Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

(d^{s + t} ÷ d^s) ÷ d^t
= $$\left( {\frac{{{{\text{d}}^{{\text{s}} + {\text{t}}}}}}{{{{\text{d}}^{\text{s}}}}}} \right)$$  ÷ d^t
= (d^{s + t - s}) ÷ d^t
= (d^t) ÷ d^t
= $$\frac{{{{\text{d}}^{\text{t}}}}}{{{{\text{d}}^{\text{t}}}}}$$
= d^{t - t}
= d⁰
= 1

(3³³ + 3³³ + 3³³)(2³³ + 2³³) = 6^x
(3.3³³)(2.2³³) = 6^x
(3³⁴)(2³⁴) = 6^x
6³⁴ = 6^x
x = 34

$$\eqalign{ & \frac{{\left( {x - \sqrt {24} } \right)\left( {\sqrt {75} + \sqrt {50} } \right)}}{{\sqrt {75} - \sqrt {50} }} = 1 \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{\sqrt {75} - \sqrt {50} }}{{\sqrt {75} + \sqrt {50} }} \times \frac{{\sqrt {75} - \sqrt {50} }}{{\sqrt {75} - \sqrt {50} }} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{{{\left( {\sqrt {75} - \sqrt {50} } \right)}^2}}}{{75 - 50}} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{75 + 50 - 2\sqrt {75} \sqrt {50} }}{{25}} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{125 - 2 \times 5\sqrt 3 \times 5\sqrt 2 }}{{25}} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{125 - 50\sqrt 6 }}{{25}} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{25\left( {5 - 2\sqrt 6 } \right)}}{{25}} \cr & \Rightarrow x - 2\sqrt 6 = 5 - 2\sqrt 6 \cr & \Rightarrow x = 5 \cr} $$

$$\eqalign{ & \frac{{3\sqrt 7 }}{{\sqrt 5 + \sqrt 2 }} - \frac{{5\sqrt 5 }}{{\sqrt 2 + \sqrt 7 }} + \frac{{2\sqrt 2 }}{{\sqrt 7 + \sqrt 5 }} \cr & = \frac{{3\sqrt 7 }}{{\sqrt 5 + \sqrt 2 }} \times \frac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }} - \frac{{5\sqrt 5 }}{{\sqrt 7 + \sqrt 2 }} \times \frac{{\sqrt 7 - \sqrt 2 }}{{\sqrt 7 - \sqrt 2 }} + \frac{{2\sqrt 2 }}{{\sqrt 7 + \sqrt 5 }} \times \frac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 - \sqrt 5 }} \cr & = \frac{{3\sqrt 7 \left( {\sqrt 5 - \sqrt 2 } \right)}}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} - \frac{{5\sqrt 5 \left( {\sqrt 7 - \sqrt 2 } \right)}}{{{{\left( {\sqrt 7 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} + \frac{{2\sqrt 2 \left( {\sqrt 7 - \sqrt 5 } \right)}}{{{{\left( {\sqrt 7 } \right)}^2} - {{\left( {\sqrt 5 } \right)}^2}}} \cr & = \sqrt {35} - \sqrt {14} - \sqrt {35} + \sqrt {10} + \sqrt {14} - \sqrt {10} \cr & = 0 \cr} $$

$$\eqalign{ & \frac{1}{{y + 1}} + \frac{{2y + 1}}{{{y^2} - 1}} \cr & = \frac{{y - 1 + 2y + 1}}{{{y^2} - 1}} \cr & = \frac{{3y}}{{{y^2} - 1}} \cr & = \frac{{3\left( {\frac{{2 - x}}{{1 + x}}} \right)}}{{{{\left( {\frac{{2 - x}}{{1 + x}}} \right)}^2} - 1}} \cr & = \frac{{3\left( {2 - x} \right) \times \left( {1 + x} \right)}}{{4 + {x^2} - 4x - 1 - {x^2} - 2x}} \cr & = \frac{{3\left( {2 - x} \right) \times \left( {1 + x} \right)}}{{ - 6x + 3}} \cr & = \frac{{3\left( {2 - x} \right) \times \left( {1 + x} \right)}}{{3\left( {1 - 2x} \right)}} \cr & = \frac{{\left( {2 - x} \right)\left( {1 + x} \right)}}{{\left( {1 - 2x} \right)}} \cr & \cr & {\bf{Alternate}}\,{\bf{solution:}} \cr & y = \frac{{2 - x}}{{1 + x}} \cr & {\text{Put }}x = 0, \cr & y = 2 \cr & \frac{1}{{y + 1}} + \frac{{2y + 1}}{{{y^2} - 1}} \cr & = \frac{1}{3} + \frac{5}{3} \cr & = \frac{6}{3} \cr & = 2 \cr & {\text{Go through option}} \cr & \frac{{\left( {1 + x} \right)\left( {2 - x} \right)}}{{\left( {1 - 2x} \right)}} \cr & = \frac{{1 \times 2}}{1} \cr & = 2 \cr} $$

I. (0.7)² + (0.07)² + (11.1)² > 123.8
L.H.S. = 0.49 + 0.0049 + 123.21
= 123.7049 $${\not > }$$ R.H.S.
⇒ Statement I is incorrect
II. (1.12)² + (10.3)² + (1.05)² > 108.3
L.H.S. = 1.2544 + 106.09 + 1.1025
= 108.4469 > R.H.S.
⇒ Statement II is true

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

(65)^1/6
(17)^1/4
(12)^1/3
LCM of 4, 6, 3 = 12
∴ (65²)^1/12 = (4225)^1/12 . . . . . . (I)
(17³)^1/12 = (4913)^1/12 . . . . . . (II)
(12⁴)^1/12 = (20736)^1/12 . . . . . . (III)
From equation (I), (II), (III) we can see that only (III) statement follows.

$$\eqalign{ & \frac{1}{{\root 3 \of {12} }},\,\frac{1}{{\root 4 \of {29} }},\,\frac{1}{{\sqrt 5 }} \cr & {\text{Take LCM of roots}} = 12 \cr & \frac{1}{{{{12}^{\frac{1}{3}}}}},\,\frac{1}{{{{29}^{\frac{1}{4}}}}},\,\frac{1}{{{5^{\frac{1}{2}}}}} \cr & {\text{By multiplying by 12}} \cr & \frac{1}{{{{12}^4}}},\,\frac{1}{{{{29}^3}}},\,\frac{1}{{{5^6}}} \cr & \frac{1}{{20736}},\,\frac{1}{{24389}},\,\frac{1}{{15625}} \cr & {\text{Divisible by largest number gives lowest quotient}}{\text{.}} \cr} $$

mⁿ = 169
mⁿ = 13²
⇒ m = 13, n = 2
∴ (m + 1)^{(n - 1)}
= (13 + 1)^{(2 - 1)}
= 14