Addition and subtraction
Addition of 5-digit and 6-digit numbers
Step-by-Step Guide to add numbers:
1. Write the numbers one below the other:
– Make sure to align the numbers by place value (ones, tens, hundreds, thousands, etc.).
– Place the digits in the correct columns: Ones under ones, tens under tens, and so on.
2. Start adding from the right (ones column):
– Always begin adding the numbers from the ones column, then move to the left.
3. Add each column:
– Add the digits in the ones column first.
– If the sum of the digits in any column is 10 or more, you need to carry over to the next column.
4. Carry over if necessary:
– If you get a two-digit number as the sum in any column, write the ones digit in that column and carry the tens digit to the next column.
5. Continue until you’ve added all columns:
– Keep moving to the left, adding digits in each column and carrying over if needed.
6. Write the final sum:
– Once you’ve added all the columns, write down the sum of the numbers.
Example: Adding 5-digit and 6-digit numbers
Let’s add 43,567 (5-digit number) and 123,456 (6-digit number).
123,456
+ 43,567
—————-
1. Start from the right (ones place):
– 6 + 7 = 13 (write 3 in the ones place, carry over 1 to the tens place).
2. Tens place:
– 5 + 6 = 11 (plus 1 carried over from the previous step, 11 + 1 = 12)
– Write 2 in the tens place, and carry over 1 to the hundreds place.
3. Hundreds place:
– 4 + 5 = 9 (plus 1 carried over, 9 + 1 = 10)
– Write 0 in the hundreds place, and carry over 1 to the thousands place.
4. Thousands place:
– 3 + 3 = 6 (plus 1 carried over, 6 + 1 = 7)
– Write 7 in the thousands place.
5. Ten thousands place:
– 2 + 4 = 6
– Write 6 in the ten thousands place.
6. Hundred thousands place:
– 1 + 0 = 1
– Write 1 in the hundred thousands place (since there’s no number to add from the second number).
So, the final sum is:
123,456
+43,567
—————
167,023
Addition without regrouping
Addition without regrouping means adding numbers where the sum of the digits in each place value (ones, tens, hundreds, etc.) is less than 10. Therefore, there’s no need to carry over any values to the next column.
Step-by-Step Guide for Adding 5-Digit and 6-Digit Numbers (Without Regrouping)
1. Write the numbers one below the other:
– Align the numbers so that each digit is in the correct place value (ones, tens, hundreds, thousands, etc.).
– The 5-digit number should be placed under the 6-digit number, and it will have a “gap” in the hundred thousand place.
2. Start adding from the right (ones place):
– Begin by adding the digits in the ones column, and then move left to the tens, hundreds, thousands, and so on.
3. No regrouping needed:
– Since the sum in each column is less than 10, you won’t need to carry over any values.
4. Write down the sum for each column:
– The sum of each place value is written directly in that column.
5. Complete the addition:
– Once you’ve added all the columns, the final sum is your result.
Example: Adding a 5-Digit and 6-Digit Number Without Regrouping
Let’s add 345,672 (6-digit number) and 12,345 (5-digit number).
345,672
+ 12,345
————–
1. Start with the ones place:
– 2 + 5 = 7
– Write 7 in the ones place.
2. Move to the tens place:
– 7 + 4 = 11 → This result requires regrouping, which is not allowed in this explanation! I’ll fix the example: let’s change the numbers so that this step doesn’t require regrouping.
Instead, let’s use 345,672 and 10,023 to ensure no regrouping happens.
New Example: Adding 345,672 and 10,023
345,672
+ 10,023
————–
1. Ones place:
– 2 + 3 = 5
– Write 5 in the ones place.
2. Tens place:
– 7 + 2 = 9
– Write 9 in the tens place.
3. Hundreds place:
– 6 + 0 = 6
– Write 6 in the hundreds place.
4. Thousands place:
– 5 + 0 = 5
– Write 5 in the thousands place.
5. Ten thousands place:
– 4 + 1 = 5
– Write 5 in the ten thousands place.
6. Hundred thousands place:
– The second number has no hundred thousands digit, so just bring down the 3 from 345,672.
The final sum is:
345,672
+ 10,023
—————
355,695
Key Points to Remember:
– No carrying over is required since the sum in each column is less than 10.
– Always start adding from the ones place and move to the left (towards tens, hundreds, thousands, etc.).
– Be careful to align the numbers correctly by their place values.
– You may have a “gap” in the highest place value if one number has fewer digits.
Addition with regrouping
Addition with regrouping happens when the sum of digits in any place value (ones, tens, hundreds, etc.) is 10 or greater. When this happens, we carry over the extra value to the next place value.
For example, if two digits add up to 12, we write 2 in the current column and carry over 1 to the next column (this is called regrouping or carrying over).
Step-by-Step Guide for Addition With Regrouping
1. Write the numbers one below the other:
– Align the numbers by their place value (ones, tens, hundreds, etc.).
– The 5-digit number should be placed below the 6-digit number, and there will be a gap in the hundred thousand place for the 5-digit number.
2. Start adding from the right (ones place):
– Add the digits in the ones column first. If the sum is greater than 9, you will regroup by carrying the tens digit to the next column.
3. Move left to the next place value (tens, hundreds, etc.):
– Add the digits in the next column, including any number you carried over from the previous step.
4. Carry over (regroup) when necessary:
– If the sum of the digits is 10 or more, write down the ones digit and carry the tens digit over to the next column.
5. Continue until all columns are added:
– Continue adding each column and regrouping as needed, moving from right to left.
6. Write down the final sum.
Example: Adding 5-Digit and 6-Digit Numbers With Regrouping
Let’s add 456,789 (6-digit number) and 34,567 (5-digit number).
456,789
+ 34,567
—————
1. Start with the ones place:
– 9 + 7 = 16 → Write 6 in the ones place and carry over 1 to the tens place.
2. Move to the tens place:
– 8 + 6 = 14, plus the 1 carried over makes 15.
– Write 5 in the tens place and carry over 1 to the hundreds place.
3. Move to the hundreds place:
– 7 + 5 = 12, plus the 1 carried over makes 13.
– Write 3 in the hundreds place and carry over 1 to the thousands place.
4. Move to the thousands place:
– 6 + 4 = 10, plus the 1 carried over makes 11.
– Write 1 in the thousands place and carry over 1 to the ten thousands place.
5. Move to the ten thousands place:
– 5 + 3 = 8, plus the 1 carried over makes 9.
– Write 9 in the ten thousands place.
6. Move to the hundred thousands place:
– 4 + 0 = 4 (no carrying over is needed here).
– Write 4 in the hundred thousands place.
The final sum is:
456,789
+ 34,567
————–
491,356
Another Example with Different Numbers
Let’s add 567,892 and 45,678.
567,892
+ 45,678
————–
1. Ones place:
– 2 + 8 = 10 → Write 0 in the ones place, carry over 1 to the tens place.
2. Tens place:
– 9 + 7 = 16, plus 1 carried over makes 17.
– Write 7 in the tens place, carry over 1 to the hundreds place.
3. Hundreds place:
– 8 + 6 = 14, plus 1 carried over makes 15.
– Write 5 in the hundreds place, carry over 1 to the thousands place.
4. Thousands place:
– 7 + 5 = 12, plus 1 carried over makes 13.
– Write 3 in the thousands place, carry over 1 to the ten thousands place.
5. Ten thousands place:
– 6 + 4 = 10, plus 1 carried over makes 11.
– Write 1 in the ten thousands place, carry over 1 to the hundred thousands place.
6. Hundred thousands place:
– 5 + 0 = 5, plus 1 carried over makes 6.
– Write 6 in the hundred thousands place.
The final sum is:
567,892
+ 45,678
—————
613,570
Key Points to Remember:
– Always start from the ones place and move left.
– If the sum of any place is 10 or more, write the ones digit in that place and carry over the tens digit to the next column.
– Make sure the numbers are aligned by place value (ones under ones, tens under tens, etc.).
– Be careful with regrouping (carrying over) and double-check each step to avoid mistakes.
Finding the missing digits in addition sums
Example 1: Finding Missing Digits in a 5-Digit Addition Problem
Let’s say we have the following sum with missing digits:
4 _ 3 5 9
+ 2 7 _ 4 6
—————
7 9 0 0 5
We’ll go step by step from the ones place to the ten-thousands place.
Step 1: Ones Place
The ones column is the rightmost. The sum here is 9 (from the first number) + 6 (from the second number) = 15. The 5 stays in the ones place, and 1 is carried over to the tens place.
Step 2: Tens Place
Now, look at the tens place. The sum here is 5 + 4 + 1 (carryover from the ones place) = 10. The 0 goes in the tens place, and 1 is carried to the hundreds place.
Step 3: Hundreds Place
Look at the hundreds place. In the sum, it says the result should be 0. We already have a carryover of 1 from the previous step.
– The first number has 3 in the hundreds place.
– The second number is missing here.
We need to think: What + 3 + 1 (carryover) = 0 (or 10, because there’s a carryover)? The missing number must be 6. So, 6 + 3 + 1 = 10, with a carryover of 1.
Step 4: Thousands Place
Look at the thousands place. We have 7 + the missing number + 1 (carryover) = 9.
– What + 7 + 1 = 9? The missing digit is 1.
Step 5: Ten-Thousands Place
Look at the ten-thousands place. The first number has 4, and the second number has 2.
– 4 + 2 = 6. Since the sum is 7, there must be a carryover of 1 from the previous step.
Final Answer:
So the missing digits are 1 and 6, and the completed sum looks like this:
4 1 3 5 9
+ 2 7 6 4 6
—————–
7 9 0 0 5
Example 2: Finding Missing Digits in a 6-Digit Addition Problem
Now, let’s look at an example with a 6-digit number:
_ 5 8 9 _ 2
+ 2 _ 6 4 7 3
——————-
8 1 _ 3 4 5
Step 1: Ones Place
In the ones place, we have 2 + 3 = 5, which matches the sum, so no missing digit here.
Step 2: Tens Place
In the tens place, we have 9 + 7 = 16. The 6 goes in the tens place, and 1 is carried over to the hundreds place.
Step 3: Hundreds Place
In the hundreds place, the sum says 4, but there’s a carryover of 1 from the tens place.
– The first number has 8, and the second number has 6.
– 8 + 6 + 1 (carryover) = 15, so the missing digit must be **5**.
Step 4: Thousands Place
In the thousands place, we have 5 + 4 = 9, but the sum shows 3.
– This means there must be a carryover from the next place.
– The missing digit is 3 because 3 + 5 + 1 (carryover) = 9.
Step 5: Ten-Thousands Place
In the ten-thousands place, we have the first number missing, and the second number is 2.
– The sum shows 1, so we need a carryover from the next place.
– To make this work, the missing number should be 7 because 7 + 2 = 9, which gives a carryover of 1.
Step 6: Hundred-Thousands Place
In the hundred-thousands place, the first number is missing, and the second number is 2.
– We already have a carryover of 1, and the sum shows 8.
– So, 7 + 2 + 1 = 8.
Final Answer:
The missing digits are 7, 5, and 3, and the completed sum looks like this:
7 5 8 9 3 2
+ 2 3 6 4 7 3
——————-
8 1 9 3 4 5
Tips for Students:
1. Start from the ones place and move left, just like normal addition.
2. Look for carryovers—this is key when sums go over 9 in any column.
3. Use subtraction to find the missing number when needed. For example, if you know the result is 9, and one number is 5, you can subtract: 9 – 5 = 4.
4. Write the carryover above the next column to help keep track of it.
Properties of addition
Addition is a basic arithmetic operation, and it follows certain properties that make it easier to understand and work with.
1. Commutative Property of Addition
Definition: The order in which you add two numbers does not change the sum.
Explanation: If you change the order of the numbers you are adding, the result will still be the same.
– Example:
– 5 + 3 = 8
– 3 + 5 = 8
Both give the same result, 8. This means 5 + 3 = 3 + 5.
Key idea: You can “switch” the numbers when adding, and the answer will remain the same.
2. Associative Property of Addition
Definition: When adding three or more numbers, the way you group the numbers does not change the sum.
Explanation: If you are adding more than two numbers, it doesn’t matter how you group them. The sum will be the same no matter where you place the brackets.
– Example:
– (2 + 3) + 4 = 5 + 4 = 9
– 2 + (3 + 4) = 2 + 7 = 9
In both cases, the sum is 9. So, (2 + 3) + 4 = 2 + (3 + 4).
Key idea: You can group the numbers differently when adding, and the total will still be the same.
3. Identity Property of Addition
Definition: Adding 0 to any number does not change the value of the number.
Explanation: When you add 0 to a number, the number stays the same. Zero is called the additive identity because it keeps the number’s identity.
– Example:
– 6 + 0 = 6
– 0 + 12 = 12
No matter how many zeros you add, the number remains the same.
Key idea: Adding zero to a number doesn’t change the number. It’s like “doing nothing” to the number.
4. Additive Inverse Property (for understanding)
This property is more advanced, but in a simple form, it says that for every number, there is another number that can be added to it to make zero.
– Example:
– The number 5 has an inverse, which is -5, because 5 + (-5) = 0.
5. Zero Property of Addition
This property is related to the identity property and can be summarized as:
– Any number plus zero equals that number.
– Example:
– 10 + 0 = 10
6. Closure Property of Addition
Definition: When you add two whole numbers, the result is always a whole number.
Explanation: If you take any two whole numbers and add them, the answer will always be another whole number. This is called being “closed” under addition.
– Example:
– 7 + 8 = 15 (15 is also a whole number)
– 4 + 10 = 14 (14 is a whole number)
Key idea: When you add two whole numbers, the result is always a whole number. There are no fractions or decimals involved.
Real life problem on addition
Let’s explore some real-life problems and learn how to solve them step by step.
1. Population Problem
Problem:
The population of City A is 45,320, and the population of City B is 58,675. What is the total population of both cities?
Solution:
We need to add the population of City A and City B to get the total population.
– 45,320 + 58,675
Let’s add them:
45,320
+ 58,675
————–
103,995
So, the total population of both cities is 103,995.
2. School Fundraising Problem
Problem:
A school raised ₹34,560 in the first year and ₹45,720 in the second year. How much money did the school raise in both years combined?
Solution:
We add the amount raised in the first year and the second year to get the total amount.
– ₹34,560 + ₹45,720
Add the numbers:
34,560
+ 45,720
—————
80,280
The school raised a total of **₹80,280** in both years.
3. Distance Between Cities Problem**
Problem:
The distance from Delhi to Agra is 231 km, and the distance from Agra to Jaipur is 276 km. How far is it from Delhi to Jaipur if you travel through Agra?
Solution:
We add the distance from Delhi to Agra and the distance from Agra to Jaipur to get the total distance.
– 231 km + 276 km
Add the numbers:
231
+ 276
————
507
So, the total distance from Delhi to Jaipur via Agra is 507 kilometers.
Subtraction of 5-digit and 6-digit numbers
Example 1: Subtracting 5-Digit Numbers
Let’s start with a simple 5-digit subtraction problem.
Problem:
Subtract 42,538 from 67,834.
– Step 1: Align the numbers by place value**. Make sure that the digits in the ones, tens, hundreds, thousands, and ten-thousands places are correctly lined up.
67,834
– 42,538
———————
– Step 2: Subtract starting from the ones place.
1. Ones place: Subtract 8 from 4. Since 4 is smaller than 8, we need to borrow 1 from the tens place. Change 3 to 2, and add 10 to the ones place (so now it’s 14):
– 14 – 8 = 6
2. Tens place: After borrowing, the tens place has 2. Subtract 3 from 2. Since 2 is smaller than 3, borrow 1 from the hundreds place, changing 8 to 7, and adding 10 to the tens place (now it’s 12):
– 12 – 3 = 9
3. Hundreds place: After borrowing, the hundreds place has 7. Subtract 5 from 7:
– 7 – 5 = 2
4. Thousands place: Subtract 2 from 7:
– 7 – 2 = 5
5. Ten-thousands place: Subtract 4 from 6:
– 6 – 4 = 2
– Step 3: Write the answer.
67,834
– 42,538
————–
25,296
So, the result is 25,296.
Example 2: Subtracting 6-Digit Numbers
Let’s try subtracting a 6-digit number now.
Problem:
Subtract 453,678 from 832,105.
– Step 1: Align the numbers by place value.
832,105
– 453,678
—————-
– Step 2: Subtract starting from the ones place.
1. Ones place: Subtract 8 from 5. Since 5 is smaller than 8, we need to borrow from the tens place. Change the tens place from 0 to 9 (after borrowing from the hundreds), and make the ones place 15:
– 15 – 8 = 7
2. Tens place: After borrowing, the tens place has 9. Subtract 7 from 9:
– 9 – 7 = 2
3. Hundreds place: After borrowing, the hundreds place has 0. Subtract 6 from 0. Since 0 is smaller than 6, borrow 1 from the thousands place, making it 10:
– 10 – 6 = 4
4. Thousands place: After borrowing, the thousands place has 1. Subtract 3 from 1. Since 1 is smaller than 3, borrow from the ten-thousands place, making it 11:
– 11 – 3 = 8
5. Ten-thousands place: After borrowing, the ten-thousands place has 2. Subtract 5 from 2. Since 2 is smaller than 5, borrow from the hundred-thousands place, making it 12:
– 12 – 5 = 7
6. Hundred-thousands place: After borrowing, the hundred-thousands place has 7. Subtract 4 from 7:
– 7 – 4 = 3
– Step 3: Write the answer.
832,105
– 453,678
—————–
378,427
So, the result is 378,427.
Step-by-Step Explanation of Borrowing (Regrouping):
1. Identify when borrowing is needed: Borrowing happens when the top digit in any column is smaller than the bottom digit. When this happens, borrow 1 from the next column to the left.
2. Adjust the numbers: Reduce the next column by 1 and increase the column where borrowing is needed by 10.
3. Repeat if needed: You may need to borrow from more than one column if multiple digits in the top number are smaller than the bottom number.
Example 3: Subtraction Without Borrowing
Let’s look at a subtraction problem where no borrowing is required.
Problem:
Subtract 123,456 from 654,321.
654,321
– 123,456
——————–
In this case, you can directly subtract digit by digit without borrowing:
1. Ones place: 1 – 6 = 5
2. Tens place: 2 – 5 = 7
3. Hundreds place: 3 – 4 = 9
4. Thousands place: 4 – 3 = 1
5. Ten-thousands place: 5 – 2 = 3
6. Hundred-thousands place: 6 – 1 = 5
So, the answer is:
654,321
-123,456
—————-
530,865
Subtraction without regrouping (borrowing)
Steps for Subtracting without Regrouping (Same as before but with larger numbers):
1. Line up the numbers by place value: Make sure the digits of ones, tens, hundreds, thousands, etc., are lined up properly.
2. Start from the right (ones place) and subtract digit by digit.
3. No borrowing (regrouping) is needed when the top number in each column is greater than or equal to the number below it.
Example with 5-Digit Numbers:
Let’s subtract 83,462 – 42,130.
83,462
– 42,130
—————-
1. Subtract the ones place:
2 – 0 = 2
2. Subtract the tens place:
6 – 3 = 3
3. Subtract the hundreds place:
4 – 1 = 3
4. Subtract the thousands place:
3 – 2 = 1
5. Subtract the ten thousands place:
8 – 4 = 4
So, the answer is 41,332.
83,462
– 42,130
—————-
41,332
Example with 6-Digit Numbers:
Now let’s subtract 764,531 – 523,410.
764,531
– 523,410
—————–
1. Subtract the ones place:
1 – 0 = 1
2. Subtract the tens place:
3 – 1 = 2
3. Subtract the hundreds place:
5 – 4 = 1
4. Subtract the thousands place:
4 – 3 = 1
5. Subtract the ten thousands place:
6 – 2 = 4
6. Subtract the hundred thousands place:
7 – 5 = 2
So, the answer is 241,121.
764,531
– 523,410
—————–
241,121
Key Points:
– In both examples, since each top digit is greater than or equal to the bottom digit, no borrowing (regrouping) is needed.
– Always subtract digit by digit starting from the ones place and moving left.
Subtraction with regrouping (borrowing)
Subtraction with regrouping (borrowing) is needed when the number in any place value (ones, tens, hundreds, etc.) on top is smaller than the number below it. In such cases, we “borrow” 1 from the next higher place value.
Steps for Subtraction with Regrouping (Borrowing):
1. Line up the numbers by place value: Ensure the ones, tens, hundreds, etc., are lined up.
2. Start from the right (ones place) and subtract.
3. If the top number is smaller than the bottom number, borrow 1 from the next place value to the left.
4. Continue borrowing if needed, moving left to higher place values.
5. Subtract once borrowing is done.
Example with 5-Digit Numbers:
Let’s subtract 72,435 – 48,267.
72,435
– 48,267
———–
1. Subtract the ones place:
5 – 7 → Since 5 is smaller than 7, we borrow 1 from the tens place:
– The 3 in the tens place becomes 2.
– Now, the 5 in the ones place becomes 15.
– Subtract 15 – 7 = 8.
2. Subtract the tens place:
2 – 6 → Again, 2 is smaller than 6, so we borrow from the hundreds place:
– The 4 in the hundreds place becomes 3.
– Now the 2 in the tens place becomes 12.
– Subtract 12 – 6 = 6.
3. Subtract the hundreds place:
3 – 2 = 1 (No need to borrow here).
4. Subtract the thousands place:
2 – 8 → 2 is smaller than 8, so we borrow from the ten thousands place:
– The 7 in the ten thousands place becomes 6.
– Now the 2 becomes 12.
– Subtract 12 – 8 = 4.
5. Subtract the ten thousands place:
6 – 4 = 2.
The answer is 24,168.
72,435
– 48,267
—————-
24,168
Example with 6-Digit Numbers:
Now let’s subtract 501,234 – 286,987.
501,234
– 286,987
—————-
1. Subtract the ones place:
4 – 7 → 4 is smaller than 7, so we borrow from the tens place:
– The 3 in the tens place becomes 2.
– Now the 4 becomes 14.
– Subtract 14 – 7 = 7.
2. Subtract the tens place:
2 – 8 → 2 is smaller than 8, so we borrow from the hundreds place:
– The 2 in the hundreds place becomes 1.
– Now the 2 in the tens place becomes 12.
– Subtract 12 – 8 = 4.
3. Subtract the hundreds place:
1 – 9 → 1 is smaller than 9, so we borrow from the thousands place:
– The 1 in the thousands place becomes 0.
– Now the 1 becomes 11.
– Subtract 11 – 9 = 2.
4. Subtract the thousands place:
0 – 6 → 0 is smaller than 6, so we borrow from the ten thousands place:
– The 0 in the ten thousands place becomes 9.
– Now the 0 becomes 10.
– Subtract 10 – 6 = 4.
5. Subtract the ten thousands place:
9 – 8 = 1.
6. Subtract the hundred thousands place:
5 – 2 = 2.
The answer is 214,247.
501,234
– 286,987
—————–
214,247
Key Points:
– Borrow from the next higher place value when the top digit is smaller than the bottom digit.
– Adjust the numbers as needed after borrowing, then subtract.
– Always start from the rightmost place value (ones) and move left.
Finding the missing digits in subtraction sums
Steps to Find the Missing Digits in Subtraction:
1. Line up the numbers by place value: Write the numbers so that the ones, tens, hundreds, etc., are lined up correctly.
2. Look at the digits you know: Use the parts of the subtraction problem that are already filled in to help figure out the missing digits.
3. Start subtracting from the ones place: Work from right to left. If a digit is missing, think about what would make the subtraction correct. Sometimes, you might need to borrow (regroup) from the next place value.
4. Use trial and error if needed: You might need to try different digits to see which one fits the subtraction correctly.
Example with 5-Digit Numbers:
Let’s solve a missing digit problem like this:
6 8, _ 5 4
– 4 _ , 7 2 3
_____________
2 3, _ 3 1
We need to find the missing digits.
Step-by-Step:
1. Look at the ones place:
We have 4 – 3 = 1. This is correct, so no missing digit here.
2. Look at the tens place:
We have 5 – 2 = 3. This is also correct, so no missing digit here.
3. Look at the hundreds place:
We have a missing digit on top: ? – 7 = ?. From the answer below, we know the hundreds digit should be 3.
This means we need to borrow:
– Change the thousands digit from 8 to 7 (since we’re borrowing).
– Now, in the hundreds place, we have 10 – 7 = 3.
So, the missing hundreds digit is 0.
4. Look at the thousands place:
Now, we have 7 – ? = 3. The missing digit must be 4 because 7 – 4 = 3.
5. Look at the ten thousands place:
We have 6 – 4 = 2, which is correct.
So, the missing digits are:
6 8, 0 5 4
– 4 4, 7 2 3
____________
2 3, 3 3 1
Example with 6-Digit Numbers:
Let’s solve another missing digit problem like this:
7 _ 3, 5 _ 6
– 4 8 5, _ 2 9
_______________
2 8 _, 3 9 7
Step-by-Step:
1. Look at the ones place:
6 – 9 → Since 6 is smaller than 9, we need to borrow from the tens place.
– Borrowing from the tens place makes the ? – 2 = ?.
– Now we have 16 – 9 = 7, so this works.
2. Look at the tens place:
We borrowed, so now we have ? – 2 = 9. This means the missing digit must be 1 because 11 – 2 = 9.
3. Look at the hundreds place:
We have 3 – ? = 3. The missing digit must be 0 because 3 – 0 = 3.
4. Look at the thousands place:
We have ? – 5 = 8. The missing digit must be 13 (because 13 – 5 = 8), after borrowing from the ten thousands place). So, the original digit was 6.
5. Look at the ten thousands place:
After borrowing, we have 6 – 8, but we borrowed from the hundred thousands place, making it 16 – 8 = 8.
6. Look at the hundred thousands place:
Finally, 7 – 4 = 2, which is correct.
So, the missing digits are:
7 6 3, 5 1 6
– 4 8 5, 2 2 9
______________
2 8 8, 3 9 7
Key Points:
– Always start from the ones place and work left.
– If the top digit is smaller than the bottom digit, borrow from the next higher place value.
– Use clues from the digits you already know to find the missing ones.
This method helps in solving missing digit problems step by step, even when borrowing (regrouping) is needed!
Properties of subtraction
Subtraction follows certain properties . These properties help in understanding how subtraction works and how to check answers. Let’s go through these properties-
1. Order Matters (Non-Commutative Property)
In subtraction, the order of numbers matters. If you change the order of the numbers, the result will be different.
– Example for 5-digit numbers:
– 34567 – 12345 ≠ 12345 – 34567
– 34567 – 12345 = 22222
– 12345 – 34567 = -22222
– Example for 6-digit numbers:
– 567890 – 234567 ≠ 234567 – 567890
– 567890 – 234567 = 333323
– 234567 – 567890 = -333323
– Conclusion:
The order of subtraction is important, and it is not reversible like addition.
2. Zero Property of Subtraction
When you subtract zero from any number, the result is the number itself. Subtracting zero does not change the value.
– Example for 5-digit numbers:
– 45678 – 0 = 45678
– Example for 6-digit numbers:
– 654321 – 0 = 654321
– Conclusion:
Subtracting zero from a number leaves it unchanged.
3. Subtraction of a Number from Itself
When you subtract a number from itself, the result is always zero.
– Example for 5-digit numbers:
– 78965 – 78965 = 0
– Example for 6-digit numbers:
– 987654 – 987654 = 0
– Conclusion:
Any number subtracted from itself results in zero.
4. Borrowing in Subtraction
When subtracting larger numbers, sometimes borrowing is required. Borrowing happens when the digit in the top number is smaller than the digit below it.
– Example for 5-digit numbers:
– Subtract 27894 from 40032 :
40032 – 27894
= 12138
Borrowing happens in the tens and hundreds place.
– Example for 6-digit numbers:
– Subtract 348719 from 523401 :
523401 – 348719 = 174682
Borrowing happens in multiple places.
5. Checking Subtraction Using Addition (Inverse Operation)
You can check the result of subtraction by adding the difference to the smaller number. If the sum equals the larger number, the subtraction is correct.
– Example for 5-digit numbers:
– 54321 – 32145 = 22176
– Check: 32145 + 22176 = 54321
– Example for 6-digit numbers:
– 782345 – 567890 = 214455
– Check: 567890 + 214455 = 782345
Summary of Properties:
1. Non-Commutative Property: Changing the order changes the result.
2. Zero Property: Subtracting zero from any number does not change the number.
3. Subtraction from Itself: Subtracting a number from itself results in zero.
4. Borrowing: Used when the digit in the top number is smaller than the one below it.
5. Check Subtraction: Using addition to verify the answer.
Real life problems on subtraction
Here are some real-life problems based on the subtraction of 5-digit and 6-digit numbers.
Problem1: Buying a Car
A family has saved ₹78,650 to buy a car. The car they want to buy costs ₹65,450. How much money will they have left after buying the car?
Solution:
To find out how much money the family will have left, we need to subtract the cost of the car from the total amount they saved.
78,650 – 65,450 = 13,200
Answer:
The family will have ₹13,200 left after buying the car.
Problem2: Population Difference
In two nearby towns, the population of Town A is 524,768, and the population of Town B is 348,567. What is the difference in the population between the two towns?
Solution:
To find the population difference, subtract the population of Town B from the population of Town A.
524,768 – 348,567 = 176,201
Answer:
The population difference between Town A and Town B is 176,201 people.
Addition and subtraction together
Steps for Solving Addition and Subtraction Problems:
1. Follow the Order:
– Always start with the numbers given in the problem.
– Perform addition first if that part comes first, or perform subtraction first if it comes first in the problem.
– Make sure to carefully solve step by step.
2. Use Brackets (if needed):
– Sometimes brackets (parentheses) are used to show which part of the problem should be solved first.
– Solve the operations inside the brackets before doing anything else.
Example 1:
A family had ₹56,420 in their savings account. They deposited ₹12,580 more in the account but later withdrew ₹8,350 for an emergency. How much money is left in the account?
Steps to Solve:
1. Start with the initial savings: ₹56,420
2. Add the deposit: ₹56,420 + ₹12,580
56,420 + 12,580 = 69,000
3. Subtract the withdrawal: ₹69,000 – ₹8,350
69,000 – 8,350 = 60,650
Answer:
The family has ₹60,650 left in their account.
Example 2:Anita went shopping with ₹45,760. She bought a dress for ₹12,340, a pair of shoes for ₹8,230, and returned an item worth ₹3,500. How much money does she have left after all these transactions?
Steps to Solve:
1. Start with the total amount Anita had: ₹45,760
2. Add the amount she got back: ₹45,760 + ₹3,500
45,760 + 3,500 = 49,260
3. Subtract the cost of the dress and shoes:
– Subtract the dress: ₹49,260 – ₹12,340
49,260 – 12,340 = 36,920
– Subtract the shoes: ₹36,920 – ₹8,230
36,920 – 8,230 = 28,690
Answer:
Anita has ₹28,690 left after shopping.
Estimating sum and difference
Steps for Estimating Sum and Difference:
1. Round the Numbers: First, round the given numbers to the nearest ten, hundred, thousand, or even ten thousand, depending on the size of the numbers.
– For 5-digit numbers, we often round to the nearest thousand or ten thousand.
2. Add or Subtract the Rounded Numbers: After rounding, perform the addition or subtraction using the simpler rounded numbers.
3. Understand It’s an Estimate: Remember, your result will be close to the actual answer, but it won’t be exact. This is useful for quickly checking if your exact answer makes sense.
Rounding Rules:
– If the digit to the right of the place you are rounding to is 5 or more, round up.
– If the digit is less than 5, round down.
Example 1: Estimate the sum of 43,589 and 56,217 by rounding to the nearest thousand.
Steps to Solve:
1. Round the numbers to the nearest thousand:
– 43,589 rounds to 44,000.
– 56,217 rounds to 56,000.
2. Add the rounded numbers:
44,000 + 56,000 = 100,000
Answer:
The estimated sum is 100,000.
Example 2: Estimate the difference between 85,654 and 43,789 by rounding to the nearest thousand.
Steps to Solve:
1. Round the numbers to the nearest thousand:
– 85,654 rounds to 86,000.
– 43,789 rounds to 44,000.
2. Subtract the rounded numbers:
86,000 – 44,000 = 42,000
Answer:
The estimated difference is 42,000.
Example 3: Estimate the sum of 567,892 and 432,345 by rounding to the nearest ten thousand.
Steps to Solve:
1. Round the numbers to the nearest ten thousand:
– 567,892 rounds to 570,000.
– 432,345 rounds to 430,000.
2. Add the rounded numbers:
570,000 + 430,000 = 1,000,000
Answer:
The estimated sum is 1,000,000.
Example 4: Estimate the difference between 789,654 and 345,276 by rounding to the nearest hundred thousand.
Steps to Solve:
1. Round the numbers to the nearest hundred thousand:
– 789,654 rounds to 800,000.
– 345,276 rounds to 300,000.
2. Subtract the rounded numbers:
800,000 – 300,000 = 500,000
Answer:
The estimated difference is 500,000.