Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & \sqrt {0.03 \times 0.3a} = {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times \sqrt b \cr & {\text{Squaring both sides}} \cr & \Rightarrow {\text{0}}{\text{.03}} \times {\text{0}}{\text{.3a}} = {\left( {0.3} \right)^2} \times {\left( {0.3} \right)^2} \times b \cr & \Rightarrow \frac{3}{{100}} \times \frac{3}{{10}}a = \frac{9}{{100}} \times \frac{9}{{100}} \times b \cr & \Rightarrow 9a = \frac{{81}}{{10}}b \cr & \Rightarrow 90a = 81b \cr & \Rightarrow 10a = 9b \cr & \Rightarrow \frac{a}{b} = \frac{9}{{10}} \cr & \Rightarrow \frac{a}{b} = 0.9 \cr} $$

$$\eqalign{ & x*y = {\left( {x + 3} \right)^2}\left( {y - 1} \right) \cr & \Leftrightarrow 5*4 = {\left( {5 + 3} \right)^2}\left( {4 - 1} \right) \cr & \Leftrightarrow 5*4 = 64 \times 3 \cr & \Leftrightarrow 5*4 = 192 \cr} $$

$$\eqalign{ & \sqrt {1 + \frac{x}{{961}}} = \frac{{32}}{{31}} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow 1 + \frac{x}{{961}} = \frac{{1024}}{{961}} \cr & \Rightarrow \frac{{961 + x}}{{961}} = \frac{{1024}}{{961}} \cr & \Rightarrow x = 1024 - 961 \cr & \Rightarrow x = 63 \cr} $$

$$\eqalign{ & {\text{ }}\sqrt {0.04 \times 0.4 \times a} = 0.004 \times 0.4 \times \sqrt b \cr & \Rightarrow \sqrt {\frac{4}{{100}} \times \frac{4}{{10}} \times a} = \frac{4}{{1000}} \times \frac{4}{{10}} \times \sqrt b \cr & \Rightarrow \frac{4}{{10}}\sqrt {\frac{a}{{10}}} = \frac{4}{{10}} \times \frac{4}{{1000}} \times \sqrt b \cr & \Rightarrow \frac{a}{{10}} = \frac{{16}}{{1000000}} \times b \cr & \Rightarrow \frac{a}{b} = \frac{{16 \times 10}}{{1000000}} \cr & \Rightarrow \frac{a}{b} = \frac{{16}}{{100000}} \cr & \Rightarrow \frac{a}{b} = 16 \times {10^{ - 5}} \cr} $$

$$\eqalign{ & \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{3}{1} \cr & \Rightarrow \frac{{2 \times 9 + 3 \times 9 + 4 \times 9}}{{2 \times 1 + 3 \times 1 + 4 \times 1}} \cr & \Rightarrow \frac{{18 + 27 + 36}}{{2 + 3 + 4}} \cr & \Rightarrow \frac{{81}}{9} \cr & \Rightarrow 9 \cr} $$

$$\eqalign{ & x \propto \frac{1}{{{y^2} - 1}}{\text{ }}\left( {{\text{Given}}} \right) \cr & x = k \times \frac{1}{{{y^2} - 1}}\left( {k{\text{ is constant}}} \right) \cr & {\text{Now }}x = 24{\text{ when }}y = 10{\text{ given}} \cr & \Rightarrow 24 = k \times \frac{1}{{{{\left( {10} \right)}^2} - 1}} \cr & \Rightarrow 24 = \frac{k}{{99}} \cr & \Rightarrow k = 24 \times 99 \cr & x = ? \cr & y = 5 \cr & \Rightarrow x = 24 \times 99 \times \frac{1}{{25 - 1}} \cr & \Rightarrow x = 24 \times 99 \times \frac{1}{{24}} \cr & \Rightarrow x = 99 \cr} $$

$$\eqalign{ & {x^2} + {y^2} + 2x + 1 = 0 \cr & \Rightarrow {x^2} + 2x + 1 + {y^2} = 0 \cr & \Rightarrow {\left( {x + 1} \right)^2} + {y^2} = 0 \cr} $$
Hence both terms are squares and there addition is zero.
So, it can be possible only when both terms are zeros.
$$\eqalign{ & \therefore x + 1 = 0 \cr & \Rightarrow x = - 1 \cr & y = 0 \cr & \therefore {x^{31}} + {y^{35}} \cr & \Rightarrow {\left( { - 1} \right)^{31}} + {\left( 0 \right)^{35}} \cr & \Rightarrow - 1 \cr} $$

a² + b² + c² = 2(a - b - c) - 3
⇒ a² + b² + c² = 2a - 2b - 2c - 3
⇒ a² + b² + c² - 2a + 2b +2c + 1 + 1 + 1 = 0
⇒ (a² - 2a + 1) + (b² + 2b + 1) + (c² + 2c + 1) = 0
⇒ (a - 1)² + (b + 1)² + (c + 1)² = 0
a = 1
b = -1
c = -1
∴ 2a - 3b + 4c
= 2 × 1 - 3 × (-1) + 4 × (-1)
= 2 + 3 - 4
= 1

$$\eqalign{ & {\left( {3a + 1} \right)^2} + {\left( {b - 1} \right)^2} + {\left( {2c - 3} \right)^2} = 0 \cr & {\left( {3a + 1} \right)^2} = 0 \cr & \Rightarrow 3a = - 1 \cr & \Rightarrow a = - \frac{1}{3} \cr & {\left( {b - 1} \right)^2} = 0 \cr & \Rightarrow b - 1 = 0 \cr & \Rightarrow b = 1 \cr & {\left( {2c - 3} \right)^2} = 0 \cr & \Rightarrow c = \frac{3}{2} \cr & \therefore 3a + b + 2c \cr & = 3 \times - \frac{1}{3} + 1 + \frac{3}{2} \times 2 \cr & = - 1 + 1 + 3 \cr & = 3 \cr} $$

$$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$   $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$    $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$
Now,
$$ \Rightarrow \frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} \times \frac{{\left( {a - b} \right)}}{{a - b}}$$
Multiply divide by (a - b) in 1^st term
Multiply divide by (b - c) in 2^nd term
Multiply divide by (c - a) in 3^rd term
$$ \Rightarrow \frac{{{{\left( {a - b} \right)}^2}\left( {a - b} \right)}}{{\left( {b - c} \right)\left( {c - a} \right)\left( {a - b} \right)}} + $$     $$\frac{{{{\left( {b - c} \right)}^2}\left( {b - c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} + $$     $$\frac{{{{\left( {c - a} \right)}^2}\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$$
$$\eqalign{ & {\text{Let, }}a - b = x \cr & b - c = y \cr & c - a = z \cr & \therefore x + y + z = 0 \cr & \therefore {x^3} + {y^3} + {z^3} = 3xyz \cr & \therefore {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \cr & = 3\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right) \cr & \therefore \frac{{3\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} \cr & = 3 \cr} $$