Practice MCQ Questions and Answer on Simplification

Solution:

 $$\eqalign{ & {\text{Given expression ,}} \cr & \frac{{{a^2} - {b^2}}}{{a + b}} = a - b \cr & = \left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) - \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right) \cr & = 2 \times \frac{1}{{\left( {101/100} \right)}} \cr & = 2 \times \frac{{100}}{{101}} \cr & = \frac{{200}}{{101}} \cr} $$

Solution:

 $$\eqalign{ & {\text{Given xepression ,}} \cr & \frac{{\left( {x + y - z} \right)\left( {x - y + z} \right)}}{{\left( {x + y + z} \right)\left( {x + z - y} \right)}} + \frac{{\left( {y + x - z} \right)\left( {y - x + z} \right)}}{{\left( {x + y + z} \right)\left( {x + y - z} \right)}} + \frac{{\left( {z + x - y} \right)\left( {z - x + y} \right)}}{{\left( {y + z + x} \right)\left( {y + z - x} \right)}} \cr & = \frac{{\left( {x + y - z} \right)}}{{\left( {x + y + z} \right)}} + \frac{{\left( {y - x + z} \right)}}{{\left( {x + y + z} \right)}} + \frac{{\left( {x - y + z} \right)}}{{\left( {x + y + z} \right)}} \cr & = \frac{{\left( {x + y - z} \right) + \left( {y - x + z} \right) + \left( {x - y + z} \right)}}{{\left( {x + y + z} \right)}} \cr & = \frac{{x + y + z}}{{x + y + z}} \cr & = 1 \cr} $$

Solution:

 $$\eqalign{ & a + \frac{1}{b} = 1{\text{ }} \cr & \Rightarrow ab + 1 = b \cr & \Rightarrow ab - b = - 1 \cr & \Rightarrow b\left( {a - 1} \right) = - 1 \cr & \Rightarrow b = \frac{1}{{\left( {1 - a} \right)}} \cr & \cr & b + \frac{1}{c} = 1 \cr & \Rightarrow bc + 1 = c \cr & \Rightarrow bc - c = - 1 \cr & \Rightarrow c\left( {b - 1} \right) = - 1 \cr & \Rightarrow c = \frac{1}{{\left( {1 - b} \right)}} \cr & \cr & \therefore c + \frac{1}{a} = \frac{1}{{\left( {1 - b} \right)}} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{1}{{1 - \left( {\frac{1}{{1 - a}}} \right)}} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{1}{{\frac{{\left( {1 - a} \right) - 1}}{{\left( {1 - a} \right)}}}} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{{\left( {1 - a} \right)}}{{ - a}} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{{\left( {a - 1} \right)}}{a} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{{a - 1 + 1}}{a} \cr & c + \frac{1}{a} = \frac{a}{a} \cr & c + \frac{1}{a} = 1 \cr} $$

Solution:

 $$\eqalign{ & M = \frac{3}{7} \div \frac{6}{5} \times \frac{2}{3} + \frac{1}{5} \times \frac{3}{2} \cr & = \frac{3}{7} \times \frac{5}{6} \times \frac{2}{3} + \frac{3}{{10}} \cr & = \frac{5}{{21}} + \frac{3}{{10}} \cr & = \frac{{50 + 63}}{{210}} \cr & = \frac{{113}}{{210}} \cr & N = \frac{2}{5} \times \frac{5}{6} \div \frac{1}{3} + \frac{3}{5} \times \frac{2}{3} \div \frac{3}{5} \cr & = \frac{2}{5} \times \frac{5}{6} \times \frac{3}{1} + \frac{3}{5} \times \frac{2}{3} \times \frac{5}{3} \cr & = 1 + \frac{2}{3} \cr & = \frac{5}{3} \cr & \frac{M}{N} = \frac{{113}}{{210}} \div \frac{5}{3} \cr & = \frac{{113}}{{210}} \times \frac{3}{5} \cr & = \frac{{113}}{{350}} \cr} $$

Solution:

 Let the three consecutive numbers are x, x + 1, x + 2 According to question x2 + (x + 1)2 + (x + 2)2 = 110 ⇒ x2 + x2 + 1 + 2x + x2 + 4 + 4x = 110 ⇒ 3x2 + 6x + 5 = 110 ⇒ 3x2 + 6x - 105 = 0 ⇒ x2 + 2x - 35 = 0 ⇒ x2 + 7x - 5x - 35 = 0 ⇒ x(x + 7) -5(x + 7) = 0 ⇒ (x + 7)(x - 5) = 0 ⇒ x = 5 & -7 [∴ (-) value can not considered] ∴ Smallest number is = 5

Solution:

 $$\eqalign{ & {\text{Given equation is :}} \cr & \frac{{\left( {5x + 1} \right)}}{x} \times \frac{{\left( {4y + 3} \right)}}{4} = 20 \cr & \Leftrightarrow \left( {5x + 1} \right)\left( {4y + 3} \right) = 80x \cr & {\text{Clearly, }}x = 3{\text{ and}} \cr & {\text{ }}y = 3{\text{ satisfy}} \cr} $$

Solution:

 $$\eqalign{ & {\text{Given expression,}} \cr & \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + \left( {\frac{1}{{{4^2}}} - \frac{1}{{{5^2}}}} \right) . . . . + \left( {\frac{1}{{{9^2}}} - \frac{1}{{{{10}^2}}}} \right) \cr & = \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) \cr & = \left( {1 - \frac{1}{{100}}} \right) \cr & = \frac{{99}}{{100}} \cr} $$

Solution:

 $$\eqalign{ & {\text{Let,}} \cr & {\text{853 + }}x \div 17 = 1000 \cr & {\text{Then,}} \cr & {\text{853 + }}\frac{x}{{17}} = 1000 \cr & \Leftrightarrow \frac{x}{{17}} = 1000 - 853 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 147 \cr & \Leftrightarrow x = 147 \times 17 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 2499 \cr} $$

Solution:

 3.352 - (9.759 - x) - 19.64 = 7.052 ⇒ 3.352 - 9.759 + x - 19.64 - 7.052 = 0 ⇒ x = 36.451 - 3352 ∴ x = 33.099

Solution:

 $$\eqalign{ & \frac{{x + 1}}{{x - 1}}{\text{ = }}\frac{a}{b}{\text{ }} \cr & {\text{By componendo and dividendo}} \cr & \frac{{x + 1 + x - 1}}{{x + 1 - x + 1}} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow \frac{{2x}}{2} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow x = \frac{{a + b}}{{a - b}}.....(i) \cr & {\text{Again ,}}\frac{{1 - y}}{{1 + y}}{\text{ = }}\frac{b}{a}{\text{ }} \cr & \Rightarrow \frac{{1 + y}}{{1 - y}} = \frac{a}{b} \cr & \Rightarrow \frac{{1 + y + 1 - y}}{{1 + y - 1 + y}} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow \frac{2}{{2y}} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow \frac{1}{y} = \frac{{a + b}}{{a - b}} \cr & \Rightarrow y = \frac{{a - b}}{{a + b}}.....(ii) \cr & {\text{Subtracting equation (ii) from (i) we get }} \cr & \therefore x - y = \frac{{a + b}}{{a - b}} - \frac{{a - b}}{{a + b}} \cr & \left[ {\because {{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2} = 4ab\,\,\,and\,\,{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \right] \cr & = \frac{{{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}}}{{\left( {a - b} \right)\left( {a + b} \right)}} \cr & = \frac{{4ab}}{{{a^2} - {b^2}}} \cr & {\text{Multiply equation (i) and (ii) we get}} \cr & xy = \frac{{a + b}}{{a - b}} \times \frac{{a - b}}{{a + b}} = 1 \cr & \therefore {\text{Expression,}} \cr & {\text{ = }}\frac{{x - y}}{{1 + xy}} = \frac{{\frac{{4ab}}{{{a^2} - {b^2}}}}}{{1 + 1}} \cr &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4ab}}{{2\left( {{a^2} - {b^2}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2ab}}{{{a^2} - {b^2}}} \cr} $$