# Class 9 Maths Worksheet for Number System

Here we are trying to give the Class 9 Maths Worksheet for Number System with answers. It tests the basic concepts and at the same time makes the student comfortable with the questions.It contains All type of questions like True and false, Simplification, Match the column

Question 1
Find the value of each of the Following
a. $16^{1/4}$
b. $625^{-3/4}$
c. $2 \sqrt {3} + 8 \sqrt {18}$
d. $15 \sqrt {6} + \sqrt {216}$
e. $\frac {10^{5/2}}{10^{3/2}}$
f. $5^{3/2} \times 7^{3/2}$
g. $\frac {\sqrt {162}}{ \sqrt {2}}$
h. $\frac {(1+ \sqrt {2})}{(1- \sqrt {2})}$
Solutions

a. $16^{1/4} = (2^4)^{1/4} = 2$
b. $625^{-3/4} = (5^4)^{-3/4}= 5^{-3} = \frac {1}{125}$
c. $2 \sqrt {3} + 8 \sqrt {18}= 2 \sqrt {3} + 24 \sqrt {2} = \sqrt {2}( \sqrt {6} + 24)$
d. $15 \sqrt {6} + \sqrt {216} =15 \sqrt {6} + \sqrt {6^2 \times 6}= 15 \sqrt {6} + 6 \sqrt {6}=21 \sqrt {6}$
e. $\frac {10^{5/2}}{10^{3/2}} = 10^{\frac {5}{2} - \frac {3}{2}} = 10^1 = 10$
f. $5^{3/2} \times 7^{3/2} = (125)^{1/2} \times (343)^{1/2} = \sqrt {42875}$
g. $\frac {\sqrt {162}}{ \sqrt {2}} = \frac {\sqrt {81 \times 2}}{ \sqrt {2}} =\frac { 9 \times \sqrt {2}}{\sqrt {2}}= 9$
h. $\frac {(1+ \sqrt {2})}{(1- \sqrt {2})} = \frac {(1+ \sqrt {2})}{(1- \sqrt {2})} \times \frac {(1+ \sqrt {2})}{(1+ \sqrt {2})} = -( 1 + \sqrt {2})^2 = -(3+ 2\sqrt {2})$

Question 2
Simplify the below expression
a. $\frac {1+ \sqrt {2}}{1 - \sqrt {3}} + \frac {1- \sqrt {2}}{1 + \sqrt {3}}$

b.$\frac {5+ \sqrt {2}}{5 - \sqrt {2}} + \frac {5- \sqrt {2}}{5 + \sqrt {2}}$

c. $\frac {1}{1 - \sqrt {2} + \sqrt {3}}$

d. $\frac {1}{\sqrt {3} - \sqrt {2}}$

Solutions
a.$\frac {1+ \sqrt {2}}{1 - \sqrt {3}} + \frac {1- \sqrt {2}}{1 + \sqrt {3}}$
=$\frac {(1+ \sqrt {2})(1 + \sqrt {3}) + (1- \sqrt {2})(1 - \sqrt {3})}{ 1 -3}$
$= -\frac {1}{2} \times [1 + \sqrt {2} + \sqrt {3} + \sqrt {6} + 1 -\sqrt {2} + \sqrt {3} - \sqrt {6}]$
$=-\frac {1}{2} \times [2 + 2 \sqrt {3}]= -(1 + \sqrt {3})$

b.$\frac {5+ \sqrt {2}}{5 - \sqrt {2}} + \frac {5- \sqrt {2}}{5 + \sqrt {2}}$
$= \frac {(5+ \sqrt {2})(5 + \sqrt {2}) + (5- \sqrt {2})(5- \sqrt {2})}{5-2}$
$= \frac {1}{3} \times [25 + 2 + 10 \sqrt {2} + 25 + 2 -10 \sqrt {2}]=18$

c. $\frac {1}{1 - \sqrt {2} + \sqrt {3}}$
$= \frac {1}{1 - \sqrt {2} + \sqrt {3}} \times \frac {1 - \sqrt {2} - \sqrt {3}}{1 - \sqrt {2} - \sqrt {3}}$
$= \frac {1 - \sqrt {2} - \sqrt {3}}{ (1-\sqrt {2})^2 - 3}$
$= \frac {1 - \sqrt {2} - \sqrt {3}}{1 + 2 -2 \sqrt {2} -3}$
$=-\frac {1 - \sqrt {2} - \sqrt {3}}{2 \sqrt {2}}$
$=-\frac {1 - \sqrt {2} - \sqrt {3}}{2 \sqrt {2}} \times \frac {\sqrt {2}}{\sqrt {2}}$
$= -\frac {1}{4} \times [\sqrt {2} -2 - \sqrt {6}]$

d. $\frac {1}{\sqrt {3} - \sqrt {2}}$
$= \frac {1}{\sqrt {3} - \sqrt {2}} \times \frac {\sqrt {3} + \sqrt {2}}{\sqrt {3} + \sqrt {2}}$
$= \sqrt {3} + \sqrt {2}$

Question 3
Find the value of a and b such that
$\frac {5 + \sqrt {3}}{7 + 2 \sqrt {3}} = a- b \sqrt {3}$
Solution
$\frac {5 + \sqrt {3}}{7 + 2 \sqrt {3}}$
$= \frac {5 + \sqrt {3}}{7 + 2 \sqrt {3}} \times \frac {7 - 2 \sqrt {3}}{7 - 2 \sqrt {3}}$
$= \frac {1}{37} \times [35 -10 \sqrt {3} + 7 \sqrt {3} -6]$
$= \frac {1}{37} \times [29 -3 \sqrt {3}]$
$= \frac {29}{37} - \frac {3}{37} \sqrt {3}$

Now
$\frac {29}{37} - \frac {3}{37} \sqrt {3}= a- b \sqrt {3}$
so $a=\frac {29}{37}$ and $b=\frac {3}{37}$

Question 4
If $p = 1 + \sqrt {3}$
Find the value of
i. $p^2 + \frac {1}{p^2}$
ii. $p^4 + \frac {1}{p^4}$
Solution
$p = 1 + \sqrt {3}$
$p^2 = (1 + \sqrt {3})^2 = 1 + 3 + 2 \sqrt {3}= 2(2 + \sqrt {3})$
$p^4 = 4(2 + \sqrt {3})^2 = 4(4 + 3 + 4 \sqrt {3})= 4(7 + 4\sqrt {3})$
i. $p^2 + \frac {1}{p^2}$
$=2(2 + \sqrt {3}) + \frac {1}{2(2 + \sqrt {3})}$
$= 2(2 + \sqrt {3}) + \frac {1}{2(2 + \sqrt {3})} \times \frac {2 - \sqrt {3}}{2 - \sqrt {3}}$
$= 2(2 + \sqrt {3}) + \frac {2 - \sqrt {3}}{2}$
$= \frac {10 + 3 \sqrt {3}}{2}$

ii. $p^4 + \frac {1}{p^4}$
$=(p^2 + \frac {1}{p^2})^2 -2$
$=(\frac {10 + 3 \sqrt {3}}{2})^2 -2$
$= \frac {100 + 27 + 60 \sqrt {3} -8}{4}$
$= \frac {119 + 60 \sqrt {3}}{4}$

Question 5
State whether the following statements are true or false
(i) $\frac {2}{\sqrt {5}}$ is a rational number.
(ii) There are infinitely many integers between any two integers.
(iii) Number of rational numbers between 11 and 13 is finite.
(iv) There are numbers which cannot be written in the form  p/q , q ≠ 0 , p, q both   are integers.
(v) The square of an irrational number is always rational.
(vi)$\frac {\sqrt {18}}{\sqrt {2}}$ is not a rational number as $\sqrt {18}$ and $\sqrt {2}$ are not integers.
Solution
i. False
ii. False
iii. False
iv. True
v. False
vi. False

Question 6
Match the column

Solution
(p) -> (b)
As
$4^{2x-1} =64$
$4^{2x-1} =4^3$ or $2x-1=3$ or $x=2$
(q) -> (b)
As
$25^{x-1} =5^{2x-1} -100$
$\frac {5^{2x}}{25} = \frac {5^{2x}}{5} -100$
$100=\frac {5^{2x}}{5} - \frac {5^{2x}}{25}$
$100= 5^{2x} [ \frac {1}{5} - {1}{25}$
$100=25^x \times \frac {4}{25}$
$25^2=25^x$ or x=2
(r) -> (a)
$3^x=9^2$
$3^x=3^4$ or x=4
(s) -> (c)
$64.2^x =1$
$2^6 .2^x=1$
$2^{6+x}=2^0$
x=-6

## Summary

This Number System Class 9 worksheet 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.

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