Question

If \[x=0.02\overline{7}\] show that \[12x=0.\overline{3}.\]

Answer 1

Hint: We have a number in the decimal expansion with the repeated term. We will first convert x into the rational (fractional) form by cancelling the repeated term. Once we have x in the fractional form, we will multiply it by 12 and show that 12x is \[0.\overline{3}\] in the decimal expansion.

Complete step-by-step answer:
We are given that \[x=0.02\overline{7}\] and we are asked to show that \[12x=0.\overline{3}.\] We will first find the fractional form of x and then we will solve further.
Now, we have x as
\[x=0.02\overline{7}.....\left( i \right)\]
We can see that the repeating number is after the second place from the decimal. So, we will multiply equation (i) by 100.
Multiplying (i) by 100, we get,
\[\Rightarrow 100x=2.\overline{7}.....\left( ii \right)\]
As the repeating term is just 1 digit, so we multiply (ii) by 10. So, we get,
\[\Rightarrow 1000x=27.\overline{7}.....\left( iii \right)\]
Now, we get rid of the repeating part. To do so, we will subtract (ii) from (iii). So, we get,
\[\begin{align}
  & 1000x=27.\overline{7} \\
 & 100x=2.\overline{7} \\
 & \underline{-\text{ }-} \\
 & 900x=25.0 \\
\end{align}\]
Now, we have,
\[\Rightarrow 900x=25\]
Divide both the sides by 900, we get,
\[\Rightarrow \dfrac{900x}{900}=\dfrac{25}{900}\]
Simplifying further, we get,
\[\Rightarrow x=\dfrac{1}{36}\]
So, we get the fraction from \[x=0.02\overline{7}\] as \[\dfrac{1}{36}.\]
Now, we found 12x as \[x=\dfrac{1}{36}.\]
So,
\[\Rightarrow 12x=12\times \dfrac{1}{36}\]
Solving further, we get,
\[\Rightarrow 12x=\dfrac{1}{3}\]
Changing \[\dfrac{1}{3}\] to decimal form, we get,
\[\Rightarrow 12x=0.\overline{3}\]
Hence proved.

Note: We always first shift the decimal near to the repeating term and then shift the repeating term and then lastly subtract to cancel the repeating term. If we directly shift the decimal to the repeating term, then we get no solution.
As, \[x=0.02\overline{7}.....\left( i \right)\]
Multiplying both the sides by 1000 to shift the decimal to the repeating term, we get,
\[\Rightarrow 1000x=27.\overline{7}.....\left( ii \right)\]
Now, if we subtract (i) from (ii), we get no solution as there is no difference in the location of the repeating term.