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The sum of $\overline{2} .75$ and $\overline{3} .78$ is :

Option: 1

$\overline{5} .53$


Option: 2

$\overline{4} .53$


Option: 3

$\overline{1} .53$


Option: 4

$\overline{1} .03$


B

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Posted by

Karina kishor kadam

The simplification of $(0 . \overline{6}+0 . \overline{7}+0 . \overline{8}+0 . \overline{3})$ yields the result:

Option: 1

$10 / 9$


Option: 2

$20 / 9$


Option: 3

2. $\overline{35}$


Option: 4

$23 / 5$


B

 

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Posted by

Karina kishor kadam

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A massless string connects two pulley of masses ' 2 \mathrm{~kg}' and '1 \mathrm{~kg}' respectively as shown in the figure.

The heavier pulley is fixed and free to rotate about its central axis while the other is free to rotate as well as translate. Find the acceleration of the lower pulley if the system was released from the rest. [Given, g=10 \mathrm{~m} / \mathrm{s}^2]

Option: 1

\frac{4}{3} \mathrm{~gm} / \mathrm{s}^2


Option: 2

\frac{3}{2} \mathrm{~gm} / \mathrm{s}^2


Option: 3

\frac{3}{4} \mathrm{~gm} / \mathrm{s}^2


Option: 4

\frac{2}{3} \mathrm{~gm} / \mathrm{s}^2


3/4gm/s2

 

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Posted by

Guru G

Two particles of mass 5kg and 10kg respectively are attached to the two ends of a rigid rod of length 1m with negligible mass. The centre of mass of the system from the 5kg particle is nearly at a distance of:
Option: 1  33 cm  
Option: 2 50 cm
Option: 3 67 cm
Option: 4 80 cm

\text { The position of } COM \text { from } m _{1} \text { mass is } \frac{ m _{2} r }{\left( m _{1}+ m _{2}\right)}

Here, m1 = 5  kg

r = 1 m

m2 = 10 kg

By putting this value we will get 0.666667 meter

or we can write = 67 cm

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Posted by

Deependra Verma

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(i) Simplify :- (13 + 23 + 33)1/2
(ii) Simplify :- \left ( \frac{3}{5} \right )^{4}\left ( \frac{8}{5} \right )^{-12}\left ( \frac{32}{5} \right )^{6}
(iii) Simplify :- \left ( \frac{1}{27} \right )^{-\frac{2}{3}}
(iv) Simplify :- \left [ \left \{ \left ( 625 \right )^{-\frac{1}{2}} \right \}^{-\frac{1}{4}} \right ]^{2}
(v) Simplify :- \frac{9^{\frac{1}{3}}\times 27^{\frac{1}{2}}}{3^{\frac{1}{6}}\times 9^{-\frac{2}{3}}}
(vi) Simplify :-64^{-\frac{1}{3}}\left ( 64^{\frac{1}{3}}-64^{\frac{2}{3}} \right )
(vii) Simplify :- \frac{8^{\frac{1}{3}}\times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}


 

(i) Answer.          6
Solution.     (13 + 23 + 33)1/2
We know that
13 = 1.1.1 = 1
23 = 2.2.2 = 8
33 = 3.3.3 = 27
Putting these values we get
(13 + 23 + 33)1/2 = \sqrt{1+8+27}
=\sqrt{36}= 6
Hence the answer is 6

(ii) Answer.  \frac{2025}{64}       

Solution.\left ( \frac{3}{5} \right )^{4}\left ( \frac{8}{5} \right )^{-12}\left ( \frac{32}{5} \right )^{6}
 We know that
8 = 2.2.2 = 23
32 = 2.2.2.2.2 = 25
\left ( \frac{3}{5} \right )^{4}\left ( \frac{8}{5} \right )^{-12}\left ( \frac{32}{5} \right )^{6}= \left ( \frac{3}{5} \right )^{4}\left ( \frac{2^{3}}{5} \right )^{-12}\left ( \frac{2^{5}}{5} \right )^{6}
= \frac{3^{4}\left ( 2^{3} \right )^{-12}\left ( 2^{5} \right )^{6}}{5^{4}5^{-12}5^{6}}                \because \left ( \frac{a}{b} \right )^{m}= \frac{a^{m}}{b^{m}}
= \frac{3^{4}2^{-36}2^{30}}{5^{4}5^{-12}5^{6}}                            \because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}
= \frac{3^{4}\times 2^{-36+30}}{5^{4-12+6}}                        \because \left ( a \right )^{m}\left ( a \right )^{n}= \left ( a \right )^{m+n}
= \frac{3^{4}\times 2^{-6}}{5^{-2}}
= \frac{3^{4}\times 5^{2}}{2^{6}}                                   \because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}
= \frac{81\times 25}{64}
= \frac{2025}{64}
Hence the answer is \frac{2025}{64}
     
(iii) Answer. 9
Solution. Given \left ( \frac{1}{27} \right )^{-\frac{2}{3}}
We know that
27 = 3.3.3 = 33
\left ( \frac{1}{27} \right )^{-\frac{2}{3}}= \left ( \frac{1}{3^{3}} \right )^{-\frac{2}{3}}
= \left ( 3^{3} \right )^{\frac{2}{3}}            \because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}
= \left ( 3 \right )^{3\times \frac{2}{3}}         \because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}
= 32 = 9
Hence the answer is 9

(iv) Answer. 5
Solution. Given \left [ \left \{ \left ( 625 \right )^{-\frac{1}{2}} \right \}^{-\frac{1}{4}} \right ]^{2}
We know that
625= \left ( 25 \right )\left ( 25 \right )= 5\cdot 5\cdot 5\cdot 5= 5^{4}
\left [ \left \{ \left ( 625 \right )^{-\frac{1}{2}} \right \}^{-\frac{1}{4}} \right ]^{2}= \left [ \left \{ \left ( \left ( 5 \right )^{4} \right )^{-\frac{1}{2}} \right \} ^{-\frac{1}{4}}\right ]^{2}
= 5^{4\times \frac{-1}{2}\times \frac{-1}{4}\times 2}        \because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}

= 51 = 5
Hence the answer is 5

(v) Answer.      \sqrt[3]{\frac{1}{3}}
Solution. We have \frac{9^{\frac{1}{3}}\times 27^{\frac{1}{2}}}{3^{\frac{1}{6}}\times 9^{-\frac{2}{3}}}
Now we know that
9 = 3.3 = 32
27 = 3.3.3 = 33
\frac{9^{\frac{1}{3}}\times 27^{\frac{1}{2}}}{3^{\frac{1}{6}}\times 9^{-\frac{2}{3}}}= \frac{\left ( 3^{2} \right )^{\frac{1}{3}}\times \left ( 3^{3} \right )^{\frac{1}{2}}}{\left ( 3 \right )^{\frac{1}{6}}\times \left ( 3^{2} \right )^{\tfrac{-2}{3}}}
= \frac{\left ( 3\right )^{2\times \frac{1}{3}}\times \left ( 3 \right )^{3\times \frac{-1}{2}}}{\left ( 3 \right )^{\frac{1}{6}}\times \left ( 3 \right )^{\tfrac{-2}{3}}}        \because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}
= \frac{\left ( 3 \right )^{\frac{2}{3}-\frac{3}{2}}}{\left ( 3 \right )^{\frac{1}{6}-\frac{2}{3}}}              \because \left ( a \right )^{m}\left ( a \right )^{n}= \left ( a \right )^{m+n}
= \frac{3^{\frac{4-9}{6}}}{3^{\frac{1-4}{6}}}
=\frac{3^{-\frac{5}{6}}}{3^{-\frac{3}{6}}}
= 3^{\frac{-5}{6}-\left ( \frac{-3}{6} \right )}           \because \frac{\left ( a \right )^{m}}{\left ( a \right )^{n}}= \left ( a \right )^{m-n}
= 3^{-\frac{2}{6}}
= \left ( \frac{1}{3} \right )^{\frac{1}{3}}                 \because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}
= \sqrt[3]{\frac{1}{3}}
Hence the answer is \sqrt[3]{\frac{1}{3}}

(vi) Answer. – 3
Solution. We have ,64^{-\frac{1}{3}}\left ( 64^{\frac{1}{3}}-64^{\frac{2}{3}} \right )
We know that 64 =4.4.4=43
= \left ( 4^{3} \right )^{\frac{-1}{3}}\left \{ \left ( \left ( 4^{3} \right ) ^{\frac{1}{3}}-\left ( 4^{3} \right ) ^{\frac{2}{3}}\right )\right \}
= \left ( 4 \right )^{3\times \frac{-1}{3}}\left \{ \left ( \left ( 4 \right )^{3\times \frac{1}{3}}-\left ( 4 \right )^{3\times \frac{2}{3}} \right ) \right \}      \because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}
= 4^{-1}\left ( 4-4^{2} \right )
= \frac{1}{4}\left ( 4-16 \right )                                                        \because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}
= \frac{1}{4}\left ( -12 \right )
= – 3
Hence the answer is – 3
 

(vii) Answer. 16
Solution. Given ,\frac{8^{\frac{1}{3}}\times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}
We know that
8 = 2.2.2 = 23
16 = 2.2.2.2 = 24
32 = 2.2.2.2.2 = 25
\frac{8^{\frac{1}{3}}\times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}= \frac{\left ( 2^{3} \right )^{\frac{1}{3}}\times\left ( 2^{4} \right )^{\frac{1}{3}}}{\left ( 2^{5} \right )^{-\frac{1}{3}}}
= \frac{2^{3\times\frac{1}{3}}\times2^{4\times\frac{1}{3}}}{2^{5\times\frac{-1}{3}}}              \because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}       
= 2^{1+\frac{4}{3}+\frac{5}{3}}                             \because \left ( a \right )^{m}\left ( a \right )^{n}= \left ( a \right )^{m+n}  and
                                                       \because \frac{\left ( a \right )^{m}}{\left ( a \right )^{n}}= \left ( a \right )^{m-n}
= 2^{\frac{3+4+5}{3}}= 2^{\frac{12}{3}}
= 2^{4}= 16
Hence the answer is 16.

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Posted by

infoexpert27

(i) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{4}{\sqrt{3}}
(ii) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{6}{\sqrt{6}}
(iii) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}
(iv) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{2}{2+\sqrt{2}}
(v) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{1}{\sqrt{3}+\sqrt{2}}

(i) Answer.   2.3093
Solution. Given: \frac{4}{\sqrt{3}}
Rationalising,
\frac{4}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}= \frac{4\sqrt{3}}{3}
(Given that \sqrt{3}= 1.732)
= \frac{4\times 1\cdot 732}{3}
= 2.3093
Hence the answer is 2.3093

(ii) Answer.  2.449
Solution.  Given: \frac{6}{\sqrt{6}}
Rationalising,
\frac{6}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}= \frac{6\sqrt{6}}{6}
= \frac{6\times \sqrt{2}\sqrt{3}}{6}
Putting the given values,
\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732
We get :
= \sqrt{2}\cdot \sqrt{3}
= 1\cdot 414\times 1\cdot 732= 2\cdot 449
Hence the answer is 2.449

(iii) Answer.  0.462852
Solution.   Given that \frac{\sqrt{10}-\sqrt{5}}{2}
This can be written as

\frac{\sqrt{2}\times \sqrt{5}-\sqrt{5}}{2}
Now putting the given values,

\sqrt{2}= 1\cdot 414,\sqrt{5}= 2\cdot 236
We get :
\Rightarrow \frac{1\cdot 414\times 2\cdot 236-2\cdot 236}{2}
= 0.462852
  Hence the answer is 0.462852

(iv) Answer.  0.414
Solution.  Given: \frac{\sqrt{2}}{2+\sqrt{2}}
Rationalising,
\frac{\sqrt{2}}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}
Using   (a – b) (a + b) = a2 – b2
= \frac{\sqrt{2}\left ( 2-\sqrt{2} \right )}{2^{2}-\sqrt{2}^{2}}
= \frac{2\sqrt{2}-2}{4-2}
= \frac{2\left ( \sqrt{2}-1 \right )}{2}
= \sqrt{2}-1
Putting the given value of \sqrt{2}= 1\cdot 414
We get
= 1.414 – 1
= 0.414
 Hence the answer is 0.414

(v)  Answer.  0.318
Solution. Given that \frac{1}{\sqrt{3}+\sqrt{2}}
Rationalising,
\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}
Using   (a – b) (a + b) = a2 – b2
= \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}^{2}-\sqrt{2}^{2}}
= \frac{\sqrt{3}-\sqrt{2}}{3-2}
= \sqrt{3}-\sqrt{2}
Putting the given values, \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732
We get,
= 1.732 – 1.414
= 0.318
Hence the answer is 0.318

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Posted by

infoexpert27

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

(i) Find the values of a in each of the following : \frac{5+2\sqrt{3}}{7+4\sqrt{3}}= a-6\sqrt{3}
(ii) Find the values of a in the following : \frac{3-\sqrt{5}}{3+2\sqrt{5}}= a\sqrt{5}-\frac{19}{11}

(iii) Find the values of b in the following : \frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}
(iv) Find the values of a and b in the following : \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}= a+\frac{7}{11}\sqrt{5b}

(i) Answer.   a = 11
Solution.   We have, \frac{5+2\sqrt{3}}{7+4\sqrt{3}}= a-6\sqrt{3}
LHS = \frac{5+2\sqrt{3}}{7+4\sqrt{3}}
Rationalising the denominator, we get:
= \frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}
= \frac{\left ( 5+2\sqrt{3} \right )\left ( 7-4\sqrt{3} \right )}{\left ( 7+4\sqrt{3} \right )\left ( 7-4\sqrt{3} \right )}
{Using (a – b) (a + b) = a2 – b2}

= \frac{35+14\sqrt{3}-20\sqrt{3}-24}{7^{2}-\left ( 4\sqrt{3} \right )^{2}}
= \frac{11-6\sqrt{3}}{49-48}
= 11-6\sqrt{3}
Now RHS = a-6\sqrt{3}
\Rightarrow 11-6\sqrt{3}= a-6\sqrt{3}
\Rightarrow 11-6\sqrt{3}= a-6\sqrt{3}
\Rightarrow a= 11
Hence a = 11 is the required answer

(ii)Answer.   a= \frac{9}{11}
Solution.  Given that, \frac{3-\sqrt{5}}{3+2\sqrt{5}}= a\sqrt{5}-\frac{19}{11}

LHS = \frac{3-\sqrt{5}}{3+2\sqrt{5}}
Rationalising the denominator, we get:
LHS = \frac{3-\sqrt{5}}{3+2\sqrt{5}}\times \frac{3-2\sqrt{5}}{3-2\sqrt{5}}
{Using (a – b) (a + b) = a2 – b2}
= \frac{9-3\sqrt{5}-6\sqrt{5}+10}{3^{2}-\left ( 2\sqrt{5} \right )^{2}}
= \frac{19-9\sqrt{5}}{9-20}
Now RHS = a\sqrt{5}-\frac{19}{11}
\Rightarrow \frac{9\sqrt{5}}{11}-\frac{19}{11}= a\sqrt{5}-\frac{19}{11}
Comparing both , we get
\Rightarrow a= \frac{9}{11}
Hence a= \frac{9}{11}is the correct answer

(iii) Answer:b = -\frac{5 }{6}

Solution:

Given:

\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}

LHS = \frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}

Rationalize

= \frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}} \times \frac{3 \sqrt{2}+2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}

=\frac{\sqrt{2}(3 \sqrt{2}+2 \sqrt{3})+\sqrt{3}(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2})^{2}-(2 \sqrt{3})^{2}}

= \frac{6+2 \sqrt{6}+3 \sqrt{6}+6}{18-12}

= 2+\frac{5 \sqrt{6}}{6}

2+\frac{5 \sqrt{6}}{6}=2-b \sqrt{6}

b = -\frac{5 }{6}

 

(iv) Answer. a = 0, b = 1
Solution.         Given,
\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}= a+\frac{7}{11}\sqrt{5b}
LHS = \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}
=\frac{\left ( 7+\sqrt{5} \right )\times\left ( 7+\sqrt{5} \right )-\left ( 7-\sqrt{5} \right ) \times \left ( 7-\sqrt{5} \right )}{\left (7 -\sqrt{5} \right )\left (7 +\sqrt{5} \right )}
Using (a – b) (a + b) = a2 – b2
(a + b)2 = a2 + b2 + 2ab
(a - b)2 = a2 + b2 - 2ab
= \frac{\left ( 7^{2} +\sqrt{5}^{2}+2\cdot 7\cdot \sqrt{5}\right )-\left ( 7^{2} +\sqrt{5}^{2}-2\cdot 7\cdot \sqrt{5} \right )}{7^{2}-\sqrt{5}^{2}}
= \frac{\left ( 49+5+14\sqrt{5} \right )-\left ( 49+5-14\sqrt{5} \right )}{49-5}
= \frac{54+14\sqrt{5}-54+14\sqrt{5}}{44}
= \frac{28\sqrt{5}}{44}
RHS = a+\frac{7}{11}\sqrt{5b}
Now LHS = RHS
\Rightarrow \frac{28\sqrt{5}}{44}= a+\frac{7}{11}\sqrt{5b}
\Rightarrow 0+\left ( \frac{4}{4} \right )\frac{7}{11}\sqrt{5}= a+\frac{7}{11}\sqrt{5b}

\Rightarrow a = 0, b = 1
Hence the answer is a = 0, b = 1

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Posted by

infoexpert27

(i) Rationalise the denominator of the following : \frac{2}{3\sqrt{3}}
(ii)Rationalise the denominator of the following : \frac{\sqrt{40}}{\sqrt{3}}
(iii) Rationalise the denominator of the following : \frac{3+\sqrt{2}}{4\sqrt{2}}
(iv)Rationalise the denominator of the following :\frac{16}{\sqrt{41}-5}
(v) Rationalise the denominator of the following : \frac{2+\sqrt{3}}{2-\sqrt{3}}
(vi) Rationalise the denominator of the following  : \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}
(vii) Rationalise the denominator of the following : \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}

(viii) Rationalise the denominator of the following : \frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}

(ix) Rationalise the denominator of the following : \frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}


 

(i) Answer.  \frac{2\sqrt{3}}{9}  
Solution.         We have, \frac{2}{3\sqrt{3}}
Rationalising the denominator, we get:
\frac{2}{3\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}
= \frac{2\sqrt{3}}{3\sqrt{3}\sqrt{3}}
= \frac{2\sqrt{3}}{9}
Hence the answer is \frac{2\sqrt{3}}{9}
 

(ii) Answer.  \frac{2\sqrt{30}}{3}  
Solution. We have ,\frac{\sqrt{40}}{\sqrt{3}}
We know that, 40 = (2) (2) (10)
\frac{\sqrt{40}}{\sqrt{3}}= \frac{\sqrt{2\cdot 2\cdot 10}}{\sqrt{3}}= \frac{2\sqrt{10}}{\sqrt{3}}
Rationalising the denominator, we get:
= \frac{2\sqrt{10}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}
= \frac{2\sqrt{10}\sqrt{3}}{\sqrt{3}\sqrt{3}}
= \frac{2\sqrt{30}}{3}
Hence the answer is: \frac{2\sqrt{30}}{3}

(iii) Answer.  \frac{3\sqrt{2}+2}{8}
Solution.  We have \frac{3+\sqrt{2}}{4\sqrt{2}}
Rationalising the denominator, we get:
\frac{3+\sqrt{2}}{4\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}
= \frac{\left ( 3+\sqrt{2} \right )\sqrt{2}}{4\sqrt{2}\sqrt{2}}
= \frac{3\sqrt{2}+2}{8}
Hence the answer is \frac{3\sqrt{2}+2}{8}

(iv) Answer. \sqrt{41}+5
Solution. We have \frac{16}{\sqrt{41}-5}
Rationalising the denominator, we get:
\frac{16}{\sqrt{41}-5}\times \frac{\sqrt{41}+5}{\sqrt{41}+5}
= \frac{16\left ( \sqrt{41}+5 \right )}{\left ( \sqrt{41}-5 \right )\sqrt{41}+5}
Using  the identity (a – b) (a + b) = a2 – b2
We get:
= \frac{16\left ( \sqrt{41}+5 \right )}{\left ( \sqrt{41} \right )^{2}-\left ( 5 \right )^{2}}
= \frac{16\left ( \sqrt{41}+5 \right )}{41-25}
= \frac{16\left ( \sqrt{41}+5 \right )}{16}
= \sqrt{41}+5
Hence the answer is \sqrt{41}+5

(v) Answer.    7+4\sqrt{3}
Solution. We have, \frac{2+\sqrt{3}}{2-\sqrt{3}}
Rationalising the denominator, we get:
\frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2-\sqrt{3}}
= \frac{\left ( 2+\sqrt{3} \right )^{2}}{\left ( 2-\sqrt{3} \right )\left ( 2+\sqrt{3} \right )}

Using   (a – b) (a + b) = a2 – b2
and      (a + b)2 = a2 + b2 + 2ab
= \frac{2^{2}+\left ( \sqrt{3} \right )^{2}+2\cdot 2\cdot \sqrt{3}}{2^{2}-\left ( \sqrt{3} \right )^{2}}
= \frac{4+3+4\sqrt{3}}{4-3}
= 7+4\sqrt{3}
Hence the answer is 7+4\sqrt{3}

(vi)Answer.  3\sqrt{2}-2\sqrt{3}
Solution.   We have, \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}
Rationalising the denominator, we get:
= \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}
= \frac{\sqrt{6}\left ( \sqrt{3}-\sqrt{2} \right )}{\left ( \sqrt{2}+\sqrt{3} \right )\left ( \sqrt{3}-\sqrt{2} \right )}
Using  the identity (a – b) (a + b) = a2 – b2
We get:
= \frac{\sqrt{18}-\sqrt{12}}{\left ( \sqrt{3} \right )^{2}\left ( \sqrt{2} \right )^{2}}
= \frac{\sqrt{3\cdot 3\cdot 2}-\sqrt{2\cdot 2\cdot 3}}{3-2}
= \frac{3\sqrt{2}-2\sqrt{3}}{1}
= 3\sqrt{2}-2\sqrt{3}
Hence the answer 3\sqrt{2}-2\sqrt{3}

(vii) Answer. 5+2\sqrt{6}
Solution. We have, \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}

Rationalising the denominator, we get:
\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}
= \frac{\left ( \sqrt{3} +\sqrt{2}\right )^{2}}{\left ( \sqrt{3} -\sqrt{2} \right )\left ( \sqrt{3} +\sqrt{2} \right )}

Using   (a – b) (a + b) = a2 – b2
and      (a + b)2 = a2 + b2 + 2ab
= \frac{\left ( \sqrt{3} \right )^{2}+\left ( \sqrt{2} \right )^{2}+2\sqrt{3}\sqrt{2}}{\left ( \sqrt{3} \right )^{2}-\left ( \sqrt{2} \right )^{2}}
= \frac{3+2+2\sqrt{6}}{3-2}
= 5+2\sqrt{6}
Hence the answer is 5+2\sqrt{6}

(viii)

Answer: 9+2 \sqrt{15}

Solution:

We have \frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}

Rationalize

=\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{3 \sqrt{5}(\sqrt{5}+\sqrt{3})+\sqrt{3}(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}

=\frac{15+3 \sqrt{15}+\sqrt{15}+3}{5-3}=\frac{18+4 \sqrt{15}}{2}

=9+2 \sqrt{15}

(ix) Answer: \frac{9+4 \sqrt{6}}{15}

Solution:

We have \frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}

=\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}=\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{16 \times 3}+\sqrt{9 \times 2}}=\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}}

Rationalize

=\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}} \times \frac{(4 \sqrt{3}-3 \sqrt{2})}{(4 \sqrt{3}-3 \sqrt{2})}

=\frac{4 \sqrt{3}(4 \sqrt{3}-3 \sqrt{2})+5 \sqrt{2}(4 \sqrt{3}-3 \sqrt{2})}{(4 \sqrt{3})^{2}-(3 \sqrt{2})^{2}}

=\frac{48-12 \sqrt{6}+20 \sqrt{6}-30}{30}

=\frac{18+8 \sqrt{6}}{30}

=\frac{9+4 \sqrt{6}}{15}

 

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(i) Simplify the following : \sqrt{45}-3\sqrt{20}+4\sqrt{5}
(ii) Simplify the following : \frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}
(iii) Simplify the following : 4\sqrt{12}\times 7\sqrt{6}
(iv) Simplify the following : 4\sqrt{28}\div 3\sqrt{7}\div 3\sqrt{7}
(v) Simplify the following : 3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}
(vi) Simplify the following : \left ( \sqrt{3}-\sqrt{2} \right )^{2}
(vii) Simplify the following : \sqrt[4]{81}- 8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{255}
(viii) Simplify the following : \frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}
(ix) Simplify the following : \frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}






 

(i) Answer. \sqrt{5}
Solution.    \sqrt{45}-3\sqrt{20}+4\sqrt{5} 
We know that,
45 = 3\times 3\times 5
20 = 2\times 2\times 5
So we get
\sqrt{3\times3\times5 }-3\sqrt{2\times2\times5}+4\sqrt{5}
= 3\sqrt{5}-3\left ( 2\sqrt{5} \right )+4\sqrt{5}
= 3\sqrt{5}-6\sqrt{5}+4\sqrt{5}
= 7\sqrt{5}-6\sqrt{5}
= \sqrt{5}  
Hence the answer is \sqrt{5}

(ii)  Answer. \frac{7\sqrt{6}}{12}
Solution. We have, \frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}
We know that,

24= 6\times 4= 3\times 2\times 2\times 2
54= 9\times 6= 3\times 3\times 3\times 2
So we get
\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}= \frac{2\sqrt{6}}{8}+\frac{3\sqrt{6}}{9}
= \frac{\sqrt{6}}{4}+\frac{\sqrt{6}}{3}
Taking LCM (3,4) = 12

= \frac{3\sqrt{6}+4\sqrt{6}}{12}
= \frac{7\sqrt{6}}{12}

(iii) Answer.  \sqrt[28]{2^{18} \times 3^{11}}
Solution.   We have
\sqrt[4]{12}\times \sqrt[7]{6}
We know that
12 = 2\times 2\times 3
6 = 2\times 3
So we get,
=\sqrt[4]{2\times 2\times 3}\times \sqrt[7]{2\times 3}  
=2^{1 / 4} \cdot 2^{1 / 4} \cdot 3^{1 / 4} \cdot 2^{1 / 7} \cdot 3^{1 / 7}
=2^{\frac{1}{4}+\frac{1}{4}+\frac{1}{7}} \times 3^{\frac{1}{4}+\frac{1}{7}}

=2^{9 / 14} \times 3^{11 / 28}

=\sqrt[28]{2^{18} \times 3^{11}}
Hence the number is \sqrt[28]{2^{18} \times 3^{11}}.

(iv)Answer.    \frac{8}{\left ( 3\times \sqrt[3]{7} \right )}     
Solution.   We have, 4\sqrt{28}\div 3\sqrt{7}\div \sqrt[3]{7}
We know that
28 = 4\times 7
So we can write,
4\sqrt{28}\div 3\sqrt{7}\div \sqrt[3]{7}= \left [ \frac{4\sqrt{28}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}
= \left [ \frac{4\sqrt{4\times 7}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}

= \left [ \frac{4\times 2\sqrt{ 7}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}

= \frac{8}{3}\div 7^{\frac{1}{3}}
= \frac{8}{\left ( 3\times 7^{\frac{1}{3}} \right )}
= \frac{8}{\left ( 3\times \sqrt[3]{7} \right )}
  Hence the answer is \frac{8}{\left ( 3\times \sqrt[3]{7} \right )}

(v) Answer.  3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}
We know that
27 = 3\times 3\times 3
So, 3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}} = 3\sqrt{3}+2\sqrt{3\times 3\times 3}+\frac{7}{\sqrt{3}}

= 3\sqrt{3}+2\left ( 3\sqrt{3} \right )+\frac{7}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}                
(Rationalising the denominator)
= 3\sqrt{3}+6\left ( \sqrt{3} \right )+\frac{7\sqrt{3}}{3}

= \left ( 3+6+\frac{7}{3} \right )\sqrt{3}         (Taking \sqrt{3} common)
Now LCM (1,1,3) = 3
= \left ( \frac{9+18+7}{3} \right )\sqrt{3}
= \frac{34}{3}\sqrt{3}
= 19\cdot 63
Hence the answer is 19.63

(vi) Answer. 5-2\sqrt{6}
Solution. Given, \left ( \sqrt{3}-\sqrt{2} \right )^{2}
We know that (a + b)2 = a2 – 2ab + b2
Comparing the given equation with the identity, we get:

\left ( \sqrt{3}-\sqrt{2} \right )^{2}= \left ( \sqrt{3} \right )^{2}-2\left ( \sqrt{3} \right )\left ( \sqrt{2} \right )+\left ( \sqrt{2} \right )^2
= 3 + 2 – 2\sqrt{3\times 2}
= 5-2\sqrt{6}
Hence the answer is 5-2\sqrt{6}

(vii) Answer. 0
Solution. We have, \sqrt[4]{81}-8 \sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}
We know that
81 = 3\times 3\times3\times3
216 = 6\times 6\times6
32 = 2\times 2\times2\times2\times2
225 = 15\times 15
So,\sqrt[4]{81}-8 \sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}
= \sqrt[4]{3\times3\times3\times3 }-8\sqrt[3]{6\times6\times6}+15\sqrt[5]{2\times2\times2\times2\times2}+\sqrt{15\times15}
= 3 – 8 × 6 + 15 × 2 + 15
= 3 – 48 + 30 + 15
= – 45 + 45
= 0
Hence the answer is 0

(viii) Answer.   \frac{5}{2\sqrt{2}}

Solution.   We have, \frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}
  We know that, 8 =2\times 2\times 2
So,

\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2\times 2\times 2}}+\frac{1}{\sqrt{2}}

= \frac{3}{2\sqrt{2}}+\frac{1}{\sqrt{2}}

= \frac{3}{2\sqrt{2}}+\frac{2}{2\sqrt{2}}
= \frac{5}{2\sqrt{2}}
Hence the answer is \frac{5}{2\sqrt{2}}

(ix) Answer.     \frac{\sqrt{3}}{2}
Solution.     We have, \frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}
LCM (3,6) = 6

\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}= \frac{4\sqrt{3}}{6}-\frac{\sqrt{3}}{6}

= \frac{4\sqrt{3}-\sqrt{3}}{6}

= \frac{3\sqrt{3}}{6}
= \frac{\sqrt{3}}{2}
Hence the answer is \frac{\sqrt{3}}{2}.

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(i) Express the following in the form \frac{p}{q}, where p and q are integers and q\neq 0. 0.2
(ii) Express the following in the form \frac{p}{q}, where p and q are integers and q\neq 0.

0.888…….
(iii)  Express the following in the form \frac{p}{q}, where p and q are integers and q\neq 0.5.\bar{2}   
(iv) Express the following in the form \frac{p}{q}, where p and q are integers and q\neq 0.0\cdot \overline{001}
(v) Express the following in the form \frac{p}{q}, where p and q are integers and q\neq 0.0.2555……
(vii) Express the following in the form \frac{p}{q}, where p and q are integers and q\neq 0..00323232…..
(viii) Express the following in the form \frac{p}{q}, where p and q are integers and q\neq 0..404040……..

 

 


       

 

 

(i) Answer.     \frac{1}{5}
Solution.   We know that
0.2 can be written as \frac{2}{10}
Now,
  \frac{2}{10}= \frac{1}{5}
Hence the answer is \frac{1}{5}

(ii) Answer.    \frac{8}{9}
Solution.    Let x = 0.888…..        .…(i)
Multiply RHS and LHS by 10
10 x = 8.88…….         …(ii)
Subtracting equation (i) from (ii)
We get
10x – x = 8.8 – 0.8 
\Rightarrow 9x= 8
\Rightarrow x= \frac{8}{9}
Hence answer is  \frac{8}{9}

(iii) Answer.  \frac{47}{9}
Solution.  Let x = 5\cdot \bar{2}          …eq. (1)
Multiply by 10 on both sides
10x = 52\cdot \bar{2}                  …eq (2)

Subtracting equation (1) from (2)
We get
10x – x = 52\cdot \bar{2} – 5\cdot \bar{2}
\Rightarrow 9x = 47
\Rightarrow x = \frac{47}{9}
Hence the answer is \frac{47}{9}

(iv) Answer.   \frac{1}{999}
Solution.    Let x = 0\cdot \overline{001}              …. Eq. (1)
Multiply by 1000 on both sides
1000 x = 1\cdot \overline{001}            …eq.(2)
Subtracting eq. (1) from (2)
We get
1000 x – x = 1\cdot \overline{001} – 1\cdot \overline{001}

\Rightarrow 999x = 1
  \Rightarrow x = \frac{1}{999}
Hence the answer is \frac{1}{999} 

(v) Answer.   \frac{23}{90}

Solution. Let x = 0.2555 …..    …eq.(1)
Multiply by 10 on both sides
10x = 2.555…                         …eq.(2)
Multiply by 100 on both sides
100x = 25.55…                       …eq. (3)
Subtracting eq. (2) from (3),
We get
100x – 10x = 25.555… – 2.555…

\Rightarrow 90x = 23
\Rightarrow x = \frac{23}{90} 
  Hence the answer is \frac{23}{90}

(vii) Answer.    \frac{8}{2475}
Solution.   Let    x = 0.00323232…..      …eq.(1)
Multiply with 100 on both sides.
We get
100x = 0.3232…                …eq(2)
Multiply again with 100 on both sides,
10000x = 32.3232…                           …eq(3)
Equation (3) – (2)
we get,
10000 x – 100x = 32.3232… – 0.3232…
\Rightarrow  9900x = 32
 \Rightarrow x = \frac{32}{9900}
 x = \frac{8}{2475}
Hence the answer is  \frac{8}{2475}

(viii) Answer.     \frac{40}{99}
Solution. Let x = 0.404040…….           …(1)
Multiplying by 100 on both sides
we get
100x = 40.40…                       …(2)
Subtracting equation (1) from (2)
we get
100x – x = 40.40 – 0.40
\Rightarrow99x = 40
 \Rightarrow  x = \frac{40}{99} 
Hence the answer is \frac{40}{99}

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