Simplification - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option C

4376 + 3209 - 1784 + 97 = 3125 + x
⇒ 7682 - 1784 = 3125 + x
or, x = 7682 - 1784 - 3125
⇒ x = 2773
Hence,
the number is 2773

Answer: Option D

Let the missing number is a
Given,
$$\eqalign{ & 14 \times 627 \div \sqrt {\left( {1089} \right)} = {\left( a \right)^3} + 141 \cr & \frac{{14 \times 627}}{{\sqrt {1089} }} = {\left( a \right)^3} + 141 \cr & \frac{{14 \times 627}}{{33}} = {a^3} + 141 \cr & 266 = {a^3} + 141 \cr & {a^3} = 125 \cr & a = \root 3 \of {125} \cr & a = 5 \cr} $$

Answer: Option C

$$\eqalign{ & {\text{Given,}} \cr & \left( {x + \frac{1}{x}} \right) = 3{\text{ }} \cr & {\text{On squaring both sides, we get}} \cr & {\left( {x + \frac{1}{x}} \right)^2} = {3^2} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 9 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 9 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr} $$

Answer: Option B

As we know
a³ + b³ + c³ + 3abc = (a² + b² + c² - ab - bc - ca)(a + b + c)
When, a + b + c = 0
Then a³ + b³ + c³ - 3abc = 0
When, x + y + z = 0
⇒ x³ + y³ + z³ = 3xyz
⇒ x³ + y³ + z³ + 3xyz = 3xyz + 3xyz
⇒ x³ + y³ + z³ + 3xyz = 6xyz

Answer: Option B

67 × 119 + 17 - 27 ÷ 259
⇒ 67 + 119 ÷ 17 × 27 - 259
⇒ 67 + 7 × 27 - 259
⇒ 67 + 189 - 259
⇒ 256 - 259
⇒ -3

Answer: Option D

$$\eqalign{ & {\text{Given,}} \cr & \frac{{{{\left( {0.73} \right)}^3} + {{\left( {0.27} \right)}^3}}}{{{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27}} \cr & {\text{let }}a = 0.73{\text{ and }}b = 0.27 \cr} $$
 $$ = \frac{{\left( {0.73 + 0.27} \right)\left[ {{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27} \right]}}{{{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27}}$$
$$\eqalign{ & \left[ {\because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = 0.73 + 0.27 \cr & = 1 \cr} $$

Answer: Option D

$$\eqalign{ & \frac{{{{\left( {a - b} \right)}^2} - {{\left( {a + b} \right)}^2}}}{{ - 4a}}{\text{ = }}\frac{x}{y} \cr & \Rightarrow \frac{{\left( {{a^2} + {b^2} - 2ab} \right) - \left( {{a^2} + {b^2} + 2ab} \right)}}{{ - 4a}} = \frac{x}{y} \cr & \Rightarrow \frac{{ - 4ab}}{{ - 4a}} = \frac{x}{y} \cr & \Rightarrow b = \frac{x}{y} \cr & \Rightarrow x = yb \cr} $$