Here we will learn subtracting 2digit numbers with borrowing. The subtractions with borrowing are solved stepbystep in four different ways.
Workedout examples on subtracting 2digit 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
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