Practice MCQ Questions and Answer on Surds and Indices

Solution:

 $$\eqalign{ & {\text{I}}.\sqrt {64} + \sqrt {0.0064} + \sqrt {0.81} + \sqrt {0.0081} = 9.07 \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = 8 + 0.08 + 0.9 + 0.09 \cr & = 9.07 = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr & {\text{Hence, statement I is true}} \cr & {\text{II}}.\sqrt {0.010201} + \sqrt {98.01} + \sqrt {0.25} = 11.51 \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \sqrt {{{\left( {0.101} \right)}^2}} + \sqrt {{{\left( {9.9} \right)}^2}} + \sqrt {{{\left( {0.5} \right)}^2}} \cr & = 0.101 + 9.9 + 0.5 \cr & = 10.501 \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr & \Rightarrow {\text{Statement II is not true}} \cr} $$

Solution:

 $$\eqalign{ & \left( {\sqrt 3 + 1} \right)\left( {10 + \sqrt {12} } \right)\left( {\sqrt {12} - 2} \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow \left( {\sqrt 3 + 1} \right)\left( {10 + 2\sqrt 3 } \right)\left( {2\sqrt 3 - 2} \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow \left( {\sqrt 3 + 1} \right) \times 2\left( {5 + \sqrt 3 } \right) \times 2\left( {\sqrt 3 - 1} \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow 4\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)\left( {5 - \sqrt 3 } \right)\left( {5 + \sqrt 3 } \right) \cr & \Rightarrow 4\left( {3 - 1} \right)\left( {25 - 3} \right)\,\,\,\,\,\left[ {\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right] \cr & \Rightarrow 4 \times 2 \times 22 \cr & \Rightarrow 176 \cr & \therefore {\text{The required answer is }}176 \cr} $$

Solution:

 3200, 2300, 7100 9100, 8100, 7100 It is clear that 3200 is largest

Solution:

 $$\eqalign{ & {\text{According to question,}} \cr & \Rightarrow {{\text{3}}^{x + y}}{\text{ = 81}}\,{\text{and}}\,{\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr & \Rightarrow {{\text{3}}^{x + y}}{\text{ = (3}}{{\text{)}}^4}\,{\text{and}}\,{\left( 3 \right)^{4(}}^{x - y)}{\text{ = }}{{\text{3}}^1} \cr & \Rightarrow x + y = 4\,{\text{and}}\,x - y = \frac{1}{4} \cr & x + y = 4......{\text{(i)}} \cr & {\text{ }}x - y = \frac{1}{4}.....(ii) \cr & {\text{Solve the equation of (i) and (ii)}} \cr & x = \frac{{17}}{8}, \cr & y = \frac{{15}}{8}, \cr & \Rightarrow \frac{x}{y} = \frac{{17}}{{15}} \cr} $$

Solution:

 $$\eqalign{ & \sqrt {121} + \sqrt {12321} + \sqrt {1234321} + \sqrt {123454321} \cr & = 11 + 111 + 1111 + 11111 \cr & = 12344 \cr} $$

Solution:

 $$\eqalign{ & \left( {\frac{{{9^2} \times {{18}^4}}}{{{3^{16}}}}} \right) \cr & = \frac{{{9^2} \times {{\left( {9 \times 2} \right)}^4}}}{{{3^{16}}}} \cr & = \frac{{{{\left( {{3^2}} \right)}^2} \times {{\left( {{3^2}} \right)}^4} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^4} \times {3^8} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^{\left( {4 + 8} \right)}} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^{12}} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{2^4}}}{{{3^{\left( {16 - 12} \right)}}}} \cr & = \frac{{{2^4}}}{{{3^4}}} \cr & = \frac{{16}}{{81}} \cr} $$

Solution:

 $$\eqalign{ & {x^z}{\text{ = }}{y^2}{\text{ }} \cr & \Leftrightarrow {\left( {{{10}^{0.48}}} \right)^z} = {\left( {{{10}^{0.70}}} \right)^2} \cr & \Leftrightarrow {10^{\left( {0.48z} \right)}} = {10^{\left( {2 \times 0.70} \right)}} = {10^{1.40}} \cr & \Leftrightarrow 0.48z = 1.40 \cr & \Leftrightarrow z = \frac{{140}}{{48}} = \frac{{35}}{{12}} \cr & \Leftrightarrow z = 2.9\left( {{\text{approx}}} \right) \cr} $$

Solution:

 $$\eqalign{ & {\left( {10} \right)^{150}} \div {\left( {10} \right)^{146}} = \frac{{{{10}^{150}}}}{{{{10}^{146}}}} \cr & = {10^{150 - 146}} \cr & = {10^4} \cr & = 10000 \cr} $$

Solution:

Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{Let }}{25^{2.7}} \times {5^{4.2}} \div {5^{5.4}} = {25^x} \cr & {\text{Then, }}{25^{2.7}} \times {5^{(4.2 - 5.4)}} = {25^x} \cr & \Leftrightarrow {25^{2.7}} \times {5^{( - 1.2)}} = {25^x} \cr & \Leftrightarrow {25^{2.7}} \times \frac{1}{{{5^{1.2}}}} = {25^x} \cr & \Leftrightarrow \frac{{{{25}^{2.7}}}}{{{{\left( {{5^2}} \right)}^{0.6}}}} = {25^x} \cr & \Leftrightarrow \frac{{{{\left( {25} \right)}^{2.7}}}}{{{{\left( {25} \right)}^{0.6}}}} = {25^x} \cr & \Leftrightarrow {25^x} = {25^{\left( {2.7 - 0.6} \right)}} = {25^{2.1}} \cr & \Leftrightarrow x = 2.1 \cr} $$

Solution:

 $$\eqalign{ & \left( {4 + \sqrt 7 } \right) \cr & = \frac{7}{2} + \frac{1}{2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr & = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr & = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right)^2} \cr & = \frac{1}{2}{\left( {\sqrt 7 + 1} \right)^2} \cr} $$