Here we are trying to give the number system class 9 important extra questions along with answer. It tests the basic concepts and at the same makes the student comfortable with the questions
Rational number between two number can be find using mean method
(i) Mean of 0.75 and 1.2 = $\frac {.75 + 1.2}{2}= .975$
So .975 is the rational lying between 0.75 and 1.2
(ii) Mean of -3/4 and -2/5 = $\frac {(-3/4) + (-2/5)}{2}= \frac {-23}{40}$
So $\frac {-23}{40}$ is the rational number
3.1,3.2,.3.3,3.4,3.5,3.6
2.101
2.102
2.103
2.104
2.105
2.106
2.107
2.108
2.109
2.110
2.111
2.112
2.113
2.114
2.115
2.116
$0.\overline{9}=0.99999..$
Let $x = 0.9999...$
$10x = 9.9999...$
$10x = 9 + x$
$9x = 9$
$x = 1$
(i) x=√ 11, So it is irrational Number
(ii) y=6, So it is rational number
(iii) z==.2, So it is rational number
(iv) u = √19/3, So it is rational number
$\frac {1}{7}=.142857142857.....= 0.\overline{142857}$
$\frac {2}{7}=.285714285714.....= 0.\overline{285714}$
Now we know that a irrational number is a non-terminating and non repeating number, So irrational number between these can be
.162857147647222977....
or
.17285716786222977....
Answer is (i)
(i)$0.\overline{36}=0.363636..$
Let $x = 0.363636.$
$100x = 36.363636.$
$100x = 36 + x$
$99x = 36$
$x = \frac {36}{99}$
(ii)$0.5\overline{6}=0.56666..$
Let $x = 0.56666..$
$10x = 5.666666....= 5 + 0.\overline{6}$
Now let $y=0.\overline{6}$
$10y=6.6666..=6+y$
or $y=\frac {6}{9}$
So $10x = 5 + \frac {6}{9}$
$x=\frac {51}{90}= \frac {17}{30}$
1.Irrational
2.Integers
3.recurring
4.Rational
5.Archimedes
6. Pythagoras
7. two
8. Natural
$\frac {\sqrt 2 -1}{\sqrt + 1} = \frac {\sqrt 2 -1}{\sqrt 2 + 1} \times \frac {\sqrt 2 -1}{\sqrt 2 - 1}=(\sqrt 2 -1)^2=3 -2 \sqrt 2$
Therefore p=3 and q=2
$x + y = \frac {2 + \sqrt 3}{ \sqrt 2 +1} + \frac {2 - \sqrt 3}{ \sqrt 2 -1} = (2+ \sqrt 3}{(\sqrt 2 -1) + (2 - \sqrt 3}{(\sqrt 2 +1)$
$=2 \sqrt 2 -2 + \sqrt 6 - \sqrt 3 + 2 \sqrt 2 +2 - \sqrt 6 -\sqrt 3=4 \sqrt 2 -2 \sqrt 3= 2( 2\sqrt 2 - \sqrt 3)$
$ x - y = \frac {2 + \sqrt 3}{ \sqrt 2 +1} - \frac {2 - \sqrt 3}{ \sqrt 2 -1} = (2+ \sqrt 3}{(\sqrt 2 -1) - (2 - \sqrt 3}{(\sqrt 2 +1)$
$=2 \sqrt 2 -2 + \sqrt 6 - \sqrt 3 - 2 \sqrt 2 -2 + \sqrt 6 +\sqrt 3=2 \sqrt 6 -4= 2( \sqrt 6 - \2)$
(p) -> (a) as $(4 -\sqrt 3) (4 + \sqrt 3)= 16-3=13$
(q) -> (a)
(r) - >(b)
(s) -> (b)
(i) $ a^3 + \frac {1}{a^3}= (\sqrt 2 -1) + \frac {1}{\sqrt 2 -1}= (\sqrt 2 - 1) + \frac {1}{\sqrt 2 -1} \times \frac {\sqrt 2 +1}{\sqrt 2 +1}= (\sqrt 2 - 1) + (\sqrt 2 + 1)=2 \sqrt 2$
(ii) $ a^3 - \frac {1}{a^3}= (\sqrt 2 -1) - \frac {1}{\sqrt 2 -1}= (\sqrt 2 - 1) - \frac {1}{\sqrt 2 -1} \times \frac {\sqrt 2 +1}{\sqrt 2 +1}= (\sqrt 2 - 1) - (\sqrt 2 + 1)=-2$
This Number System Class 9 Important questions with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.