$(0.34\overline{67} + 0.13\overline{33})$ is equal to

Concept & Formula When adding mixed recurring decimals, direct vertical addition can be tricky due to carryovers cascading infinitely. The most foolproof method is converting them into fractions, summing them, and converting the result back to a mixed recurring decimal. $$ \text{Fraction} = \frac{\text{Entire Number} - \text{Non-repeating part}}{\text{9s (repeating count)} \text{0s (non-repeating count)}} $$ Step-by-Step Solution Given expression: $(0.34\overline{67} + 0.13\overline{33})$ 1. Convert both decimals to fractions: $$ 0.34\overline{67} = \frac{3467 - 34}{9900} = \frac{3433}{9900} $$ $$ 0.13\overline{33} = \frac{1333 - 13}{9900} = \frac{1320}{9900} $$ 2. Add the fractions: $$ \frac{3433}{9900} + \frac{1320}{9900} = \frac{4753}{9900} $$ 3. Convert the resulting fraction back to a decimal: Because the denominator is $9900$, the final decimal has two non-repeating digits and two repeating digits, in the format $0.ab\overline{cd}$. Using the formula backwards: $$ \frac{abcd - ab}{9900} = \frac{4753}{9900} $$ $$ abcd - ab = 4753 $$ Let's test $ab = 48$ (since the complete number is close to $4753$): $$ 48cd - 48 = 4753 $$ $$ 48cd = 4753 + 48 $$ $$ 48cd = 4801 $$ This means the complete four-digit sequence is $4801$. Thus, $ab = 48$ and $cd = 01$. 4. Write the final decimal: $$ 0.48\overline{01} $$ Exam Strategy & Shortcut You can use vertical addition if you expand carefully and watch for the carry-over from the repeating block. $$ \quad 0.34676767... $$ $$ + 0.13333333... $$ Notice $67 + 33 = 100$. This causes a carry of $1$ into the next block. $$ \quad 0.48010100... $$ The pattern $01$ repeats infinitely, yielding $0.48\overline{01}$. Common Pitfall A major pitfall is adding $3467 + 1333$ to get $4800$ and incorrectly concluding the answer is $0.48\overline{00}$ or just $0.48$ (Option d). This ignores the fact that the infinite sequence $6767... + 3333...$ continuously generates carry-overs. Final Answer Therefore, the correct answer is $0.48\overline{01}$.