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

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

Answer: Option B

$$\eqalign{ & a*b = a + b + ab \cr & 3*4 \cr & = 3 + 4 + 3 \times 4 \cr & = 19 \cr & 2*3 \cr & = 2 + 3 + 2 \times 3 \cr & = 11 \cr & \therefore 3*4 - 2*3{\text{ }} \cr & = 19 - 11 \cr & = 8 \cr} $$

Answer: Option B

$$\eqalign{ & a:b = 2:3{\text{ and }}b:c = 4:5 \cr & a{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}b{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ }}c \cr & {2_{ \times \left( 4 \right)}}\,\,\,\,{\text{ }}{3_{ \times \left( 4 \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{4_{ \times \left( 3 \right)}}\,\,\,\,\,\,\,\,{\text{ }}{5_{ \times \left( 3 \right)}} \cr & \overline {\underline {\,\,8{\text{ }}\,\,\,\,\,\,\,\,\,\,{\text{ }}12{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}15{\text{ }}\,\,{\text{ }}} } \cr & \therefore {a^2}:{b^2}:bc \cr & \Rightarrow {\left( 8 \right)^2}:{\left( {12} \right)^2}:\left( {12 \times 15} \right) \cr & \Rightarrow 64:144:180 \cr & \Rightarrow 16:36:45 \cr} $$

Answer: Option C

$$\eqalign{ & {\text{A}}:{\text{B}} = \frac{1}{2}:\frac{3}{8} \cr & \Rightarrow {\text{A}}:{\text{B}} = 8:6 \cr & \Rightarrow {\text{A}}:{\text{B}} = 4:3 \cr & {\text{B}}:{\text{C}} = \frac{1}{3}:\frac{5}{9} \cr & \Rightarrow {\text{B}}:{\text{C}} = 9:15 \cr & \Rightarrow {\text{B}}:{\text{C}} = 3:5 \cr & {\text{C}}:{\text{D}} = \frac{5}{6}:\frac{3}{4} \cr & \Rightarrow {\text{C}}:{\text{D}} = 20:18 \cr & \Rightarrow {\text{C}}:{\text{D}} = 10:9 \cr & {\text{A}}:{\text{ B }}:{\text{C }}:{\text{D}} \cr & 4{\text{ }}:{\text{ }}3 \cr & \,\,\,\,\,\,\,\,\,\,3{\text{ }}:{\text{ }}5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{\text{ }}:{\text{ }}9 \cr & \overline {\underline {{\text{ }}8{\text{ }}:6{\text{ }}:10{\text{ }}:{\text{ }}9{\text{ }}} } \cr} $$

Answer: Option C

$$\eqalign{ & {\text{A}}:{\text{B}}:{\text{C}} \cr & {\text{ }}2:\,\,3\,\,:4 \cr & \therefore \frac{{\text{A}}}{{\text{B}}}:\frac{{\text{B}}}{{\text{C}}}:\frac{{\text{C}}}{{\text{A}}} \cr & \left( {{\text{multiply with ABC}}} \right) \cr & \therefore \frac{{{\text{A}} \times {\text{ABC}}}}{{\text{B}}}{\text{:}}\frac{{{\text{B}} \times {\text{ABC}}}}{{\text{C}}}{\text{:}}\frac{{{\text{C}} \times {\text{ABC}}}}{{\text{A}}} \cr & \Rightarrow {{\text{A}}^2}{\text{C}}:{{\text{B}}^2}{\text{A}}:{\text{B}}{{\text{C}}^2} \cr & \Rightarrow {\left( 2 \right)^2} \times 4:{\left( 3 \right)^2} \times 2:{\left( 4 \right)^2} \times 3 \cr & \Rightarrow 16:18:48 \cr & \Rightarrow 8:9:24 \cr & \cr & {\bf{Alternate:}} \cr & \frac{{\text{A}}}{{\text{B}}}:\frac{{\text{B}}}{{\text{C}}}:\frac{{\text{C}}}{{\text{A}}} = \frac{2}{3}:\frac{3}{4}:\frac{4}{2} \cr & \Rightarrow \frac{2}{3} \times 12:\frac{3}{4} \times 12:\frac{4}{2} \times 12 \cr & \Rightarrow 8:9:24 \cr} $$

Answer: Option D

$$\eqalign{ & \frac{{144}}{{0.144}} = \frac{{14.4}}{x} \cr & \Rightarrow \frac{{144 \times 1000}}{{144}} = \frac{{144}}{{x \times 10}} \cr & \Rightarrow 1000 = \frac{{144}}{{10x}} \cr & \Rightarrow x = \frac{{144}}{{1000 \times 10}} \cr & \Rightarrow x = \frac{{144}}{{10000}} \cr & \Rightarrow x = 0.0144 \cr} $$