Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

Here,
Base = 3 units
Height = 4 units
Area of Δ = $$\frac{1}{2}$$ × b × h
= $$\frac{1}{2}$$ × 3 × 4
= 6 sq. units

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{If }}x = 5 \cr & \therefore {x^2} - 2 + \frac{1}{{{x^2}}} \cr & = {\left( {x - \frac{1}{x}} \right)^2} \cr & = {\left( {5 - \frac{1}{5}} \right)^2} \cr & = {\left( {\frac{{24}}{5}} \right)^2} \cr & = \frac{{576}}{{25}} \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & x = 2,{\text{ }}y = 1,{\text{ }}z = - 3 \cr & {x^3} + {y^3} + {z^3} - 3xyz = ? \cr & {\text{As we know that}} \cr & a + b + c = 0{\text{ then }}{a^3} + {b^3} + {c^3} - 3abc = 0 \cr & 2 + 1 - 3 = 0 \cr & \therefore {x^3} + {y^3} + {z^3} - 3xyz = 0 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & \frac{1}{{x + y}} + \frac{1}{{x - y}} \cr & = \frac{{x - y + x + y}}{{{x^2} - {y^2}}} \cr & = \frac{{2x}}{{{x^2} - {y^2}}} \cr} $$

$$\eqalign{ & {\text{Given, }}x + y = 2a \cr & {\text{Find, }}\frac{a}{{x - a}} + \frac{a}{{y - a}}{\text{ = ?}} \cr & \mathop {\mathop x\limits_ \downarrow \,}\limits_3 + \mathop {\mathop y\limits_ \downarrow }\limits_1 = \mathop {\mathop {2a}\limits_ \downarrow }\limits_2 \cr & {\text{Let }}x = 3,{\text{ }}y = 1,{\text{ }}a = 2 \cr & \therefore \frac{a}{{x - a}} + \frac{a}{{y - a}} \cr & = \frac{2}{{\left( {3 - 2} \right)}} + \frac{2}{{\left( {1 - 2} \right)}} \cr & = \frac{2}{1} + \frac{2}{{ - 1}} \cr & = 0 \cr} $$

$$\eqalign{ & {\text{Given}} \cr & x + y + z = 6 \cr & xy + yz + zx = 10 \cr & {\text{To find }}{x^3} + {y^3} + {z^3} - 3xyz{\text{ = ?}} \cr & \Rightarrow {\text{ Using formula,}} \cr} $$
  $$ \Rightarrow {\left( {x + y + z} \right)^2} = $$   $${x^2} \,+$$ $$\, {y^2} \,+ $$ $$\, {z^2} \,+ $$ $$\, 2\left( {xy + yz + zx} \right)$$
$$\eqalign{ & \Rightarrow {6^2} = {x^2} + {y^2} + {z^2} + 2 \times 10 \cr & \Rightarrow 36 = {x^2} + {y^2} + {z^2} + 20 \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 16 \cr} $$
  $$ \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = $$     $$\left( {x + y + z} \right)$$  $$\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right)$$
$$\eqalign{ & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 6\left[ {16 - 10} \right] \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 6 \times 6 \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 36 \cr} $$

$$\eqalign{ & {\text{Given ,}} \cr & \frac{{x + 1}}{{x - 1}} = \frac{a}{b} \cr & \left( {{\text{Using componendo & dividendo}}} \right) \cr & \Leftrightarrow \frac{x}{1} = \frac{{a + b}}{{a - b}} \cr & \Leftrightarrow x = \frac{{a + b}}{{a - b}}\,.....(i) \cr & {\text{Again,}}\frac{{1 - y}}{{1 + y}} = \frac{b}{a} \cr & \Leftrightarrow \frac{{1 + y}}{{1 - y}} = \frac{a}{b} \cr & \Leftrightarrow \frac{1}{y} = \frac{{a + b}}{{a - b}} \cr & \Leftrightarrow y = \frac{{a - b}}{{a + b}}\,.....(ii) \cr & {\text{From question,}} \cr & \frac{{x - y}}{{1 + xy}} \cr & \Rightarrow \frac{{\frac{{a + b}}{{a - b}} - \frac{{a - b}}{{a + b}}}}{{1 + \left( {\frac{{a + b}}{{a - b}}} \right)\left( {\frac{{a - b}}{{a + b}}} \right)}} \cr & \Rightarrow \frac{{{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}}}{{\left( {{a^2} - {b^2}} \right)\left( {1 + 1} \right)}} \cr & \Rightarrow \frac{{4ab}}{{2\left( {{a^2} - {b^2}} \right)}} \cr & \Rightarrow \frac{{2ab}}{{{a^2} - {b^2}}} \cr} $$

$$\eqalign{ & {\text{Given, }} \cr & {a^2} + {b^2} + {c^2} - ab - bc - ca = 0 \cr & {\text{Find, }}a:b:c = ? \cr & {\text{According to the question, }} \cr & {a^2} + {b^2} + {c^2} - ab - bc - ca = 0 \cr & \Rightarrow {a^2} + {b^2} + {c^2} = ab + bc + ca \cr & {\text{Let }}a = b = c = 1 \cr & \Rightarrow {1^2} + {1^2} + {1^2} = 1 \times 1 + \left( {1 \times 1} \right) + \left( {1 \times 1} \right) \cr & \Rightarrow 3 = 3 \cr & \Rightarrow {\text{So, ratio of }}a:b:c = 1:1:1 \cr} $$

$$\eqalign{ & x - \frac{1}{x} = 2{\text{ to find }}{x^3} - \frac{1}{{{x^3}}} \cr & \Rightarrow x - \frac{1}{x} = 2 \cr & \left[ {{\text{Cubing both sides}}} \right] \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {\left( 2 \right)^3} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times x \times \frac{1}{x}\left( {x - \frac{1}{x}} \right) = 8 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times \left( 2 \right) = 8 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 14 \cr} $$

$$\eqalign{ & x = a\left( {b - c} \right) \cr & y = b\left( {c - a} \right) \cr & z = c\left( {a - b} \right) \cr & {\text{Let, }} \cr & \frac{x}{a} = b - c{\text{ }}\,\,\,\,\,\,\,\,\frac{x}{a} = {\text{A}} \cr & \frac{y}{b} = c - a{\text{ }}\,\,\,\,\,\,\,\,\frac{y}{b} = {\text{B}} \cr & \frac{z}{c} = a - b{\text{ }}\,\,\,\,\,\,\,\,\frac{z}{c} = {\text{C}} \cr & \therefore {\text{A}} + {\text{B}} + {\text{C}} \cr & = b - c + c - a + a - b \cr & = 0 \cr & \therefore {{\text{A}}^3} + {{\text{B}}^3} + {{\text{C}}^3} = {\text{3ABC}} \cr & \therefore {\left( {\frac{x}{a}} \right)^3}{\text{ + }}{\left( {\frac{y}{b}} \right)^3} + {\left( {\frac{z}{c}} \right)^3} \cr & = 3 \times \frac{x}{a} \times \frac{y}{b} \times \frac{z}{c} \cr & = \frac{{3xyz}}{{abc}} \cr} $$