Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

Answer: Option A

$$\eqalign{ & 3*5 + 5*3 \cr & \Rightarrow 3*5 = 2 \times 3 - 3 \times 5 + 3 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6 - 15 + 15 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6 \cr & \Rightarrow 5*3 = 2 \times 5 - 3 \times 3 + 3 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 10 - 9 + 15 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 16 \cr & \therefore 3*5 + 5*3 \cr & \Rightarrow 6 + 16 = 22 \cr} $$

Answer: Option C

$$\eqalign{ & {\text{8}} \times {\text{2}} \cr & = 8 + 2 + \frac{8}{2} \cr & = 10 + 4 \cr & = 14 \cr} $$

Answer: Option D

$$\left( {{\text{1 + }}\frac{1}{x}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 1}}} \right)$$  $$\left( {{\text{1 + }}\frac{1}{{x + 2}}} \right)$$  $$\left( {{\text{1 + }}\frac{1}{{x + 3}}} \right)$$
Taking L.C.M of each term
$$ \Rightarrow \left( {\frac{{x + 1}}{x}} \right)$$ $$\left( {\frac{{x + 1 + 1}}{{x + 1}}} \right)$$  $$\left( {\frac{{x + 2 + 1}}{{x + 2}}} \right)$$  $$\left( {\frac{{x + 3 + 1}}{{x + 3}}} \right)$$
$$\eqalign{ & \Rightarrow \frac{1}{x} \times \left( {x + 4} \right) \cr & \Rightarrow \frac{{x + 4}}{x} \cr} $$

Answer: Option A

$$\eqalign{ & \frac{a}{b}{\text{ = }}\frac{2}{3}{\text{ and }}\frac{b}{c}{\text{ = }}\frac{4}{5}\,\,\left( {{\text{Given}}} \right) \cr & or\,\frac{c}{b} = \frac{5}{4} \cr & \frac{{a + b}}{{b + c}} \cr & = \frac{{b\left( {\frac{a}{b} + 1} \right)}}{{b\left( {\frac{c}{b} + 1} \right)}} \cr & = \frac{{\frac{a}{b} + 1}}{{\frac{c}{b} + 1}} \cr & = \frac{{\left( {\frac{2}{3} + 1} \right)}}{{\left( {\frac{5}{4} + 1} \right)}} \cr & = \frac{{\frac{2 + 3}{3}}}{{\frac{{5 + 4}}{4}}} \cr & = \frac{{5 \times 4}}{{3 \times 9}} \cr & = \frac{{20}}{{27}} \cr & \therefore \frac{{a + b}}{{b + c}} = \frac{{20}}{{27}} \cr & {\bf{Alternate:}} \cr & a{\text{ }}\,\,\,{\text{ }}\,{\text{ }}\,\,\,\,{\text{ }}b{\text{ }}\,\,\,\,{\text{ }}\,\,\,\,{\text{ }}\,\,\,{\text{ }}c \cr & {2_{ \times \left( 4 \right)}}\,\,\,\,\,{\text{ }}{3_{ \times \left( 4 \right)}} \cr & \,{\text{ }}\,{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,{4_{ \times \left( 3 \right)}}{\text{ }}\,\,\,\,\,\,{5_{ \times \left( 3 \right)}} \cr & \overline {\underline {8{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}12{\text{ }}\,\,\,\,\,\,\,\,\,\,{\text{ }}15\,\,\,\,} } \cr & \therefore \frac{{a + b}}{{b + c}} = \frac{{8 + 12}}{{12 + 15}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{{27}} \cr} $$

Answer: Option C

$$\eqalign{ & \frac{{2a + b}}{{a + 4b}} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \Rightarrow 2a + b = 3\left( {a + 4b} \right) \cr & \Rightarrow 2a + b = 3a + 12b \cr & \Rightarrow - a = 11b \cr & \Rightarrow a = - 11b \cr & \therefore \frac{{a + b}}{{a + 2b}} \cr & \Rightarrow \frac{{ - 11b + b}}{{ - 11b + 2b}} \cr & \Rightarrow \frac{{ - 10b}}{{ - 9b}} \cr & \Rightarrow \frac{{10}}{9} \cr} $$