41.
If x = 2 then the value of x3 + 27x2 + 243x + 631 = ?
(A) 1321
(B) 1233
(C) 1231
(D) 1211
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Solution:
$$\eqalign{ & {\text{Given, }}x = 2 \cr & {\text{Find }}{x^3} + 27{x^2} + 243x + 631 \cr & {\text{To put value }}x = 2{\text{ }} \cr & \Rightarrow {2^3} + 27 \times {2^2} + \left( {243 \times 2} \right) + 631 \cr & \Rightarrow 8 + 108 + 486 + 631 \cr & \Rightarrow 1233 \cr} $$
42.
If a2 + b2 = 5ab, then the value of $$\left( {\frac{{{a^2}}}{{{b^2}}}{\text{ + }}\frac{{{b^2}}}{{{a^2}}}} \right)$$ ÃÂÃÂ is?
(A) 32
(B) 16
(C) 23
(D) -23
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Solution:
$$\eqalign{ & {a^2} + {b^2} = 5ab \cr & \Rightarrow \frac{{{a^2}}}{{ab}} + \frac{{{b^2}}}{{ab}} = 5 \cr & \Rightarrow \frac{a}{b}{\text{ + }}\frac{b}{a}{\text{ = 5}} \cr & {\text{Squaring the both sides}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^2}{\text{ + }}{\left( {\frac{b}{a}} \right)^2} + 2 \times \frac{a}{b} \times \frac{b}{a} = 25 \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}}{\text{ + }}\frac{{{b^2}}}{{{a^2}}} = 25 - 2 \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}}{\text{ + }}\frac{{{b^2}}}{{{a^2}}} = 23 \cr} $$
43.
$${\text{If }}\frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {48} + \sqrt {18} }} = a + b\sqrt 6 {\text{,}}$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ then the value of a and b are respectively?
(A) $$\frac{9}{{15}}, - \frac{4}{{15}}$$
(B) $$\frac{3}{{11}},\frac{4}{{33}}$$
(C) $$\frac{9}{{10}},\frac{2}{5}$$
(D) $$\frac{3}{5},\frac{4}{{15}}$$
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Solution:
$$\eqalign{ & \frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {48} + \sqrt {18} }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {16 \times 3} + \sqrt {9 \times 2} }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} \times \frac{{4\sqrt 3 - 3\sqrt 2 }}{{4\sqrt 3 - 3\sqrt 2 }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{\left( {4\sqrt 3 + 5\sqrt 2 } \right)\left( {4\sqrt 3 - 3\sqrt 2 } \right)}}{{48 - 18}} = a + b\sqrt 6 \cr & \Rightarrow \frac{{8\sqrt 6 + 18}}{{30}} = a + b\sqrt 6 \cr & \Rightarrow \frac{{8\sqrt 6 }}{{30}} + \frac{{18}}{{30}} = a + b\sqrt 6 \cr & \Rightarrow \frac{4}{{15}}\sqrt 6 + \frac{3}{5} = a + b\sqrt 6 \cr & \Rightarrow \frac{3}{5} + \frac{4}{{15}}\sqrt 6 = a + b\sqrt 6 \cr} $$ By comparing coefficients of rational and irrational parts. $$\eqalign{ & \Rightarrow a = \frac{3}{5}{\text{ , }}b = \frac{4}{{15}} \cr & \therefore \left( {\frac{3}{5},\frac{4}{{15}}} \right) \cr} $$
44.
If 4pxy = (x + 2y)2 - (x - 2y)2 , then what will be the value of p?
(A) 0.5
(B) 0.25
(C) 4
(D) 2
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Solution:
4pxy = (x + 2y)2 - (x - 2y)2 ⇒ 4pxy = x2 + 4y2 + 4xy - x2 - 4y2 + 4xy ⇒ 4pxy = 8xy ⇒ $$p = \frac{{8xy}}{{4xy}} = 2$$ Alternate: (a + b)2 - (a - b)2 = 4ab 4pxy = 4 × x × 2y 4pxy = 8xy p = 2
45.
If 2, 0 is a solution of the linear equation 2x + 3y = k, then the value of k is?
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Solution:
$$\eqalign{ & 2x + 3y = k\left( {2 = x,{\text{ }}0 = y} \right) \cr & \therefore 2 \times 2 + 3 \times 0 = k \cr & \Leftrightarrow k = 4 \cr} $$
46.
If $$A = \frac{{0.216 + 0.008}}{{0.36 + 0.04 - 0.12}}$$ ÃÂÃÂ ÃÂÃÂ and $$B = \frac{{0.729 - 0.027}}{{0.81 + 0.09 + 0.27}},$$ ÃÂÃÂ ÃÂÃÂ then what is the value of (A2 + B2 )2 ?
(A) 0.8
(B) 1
(C) 1.4
(D) 2.2
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Solution:
$$\eqalign{ & A = \frac{{0.216 + 0.008}}{{0.36 + 0.04 - 0.12}} \cr & = \frac{{{{\left( {0.6} \right)}^3} + {{\left( {0.2} \right)}^3}}}{{{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)}} \cr & = \frac{{\left( {0.6 + 0.2} \right)\left\{ {{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)} \right\}}}{{\left\{ {{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)} \right\}}} \cr & = 0.8 \cr & B = \frac{{0.729 - 0.027}}{{0.81 + 0.09 + 0.27}} \cr & = \frac{{{{\left( {0.9} \right)}^3} - {{\left( {0.3} \right)}^3}}}{{{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)}} \cr & = \frac{{\left( {0.9 - 0.3} \right)\left\{ {{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)} \right\}}}{{\left\{ {{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)} \right\}}} \cr & = 0.6 \cr & \therefore \,{\left( {{A^2} + {B^2}} \right)^2} \cr & = {\left[ {{{\left( {0.8} \right)}^2} + {{\left( {0.6} \right)}^2}} \right]^2} \cr & = {\left( {0.64 + 0.36} \right)^2} \cr & = {\left( 1 \right)^2} \cr & = 1 \cr} $$
47.
The value of $$\left( {{\text{1 + }}\frac{1}{x}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 1}}} \right)$$ÃÂÃÂ $$\left( {{\text{1 + }}\frac{1}{{x + 2}}} \right)$$ÃÂÃÂ $$\left( {{\text{1 + }}\frac{1}{{x + 3}}} \right)$$ ÃÂÃÂ is?
(A) $$1 + \frac{1}{{x + 4}}$$
(B) x + 4
(C) $$\frac{1}{x}$$
(D) $$\frac{{x + 4}}{x}$$
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Solution:
$$\left( {{\text{1 + }}\frac{1}{x}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 1}}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 2}}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 3}}} \right)$$ Taking L.C.M of each term $$ \Rightarrow \left( {\frac{{x + 1}}{x}} \right)$$ $$\left( {\frac{{x + 1 + 1}}{{x + 1}}} \right)$$ $$\left( {\frac{{x + 2 + 1}}{{x + 2}}} \right)$$ $$\left( {\frac{{x + 3 + 1}}{{x + 3}}} \right)$$ $$\eqalign{ & \Rightarrow \frac{1}{x} \times \left( {x + 4} \right) \cr & \Rightarrow \frac{{x + 4}}{x} \cr} $$
48.
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ ÃÂÃÂÃÂÃÂ ÃÂÃÂÃÂÃÂ then the value of 8 ÃÂÃÂÃÂÃÂÃÂÃÂ 2 is?
(A) 6
(B) 10
(C) 14
(D) 16
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Solution:
$$\eqalign{ & {\text{8}} \times {\text{2}} \cr & = 8 + 2 + \frac{8}{2} \cr & = 10 + 4 \cr & = 14 \cr} $$
49.
If $$x + \frac{1}{{9x}} = 4{\text{,}}$$ ÃÂÃÂ then $${\text{9}}{x^2} + \frac{1}{{9{x^2}}}$$ ÃÂÃÂ is?
(A) 140
(B) 142
(C) 144
(D) 146
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Solution:
$$\eqalign{ & {\text{ }}x + \frac{1}{{9x}} = 4 \cr & {\text{Multiply by 3 both side}} \cr & \Rightarrow {\text{3}}x + \frac{1}{{3x}} = 12 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {\text{9}}{x^2} + \frac{1}{{9{x^2}}} + 2 \times 3x \times \frac{1}{{3x}} = 144 \cr & \Rightarrow {\text{9}}{x^2} + \frac{1}{{9{x^2}}} + 2 = 144 \cr & \Rightarrow {\text{9}}{x^2} + \frac{1}{{9{x^2}}} = 142 \cr} $$
50.
If x + y + z = 19, x2 + y2 + z2 = 133 and xz = y2 , then the difference between z and x is:
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Solution:
Given: x + y + z = 19, x2 + y2 + z2 = 133 and xz = y2, Formula used: (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) Calculation: (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) ⇒ (19)2 = 133 + 2(xy + yz + y2) ⇒ 133 + 2[y(x + y + z)] = 361 ⇒ 2y(19) = 361 - 133 ⇒ y = 6 x + y + z = 19 ⇒ x + z = 13 The possible value of x and z is 9 and 4 x - 4 ⇒ 9 - 4 ⇒ 5 ∴ The value is 5