Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & x + \frac{1}{x} = 99 \cr & \therefore {x^2} + 1 = 99x \cr & \Rightarrow 2\left( {{x^2} + 1} \right) = 2 \times 99x \cr & \Rightarrow 2{x^2} + 2 = 198x \cr & = \frac{{100x}}{{2{x^2} + 2 + 102x}} \cr & = {\text{ }}\frac{{100x}}{{198x + 102x}} \cr & = \frac{{100x}}{{300x}} \cr & = \frac{1}{3} \cr} $$

$$\eqalign{ & \frac{{4x - 3}}{x} + \frac{{4y - 3}}{y} + \frac{{4z - 3}}{z} = 0 \cr & \Rightarrow \frac{{4x}}{x} - \frac{3}{x} + \frac{{4y}}{y} - \frac{3}{y} + \frac{{4z}}{z} - \frac{3}{z} = 0 \cr & \Rightarrow 4 - \frac{3}{x} + 4 - \frac{3}{y} + 4 - \frac{3}{z} = 0 \cr & \Rightarrow 12 - 3\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = 0 \cr & \Rightarrow - 3\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = - 12 \cr & \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 \cr} $$

$$\eqalign{ & \frac{{xy}}{{x + y}} = a,\,\frac{{xz}}{{x + z}} = b,\,\frac{{yz}}{{y + z}} = c{\text{ }} \cr & {\text{Now,}} \cr & \frac{{x + y}}{{xy}} = \frac{1}{a} \cr & \frac{{x + z}}{{xz}} = \frac{1}{b} \cr & \frac{{y + z}}{{yz}} = \frac{1}{c} \cr & \Rightarrow \frac{1}{y} + \frac{1}{x} = \frac{1}{a},\frac{1}{z} + \frac{1}{x} = \frac{1}{b},\frac{1}{x} + \frac{1}{y} = \frac{1}{c} \cr & {\text{Now we have to find the value of }}x \cr & \therefore \frac{1}{a} + \frac{1}{b} - \frac{1}{c} = \frac{1}{y} + \frac{1}{x} + \frac{1}{z} + \frac{1}{x} - \frac{1}{y} - \frac{1}{z} \cr & \therefore \frac{1}{a} + \frac{1}{b} - \frac{1}{c} = \frac{2}{x} \cr & \Rightarrow \frac{{bc + ac - ab}}{{abc}} = \frac{2}{x} \cr & \Rightarrow x = \frac{{2abc}}{{bc + ac - ab}} \cr} $$

$$\eqalign{ & xy = 8{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{So, }}\left( {x,y} \right) = \left( {1,8} \right) \cr} $$
We have to question the options and check them.
$$\eqalign{ & \left( {8,1} \right) \cr & \left( {2,4} \right) \cr & \left( {4,2} \right) \cr & \therefore {\text{ }}2x + y \cr & = 2 \times 1 + 8 \cr & = 10 \cr & 2 \times 8 + 1 = 11 \cr & 2 \times 2 + 4 = 8{\text{ }}\left( {{\text{Minimum}}} \right) \cr & 2 \times 4 + 2 = 10 \cr} $$
Hence, in this question we have all the options.
So, take all the positive factor otherwise we should have to take - ve(negative) values also.
$$\eqalign{ & \left( {x,y} \right) = \left( {1,8} \right) \cr & {\text{ }}\left( {8,1} \right) \cr & {\text{ }}\left( {2,4} \right) \cr & {\text{ }}\left( {4,2} \right) \cr & {\text{ }}\left( { - 1, - 8} \right) \cr & {\text{ }}\left( { - 8, - 1} \right) \cr & {\text{ }}\left( { - 2, - 4} \right) \cr & {\text{ }}\left( { - 4, - 2} \right) \cr} $$

$$\eqalign{ & {a^2} - 4a - 1 = 0 \cr & {a^2} - 1 = 4a \cr & a - \frac{1}{a} = 4 \cr & {\text{Squring both sides}} \cr & {a^2} + \frac{1}{{{a^2}}} - 2 = 16 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} = 18 \cr & \therefore {a^2} + \frac{1}{{{a^2}}} + 3a - \frac{3}{a} \cr & = {a^2} + \frac{1}{{{a^2}}} + 3\left( {a - \frac{1}{a}} \right) \cr & = 18 + 3 \times 4 \cr & = 18 + 12 \cr & = 30 \cr} $$

$$\eqalign{ & \left( {x - 2} \right)\left( {x - 9} \right) \cr & = {x^2} - 9x - 2x + 18 \cr & = {x^2} - 11x + 18 \cr & = a{x^2} + bx + c = 0 \cr & {\text{For minimum value}} \cr & = \frac{{4ac - {b^2}}}{{4a}} \cr & = \frac{{4 \times 1 \times 18 - {{\left( { - 11} \right)}^2}}}{{4 \times 1}} \cr & = \frac{{72 - 121}}{4} \cr & = \frac{{ - 49}}{4} \cr & = - \frac{{49}}{4} \cr} $$

$$\eqalign{ & \sqrt x = \sqrt 3 - \sqrt 5 \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow x = 3 + 5 - 2.\sqrt {3.} \sqrt 5 \cr & \Rightarrow x = 8 - 2\sqrt {15} \cr & \Rightarrow x - 8 = - 2\sqrt {15} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {x^2} + 64 - 16x = 60 \cr & \Rightarrow {x^2} + 4 - 16x = 0 \cr & \Rightarrow {x^2} + 6 - 16x = 2 \cr & \Rightarrow {x^2} - 16x + 6 = 2 \cr} $$

$$\eqalign{ & {x^2} = y + z \cr & {y^2} = z + x \cr & {z^2} = x + y \cr & \Rightarrow {x^2} + x = x + y + z \cr & {\text{Adding }}x{\text{ on both sides }} \cr & x\left( {x + 1} \right) = x + y + z \cr & \frac{1}{{\left( {x + 1} \right)}} = \frac{x}{{x + y + z}} \cr & {\text{Similarly,}} \cr & \frac{1}{{\left( {y + 1} \right)}} = \frac{y}{{x + y + z}} \cr & \frac{1}{{\left( {z + 1} \right)}} = \frac{z}{{x + y + z}} \cr & \therefore \frac{1}{{\left( {x + 1} \right)}} + \frac{1}{{\left( {y + 1} \right)}} + \frac{1}{{\left( {z + 1} \right)}} \cr & = \frac{x}{{x + y + z}} + \frac{y}{{x + y + z}} + \frac{z}{{x + y + z}} \cr & = \frac{{x + y + z}}{{x + y + z}} \cr & = 1 \cr} $$

a + b + c = 0
Have values
a = 1
b = 2
c = - 3
$$ \Rightarrow \left( {\frac{{a + b}}{c} + \frac{{b + c}}{a} + \frac{{c + a}}{b}} \right)$$     $$\left( {\frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}}} \right)$$
$$ \Rightarrow \left( {\frac{{1 + 2}}{{ - 3}} + \frac{{2 - 3}}{1} + \frac{{ - 3 + 1}}{2}} \right)$$     $$\left( {\frac{1}{{2 - 3}} + \frac{2}{{ - 3 + 1}} + \frac{{ - 3}}{{1 + 2}}} \right)$$
$$\eqalign{ & \Rightarrow \left( { - 1 - 1 - 1} \right)\left( { - 1 - 1 - 1} \right) \cr & \Rightarrow - 3 \times - 3 \cr & \Rightarrow 9 \cr} $$

$$\eqalign{ & a + \frac{1}{b} = 1,{\text{ }}b + \frac{1}{c} = 1 \cr & {\text{Values of }}a,{\text{ }}b,{\text{ }}c{\text{ assume}} \cr & a = \frac{1}{2} \cr & b = 2 \cr & c = - 1 \cr & \therefore abc \cr & = \frac{1}{2} \times 2 \times - 1 \cr & = - 1 \cr} $$