Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{2}}\sqrt {50} {\text{ + }}\sqrt {18} - \sqrt {72} \cr & \Rightarrow {\text{2}} \times {\text{5}}\sqrt 2 {\text{ + 3}}\sqrt 2 - 6\sqrt 2 \cr & \Rightarrow 13\sqrt 2 - 6\sqrt 2 \cr & \Rightarrow 7\sqrt 2 \cr & \Rightarrow 7 \times 1.414 \cr & \Rightarrow 9.898 \cr} $$

$$\eqalign{ & {\text{Let }}a = 55, \cr & \,\,\,\,\,\,\,\,\,\,\,b = 17, \cr & \,\,\,\,\,\,\,\,\,\,\,c = - 72 \cr & a + b + c \cr & = 55 + 17 - 72 \cr & = 0 \cr & \therefore {a^3} + {b^3} + {c^3} - 3abc = 0 \cr & \left( {a + b + c} \right) = 0 \cr & {\text{Answer is }}0. \cr} $$

$$\eqalign{ & \left( {\sqrt 3 + 1} \right)\left( {10 + \sqrt {12} } \right)\left( {\sqrt {12} - 2} \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow \left( {\sqrt 3 + 1} \right)\left( {10 + 2\sqrt 3 } \right)\left( {2\sqrt 3 - 2} \right)\left( {5 - \sqrt 3 } \right) \cr} $$
$$ \Rightarrow \left( {\sqrt 3 + 1} \right) \times $$   $$2\left( {5 + \sqrt 3 } \right) \times $$   $$2\left( {\sqrt 3 - 1} \right)$$  $$\left( {5 - \sqrt 3 } \right)$$
$$\eqalign{ & \Rightarrow 4\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)\left( {5 + \sqrt 3 } \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow 4\left[ {{{\left( {\sqrt 3 } \right)}^2} - {1^2}} \right]\left[ {{{\left( 5 \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} \right] \cr & \Rightarrow 4 \times 2 \times 22 \cr & \Rightarrow 176 \cr} $$

$$\eqalign{ & \frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} \cr & = \frac{1}{{1 + \frac{{{a^n}}}{{{a^m}}}}} + \frac{1}{{1 + \frac{{{a^m}}}{{{a^n}}}}} \cr & = \frac{{{a^m}}}{{{a^m} + {a^n}}} + \frac{{{a^n}}}{{{a^m} + {a^n}}} \cr & = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{Given expression, }} \cr & \frac{1}{{1 + \frac{{{x^b}}}{{{x^a}}} + \frac{{{x^c}}}{{{x^a}}}}} + \frac{1}{{1 + \frac{{{x^a}}}{{{x^b}}} + \frac{{{x^c}}}{{{x^b}}}}} + \frac{1}{{1 + \frac{{{x^b}}}{{{x^c}}} + \frac{{{x^a}}}{{{x^c}}}}} \cr} $$
  $$ = \frac{{{x^a}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$    $$\frac{{{x^b}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$   $$\frac{{{x^c}}}{{\left( {{x^a} + {x^{b}} + {x^c}} \right)}}$$
$$\eqalign{ & = \frac{{\left( {{x^a} + {x^b} + {x^c}} \right)}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} \cr & = 1 \cr} $$

$${x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}}$$
  $$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$    $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$   $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$
$$\eqalign{ & = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}} \cr & = \left( {{x^0} \times {x^0}} \right) \cr & = \left( {1 \times 1} \right) \cr & = 1 \cr} $$

$$\eqalign{ & {\text{ }}{{\text{2}}^{n - 1}}{\text{ + }}{{\text{2}}^{n + 1}}{\text{ = 320}} \cr & \Rightarrow {\text{ }}{{\text{2}}^{n - 1}}\left( {1 + {2^2}} \right){\text{ = 320}} \cr & \Rightarrow {\text{ }}{{\text{2}}^{n - 1}} \times {\text{5 = 320}} \cr & \Rightarrow {\text{ }}{{\text{2}}^{n - 1}}{\text{ = }}\frac{{320}}{5}{\text{ = 64}} \cr & \Rightarrow {\left( 2 \right)^{n - 1}} = {\left( 2 \right)^6} \cr & \Rightarrow n - 1 = 6 \cr & \Rightarrow n = 7 \cr} $$

$$\eqalign{ & {4^{61}} + {4^{62}} + {4^{63}} + {4^{64}} \cr & = {4^{61}}\left( {{4^0} + {4^1} + {4^2} + {4^3}} \right) \cr & = {4^{61}} \times 85 \cr} $$
Now check with option 85 is divisible by 17.

$$\eqalign{ & {x^{\frac{{b + c}}{{\left( {a - b} \right)\left( {c - a} \right)}}}}.{x^{\frac{{c + a}}{{\left( {a - b} \right)\left( {b - c} \right)}}}}.{x^{\frac{{a + b}}{{\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^{\frac{{\left( {b + c} \right)\left( {b - c} \right) + \left( {c + a} \right)\left( {c - a} \right) + \left( {a + b} \right)\left( {a - b} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^{\frac{{\left( {{b^2} - {c^2}} \right) + \left( {{c^2} - {a^2}} \right) + \left( {{a^2} - {b^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^0} \cr & = 1 \cr} $$

$$\eqalign{ & {a^1} \cr & = {c^z} \cr & = {\left( {{b^y}} \right)^z} \cr & = {b^{yz}} \cr & = {\left( {{a^x}} \right)^{yz}} \cr & = {a^{xyz}} \cr & \Rightarrow xyz = 1 \cr} $$