Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

LCM of 2, 3, 4, 6 is 12
$$\sqrt 2 = $$  $${2^{\frac{1}{2}}} = $$  $${2^{\left( {\frac{1}{2} \times \frac{6}{6}} \right)}} = $$   $${2^{\frac{6}{{12}}}} = $$  $${\left( {{2^6}} \right)^{\frac{1}{{12}}}} = $$  $${\left( {64} \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of {64} $$
$$\root 6 \of 3 = $$  $${3^{\frac{1}{6}}} = $$  $${3^{\left( {\frac{1}{6} \times \frac{2}{2}} \right)}} = $$   $${3^{\frac{2}{{12}}}} = $$  $${\left( {{3^2}} \right)^{\frac{1}{{12}}}} = $$  $${\left( 9 \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of 9 $$
$$\root 3 \of 4 = $$  $${4^{\frac{1}{3}}} = $$  $${4^{\left( {\frac{1}{3} \times \frac{4}{4}} \right)}} = $$  $${4^{\frac{4}{{12}}}} = $$  $${\left( {{4^4}} \right)^{\frac{1}{{12}}}} = $$  $${\left( {256} \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of {256} $$
$$\root 4 \of 5 = $$  $${5^{\frac{1}{4}}} = $$  $${5^{\left( {\frac{1}{4} \times \frac{3}{3}} \right)}} = $$   $${5^{\frac{3}{{12}}}} = $$  $${\left( {{5^3}} \right)^{\frac{1}{{12}}}} = $$  $${\left( {125} \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of {125} $$
Clearly $$\root {12} \of {256} \,\,i.e.,\root 3 \of 4 $$     is the greatest.

$$\eqalign{ & {\text{I}}.\sqrt {121} + \sqrt {12321} + \sqrt {1234321} = 1233 \cr & 11 + 111 + 1111 = 1233 \cr & 1233 = 1233 \cr & {\text{II}}.\sqrt {0.64} + \sqrt {64} + \sqrt {36} + \sqrt {0.36} > 15 \cr & \frac{8}{{10}} + 8 + 6 + \frac{6}{{10}} > 15 \cr & 15.4 > 15 \cr & {\text{Hence statement II is also true}}{\text{.}} \cr} $$

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

$$\eqalign{ & \sqrt {10 + 2\left( {\sqrt 6 - \sqrt {15} - \sqrt {10} } \right)} \cr & = \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} + {{\left( { - \sqrt 5 } \right)}^2} + 2\left( {\sqrt 2 \times \sqrt 3 - \sqrt 3 \times \sqrt 5 - \sqrt 5 \times \sqrt 2 } \right)} \cr & = \sqrt {{{\left( {\sqrt 2 + \sqrt 3 - \sqrt 5 } \right)}^2}} \cr & = \sqrt 2 + \sqrt 3 - \sqrt 5 \cr} $$

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

$$\eqalign{ & \frac{1}{{0.4}} + \frac{1}{{0.04}} + \frac{1}{{0.004}} + ....{\text{ upto 8 terms}} \cr & = \frac{{10}}{4} + \frac{{100}}{4} + \frac{{1000}}{4} + ....{\text{ upto 8 terms}} \cr & = \frac{1}{4}\left[ {10 + 100 + 1000 + ....{\text{ upto 8 terms}}} \right] \cr & = \frac{1}{4}\left[ {\frac{{10\left( {{{10}^8} - 1} \right)}}{{10 - 1}}} \right]\,\,\,\,\,\,\,\,\,\,\left( {\because {S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}} \right) \cr & = \frac{1}{{4 \times 9}} \times 10\left( {{{10}^4} + 1} \right)\left( {{{10}^4} - 1} \right) \cr & = \frac{{10}}{{4 \times 9}} \times \left( {{{10}^4} + 1} \right)\left( {{{10}^2} + 1} \right)\left( {10 + 1} \right)\left( {10 - 1} \right) \cr & = \frac{{10}}{{4 \times 9}} \times \left( {{{10}^4} + 1} \right)\left( {{{10}^2} + 1} \right)\left( {11} \right)\left( 9 \right) \cr & = \frac{{10 \times 10001 \times 101 \times 11}}{4} \cr & = 27777777.5 \cr} $$

$$\eqalign{ & \sqrt {28 + 10\sqrt 3 } - \sqrt {7 - 4\sqrt 3 } \cr & = \sqrt {{{\left( {5 + \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} \cr & = 5 + \sqrt 3 - 2 + \sqrt 3 \cr & = 3 + 2\sqrt 3 \cr & = 3 + 2 \times 1.73 \cr & = 3 + 3.46 \cr & = 6.5 \cr} $$

$$\eqalign{ & x = \root 6 \of {27} - \sqrt {6\frac{3}{4}} \cr & {x^2} = {\left( {{{27}^{\frac{1}{6}}} - \frac{{{{27}^{\frac{1}{2}}}}}{2}} \right)^2} \cr & {x^2} = {27^{\frac{2}{6}}} + \frac{{27}}{4} - \frac{{2 \times {{27}^{\frac{1}{6}}} \times {{27}^{\frac{1}{2}}}}}{2} \cr & {x^2} = 3 + \frac{{27}}{4} - 9 \cr & {x^2} = \frac{3}{4} \cr & y = \frac{{\sqrt {45} + \sqrt {605} + \sqrt {245} }}{{\sqrt {80} + \sqrt {125} }} \cr & y = \frac{{3\sqrt 5 + 11\sqrt 5 + 7\sqrt 5 }}{{4\sqrt 4 + 5\sqrt 5 }} \cr & y = \frac{7}{3} \cr & {y^2} = \frac{{49}}{9} \cr & {x^2} + {y^2} = \frac{3}{4} + \frac{{49}}{9} = \boxed{\frac{{223}}{{36}}} \cr} $$

847 × 385 × 675 × 3025 = 3^a × 5^b × 7^c × 11^d
LHS ⇒ 847 × 385 × 675 × 3025
= (7 × 11 × 11) × (5 × 7 × 11) × (3 × 3 × 3 × 5 × 5) × (5 × 5 × 11 × 11)
= 3³ × 5⁵ × 7² × 11⁵
= 3^a × 5^b × 7^c × 11^d
Compare value
a = 3; b = 5; c = 2; d = 5
⇒ ab - cd = 3 × 5 - 2 × 5 = 15 - 10 = 5