# Class 9 Maths Problems for Number System

Here we are trying to give the Class 9 Maths Problems for Number System. It tests the basic concepts and at the same time makes the student comfortable with 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
i. .19 ( Terminating decimals)
ii. $.\overline {33333}$ ( Non Terminating repeating decimals)
iii. $.91 \bar {6}$ ( Non Terminating repeating decimals)
iv. $.0769 \overline{230769}$ ( Non Terminating repeating decimals)
v. $. \overline {230769}$ ( Non Terminating repeating decimals)
vi. .2775 ( Terminating decimals)