- Natural and Whole Number
- |
- Integers
- |
- Rational and Irrational Numbers
- |
- Real Numbers
- |
- Laws of exponents
- |
- What is Number Line
- |
- successive Magnification

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

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. $\frac {(8+\sqrt {3})}{(8-\sqrt {3})}$

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})} $

$= \frac {1}{61} [64 + 3 + 16 \sqrt {3}]=\frac {1}{61} (67 + 16 \sqrt {3})$

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}}$

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$

Find the value of p and q such that

$ \frac {5 + \sqrt {3}}{7 + 2 \sqrt {3}} = p + q \sqrt {3}$

$ \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 $q= - \frac {3}{37}$

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. $x-y$

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

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}$

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)

Given below are the links of some of the reference books for class 9 Math.

- Mathematics for Class 9 by R D Sharma One of the best book for studying class 9 level mathematics. It has lot of problems to be solved.
- Secondary School Mathematics for Class 9 by R S Aggarwal This is also as good as R.D. Sharma. Either this or the book by R.D. Sharma will do. I find book R.S. Aggarwal little bit more challenging than the one by R.D. Sharma.
- Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
- Pearson IIT Foundation Physics, Chemistry & Maths combo for Class 9 Only buy if you are prepared to study extra topics and want to take your studies a step further. You might need help to understand topics in these books.

You can use above books for extra knowledge and practicing different questions.

Class 9 Maths Class 9 Science

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