The addition may appear simple for you in the earlier stages but you may feel it complex with each passing grade due to the complexity. Fourth Grade Addition Worksheets combine both fun and learning so that you won’t have any difficulty while solving the problems. Addition Problems available can be a great resource for 4th grade math children and they can rely on them to solve questions. Try your hand in the various problems solving on addition with and without regrouping, adding numbers in columns, finding the missing addend, etc.

Do Refer:

1. 5684 and 3218
2. 6827 and 1273
3. 1072 and 8003
4. 9362 and 2168

Solution:

1.
1. Start adding with one’s place digits.
4+8=12
Here the sum is 12. The tens digit of the sum i.e.1 will be carried to the tens place.
2. Add tens place digits along with the carryover 1.
= 1(carry) + 8 + 1 = 10.
Here the sum is 10. The tens digit of a sum (i.e., 1) will be carried to the hundreds place.
3. Now add the digit of hundreds place along with the carryover digit 1.
= 1(carry) + 6 + 2 = 9.
Here the sum is 9. Write 9 under the hundreds place.
=5+3=8
write 8 under thousands place.
Therefore, the addition of 5684 and 3218 is 8902.
2.Â
1. Start adding with one’s place digits.
7+3=10
Here the sum is 10. The tens digit of the sum i.e.1 will be carried to the tens place.
2. Add tens place digits along with the carryover 1.
= 1(carry) + 2 + 7 = 10.
Here the sum is 10. The tens digit of a sum (i.e., 1) will be carried to the hundreds place.
3. Now add the digit of hundreds place along with the carryover digit 1.
= 1(carry) + 8 + 2 = 11.
Here the sum is 11. Write 11 under the hundreds place.
4. Add thousand’s place digits along with carryover 1.
=1+6+1=8
write 8 under thousands place.
Therefore, the addition of 6827 and 1273 is 8100.
3.
1.Start adding with one’s place digits.
2+3=5
Here the sum is 5. write the sum under one’s place.
= 7 + 0 = 7.
Here the sum is 7. write the sum under ten’s place.
3. Now add the digit of hundreds place.
= 0+ 0 = 0.
Here the sum is 0. Write 0 under the hundreds place.
=1+8=9
write 9 under thousands place.
Therefore, the addition of 6827 and 1273 is 9075.
4.
1. Start adding with one’s place digits.
2+8=10
Here the sum is 10. The tens digit of the sum i.e.1 will be carried to the tens place.
2. Add tens place digits along with the carryover 1.
= 1(carry) + 6 + 6 = 13.
Here the sum is 13. The tens digit of a sum (i.e., 1) will be carried to the hundreds place.
3. Now add the digit of hundreds place along with the carryover digit 1.
= 1(carry) + 3 + 1 = 5.
Here the sum is 5. Write 5 under the hundreds place.
=9+2=11
Here the sum is 11. The tens digit of the sum i.e.1 will be carried to the ten thousand’s place.
5. Add ten thousand places. There is only carry over 1.
1+0=1
Write 1 under ten thousand’s place.
Therefore, the addition of 9362 and 2168 is 8100.

II. What is 4285 more than 2653?

Solution:

1. 1.Start adding with one’s place digits.
5+3=8
Here the sum is 8. write the sum under one’s place.
= 8 + 5 = 13.
Here the sum is 13. The tens digit of the sum i.e.1 will be carried to the hundred’s place.
3. Now add the digit of hundreds place along with carryover.
=1+ 2+ 6 = 9.
Here the sum is 9. Write 9 under the hundreds place.
=4+2=6
write 6 under thousands place.
Therefore, the addition of 4285 and 2653 is 6938.

III.Â  a)
b)
c)

Solution:

a) In one’s place, by adding 2 and 1 we get 3.
In tens place, the sum is given as 5. One addend is given as 1 and the other is missing. So the missing addend is (5-1=4) 4.
In hundred’s place, by adding 2 and 3 we get 5.
In thousand’s place, the sum is given as 8. One addend is 5 and the other is missing. So the missing addend is(8-5=3) 3.
b) In one’s place, the sum is given as 7.one addend is 2 and the other addend is missing. So the missing addend is (7-2=5)5.
In tens place, the sum is given as 4. One addend is given as 1 and the other is missing. So the missing addend is (4-1=3) 3.
In hundred’s place, by adding 2 and 6 we get 8.
In thousand’s place, the sum is given as 7. One addend is 5 and the other is missing. So the missing addend is(7-5=2) 2.
c) In one’s place, the sum is given as 8.one addend is 2 and the other addend is missing. So the missing addend is (8-2=6)6.
In tens place, the sum is given as 9. One addend is given as 8 and the other is missing. So the missing addend is (9-8=1) 1.
In hundred’s place, by adding 2 and 3 we get 5.
In thousand’s place, the sum is given as 6. One addend is 5 and the other is missing. So the missing addend is(6-5=1) 1.

IV. Write the next two numbers in the series
1. 2350,2400,2450,2500,_________ ,__________ .
2. 100,200,300,400,500,__________, ___________.
3. 10,12,14,16,18,_____ ,______ .
4. 2120,2140,2160,2180,2200,_____ ,______.

Solution:

1. In the given series of numbers 50 is added to every number to get the other number.
So the next two numbers are 2550,2600.
2. In the given series of numbers 100 is added to every number to get the other number.
So the next two numbers are 600,700.
3. In the given series of numbers 2 is added to every number to get the other number.
So the next two numbers are 20,22.
4. In the given series of numbers 20 is added to every number to get the other number.
So the next two numbers are 2220,2240.

V. Find the estimated sum by rounding to the nearest 10s and compare with the exact sum.
1.

S.No Question Rounded numbers Estimated Sum Exact Sum Which is Bigger
1 238 +159 Nearest 10’s
2 771 + 348 Nearest 10’s
3 626 + 298 Nearest 10’s
Solution:

S.No Question Rounded numbers Estimated Sum Exact Sum Which is Bigger
1 238 +159 Nearest 10’s 240 +160=400 397 estimated sum
2 771 + 348 Nearest 10’s 770 +350=1120 1119 estimated sum
3 626 + 298 Nearest 10’s 630+300=930 524 estimated sum

VI. Find the estimated sum by rounding to the nearest 100s and compare with the exact sum.

S.No Question Rounded numbers Estimated Sum Exact Sum Which is Bigger
1 3580 + 12280 Nearest 100’s
2 1721 + 2388 Nearest 100’s
3 9646 + 6298 Nearest 100’s
Solution:

S.No Question Rounded numbers Estimated Sum Exact Sum Which is Bigger
1 3582 + 12279 Nearest 100’s 3580 +12280=15860 15861 exact sum
2 1721 + 2388 Nearest 100’s 1720 + 2390=4110 4109 estimated sum
3 9646 + 6298 Nearest 100’s 9650 +6300=15950 15944 estimated sum

VII. Rajesh scored 23 marks in English and 24 marks in Maths. How many marks did he get in both subjects?

Solution:

Given,
Rajesh scored marks in english=23
Rajesh scored marks in Maths=24
Rajesh got marks in both the subjects=23+24=47 marks
Therefore, Rajesh got 47 marks in both subjects.

VIII. Ram saved the money of Rs 2000. His father gave him Rs 3000. Find the amount of money Ram has?

Solution:

Ram has money=Rs 2000
Father gave money=Rs 3000
Total amount of money Ram has=Rs 2000 + Rs 3000=Rs 5000
Hence,Ram has money=Rs 5000.

1. 1285 + 2397=
a) 3582
b) 3682
c) 3480
d) 3282
2. 121 + 225 is the same as
a) 121
b)225
c) 225 + 121
d) All the above
a) number itself
b) zero
c) Null
d)None of the above
4. Adding 1 to a number gives _______ of the number
a) predecessor
b) Successor
c) No change
d) 0

1. b
2. c
3. a
4. b

X. Match the following
1. 9289 + 2189Â  Â  Â  Â  Â  Â  a) 1114
2. 287 + 827Â  Â  Â  Â  Â  Â  Â  Â  b)Â  1700
3. 1543 + 2315Â  Â  Â  Â  Â  Â  Â c)Â  11478
4. 1200 + 500Â  Â  Â  Â  Â  Â  Â  Â  d) 3858