Calculating the Sum of Recurring Decimals

The problem asks us to find the sum of three numbers, \(A\), \(B\), and \(C\), which are given in the form of recurring decimals. To add these numbers, it is easiest to convert each recurring decimal into a fraction.

Converting Recurring Decimals to Fractions

A recurring decimal is a decimal representation of a number whose digits after a certain point are periodic. These numbers are also rational numbers and can be expressed in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).

The general method to convert a recurring decimal to a fraction involves setting the decimal equal to a variable (say, \(x\)), multiplying by appropriate powers of 10 to align the repeating part, and then subtracting the equations to eliminate the repeating part.

Converting \(A = 0.3\overline{12}\) to a Fraction

Let \(A = 0.3\overline{12}\). This means \(A = 0.3121212...\)

• First, multiply by 10 to move the non-repeating part (\(3\)) to the left of the decimal point:
  \(10A = 3.121212...\)   (Equation 1)
• Next, identify the repeating block, which is \(12\). It has 2 digits. Multiply the original number by \(10^2 = 100\) to shift the decimal point two places to the right:
  \(100A = 31.21212...\)
• However, we need to align the repeating parts. Multiply Equation 1 by 100 (since there are 2 repeating digits):
  \(10 \times 100 A = 3.121212... \times 100\)
  \(1000A = 312.121212...\)    (Equation 2)
• Now, subtract Equation 1 from Equation 2 to eliminate the repeating part:
  \(1000A - 10A = (312.121212...) - (3.121212...)\)
  \(990A = 309\)
• Solve for \(A\):
  \(A = \frac{309}{990}\)
• Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 3:
  \(A = \frac{309 \div 3}{990 \div 3} = \frac{103}{330}\)

So, \(A = \frac{103}{330}\).

Converting \(B = 0.4\overline{15}\) to a Fraction

Let \(B = 0.4\overline{15}\). This means \(B = 0.4151515...\)

• Multiply by 10 to move the non-repeating part (\(4\)) to the left:
  \(10B = 4.151515...\)    (Equation 3)
• The repeating block is \(15\), which has 2 digits. Multiply Equation 3 by \(10^2 = 100\):
  \(10 \times 100 B = 4.151515... \times 100\)
  \(1000B = 415.151515...\)    (Equation 4)
• Subtract Equation 3 from Equation 4:
  \(1000B - 10B = (415.151515...) - (4.151515...)\)
  \(990B = 411\)
• Solve for \(B\):
  \(B = \frac{411}{990}\)
• Simplify the fraction by dividing the numerator and denominator by 3:
  \(B = \frac{411 \div 3}{990 \div 3} = \frac{137}{330}\)

So, \(B = \frac{137}{330}\).

Converting \(C = 0.30\overline{9}\) to a Fraction

Let \(C = 0.30\overline{9}\). This means \(C = 0.309999...\)

A recurring decimal ending in a repeating \(9\) can be simplified. For example, \(0.\overline{9} = 1\), \(0.1\overline{9} = 0.2\), \(0.30\overline{9} = 0.31\).

Using the standard method:

• Multiply by 100 to move the non-repeating part (\(30\)) to the left:
  \(100C = 30.9999...\)    (Equation 5)
• The repeating block is \(9\), which has 1 digit. Multiply Equation 5 by \(10^1 = 10\):
  \(100 \times 10 C = 30.9999... \times 10\)
  \(1000C = 309.9999...\)    (Equation 6)
• Subtract Equation 5 from Equation 6:
  \(1000C - 100C = (309.9999...) - (30.9999...)\)
  \(900C = 279\)
• Solve for \(C\):
  \(C = \frac{279}{900}\)
• Simplify the fraction by dividing the numerator and denominator by 9:
  \(C = \frac{279 \div 9}{900 \div 9} = \frac{31}{100}\)

Alternatively, recognising \(0.30\overline{9} = 0.31\), we directly get \(C = 0.31 = \frac{31}{100}\).

So, \(C = \frac{31}{100}\).

Calculating the Sum \(A + B + C\)

Now we need to add the fractions we found:

First, add the fractions with the same denominator:

Simplify this fraction:

Now, add this result to the fraction for \(C\):

To add these fractions, we need a common denominator. The least common multiple (LCM) of 11 and 100 is \(11 \times 100 = 1100\).

• Convert \(\frac{8}{11}\) to a fraction with denominator 1100:
  \(\frac{8}{11} = \frac{8 \times 100}{11 \times 100} = \frac{800}{1100}\)
• Convert \(\frac{31}{100}\) to a fraction with denominator 1100:
  \(\frac{31}{100} = \frac{31 \times 11}{100 \times 11} = \frac{341}{1100}\)
• Add the converted fractions:
  \(\frac{800}{1100} + \frac{341}{1100} = \frac{800 + 341}{1100} = \frac{1141}{1100}\)

Conclusion

The sum of \(A\), \(B\), and \(C\) is \(\frac{1141}{1100}\).

The sum \(A + B + C = \frac{1141}{1100}\).

Revision Table: Recurring Decimals and Fractions

Additional Information: Properties of Rational Numbers

Recurring decimals represent rational numbers. Rational numbers can be added, subtracted, multiplied, and divided (except by zero), and the result is always another rational number. This property is called closure. Converting recurring decimals to fractions allows us to perform arithmetic operations using the standard rules for fractions.

The value \(0.30\overline{9}\) is a specific case. The repeating digit 9 after some non-repeating digits or a finite decimal part indicates that the number is equivalent to a terminating decimal where the last non-9 digit is incremented by one. For instance, \(0.30\overline{9} = 0.31\), \(0.5\overline{9} = 0.6\), \(2.41\overline{9} = 2.42\). Understanding this equivalence can sometimes simplify the conversion process.