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Maths 3 Marks Question

Number System · Jharkhand Board (JAC) STD 9 (Jharkhand - English Medium). 13 questions.

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3 Marks Question

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13 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}}:$ $0.\bar{4}$
Answer
Let $\text{x}=0.\bar4$ Now, $\text{x}=0.\bar4=0.444\ ...(\text{i})$
Multiplying both sides of equation $(i)$ by $10$,
we get, $10\text{x}=4.444\ ...(\text{ii}) $ Subtracting equation $(i)$ by $(ii)$
$\therefore\ 10\text{x}-\text{x}=4.444\ ...-\ 0.444\ ...$
$\Rightarrow9\text{x}=4$
$\Rightarrow\text{x}=\frac{4}{9}$ Hence, $0.\bar{4}=\frac{4}{9}$
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Question 23 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}}:$ $0.\overline{54}$
Answer
Let $\text{x}=0.\overline{54}$
$\Rightarrow\text{x}=0.5454\ ...(\text{i})$ Multiplying equation $(i)$ by $100$,
We get, $100\text{x}=54.5454\ ...(\text{ii}) $ Subtracting equation $(i)$ by equation $(ii)$
$\therefore\ 100\text{x}-\text{x}=54$
$\Rightarrow99\text{x}=54$
$\Rightarrow\text{x}=\frac{54}{99}=\frac{6}{11}$
Hence, $0.\overline{54}=\frac{6}{11}$
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Question 33 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}}:$ $125.\bar{3}$
Answer
Let $\text{x}=4.\bar{7}$
$\Rightarrow\text{x}=4.77\ ...(\text{i})$ Multiplying equation $(i)$ by $10,$
$\therefore10\text{x}=47.77\ ...(\text{ii})$ Subtracting equation $(i)$ by $(ii)$
$\therefore\ 10\text{x}-\text{x}=47.77 \ ... -4.77 \ ...$
$\Rightarrow9\text{x}=43$
$\Rightarrow\text{x}=\frac{43}{9}$ Hence, $4.\bar{7}=\frac{43}{9}$
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Question 43 Marks
Explain how irrational number is differ from rational numbers?
Answer
An irrational number is a real number which can be written as a decimal but not as a fraction i.e. it cannot be expressed as a ratio of integers. It cannot be expressed as terminating or repeating decimal. For example, $0.10110100$ is an irrational number. A rational number is a real number which can be written as a fraction and as a decimal i.e. it can be expressed as a ratio of integers. It can be expressed as terminating or repeating decimal. For examples, $0.10$ and $0.\bar{4}$ both are rational numbers.
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Question 53 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}}:$
$0.\overline{47}$
Answer
$0.\overline{47}=0.4777...$
Let $\text{x}=0.4777 \ ...\text{(i)}$
$10\text{x}=4.777$
$100\text{x}=47.777 \ ...(\text{ii})$
$(ii) - (i)$ gives
$99\text{x}=43$
$\text{x}=\frac{43}{99}$
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Question 63 Marks
Give two rational numbers lying between $0.232332333233332$ and $0.212112111211112.$
Answer
Let $a = 0.212112111211112$ And, $b = 0.232332333233332...$ Clearly, $a < b$ because in the second decimal place a has digit $1$ and $b$ has digit $3$.
If we consider rational numbers in which the second decimal place has the digit $2$, then they will lie between $a$ and $b.$
$Let x = 0.22 y = 0.22112211... $
Then $a < x < y < b$
Hence, $x$ and $y$ are required rational numbers.
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Question 73 Marks
Give two rational numbers lying between $0.515115111511115$ and $0. 5353353335...$
Answer
Let, $a = 0.515115111511115...$ And, $b = 0.5353353335...$
We observe that in the second decimal place a has digit $1$ and $b$ has digit $3,$
therefore, $a < b.$
So If we consider rational numbers $x = 0.52 y = 0.52062062...$
We find that, $a < x < y < b$
Hence $x$ and $y$ are required rational numbers.
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Question 83 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}}:$ $125.\bar{3}$
Answer
Let $\text{x}=125.\bar{3}$
$\Rightarrow\text{x}=125.33\ ...(\text{i})$ Multiplying equation $(i)$ by $10$,
$\therefore100\text{x}=1253.33\ ...(\text{ii})$ Subtracting equation $(i)$ by $(ii)$
$\therefore\ 10\text{x}-\text{x}=1253.33 \ ... -125.33 \ ...$
$\Rightarrow9\text{x}=1128$
$\Rightarrow\text{x}=\frac{1128}{9}=\frac{376}{3}$ Hence, $\text{x}=\frac{376}{3}$
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Question 93 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}}:$ $0.\overline{37}$
Answer
Let $\text{x}=0.\overline{37}$ Now, $\text{x}=0.3737\ ...(\text{i})$ Multiplying equation $(i)$ by $10$.
$\therefore \ 10\text{x}=3.737\ ...(\text{ii}) $ Multiplying equation $(ii)$ by $10$.
$100\text{x}=37.3737\ ...(\text{iii}) $
Subtracting equation $(i)$ by $(iii)$
$\therefore\ 100\text{x}-\text{x}=37$
$\Rightarrow99\text{x}=37$
$\Rightarrow\text{x}=\frac{37}{99}$
Hence, $0.\overline{37}=\frac{37}{99}$
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Question 103 Marks
Find five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}.$
Answer
Given that to find out five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$ To find $5$ rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$with $\frac{6}{6}$
We have, $\frac{3}{5}\times\frac{6}{6}=\frac{18}{30}$
$\frac{4}{5}\times\frac{6}{6}=\frac{24}{30}$
We know $18 < 19 < 20 < 21 < 22 < 23 < 24$
$\frac{18}{30}<\frac{19}{30}<\frac{20}{30}<\frac{21}{30}<\frac{22}{30}<\frac{23}{30}<\frac{24}{30}$
$\frac{3}{5}<\frac{19}{30}<\frac{20}{30}<\frac{21}{30}<\frac{22}{30}<\frac{23}{30}<\frac{4}{5}$
Therefore, $5$ rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$ are $\frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30}$
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Question 113 Marks
Find one irrational number between $0.2101$ and $0.2222\ ...=0.\bar{2}$
Answer
Let, $a = 0.2101$ and, $b = 0.2222...$
We observe that in the second decimal place a has digit $1$ and $b$ has digit $2$,
therefore $a < b$ in the third decimal place a has digit $0$. So, if
we consider irrational numbers $x = 0.211011001100011...$
We find that $a < x < b$ Hence $x$ is required irrational number.
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Question 123 Marks
Find two irrational numbers between $0.5$ and $0.55.$
Answer
Let $a =0.5=0.50$ and $b =0.55$
We observe that in the second decimal place a has digit $0$ and $b$ has digit $5$ ,therefore $a<0$ so, if
we consider irrational numbers $x=0.51051005100051 \ldots y=0.530535305353530 \ldots$
We find that $a<x<y < b$
Hence $x$ and $y$ are required irrational numbers.
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Question 133 Marks
Express the following decimals in the form $\frac{\text{p}}{\text{q}}:$ $0.\overline{621}$
Answer
Let $\text{x}=0.\overline{621}$
Now, $\text{x}=0.621621\ ...(\text{i})$ Multiplying equation $(i) $ by $1000,$
$\therefore1000\text{x}=621.621621\ ...(\text{ii})$ Subtracting equation $(i)$ by $(ii)$
$\therefore\ 1000\text{x}-\text{x}=621$
$\Rightarrow999\text{x}=621$
$\Rightarrow\text{x}=\frac{621}{999}=\frac{69}{111}=\frac{23}{37}$
Hence, $0.\overline{621}=\frac{23}{37}$
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