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Maths 3 Marks Question

Number systems · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 9 questions.

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3 Marks Question

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Question 13 Marks
Look at several examples of rational numbers in the form $\frac{p}{q}(q \neq 0)$, where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Answer
Let $\frac{1}{2}$, $\frac{1}{4}$,$\frac{5}{8}$, $\frac{17}{25}$,$\frac{2}{125}$, $\frac{13}{20}$ etc which has terminating decimal expansion . For example:
$\frac{1}{2} $=$\frac{1}{2^1}$=0.5 , denominator q = 21
$\frac{1}{4}$=$\frac{1}{2^2}$ = 0.25  denominator q = 22
$\frac{5}{8}$ =$\frac{5}{2^3}$ = 0.625 , denominator q = 23
$\frac{17}{25}$= $\frac{17}{5^2}$ = 0.68 , denominator q = 52
$\frac{2}{125}$= $\frac{2}{5^3}$ = 0.016, denominator q = 53
$\frac{13}{20}$ = $\frac{13}{2^2\times 5}$ = 0.65, denominator q = 22 $\times$ 5
It can be observed that the terminating decimal can be obtained in a condition where prime factorization of the denominator (i.e. q) of the given fractions has 2 or the power of 2 or 5 or power of 5 or both.
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Question 23 Marks
What can the maximum number of digits be in the recurring block of digits in the decimal expansion of $\frac{1}{{17}}$? Perform the division to check your answer.
Answer
We need to find the number of digits in the recurring block of $\frac{1}{{17}}$
Let us perform the long division to get the recurring block of $\frac{1}{{17}}$ We need to divide 1 by 17, to get

We can observe that while dividing 1 by 17 we got the remainder as 1, which will continue to be 1 after carrying out 16 continuous divisions. Therefore, we conclude that
$\frac{1}{{17}} = 0.{\text{0588235294117647}}.....{\text{ or }}$$\frac{1}{{17}} = 0.\overline {{\text{0588235294117647}}}$, which is a non-terminating decimal and recurring decimal.
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Question 33 Marks
You know that $\frac{1}{7}=0 . \overline{142857}$. Can you predict what the decimal expansions of  $\frac { 2 } { 7 } , \frac { 3 } { 7 } , \frac { 4 } { 7 } , \frac { 5 } { 7 } , \frac { 6 } { 7 }$ are, without actually doing the long division? If so, how?
Answer
Yes, We can predict the decimal expansions of $\frac { 2 } { 7 } , \frac { 3 } { 7 } , \frac { 4 } { 7 } , \frac { 5 } { 7 } , \frac { 6 } { 7 }$, without actually doing the long division as follows :
$\frac { 2 } { 7 } = 2 \times \frac { 1 } { 7 } = 2 \times 0 . \overline { 142857 } = 0 . \overline { 285714 }$
$\frac { 3 } { 7 } = 3 \times \frac { 1 } { 7 } = 3 \times 0 . \overline { 142857 } = 0 . \overline { 428571 }$
$\frac { 4 } { 7 } = 4 \times \frac { 1 } { 7 } = 4 \times 0 . \overline { 142857 } = 0 . \overline { 571428 }$
$\frac { 5 } { 7 } = 5 \times \frac { 1 } { 7 } = 5 \times 0 . \overline { 142857 } = \overline{0.714285}$
$\frac { 6 } { 7 } = 6 \times \frac { 1 } { 7 } = 6 \times 0 . \overline { 142857 } = 0 . \overline { 857142 }$
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Question 53 Marks
Find five rational numbers between $\frac{3}{5}{\text{ and }}\frac{4}{5}$
Answer
We know that there are infinite rational numbers between any two numbers. A rational number is the one that can be written in the form of $\frac{p}{q}$ where p and q are Integers and $q \ne 0$ We know that the numbers $\frac{3}{5}{\text{ and }}\frac{4}{5}$ can also be written as.0.6 and 0.8 or
$\eqalign{ & {3 \over 5} = {3 \over 5} \times {{20} \over {20}} = {{60} \over {100}} \cr & {4 \over 5} = {4 \over 5} \times {{20} \over {20}} = {{80} \over {100}} \cr}$
We can conclude that the numbers all lie between We need to rewrite the numbers $0.61,0.62,0.63,0.64{\text{ and }}0.65$ in $\frac{p}{q}$ form to get the rational numbers between ​$\frac{3}{5}{\text{ and }}\frac{4}{5}$​ . So, after converting, we get $\frac{{61}}{{100}},\frac{{62}}{{100}},\frac{{63}}{{100}},\frac{{64}}{{100}}{\text{ and }}\frac{{65}}{{100}}$ We can further convert the rational numbers $\frac{{62}}{{100}},\frac{{64}}{{100}}{\text{ and }}\frac{{65}}{{100}}$ into lowest fractions.On converting the fractions, we get $\frac{{31}}{{50}},\frac{{16}}{{25}}{\text{ and }}\frac{{13}}{{20}}$ Therefore, five rational numbers between $\frac{3}{5}{\text{ and }}\frac{4}{5}$are $\frac{{61}}{{100}},\frac{{31}}{{50}},\frac{{63}}{{100}},\frac{{16}}{{25}}{\text{ and }}\frac{{13}}{{50}}$
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Question 63 Marks
Show that 1.272727 = $1. \overline {27}$ can be expressed in the form $\frac{p}{q}$, where p and q are integers and q $\ne$ 0.
Answer
The given number= 1.272727....= $1. \overline {27}$
Let x = $1. \overline {27}$. Then,
x = 1.272727.......(i)
Multiplying both sides by 100 we get
100 x = 127.272727......(ii)
On subtracting (i) from (ii), we getLet x = $1. \overline {27}$. Then,
99 x = (127.272727...) - (1.272727....) =127-1
99 x = 126
 $\begin{array}{l}\mathrm x\;=\frac{126}{99}=\frac{14}{11}\\\end{array}$
Therefore,  $1. \overline {27}$ = 1.272727..=$\frac {14}{11} = \frac pq$
Hence $1. \overline {27}$ can be expressed in $\frac pq$ form where p=14 and q=11
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Question 73 Marks
Find the decimal expansions of $\frac{10}{3},\frac{7}{8}$ and $ \frac{1}{7}$.
Answer


$\frac{7}{8}$, we found that the remainder becomes zero and the decimal expansion of $\frac{7}{8}$ = 0.875.
We call the decimal expansion of such numbers terminating.
$\frac{10}{3}$ and $\frac{1}{7}$ , we notice that the remainders repeat after a certain stage forcing the decimal expansion to go on for ever. In other words, we have a repeating block of digits in the quotient. We say that this expansion is non-terminating recurring.
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Question 83 Marks
Locate $\sqrt{2}$ on the number line.
Answer
Consider a square OABC, with each side 1 unit in length (see Fig.).

Then you can see by the Pythagoras theorem that OB = $\sqrt{1^{2}+1^{2}}=\sqrt{2}$.
Transfer Fig. onto the number line making sure that the vertex O coincides with zero (see Fig.)

We have just seen that OB = $\sqrt{2}$.
Using a compass with centre O and radius OB, draw an arc intersecting the number line at the point P.
Then P corresponds to $\sqrt{2}$ on the number line.
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Question 93 Marks
Find the irrational number between $\frac{1}{7}$and $\frac{2}{7}.$
Answer
Given numbers are $\frac{1}{7}$and $\frac{2}{7}.$
$\frac{1}{7}=0.142857142857 \ldots$
$\Rightarrow \frac{1}{7}=0 . \overline{142857}$
$\frac{2}{7}=2 \times \frac{1}{7}=0.285714285714 \ldots$
$\Rightarrow \quad \frac{2}{7}=0 . \overline{285714}$
To find irrational numbers between $\frac{1}{7}$and $\frac{2}{7}$, we find numbers which are non-terminating and non-repeating.
There are infinitely many such numbers lying between $\frac{1}{7}$and $\frac{2}{7}$. 
Examples: 
  1. 0.150150015000150000 ...
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