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Maths 3 Mark Question

Number System · Maharashtra Board STD 9 (MSBSHSE - English Medium). 13 questions.

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3 Mark 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...
y = 0.530535305353530...
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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