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Class 9 Maths Problems for Number System

Here we are trying to give the Class 9 Maths Problems for Number System. This contains tough/difficult questions on number system class 9. I hope you like the questions

Question 1
Find the value of each of the Following
a. $2^{1/4} \times 2^{1/5}$
b. $16^{-3/4}$
c. $\sqrt[3]{2}.\sqrt[4]{2} \sqrt[12]{32}$
d. $15\sqrt {6} + \sqrt {216}$
e. $ \frac {11^{5/2} }{ 11^{3/2}}$
f. $5^{3/2} 7^{3/2}$
g. $ \frac {\sqrt {162}}{ \sqrt {2}}$
h. $\frac {(8+\sqrt {3})}{(8-\sqrt {3})}$

Solution
a. $2^{1/4} \times 2^{1/5} = 2^{ \frac {1}{4} + \frac {1}{5}} = 2^{9/20}$
b. $16^{-3/4} = (2^4)^{-3/4} = \frac {1}{8}$
c. $\sqrt[3]{2}.\sqrt[4]{2}. \sqrt[12]{32} = 2^{1/3} . 2^{1/4} .2^{5/12} = 2^{ \frac {1}{3} + \frac {1}{4} + \frac {5}{12}}= 2$
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 {11^{5/2} }{ 11^{3/2}} = 11^ {\frac {5}{2} - \frac {3}{2}} = 11$
f. $5^{3/2} 7^{3/2} = (5 \times 7) ^{3/2} = (35)^{3/2}$
g. $ \frac {\sqrt {162}}{ \sqrt {2}} = \frac {9 \sqrt {2}} {\sqrt {2}}=9$
h. $\frac {(8+\sqrt {3})}{(8-\sqrt {3})} = \frac {(8+\sqrt {3})}{(8-\sqrt {3})} \times \frac {(8+\sqrt {3})}{(8+\sqrt {3})} $
$= \frac {1}{61} [64 + 3 + 16 \sqrt {3}]=\frac {1}{61} (67 + 16 \sqrt {3})$

Question 2
Simplify the below expression
a. $ \sqrt[4]{81} - 8 \sqrt[3]{216} + 15 \sqrt[5]{32} + \sqrt {225}$
b. $ 64^{-1/3} [64^{1/3} - 64^{2/3}]$
c. $ \frac {4}{216^{-2/3}} + \frac {1}{256^{-3/4}} + \frac {2}{243^{-1/5}}$

Solution
a.$ \sqrt[4]{81} - 8 \sqrt[3]{216} + 15 \sqrt[5]{32} + \sqrt {225}= \sqrt[4]{3^4} - 8 \sqrt[3]{6^3} + 15 \sqrt[5]{2^5} + \sqrt {25^2}$
$= 3 -48 + 30 + 25 =10$

b. $ 64^{-1/3} [64^{1/3} - 64^{2/3}]= (4^3)^{-1/3} [ (4^3)^{1/3} - (4^3)^{2/3}]= 4^{-1} [ 4 - 16]=-2$

c.$ \frac {4}{216^{-2/3}} + \frac {1}{256^{-3/4}} + \frac {2}{243^{-1/5}} = \frac {4}{(6^3)^{-2/3}} + \frac {1}{(4^4)^{-3/4}} + \frac {2}{(3^5)^{-1/5}}$
$= \frac {4}{6^{-2}} + \frac {1}{4^{-3}} + \frac {2}{3^{-1}}$
$ = 144 + 64 + 6 = 214$

Question 3
Find the value of p and q such that
$ \frac {5 + \sqrt {3}}{7 + 2 \sqrt {3}} = p + q \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
$\frac {29}{37} - \frac {3}{37} \sqrt {3}= p + q \sqrt {3}$
so $p=\frac {29}{37}$ and $q= - \frac {3}{37}$

Question 4
If $ x = 1+\sqrt {3} $
and
$y= \frac {1}{x}$
Find the value of
i. $x^2 + y^2$
ii. $x^2 - y^2$
iii. $(x+y)^2$
iv. $x-y$

Solution
$ x = 1+\sqrt {3} $
$x^2 = (1+\sqrt {3})^2 = 2(2 + \sqrt {3})$
$y= \frac {1}{x} = \frac {1}{1+\sqrt {3}} = \frac {1}{2} (\sqrt {3} -1)$
$ y^2 = \frac {1}{2} (2 - \sqrt {3})$
Now,
i. $x^2 + y^2 = 2(2 + \sqrt {3}) + \frac {1}{2} (2 - \sqrt {3})$
$=\frac {1}{2} (10 + 3 \sqrt {3})$

ii. $x^2 - y^2 = 2(2 + \sqrt {3}) - \frac {1}{2} (2 - \sqrt {3})$
$=\frac {1}{2} (8 + 5 \sqrt {3})$

iii. $(x+y)^2 = (\frac {1 + 3 \sqrt {3}}{2})^2 = \frac {1}{4} [28 + 6 \sqrt {3}] =\frac {1}{2}( 14 + 3 \sqrt {3})$

iv. $x-y = \frac {3 + \sqrt {3}}{2}$

Question 5
Write the following in decimal form and say what kind of decimal expansion each has:
(i) $ \frac {19}{100}$
(ii) $ \frac {1}{3}$
(iii) $ \frac {11}{12}$
(iv) $ \frac {1}{13}$
(v) $ \frac {3}{13}$
(vi) $ \frac {111}{400}$

Solution
Class 9 Maths Problems for Number System

Question 6
Find the value of
$\frac {1}{1 + \sqrt 2} + \frac {1}{\sqrt 2 + \sqrt 3} + \frac {1}{\sqrt 3 + \sqrt 4} + ... + \frac {1}{\sqrt 8 + \sqrt 9}$
Solution
$\frac {1}{1 + \sqrt 2} + \frac {1}{\sqrt 2 + \sqrt 3} + \frac {1}{\sqrt 3 + \sqrt 4} + ... + \frac {1}{\sqrt 8 + \sqrt 9}$
$=\frac {1}{\sqrt 2 + 1 } + \frac {1}{\sqrt 3 + \sqrt 2} + \frac {1}{\sqrt 4 + \sqrt 3} + ... + \frac {1}{\sqrt 9 + \sqrt 8}$
$= \frac {1}{\sqrt 2 + 1} \times \frac {\sqrt 2 - 1}{\sqrt 2 - 1} + \frac {1}{\sqrt 3 + \sqrt 2} \times \frac {\sqrt 3 - \sqrt 2}{\sqrt 3 - \sqrt 2} + \frac {1}{\sqrt 4 + \sqrt 3} \times \frac {\sqrt 4 - \sqrt 3}{\sqrt 4 - \sqrt 3} + ... + \frac {1}{\sqrt 9 + \sqrt 8} \times \frac {\sqrt 9 - \sqrt 8}{\sqrt 9 - \sqrt 8}$
$= (\sqrt 2 - 1) + (\sqrt 3 - \sqrt 2) + (\sqrt 4 - \sqrt 3) ..... + (\sqrt 9 - \sqrt 8)= 2$

Question 7
Find the value of
$ \frac { \sqrt { 3 + 2 \sqrt 2} + \sqrt { 3 - 2 \sqrt 2}}{\sqrt {\sqrt 3 + 1}} - 2 \sqrt {\sqrt 3 - 1}$
Solution
$= \frac { \sqrt {(\sqrt 2 + 1)^2} + \sqrt {(\sqrt 2 -1)^2}}{ \sqrt {\sqrt 3 + 1}} - 2 \sqrt {\sqrt 3 - 1}$
$=\frac { \sqrt 2 + 1 + \sqrt 2 -1 }{ \sqrt {\sqrt 3 + 1}} - 2 \sqrt {\sqrt 3 - 1}$
$= \frac {2 \sqrt 2}{\sqrt {\sqrt 3 + 1}} - 2 \sqrt {\sqrt 3 - 1}$
$=\frac {2 \sqrt 2- 2 \sqrt 2}{\sqrt {\sqrt 3 + 1}}$
=0

Question 8
Find the value of
$\frac {1}{2 + \sqrt 3} + \frac {2}{\sqrt 5 - \sqrt 3} + \frac {1}{2 -\sqrt 5}$
Solution
$= 2 -\sqrt 3 + \sqrt 5 + \sqrt 3 - 2 + \sqrt 5=0$

Summary

This Class 9 Maths Problems for Number System 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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