The problem requires converting the repeating decimal $0.5\overline{32}$ into an equivalent fraction. The digits '32' repeat indefinitely.
Let the repeating decimal be represented by $x$.
$x = 0.5323232...$
Multiply $x$ by 10 to shift the decimal point just before the repeating block.
$10x = 5.323232...$ (Equation 1)
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)
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 $990x = 527$.
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.
Simplify the given expression.
$9 \times 0.9 \times 0.09 \times 0.009 \times \frac{1}{0.3} \times \frac{1}{0.03} \times \frac{1}{0.003}$
Arrange the following numbers in their increasing order.
(1) $-0.96$
(2) $0.83$
(3) $0.24$
(4) $-0.64$
(5) $0.58$
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The value of \(0.4\overline 6 + 0.7\overline {23} - 0.3\overline 9 \times 0.\overline 7 \) is:
The value of \(\frac{48.3\times[(4.95)^2+4.95\times13.25]}{[(12.55)^2-(5.65)^2]\times19.8} \) is:
Find the value of (1.6) 3 - (0.9) 3 - (0.7) 3.
What is the value of x, if \(5\left( {1 - \frac{x}{5}} \right) - (5 - x) - \frac{1}{{200}}{\rm{of (20 - x) = 0}}{\rm{.08}}\) ?