# Subtracting 2-digit Numbers with Borrowing (Regrouping)

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 = 8We write 8 in ones place.Subtract 3 tens from 5 tens.5 tens - 3 tens = 2 tensWrite 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 = _____

(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 = _____

(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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