
If $3.0\overline {65} $ is $\dfrac{p}{q}$ and according to Euclid’s Division Lemma, $p = bq + r$, then value of $\dfrac{r}{b}$ is
A. $2$
B. $5$
C. $\dfrac{{13}}{3}$
D. $1$
Answer
533.4k+ views
Hint: When a number can be written in the form of $\dfrac{p}{q}$ then the number is a rational number. We have to first convert $3.0\overline {65} $ into $\dfrac{p}{q}$ and find the value of $p$ and $q$. Using these values we can divide $p$ by $q$ and find the value of $r$ and $b$. Euclid’s Division Lemma states that for any two integers $p$ and $q$, we can find two integers $b$ and $r$ such that $p = bq + r$. Here $0 \leqslant r \leqslant q$.
Complete step by step solution:
We have been given that $3.0\overline {65} $ can be written in the form of $\dfrac{p}{q}$. We can find the value of this fraction and from this we can find the values of $p$ and $q$.
We can assume that $3.0\overline {65} = x$.
We multiply both sides by $10$. We get,
$
3.0\overline {65} \times 10 = x \times 10 \\
\Rightarrow 30.\overline {65} = 10x\;\;\;\;\;\;...\left( 1 \right) \\
$
We can also multiply both sides by $1000$ to get,
$
3.0\overline {65} \times 1000 = x \times 1000 \\
\Rightarrow 3065.\overline {65} = 1000x\;\;\;\;\;\;...\left( 2 \right) \\
$
We can subtract equation $\left( 1 \right)$ from equation $\left( 2 \right)$ to get,
$
\Rightarrow 3065.\overline {65} - 30.\overline {65} = 1000x - 10x \\
\Rightarrow 3035 = 990x \\
\Rightarrow x = \dfrac{{3035}}{{990}} = \dfrac{{607}}{{198}} \\
$
Thus, $3.0\overline {65} = \dfrac{{607}}{{198}}$
Thus, $p = 607$ and $q = 198$.
By Euclid’s division lemma, we can write,
$607 = 198b + r$
To find the value of $b$ and $r$ we divide $607$ by $198$.
We can write, $607 = 198 \times 3 + 13$
We get the value of $b = 3$ and $r = 13$.
We can evaluate $\dfrac{r}{b} = \dfrac{{13}}{3}$.
Hence, option (C) is correct.
So, the correct answer is “Option C”.
Note: A decimal number with repetitive decimals is a rational number because it can be written in the form of $\dfrac{p}{q}$. When we write a number in the form of $\dfrac{p}{q}$ we have to make sure that the HCF of $p$ and $q$ is $1$. For decimal numbers less than $1$ we will get $p < q$, then the quotient will become zero and the ratio $\dfrac{r}{b}$ will not be defined.
Complete step by step solution:
We have been given that $3.0\overline {65} $ can be written in the form of $\dfrac{p}{q}$. We can find the value of this fraction and from this we can find the values of $p$ and $q$.
We can assume that $3.0\overline {65} = x$.
We multiply both sides by $10$. We get,
$
3.0\overline {65} \times 10 = x \times 10 \\
\Rightarrow 30.\overline {65} = 10x\;\;\;\;\;\;...\left( 1 \right) \\
$
We can also multiply both sides by $1000$ to get,
$
3.0\overline {65} \times 1000 = x \times 1000 \\
\Rightarrow 3065.\overline {65} = 1000x\;\;\;\;\;\;...\left( 2 \right) \\
$
We can subtract equation $\left( 1 \right)$ from equation $\left( 2 \right)$ to get,
$
\Rightarrow 3065.\overline {65} - 30.\overline {65} = 1000x - 10x \\
\Rightarrow 3035 = 990x \\
\Rightarrow x = \dfrac{{3035}}{{990}} = \dfrac{{607}}{{198}} \\
$
Thus, $3.0\overline {65} = \dfrac{{607}}{{198}}$
Thus, $p = 607$ and $q = 198$.
By Euclid’s division lemma, we can write,
$607 = 198b + r$
To find the value of $b$ and $r$ we divide $607$ by $198$.
We can write, $607 = 198 \times 3 + 13$
We get the value of $b = 3$ and $r = 13$.
We can evaluate $\dfrac{r}{b} = \dfrac{{13}}{3}$.
Hence, option (C) is correct.
So, the correct answer is “Option C”.
Note: A decimal number with repetitive decimals is a rational number because it can be written in the form of $\dfrac{p}{q}$. When we write a number in the form of $\dfrac{p}{q}$ we have to make sure that the HCF of $p$ and $q$ is $1$. For decimal numbers less than $1$ we will get $p < q$, then the quotient will become zero and the ratio $\dfrac{r}{b}$ will not be defined.
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