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Number Systems : Exercise 1.6 (Mathematics NCERT Class 9th)


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Q.1     Find
              (i) {64^{{1 \over 2}}}   (ii) {32^{{1 \over 5}}}   (iii){125^{{1 \over 3}}}
Sol.

(i)    {64^{{1 \over 2}}}  = {\left( {8 \times 8} \right)^{{1 \over 2}}}  = {\left( {{8^2}} \right)^{{1 \over 2}}}  = {8^{2 \times {1 \over 2}}} = {8^1} = 8
(ii)   {32^{{1 \over 5}}}  = {\left( {2 \times 2 \times 2 \times 2 \times 2} \right)^{{1 \over 5}}} = {\left( {{2^5}} \right)^{{1 \over 5}}}  = {2^{5 \times {1 \over 5}}} = {2^1} = 2
(iii)  {125^{{1 \over 3}}}  = {\left( {5 \times 5 \times 5} \right)^{{1 \over 3}}} = {\left( {{5^3}} \right)^{{1 \over 3}}}  = {5^{3 \times {1 \over 3}}} = {5^1} = 5


Q.2       Find :
            
(i) {9^{{3 \over 2}}}   (ii) {32^{{2 \over 5}}}  
(iii) {16^{{3 \over 4}}}   (iv) {125^{{{ - 1} \over 3}}}
Sol.

(i)    {9^{{3 \over 2}}}  = {\left( {3 \times 3} \right)^{{3 \over 2}}} = {\left( {{3^2}} \right)^{{3 \over 2}}} = {3^{2 \times {3 \over 2}}}  = {3^3} = 3 \times 3 \times 3 = 27
(ii)   {32^{{2 \over 5}}} = {\left( {2 \times 2 \times 2 \times 2 \times 2} \right)^{{2 \over 5}}} = {\left( {{2^5}} \right)^{{2 \over 5}}} {2^{5\, \times {2 \over 5}}} = {2^2} = 2 \times 2 = 4
(iii)  {16^{{3 \over 4}}} = {\left( {2 \times 2 \times 2 \times 2} \right)^{{3 \over 4}}} = {\left( {{2^4}} \right)^{{3 \over 4}}}  = {2^{4 \times {3 \over 4}}} = {2^3} = 2 \times 2 \times 2 = 8
(iv)    {125^{ - {1 \over 3}}} = {1 \over {{{125}^{{1 \over 3}}}}}  ={1 \over {{{\left( {5 \times 5 \times 5} \right)}^{{1 \over 3}}}}} = {1 \over 5}


Q.3      Simplify :
           
(i) {2^{{2 \over 3}}}{.2^{{1 \over 5}}}    
           
(ii) {\left( {{1 \over {{3^3}}}} \right)^7}
           
(iii) {{{{11}^{{1 \over 2}}}} \over {{{11}^{{1 \over 4}}}}}
           
(iv) {7^{{1 \over 2}}}{.8^{{1 \over 2}}}
Sol.

(i)    {2^{{2 \over 3}}}{.2^{{1 \over 5}}} = {2^{{2 \over 3} + {1 \over 5}}} = {2^{{{10 + 3} \over {15}}}} = {2^{{{13} \over {15}}}}
(ii)   {\left( {{1 \over {{3^2}}}} \right)^7} = {{{{\left( 1 \right)}^7}} \over {{3^{2 \times 7}}}} = {1 \over {{3^{21}}}}
(iii)  {{{{11}^{{1 \over 2}}}} \over {{{11}^{{1 \over 4}}}}} = {11^{{1 \over 2} - {1 \over 4}}} = {11^{{1 \over 4}}}
(iv)   {7^{{1 \over 2}}}{.8^{{1 \over 2}}} = {\left( {7 \times 8} \right)^{{1 \over 2}}} = {56^{{1 \over 2}}} = {\left( {4 \times 14} \right)^{{1 \over 2}}} = {\left( {56} \right)^{{1 \over 2}}}



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