111.
If x = 332, y = 333, z = 335, then the value of x3 + y3 + z3 - 3xyz is?
(A) 7000
(B) 8000
(C) 9000
(D) 10000
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Solution:
Here, x = 332, y = 333, z = 335 Find x3 + y3 + z3 - 3xyz $$ = \frac{1}{2}\left( {x + y + z} \right)$$ $$\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]$$ $$ = \left( {\frac{{332 + 333 + 335}}{2}} \right)$$ $$\left[ {{{\left( {332 - 333} \right)}^2} + {{\left( {333 - 335} \right)}^2} + {{\left( {332 - 335} \right)}^2}} \right]$$ $$\eqalign{ & = \frac{{1000}}{2}\left[ {{1^2} + {2^2} + {3^2}} \right] \cr & = \frac{{1000}}{2}\left( {14} \right) \cr & = 7000{\text{ }} \cr} $$
112.
If $$2x + \frac{1}{{4x}} = 1{\text{,}}$$ ÃÂÃÂ then the value of $${x^2} + \frac{1}{{64{x^2}}}$$ ÃÂÃÂ is?
(A) 0
(B) 1
(C) $$\frac{1}{4}$$
(D) 2
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Solution:
$$\eqalign{ & 2x + \frac{1}{{4x}} = 1 \cr & {\text{Dividing by 2 both side}} \cr & \Rightarrow x + \frac{1}{{8x}} = \frac{1}{2} \cr & {\text{Squaring both side }} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} + 2 \times x \times \frac{1}{{8x}} = \frac{1}{4} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} + \frac{1}{4} = \frac{1}{4} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} = \frac{1}{4} - \frac{1}{4} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} = 0 \cr} $$
113.
Simplify $$\frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}}{\text{if }}x + \frac{{{y^2}}}{x} = 5.$$
(A) $$\frac{5}{{{y^2}}}$$
(B) $$\frac{7}{{{y^2}}}$$
(C) $$ - \frac{5}{{{y^2}}}$$
(D) $$ - \frac{7}{{{y^2}}}$$
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Solution:
$$\eqalign{ & {\text{Given:}} \cr & x + \frac{{{y^2}}}{x} = 5 \cr & {\text{Put }}y = 2,\,x = 1 \cr & {\text{Now,}} \cr & \frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}} \cr & = \frac{{{{\left( 1 \right)}^2} + 2 \times 1 + {{\left( 2 \right)}^2}}}{{{{\left( 1 \right)}^3} - 5{{\left( 1 \right)}^2}}} \cr & = \frac{7}{{ - 4}} \cr & {\text{Put }}y = 2{\text{ in option and option D will satisfy the condition}} \cr & \cr & {\bf{Alternate \,solution:}} \cr & x + \frac{{{y^2}}}{x} = 5 \cr & {x^2} + {y^2} = 5x \cr & {x^2} - 5x = - {y^2} \cr & {\text{Now, }}\frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}} \cr & = \frac{{5x + 2x}}{{x\left( {{x^2} - 5x} \right)}} \cr & = \frac{{7x}}{{x\left( { - {y^2}} \right)}} \cr & = - \frac{7}{{{y^2}}} \cr} $$
114.
If (x - 3)2 + (y - 5)2 + (z - 4)2 = 0, then the value of $$\frac{{{x^2}}}{9}{\text{ + }}\frac{{{y^2}}}{{25}}{\text{ + }}\frac{{{z^2}}}{{16}}$$ ÃÂÃÂ ÃÂÃÂ is?
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Solution:
$$\eqalign{ & {\left( {x - 3} \right)^2}{\text{ + }}{\left( {y - 5} \right)^2}{\text{ + }}{\left( {z - 4} \right)^2} = 0 \cr & \therefore {\left( {x - 3} \right)^2} = 0{\text{ }}x = 3 \cr & {\left( {y - 5} \right)^2} = 0{\text{ }}y = 5 \cr & {\left( {z - 4} \right)^2} = 0{\text{ }}z = 4{\text{ }} \cr & \therefore \frac{{{x^2}}}{9}{\text{ + }}\frac{{{y^2}}}{{25}}{\text{ + }}\frac{{{z^2}}}{{16}} \cr & \Rightarrow \frac{9}{9} + \frac{{25}}{{25}} + \frac{{16}}{{16}} \cr & \Rightarrow 3 \cr} $$
115.
If $${p^2} + {q^2} = 7pq{\text{,}}$$ ÃÂÃÂ ÃÂÃÂ then the value of $$\frac{p}{q}{\text{ + }}\frac{q}{p}$$ ÃÂÃÂ is equal to?
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Solution:
$$\eqalign{ & {p^2} + {q^2} = 7pq \cr & \frac{p}{q}{\text{ + }}\frac{q}{p} = 7 \cr} $$ (Divide whole equation by pq)
116.
If $$x + \frac{9}{x} = 6{\text{,}}$$ ÃÂÃÂ then $$\left( {{x^2} + \frac{9}{{{x^2}}}} \right)$$ ÃÂÃÂ is equal to?
(A) 8
(B) 9
(C) 10
(D) 12
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Solution:
$$\eqalign{ & x + \frac{9}{x} = 6 \cr & {\text{Take value of x}} \cr & {\text{Let }}x = 3 \cr & 3 + \frac{9}{3} = 6{\text{ }}\left( {{\text{Proved}}} \right) \cr & {\text{So, }}x = 3 \cr & \therefore {x^2} + \frac{9}{{{x^2}}} \cr & = 9 + \frac{9}{9} \cr & = 10 \cr & \cr & {\bf{Alternate:}} \cr & x + \frac{9}{x} = 6 \cr & {\text{On squaring,}} \cr & \Rightarrow {\left( {x + \frac{9}{x}} \right)^2} = 36 \cr & \Rightarrow {x^2} + \frac{{81}}{{{x^2}}} + 2 \times x \times \frac{9}{x} = 36 \cr & \Rightarrow {x^2} + \frac{{81}}{{{x^2}}} - 18 = 0 \cr & \Rightarrow {\left( {x - \frac{9}{x}} \right)^2} = 0 \cr & \Rightarrow x = \frac{9}{x} \cr & \Rightarrow {x^2} = 9 \cr & {\text{Hence, }}\left( {{x^2} + \frac{9}{{{x^2}}}} \right) \cr & = \left( {9 + \frac{9}{9}} \right) \cr & = 10 \cr} $$
117.
If $$x + \frac{1}{x} = 3{\text{,}}$$ ÃÂÃÂ then the value of $$\frac{{3{x^2} - 4x + 3}}{{{x^2} - x + 1}}$$ ÃÂÃÂ is?
(A) $$\frac{4}{3}$$
(B) $$\frac{3}{2}$$
(C) $$\frac{5}{2}$$
(D) $$\frac{5}{3}$$
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Solution:
$$\eqalign{ & x + \frac{1}{x} = 3 \cr & \frac{{3{x^2} - 4x + 3}}{{{x^2} - x + 1}} \cr & = \frac{{\frac{{3{x^2}}}{x} - \frac{{4x}}{x} + \frac{3}{x}}}{{\frac{{{x^2}}}{x} - \frac{x}{x} + \frac{1}{x}}} \cr & = \frac{{3\left( {x + \frac{1}{x}} \right) - 4}}{{\left( {x + \frac{1}{x}} \right) - 1}} \cr & = \frac{{3 \times 3 - 4}}{{3 - 1}} \cr & = \frac{{9 - 4}}{2} \cr & = \frac{5}{2} \cr} $$
118.
If the sum of square of two real numbers is 41, and their sum is 9. Then the sum of cubes of these two numbers is ?
(A) 169
(B) 209
(C) 189
(D) 198
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Solution:
$$\eqalign{ & {\text{Let the numbers are a, b}} \cr & {a^2} + {b^2} = 41 \cr & a + b = 9 \cr & \Rightarrow {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \cr & \Rightarrow {9^2} = 41 + 2ab \cr & \Rightarrow 81 - 41 = 2ab \cr & \Rightarrow ab = 20 \cr & {\text{Take }} \cr & a = 5 \cr & b = 4 \cr & \Rightarrow {a^3} + {b^3} = {5^3} + {4^3} \cr & \Rightarrow {a^3} + {b^3} = 125 + 64 \cr & \Rightarrow {a^3} + {b^3} = 189 \cr} $$
119.
If x : y = 7 : 3 then the value of $$\frac{{xy + {y^2}}}{{{x^2} - {y^2}}}{\text{ is?}}$$
(A) $$\frac{3}{4}$$
(B) $$\frac{4}{3}$$
(C) $$\frac{3}{7}$$
(D) $$\frac{7}{3}$$
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Solution:
$$\eqalign{ & x:y \cr & 7:3{\text{ }} \cr & \therefore {\text{ }}\frac{{xy + {y^2}}}{{{x^2} - {y^2}}} \cr & = \frac{{21 + 9}}{{49 - 9}} \cr & = \frac{{30}}{{40}} \cr & = \frac{3}{4} \cr} $$
120.
If $$3{a^2} = {b^2} \ne 0{\text{,}}$$ ÃÂÃÂ then the value of $$\frac{{{{\left( {a + b} \right)}^3} - {{\left( {a - b} \right)}^3}}}{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}}$$ ÃÂÃÂ ÃÂÃÂ is?
(A) $$\frac{{3b}}{2}$$
(B) b
(C) $$\frac{b}{2}$$
(D) $$\frac{{2b}}{3}$$
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Solution:
$$\eqalign{ & 3{a^2} = {b^2}{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{ }}\frac{{{{\left( {a + b} \right)}^3} - {{\left( {a - b} \right)}^3}}}{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}} \cr} $$ $$ = \frac{{{a^3} + {b^3} + 3ab\left( {a + b} \right)\, - \,\left( {{a^3} - {b^3} - 3ab\left( {a - b} \right)} \right){\text{ }}}}{{{a^2} + {b^2} + 2ab + {\text{ }}{a^2} + {b^2} - 2ab}}$$ $$\eqalign{ & = \frac{{2{b^3}{\text{ + 6}}{{\text{a}}^2}{\text{b }}}}{{2{a^2} + 2{b^2}{\text{ }}}} \cr & = \frac{{{b^3}{\text{ + 3}}{{\text{a}}^2}{\text{b }}}}{{{a^2} + {b^2}{\text{ }}}} \cr & = \frac{{{b^3} + {b^3}{\text{ }}}}{{\frac{{{b^2}}}{3} + {b^2}{\text{ }}}} \cr & = \frac{{2{b^3}}}{{{b^2}\left( {\frac{1}{3} + 1} \right)}} \cr & = \frac{{2b}}{{\frac{4}{3}}} \cr & = \frac{{3b}}{2} \cr} $$