Question

Find the sum of $\dfrac{{0.3}}{{0.5}} + \dfrac{{0.33}}{{0.55}} + \dfrac{{0.333}}{{0.555}} + ...$ to $15$ terms.
A.$10$
B.$9$
C.$3$
D.$5$

Answer 1

Hint: First, remove the decimal point from the terms series by multiplying each term with the $10,100,1000$…up to $15$ terms respectively. Then simplify the series by taking $\dfrac{3}{5}$ common from the terms. Simplify the terms inside the brackets and multiply the obtained number with the number outside the bracket.

Complete step-by-step answer:
We have to find the sum of $\dfrac{{0.3}}{{0.5}} + \dfrac{{0.33}}{{0.55}} + \dfrac{{0.333}}{{0.555}} + ...$to $15$ terms
First, we will remove the decimal point from the numerator and denominator in the series by multiplying each term with the $10,100,1000$…up to $15$ terms respectively.
$ \Rightarrow \dfrac{{0.3 \times 10}}{{0.5 \times 10}} + \dfrac{{0.33 \times 100}}{{0.55 \times 100}} + \dfrac{{0.333 \times 1000}}{{0.555 \times 1000}} + ...$ to $15$ terms
Then on multiplication, we get-
$ \Rightarrow \dfrac{3}{5} + \dfrac{{33}}{{55}} + \dfrac{{333}}{{555}} + ...$ to $15$ terms
Now on taking $\dfrac{3}{5}$ common from the terms, we get-
 $ \Rightarrow \dfrac{3}{5}\left\{ {\dfrac{1}{1} + \dfrac{{11}}{{11}} + \dfrac{{111}}{{111}} + ...{\text{to 15 terms}}} \right\}$
Now here we see that the numbers of the numerator and denominator of each term are the same so they will get cancelled and we will get-
$ \Rightarrow \dfrac{3}{5}\left\{ {1 + 1 + 1 + ...15{\text{times}}} \right\}$
Now on adding the number$1$total$15$ times we get-
$ \Rightarrow \dfrac{3}{5}\left\{ {1 \times 15} \right\}$
On solving, we get-
$ \Rightarrow \dfrac{3}{5} \times 15$
Here $15$ is divisible by $5$ as its unit digit is $5$ so we will divide the number by $5$.
Then we get-
$ \Rightarrow 3 \times 3$
On multiplication, we get-
$ \Rightarrow 9$
Hence the sum of the series is $9$.

Note: Here you can also solve this question this way-
After we obtain this $ \Rightarrow \dfrac{3}{5} + \dfrac{{33}}{{55}} + \dfrac{{333}}{{555}} + ...$ to $15$ terms
We can divide the numerator and denominator of the second term to $15th$ term by the highest common factor which is $11,111,...$ respectively up to $14$ terms for each term. The first term will be written as the same. So we get-
$ \Rightarrow \dfrac{3}{5} + \dfrac{3}{5} + \dfrac{3}{5} + ...15terms$
Since there are $15$ terms, we will have to add the number$\dfrac{3}{5}$ $15$ times. Then we can simply write it as-
$ \Rightarrow \dfrac{3}{5} \times 15$
On solving this, we will get the sum of the series=$9$