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Question

$0.5\overline{32}$ is equivalent to the fraction:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$\frac{527}{990}$

Convert Repeating Decimal to Fraction

The problem requires converting the repeating decimal $0.5\overline{32}$ into an equivalent fraction. The digits '32' repeat indefinitely.

Decimal to Fraction Conversion Steps

  1. Let the repeating decimal be represented by $x$.

    $x = 0.5323232...$

  2. Multiply $x$ by 10 to shift the decimal point just before the repeating block.

    $10x = 5.323232...$ (Equation 1)

  3. Multiply $x$ by $1000$ (10 times $10^2$, where 2 is the number of repeating digits) to shift the decimal point to the end of the first repeating block.

    $1000x = 532.323232...$ (Equation 2)

  4. Subtract Equation 1 from Equation 2 to eliminate the repeating decimal part.

    $ \begin{array}{rcrcr} 1000x & = & 532 & . & 323232... \\ - 10x & = & 5 & . & 323232... \\ \hline 990x & = & 527 & . & 000000... \end{array} $

    This simplifies to $990x = 527$.

  5. Solve for $x$ by dividing both sides by 990.

    $x = \frac{527}{990}$

The fraction equivalent to $0.5\overline{32}$ is $\frac{527}{990}$. This corresponds to Option 4.

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Important Questions from Decimals

  1. The value of \(\frac{1}{4} + \frac{{[{{(20.35)}^2} - {{(8.35)}^2}] \times 0.0175}}{{{{(1.05)}^2} + (1.05)(27.65)}}\)  is:

  2. The value of \(0.4\overline 6 + 0.7\overline {23} - 0.3\overline 9 \times 0.\overline 7 \)  is:

  3. The value of \(\frac{48.3\times[(4.95)^2+4.95\times13.25]}{[(12.55)^2-(5.65)^2]\times19.8} \)  is:

  4. Find the value of (1.6) 3 - (0.9) 3 - (0.7) 3.

  5. What is the value of x, if \(5\left( {1 - \frac{x}{5}} \right) - (5 - x) - \frac{1}{{200}}{\rm{of (20 - x) = 0}}{\rm{.08}}\) ?

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