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In the following problem, by adding $0.141414$..., $0.414141..., 0.151515...$, and $0.515151...$, I get $1.111....$ Then the expression becomes square root ($11 \times 1.1111$....). My answer is $11 \times \sqrt{0.1010101...}$, which is different from the answer sheet. Any help is highly appreciated.

What is the value of the following expression? Express your answer as a common fraction.

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M47145
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  • Do you know how to convert repeating decimals into fractions? – Fimpellizzeri May 18 '16 at 00:45
  • Without going into the repeating decimal part (and converting it to a fraction), should $0.\overline{14} + 0.\overline{41} = 0.\overline{5}$ and $0.\overline{15} + 0.\overline{51} = 0.\overline{6}$. Shouldn't you get $1.\overline{2}$? – Jared May 18 '16 at 00:54
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    $x = .4141....$ so $100x - 41.4141....$ Now subtract. – ncmathsadist May 18 '16 at 00:56

3 Answers3

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Resolving the radicand, you have $$11\cdot(14/99 + 41/99 + 15/99 + 51/99) = 1331/99 = 121/9$$ Taking the square root of this you are left with $11/3$.

ncmathsadist
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Notice that $0.\overline{14}+0.\overline{41}+0.\overline{15}+0.\overline{51}=1.\overline{2} = 1+\frac{2}{9}$.

Now $\sqrt{11(1+\frac{2}{9})}=\sqrt{\frac{99}{9}+\frac{22}{9}}=\sqrt{\frac{121}{9}}=\frac{11}{3}=3\frac{2}{3}$

M47145
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2

I like @ncmathsadist's suggestion. If you can reason that $0.\overline{14} + 0.\overline{41} = 0.\overline{5}$ and $0.\overline{15} + 0.\overline{51} = 0.\overline{6}$ Then think about it:

\begin{align} &0.555 \\ +&0.666 \\\hline &1.221 \end{align}

Clearly if it repeats forever, $0.\overline{5} + 0.\overline{6} = 1.\overline{2}$. Then we can solve by doing the following:

$$ x = 1.\overline{2} \\ 10x = 12.\overline{2}\\ 10x - x = 9x \rightarrow 9x = 12 - 1 = 11 $$

Therefore $x = 1.\overline{2} = \frac{11}{9}$. Then we have:

$$ \sqrt{11\cdot\frac{11}{9}} = \sqrt{\frac{11^2}{3^2}} = \frac{11}{3} = 3\frac{2}{3} $$

Jared
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