Practice MCQ Questions and Answer on Simplification

Solution:

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

Solution:

 $$\eqalign{ & \frac{{40 + \frac{3}{4}{\text{ of }}32}}{{37 + \frac{3}{4}{\text{ of }}\left( {34 - 6} \right)}} \cr & = \frac{{40 + 24}}{{37 + 21}} \cr & = \frac{{64}}{{58}} \cr & = \frac{{32}}{{29}} \cr & = 1\frac{3}{{29}} \cr} $$

Solution:

 $$\eqalign{ & {\text{Number of bananas required daily}} \cr & {\text{ = }}\frac{{798}}{7} \cr & = 144 \cr & \therefore \,\,\,{\text{Required number}} \cr & {\text{ = }}\left[ {114 \times \left( {\mathop {31}\limits^{{\text{Jan}}} + \mathop {29}\limits^{{\text{Feb}}} + \mathop {31}\limits^{{\text{Mar}}} } \right)} \right] \cr & = \left( {114 \times 91} \right) \cr & = 10374 \cr} $$

Solution:

 $$\eqalign{ & {\text{According to question,}} \cr & \sqrt {3 + \frac{{33}}{{64}}} \div \sqrt {9 + \frac{1}{7}} \times 2\sqrt {3\frac{1}{9}} \cr & = \sqrt {\frac{{225}}{{64}}} \div \sqrt {\frac{{64}}{7}} \times 2\sqrt {\frac{{28}}{9}} \cr & = \sqrt {\frac{{225}}{{64}} \times \frac{7}{{64}} \times \frac{{28}}{9}} \times 2 \cr & = \frac{{5 \times 7}}{{8 \times 4}} \times 2 \cr & = \frac{{35}}{{16}} \cr & = 2\frac{3}{{16}} \cr} $$

Solution:

 Given expression, $$\left[ {\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - x.\frac{1}{x}} \right)} \right]$$     $$\left[ {\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + x.\frac{1}{x}} \right)} \right]$$ $$\eqalign{ & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} - \frac{1}{{{x^3}}}} \right) \cr & = {x^6} - \frac{1}{{{x^6}}} \cr} $$

Solution:

 $$\eqalign{ & a + b + c = 0 \cr & \Rightarrow a = - \left( {b - c} \right) \cr & \Rightarrow {a^2} = {\left( {b + c} \right)^2} \cr & \therefore \frac{{{a^2}}}{{\left( {{a^2} - bc} \right)}} + \frac{{{b^2}}}{{\left( {{b^2} - ca} \right)}} + \frac{{{c^2}}}{{\left( {{c^2} - ab} \right)}} \cr} $$ $$ = \frac{{{{\left( {b + c} \right)}^2}}}{{{{(b + c)}^2} - bc}} + \frac{{{b^2}}}{{{b^2} + c\left( {b + c} \right)}} + \frac{{{c^2}}}{{{c^2} + b\left( {b + c} \right)}}$$ $$ = \frac{{{{\left( {b + c} \right)}^2}}}{{{b^2} + {c^2} + bc}} + \frac{{{b^2}}}{{{b^2} + {c^2} + bc}} + \frac{{{c^2}}}{{{b^2} + {c^2} + bc}}$$ $$\eqalign{ & = \frac{{{b^2} + {c^2} + 2bc + {b^2} + {c^2}}}{{{b^2} + {c^2} + bc}} \cr & = \frac{{2\left( {{b^2} + {c^2} + bc} \right)}}{{{b^2} + {c^2} + bc}} \cr & = 2 \cr} $$

Solution:

 $$\eqalign{ & {\text{According to question,}} \cr & \frac{5}{{3 + \frac{3}{{1 - \frac{2}{3}}}}}\,\, \cr & \Rightarrow \frac{5}{{3 + \frac{3}{{\frac{1}{3}}}}}\, \cr & \Rightarrow \frac{5}{{3 + 9}} \cr & \Rightarrow \frac{5}{{12}} \cr} $$

Solution:

 $$\eqalign{ & {\text{Given expression,}} \cr & \frac{{{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}}}{{a b}} \cr & = \frac{{4ab}}{{ab}} \cr & = 4\,\,\,\,\,\left\{ {{\text{where }}a = 469,b = 174} \right\} \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & 309 \div \left[ {\left( {\frac{3}{2}} \right){\text{of}}\left( {25 + 35} \right) - 12\frac{3}{4}} \right] \cr & = 309 \div \left[ {\frac{3}{2} \times 60 - \frac{{51}}{4}} \right] \cr & = 309 \div \left[ {90 - \frac{{51}}{4}} \right] \cr & = 309 \times \frac{4}{{360 - 51}} \cr & = 4 \cr} $$