi) 53 - 29 = _____

**Answers:**

(i) 63

(ii) 9

(iii) 37

(iv) 28

(v) 59

(vi) 18

(vii) 26

(viii) 25

(ix) 16

(x) 48

(xi) 18

(xii) 29

(xiii) 38

(xiv) 7

(xv) 25

(xvi) 24

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