Simplification - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{\sqrt {32} + \sqrt {48} }}{{\sqrt 8 + \sqrt {12} }} \cr & = \frac{{\sqrt {4 \times 4 \times 2} + \sqrt {4 \times 4 \times 3} }}{{\sqrt {2 \times 2 \times 2} + \sqrt {2 \times 2 \times 3} }} \cr & = \frac{{4\sqrt 2 + 4\sqrt 3 }}{{2\sqrt 2 + 2\sqrt 3 }} \cr & = \frac{{4\left( {\sqrt 2 + \sqrt 3 } \right)}}{{2\left( {\sqrt 2 + \sqrt 3 } \right)}} \cr & = \frac{4}{2} \cr & = 2 \cr} $$

Number of digits in 62478078 = 8
∴ Number of digits in its square root = 4

$$\eqalign{ & {\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right){\text{be a perfect square}} \cr & r = 2t = \pm 1 \cr & \left( {{\text{If }}t = 1} \right)\,\,{n^2} - n + \frac{1}{4} \cr & = {n^2} - 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n - \frac{1}{2}} \right)^2} \cr & \left( {{\text{If }}t = - 1} \right)\,\,{n^2} + n + \frac{1}{4} \cr & = {n^2} + 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n + \frac{1}{2}} \right)^2} \cr} $$

$$\eqalign{ & \frac{{{x^2} + {y^2} + {z^2} - yz - zx - xy}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{\text{ }} \cr & = \frac{{{{\left( {a + m} \right)}^2} + {{\left( {b + m} \right)}^2} + {{\left( {c + m} \right)}^2} - \left( {b + m} \right)\left( {c + m} \right) - \left( {c + m} \right)\left( {a + m} \right) - \left( {a + m} \right)\left( {b + m} \right)}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & = \frac{{{a^2} + {m^2} + 2am + {b^2} + {m^2} + 2bm + {c^2} + {m^2} + 2cm - bc - bm - cm - {m^2} - ca - cm - am - {m^2} - ab - am - bm - {m^2}}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & = \frac{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \left( {\frac{{{a^3} + {b^3}}}{{{a^2} + {b^2} - ab}}} \right){\text{ = }}\left( {a + b} \right) \cr & \left[ {{\text{where}}\,a = 785,b = 345} \right] \cr & = 785 + 345 = 1220 \cr} $$

$$\eqalign{ & {\text{Given experssion ,}} \cr & \frac{{{a^3} + {b^3} + {c^3} - 3abc}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & \left( {Let,\,38 = a,\,34 = b,\,28 = c} \right) \cr & = a + b + c \cr & = 38 + 34 + 28 \cr & = 100 \cr} $$

$$\eqalign{ & {\text{Greatest 4 digit number}} \cr & {\text{ = 9999 }} \cr & \,\,\,\,\,\,\,\,{\text{9| }}\,{\text{9999 | 99}} \cr & \,\,\,\,\,\,\,\,\,\,\,|\,\,\,81 \cr & \overline {189\,\,|\,\,\,1899} \cr & \,\,\,\,\,\,\,\,\,\,\,|\,\,\,1701 \cr & \,\,\,\,\,\,\,\,\,\,\,\overline {|\,\,\,\,\,198} \cr} $$
Greatest four digit number is a perfect square
= 999 - 198
= 9801

$$\eqalign{ & {\left( {a - b} \right)^2} = {a^2} + {b^2} + 2ab \cr & 16{a^2} - {\text{12}}a \cr & = {(4a)^2} - 2 \times 4a \times \frac{3}{2} + {\left( {\frac{3}{2}} \right)^2} \cr & {\text{Number be added}} \cr & {\text{ = }}{\left( {\frac{3}{2}} \right)^2} \cr & = \frac{9}{4} \cr} $$

Let the number of students in a row be 'x'
According to question
⇒ x × x = 1369
⇒ x² = 1369
⇒ x = 37

$$\eqalign{ & {\text{Since }}\left( {x - y} \right) + \left( {y - z} \right) + \left( {z - x} \right) = 0 \cr & {\text{So,}}{\left( {x - y} \right)^3} + {\left( {y - z} \right)^3} + {\left( {z - x} \right)^3} \cr & = 3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right) \cr & {\text{Since}}\left( {{x^2} - {y^2}} \right) + \left( {{y^2} - {z^2}} \right) + \left( {{z^2} - {x^2}} \right) = 0 \cr & {\text{So,}}\left( {{x^2} - {y^2}} \right) + \left( {{y^2} - {z^2}} \right) + \left( {{z^2} - {x^2}} \right) \cr & = 3\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right) \cr & \therefore {\text{Given expression }} \cr & {\text{ = }}\frac{{3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}}{{3\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right)}} \cr & = \frac{1}{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}} \cr & = {\left[ {\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)} \right]^{ - 1}} \cr} $$