Solving Arithmetic Problems Addition, Subtraction, And Multiplication With Percentage Explanation

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In this comprehensive guide, we will delve into basic mathematical operations with detailed examples. This guide aims to clarify the steps involved in solving various arithmetic problems, including addition, subtraction, and multiplication. Whether you're a student looking to improve your math skills or simply refreshing your knowledge, this article provides step-by-step explanations to help you master these fundamental concepts. Let's dive in and explore each operation with practical examples.

Addition Problems

Problem 42: Adding Multiple Decimal Numbers

The first addition problem involves summing several decimal numbers: R 307.47 + R 40.23 + R 11.99 + R 169.35. Addition is a fundamental operation in mathematics, and when dealing with decimal numbers, it’s crucial to align the decimal points correctly. This ensures that you are adding the corresponding place values (ones with ones, tenths with tenths, hundredths with hundredths, and so on). This meticulous alignment is the cornerstone of accurate decimal addition.

To begin, write the numbers vertically, aligning the decimal points:

  307.47
   40.23
   11.99
+ 169.35
-------

Now, start adding from the rightmost column (the hundredths place). Add 7, 3, 9, and 5, which totals 24. Write down the 4 and carry over the 2 to the tenths column. This carry-over is a crucial step in addition, representing the overflow from one place value to the next, ensuring the sum remains accurate.

Next, add the numbers in the tenths column, including the carry-over: 2 (carry-over) + 4 + 2 + 9 + 3 = 20. Write down the 0 and carry over the 2 to the ones column. The process of carrying over is a testament to the place value system, where each digit's position determines its magnitude, and overflows must be accounted for in the adjacent higher place value.

In the ones column, add the numbers along with the carry-over: 2 (carry-over) + 7 + 0 + 1 + 9 = 19. Write down the 9 and carry over the 1 to the tens column. This step reinforces the concept of place value, as the carry-over from the ones column directly impacts the sum in the tens column.

Now, add the numbers in the tens column, including the carry-over: 1 (carry-over) + 0 + 4 + 1 + 6 = 12. Write down the 2 and carry over the 1 to the hundreds column. The tens column represents the multiples of ten, and the carry-over ensures that the sum accurately reflects the total number of tens.

Finally, add the numbers in the hundreds column, including the carry-over: 1 (carry-over) + 3 + 1 = 5. Write down the 5. The hundreds column is the final column in this particular problem, and its accurate calculation is vital to the overall sum.

Combine the results to get the final answer: R 529.04. Therefore, R 307.47 + R 40.23 + R 11.99 + R 169.35 = R 529.04. This final result is the cumulative total of all the individual amounts, achieved through a systematic process of aligning decimal points and carrying over excess values from one place value to the next.

Problem 43: Summing Multiple Amounts Including Rand and Cents

Problem 43 involves adding R 1,203.66 + R 11,698.01 + R 500.00 + R 0.36 + R 3,552.75. This addition problem requires careful alignment of decimal points to ensure accurate summation of both the Rand and cents values. Aligning the numbers correctly is crucial because it ensures that corresponding place values are added together—thousands with thousands, hundreds with hundreds, ones with ones, tenths with tenths, hundredths with hundredths, and so on. This alignment is fundamental to the accuracy of the final sum.

Set up the problem by writing the numbers vertically, ensuring the decimal points are aligned:

  1,203.66
 11,698.01
   500.00
     0.36
+ 3,552.75
---------

Start by adding the hundredths column: 6 + 1 + 0 + 6 + 5 = 18. Write down the 8 and carry over the 1 to the tenths column. The hundredths column represents the smallest monetary unit (cents in this case), and accurate summation here is essential for the final total.

Add the tenths column, including the carry-over: 1 (carry-over) + 6 + 0 + 0 + 3 + 7 = 17. Write down the 7 and carry over the 1 to the ones column. The tenths column represents the first decimal place, and including the carry-over from the hundredths column ensures precision.

Now, add the ones column, including the carry-over: 1 (carry-over) + 3 + 8 + 0 + 0 + 2 = 14. Write down the 4 and carry over the 1 to the tens column. The ones column is the anchor for the whole number part of the amounts, and accurate summation here is vital.

Add the tens column, including the carry-over: 1 (carry-over) + 0 + 9 + 0 + 0 + 5 = 15. Write down the 5 and carry over the 1 to the hundreds column. The tens column represents multiples of ten, and the carry-over reflects the overflow into the next higher place value.

Next, add the hundreds column, including the carry-over: 1 (carry-over) + 2 + 6 + 5 + 0 + 5 = 19. Write down the 9 and carry over the 1 to the thousands column. The hundreds column represents multiples of one hundred, and accurate summation ensures the total hundreds are correctly accounted for.

Add the thousands column, including the carry-over: 1 (carry-over) + 1 + 1 + 0 + 3 = 6. Write down the 6. The thousands column is a significant component of these larger amounts, and accurate summation is crucial.

Finally, add the ten-thousands column: 1. Write down the 1. This final column completes the summation of the largest place values in the problem.

Combine the results to get the final answer: R 16,954.78. Thus, R 1,203.66 + R 11,698.01 + R 500.00 + R 0.36 + R 3,552.75 = R 16,954.78. This total represents the cumulative sum of all the individual amounts, derived from meticulously adding each place value and carrying over where necessary.

Subtraction Problems

Problem 44: Subtracting Decimal Numbers

The subtraction problem presented is R 336.98 - R 26.35. Subtraction, the inverse operation of addition, involves finding the difference between two numbers. When dealing with decimals, aligning the decimal points is essential to ensure accurate subtraction of corresponding place values. Aligning decimal points allows for the direct subtraction of hundredths from hundredths, tenths from tenths, ones from ones, and so forth, maintaining the integrity of the decimal system.

Begin by writing the numbers vertically, aligning the decimal points:

  336.98
-  26.35
-------

Start subtracting from the rightmost column, the hundredths place: 8 - 5 = 3. Write down the 3. This straightforward subtraction in the hundredths place provides the first digit of the difference.

Next, subtract the tenths column: 9 - 3 = 6. Write down the 6. Subtracting the tenths ensures the accuracy of the first decimal place in the result.

Now, move to the ones column: 6 - 6 = 0. Write down the 0. Subtracting the ones provides the units digit of the difference.

In the tens column, subtract: 3 - 2 = 1. Write down the 1. The tens column subtraction gives the tens digit in the result.

Finally, in the hundreds column, there is only 3, so write down 3. This completes the subtraction, providing the hundreds digit of the difference.

Combine the results to get the final answer: R 310.63. Therefore, R 336.98 - R 26.35 = R 310.63. This final amount is the difference between the two original amounts, calculated by systematically subtracting each place value.

Problem 45: Subtraction with Borrowing

Problem 45 is R 1,365.23 - R 966.89. This subtraction problem involves borrowing, a critical technique when a digit in the subtrahend (the number being subtracted) is larger than the corresponding digit in the minuend (the number from which we are subtracting). Borrowing involves reallocating values from higher place values to lower ones, enabling the subtraction to proceed smoothly. This process ensures that the difference between the two numbers is accurately calculated, even when individual digits in the minuend are smaller than their counterparts in the subtrahend.

Write the numbers vertically, aligning the decimal points:

  1,365.23
-  966.89
-------

Start with the hundredths column: 3 - 9. Since 3 is less than 9, we need to borrow 1 from the tenths place. This makes the 2 in the tenths place a 1, and the 3 in the hundredths place becomes 13. Now, subtract: 13 - 9 = 4. Write down the 4. Borrowing allows us to perform subtraction even when the digit in the minuend is smaller, which is a fundamental technique in arithmetic.

Move to the tenths column: 1 - 8. Again, 1 is less than 8, so we borrow 1 from the ones place. This makes the 5 in the ones place a 4, and the 1 in the tenths place becomes 11. Subtract: 11 - 8 = 3. Write down the 3. Borrowing from the ones place is necessary to continue the subtraction process accurately.

Now, look at the ones column: 4 - 6. Since 4 is less than 6, borrow 1 from the tens place. The 6 in the tens place becomes 5, and the 4 in the ones place becomes 14. Subtract: 14 - 6 = 8. Write down the 8. Borrowing from the tens place allows for the accurate calculation of the difference in the ones place.

In the tens column, subtract: 5 - 6. Again, 5 is less than 6, so we borrow 1 from the hundreds place. The 3 in the hundreds place becomes 2, and the 5 in the tens place becomes 15. Subtract: 15 - 6 = 9. Write down the 9. Borrowing from the hundreds place ensures the accurate calculation of the tens digit in the difference.

For the hundreds column, subtract: 2 - 9. Since 2 is less than 9, borrow 1 from the thousands place. The 1 in the thousands place becomes 0, and the 2 in the hundreds place becomes 12. Subtract: 12 - 9 = 3. Write down the 3. This step completes the borrowing process, ensuring all necessary adjustments are made for accurate subtraction.

Finally, in the thousands column, we have 0 left. Write down 0. The thousands place calculation is completed, leaving us with the final result.

Combine the results to get the final answer: R 398.34. Therefore, R 1,365.23 - R 966.89 = R 398.34. This is the final difference between the two amounts, achieved through careful borrowing and subtraction of each place value.

Multiplication Problems

Problem 46: Multiplying Two-Digit Numbers

The multiplication problem presented is 63 Ă— 15. Multiplication is a fundamental arithmetic operation that involves combining groups of equal sizes. This problem requires multiplying two two-digit numbers, which can be efficiently done using the standard multiplication algorithm. This algorithm breaks down the multiplication into simpler steps, multiplying each digit of one number by each digit of the other number, and then summing the results. This systematic approach ensures accuracy and clarity in the multiplication process.

Set up the problem:

   63
Ă—  15
----

First, multiply 63 by the ones digit of 15, which is 5:

  • 5 Ă— 3 = 15. Write down 5 and carry over 1.
  • 5 Ă— 6 = 30. Add the carry-over 1 to get 31. Write down 31.

The first partial product is 315.

Next, multiply 63 by the tens digit of 15, which is 1. Since we are multiplying by the tens digit, we add a 0 as a placeholder in the ones place of the second partial product:

  • 1 Ă— 3 = 3. Write down 3 in the tens place.
  • 1 Ă— 6 = 6. Write down 6 in the hundreds place.

The second partial product is 630.

Now, add the two partial products:

   315
+ 630
----

Add the columns:

  • 5 + 0 = 5
  • 1 + 3 = 4
  • 3 + 6 = 9

The result is 945. Therefore, 63 Ă— 15 = 945. This final result is the product of the two numbers, achieved through the systematic multiplication of digits and subsequent addition of partial products.

Problem 47: Multiplying Two-Digit Numbers

Problem 47 is 45 Ă— 23. This multiplication problem, similar to the previous one, involves multiplying two two-digit numbers. The same standard multiplication algorithm applies here, breaking down the multiplication into manageable steps. This involves multiplying each digit of one number by each digit of the other number and then summing the resulting partial products. This methodical approach ensures that the multiplication is performed accurately and efficiently.

Set up the problem:

   45
Ă—  23
----

First, multiply 45 by the ones digit of 23, which is 3:

  • 3 Ă— 5 = 15. Write down 5 and carry over 1.
  • 3 Ă— 4 = 12. Add the carry-over 1 to get 13. Write down 13.

The first partial product is 135.

Next, multiply 45 by the tens digit of 23, which is 2. Add a 0 as a placeholder in the ones place of the second partial product:

  • 2 Ă— 5 = 10. Write down 0 and carry over 1.
  • 2 Ă— 4 = 8. Add the carry-over 1 to get 9. Write down 9.

The second partial product is 900.

Now, add the two partial products:

   135
+ 900
----

Add the columns:

  • 5 + 0 = 5
  • 3 + 0 = 3
  • 1 + 9 = 10. Write down 0 and carry over 1.
  • Write down the carry-over 1.

The result is 1035. Therefore, 45 Ă— 23 = 1035. This is the product of 45 and 23, calculated by multiplying each digit and summing the partial products in a systematic manner.

Discussion Category

Problem 48: Understanding Percentages

Problem 48 mentions “10 percent.” This is an introduction to the concept of percentages, which are a fundamental part of mathematics and everyday life. A percentage is a way of expressing a number as a fraction of 100. The term