Recurring decimals to fraction

Question

1. $.\overline{36}=\frac{36}{99}$
2. $2.1\overline{36}=2\frac3{22}$

The part I do not understand however, is "you could used 1) to speed up the working of 2)" which is written in the book.

How would I use 1) to help me work out 2)?

Thanks.

Answer 1

$$2.1\overline{39}=2+0.1+\frac1{10}\cdot0.\overline{39}$$

So, $$2.1\overline{39}=2+\frac1{10}+\frac1{10}\cdot\frac{36}{99}=2+\frac1{10}\cdot\left(1+\frac4{11}\right)=2+\frac1{10}\cdot\frac{15}{11}=2+\frac3{22}$$

Answer 2

1. $.\overline{36}=\frac{36}{99}$
2. $2.1\overline{36}=2\frac3{22}$

part 1.

suppose $$\begin {equation}\dfrac pq=0.\overline{36}\end {equation}$$ this is eqn (1).

mulipltly this eqn with 100 $$\begin {equation}100\dfrac pq=36.\overline{36}\end {equation}$$ this is eqn (2).Now subtract (1) from (2). $$\begin {equation}100\dfrac pq-\frac pq=36.\overline{36}-0.\overline{36}\end {equation}$$ $$99\dfrac pq=36.0$$ $$\dfrac pq=\frac {36}{99}$$

part 2

$2.1\overline{36}=2+0.1\overline{36}$ $$\dfrac pq=0.1\overline{36}\implies 10\dfrac pq=1.\overline{36}$$this is eqn (1).

multilpy with 100 in eqn(1) $$1000\dfrac pq=136.\overline{36}$$ then subtract (1) from (2). $$990\dfrac pq=135.0$$ $$\dfrac pq=\dfrac {135}{990}\implies\dfrac {3}{22}$$

so $2.1\overline{36}\implies 2+0.1\overline{36}\implies 2+\dfrac{3}{22}\implies 2\dfrac3{22}$