141.
x varies inversely as square of y. Given that y = 2 for x = 1, the value of x for y = 6 will equal to?
(A) 3
(B) 9
(C) $$\frac{1}{3}$$
(D) $$\frac{1}{9}$$
Show Answer
Solution:
$$\eqalign{ & x \propto \frac{1}{{{y^2}}} \cr & \left( {{\text{Inversely proportional}}} \right) \cr & x = \frac{k}{{{y^2}}} \cr & \left( {{\text{Given}}} \right), \cr & \left( {y = 2} \right){\text{ for }}\left( {x = 1} \right) \cr & \therefore 1 = \frac{k}{{{{\left( 2 \right)}^2}}} \cr & \Rightarrow 1 = \frac{k}{4} \cr & \Rightarrow k = 4 \cr & \therefore {\text{For }}y = 6 \cr & x = \frac{4}{{{{\left( 6 \right)}^2}}} \cr & x = \frac{4}{{36}} \cr & x = \frac{1}{9} \cr} $$
142.
If $$a\left( {2 + \sqrt 3 } \right)$$ ÃÂÃÂ = $$b\left( {2 - \sqrt 3 } \right)$$ ÃÂÃÂ = 1, then the value of $$\frac{1}{{{a^2} + 1}}$$ÃÂÃÂ + $$\frac{1}{{{b^2} + 1}}$$ ÃÂÃÂ = ?
Show Answer
Solution:
$$\eqalign{ & a\left( {2 + \sqrt 3 } \right) = b\left( {2 - \sqrt 3 } \right) = 1 \cr & a = \frac{1}{{\left( {2 + \sqrt 3 } \right)}} \cr & b = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \cr & \Rightarrow a = \frac{1}{b} \cr & \Rightarrow \frac{1}{{{a^2} + 1}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{1}{{\frac{1}{{{b^2}}} + 1}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{1}{{\frac{{1 + {b^2}}}{{{b^2}}}}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{{{b^2}}}{{{b^2} + 1}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{{{b^2} + 1}}{{{b^2} + 1}} \cr & \Rightarrow 1 \cr} $$
143.
If 4x2 - 6x + 1 = 0, then the value of $$8{x^3} + \frac{1}{{8{x^3}}}$$ ÃÂÃÂ is:
(A) 36
(B) 13
(C) 18
(D) 11
Show Answer
Solution:
$$\eqalign{ & 4{x^2} - 6x + 1 = 0 \cr & 2x\left( {2x - 3 + \frac{1}{{2x}}} \right) = 0 \cr & 2x + \frac{1}{{2x}} = 3 \cr & 8{x^3} + \frac{1}{{8{x^3}}} = 27 - 3 \times 3 \cr & 8{x^3} + \frac{1}{{8{x^3}}} = 18 \cr} $$
144.
If a = 2.361, b = 3.263, and c = 5.624, then the value of a3 + b3 - c3 + 3abc is?
(A) (p- q)(q - r)3 + (r - p)3
(B) 3(p - q)(q - r)(r - p)
(C) 0
(D) 1
Show Answer
Solution:
$$\eqalign{ & a = 2.361 \cr & b = 3.263 \cr & c = 5.624 \cr & a + b - c = 0 \cr & 2.361 + 3.263 - 5.624 = 0 \cr & 0 = 0 \cr & \therefore {a^3} + {b^3} - {c^3} + 3abc \cr & \Rightarrow 0 \cr} $$
145.
If x(x - 3) = -1, then the value of x3 (x3 - 18) is?
Show Answer
Solution:
$$\eqalign{ & x\left( {x - 3} \right) = - 1 \cr & \Rightarrow \left( {x - 3} \right) = \frac{{ - 1}}{x} \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {x - 3} \right)^3} = {\left( {\frac{{ - 1}}{x}} \right)^3} \cr & \Rightarrow {x^3} - 27 - 9.x.\left( {x - 3} \right) = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3} - 27 - 9 \times - 1 = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3} - 27 + 9 = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3} - 18 = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3}\left( {{x^3} - 18} \right) = - 1 \cr} $$
146.
The graphs of the equations 3x - 20y - 2 = 0 and 11x - 5y + 61 = 0 intersect at P(a, b). What is the value of (a2 + b2 - ab)(a2 - b2 + ab)?
(A) $$\frac{{37}}{{35}}$$
(B) $$\frac{5}{7}$$
(C) $$\frac{{31}}{{41}}$$
(D) $$\frac{{41}}{{31}}$$
Show Answer
Solution:
It intersect at point P(a, b) = (x, y) 3x - 20y = 2 11x - 5y = -61 44x - 20y = -244 3x - 20y = 2 $$\overline {41{\text{x}}\,\,\,\,\,\,\,\,\, = - 246\,\,} $$ x = -6 y = -1 P(a, b) = (-6, -1) $$\eqalign{ & \frac{{{a^2} + {b^2} - ab}}{{{a^2} - {b^2} + ab}} \cr & = \frac{{36 + 1 - 6}}{{36 - 1 + 6}} \cr & = \frac{{31}}{{41}} \cr} $$
147.
If $$x + \frac{{16}}{x} = 8,$$ ÃÂÃÂ then the value of $${x^2} + \frac{{32}}{{{x^2}}}$$ ÃÂÃÂ is:
(A) 20
(B) 24
(C) 16
(D) 18
Show Answer
Solution:
$$\eqalign{ & x + \frac{{16}}{x} = 8 \cr & {\text{at }}x = 4{\text{ it satisfy}} \cr & {x^2} + \frac{{32}}{{{x^2}}} = {4^2} + \frac{{32}}{{{4^2}}} \cr & = 16 + 2 \cr & = 18 \cr} $$
148.
If x = 2 - p, then x3 + 6xp + p3 is equal to:
Show Answer
Solution:
x = 2 - p x + p = 2 Cube both side (x + p)3 = 23 x3 + p3 + 3xp(x + p) = 8 x3 + p3 + 2xp × 2 = 8 x3 + 6xp + p3 = 8
149.
If $$\frac{{22\sqrt 2 }}{{4\sqrt 2 - \sqrt {3 + \sqrt 5 } }} = a + \sqrt 5 b$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ with a, b > 0, then what is the value of (ab) : (a + b)?
(A) 7 : 4
(B) 7 : 8
(C) 4 : 7
(D) 8 : 7
Show Answer
Solution:
$$\eqalign{ & \frac{{22\sqrt 2 }}{{4\sqrt 2 - \sqrt {3 + \sqrt 5 } }} = a + \sqrt 5 b \cr & {\text{LHS}} = \frac{{22\sqrt 2 }}{{4\sqrt 2 - \sqrt {3 + \sqrt 5 } }} \cr & = \frac{{22\sqrt 2 }}{{4\sqrt 2 - \sqrt {\frac{{6 + 2\sqrt 5 }}{2}} }} \cr & = \frac{{22\sqrt 2 }}{{4\sqrt 2 - \frac{{\sqrt 5 + 1}}{{\sqrt 2 }}}} \cr & = \frac{{22\sqrt 2 }}{{\frac{{8 - \sqrt 5 - 1}}{{\sqrt 2 }}}} \cr & = \frac{{22 \times 2}}{{7 - \sqrt 5 }} \times \frac{{7 + \sqrt 5 }}{{7 + \sqrt 5 }} \cr & = \frac{{44\left( {7 + \sqrt 5 } \right)}}{{{7^2} - {{\left( {\sqrt 5 } \right)}^2}}} \cr & = \frac{{44\left( {7 + \sqrt 5 } \right)}}{{44}} \cr & = 7 + \sqrt 5 \cr & {\text{Compare LHS and RHS}} \cr & 7 + \sqrt 5 = a + \sqrt 5 b \cr & a = 7;\,b = 1 \cr & \left( {ab} \right):\left( {a + b} \right) = \left( {7 \times 1} \right):\left( {7 + 1} \right) = 7:8 \cr} $$
150.
If $$\frac{{x + y}}{z} = 2,$$ ÃÂÃÂ then what is the value of $$\left[ {\frac{y}{{y - z}}} \right] + \left[ {\frac{x}{{x - z}}} \right]?$$
Show Answer
Solution:
Take x = 3, y = 1, z = 2, satisfy the equation $$\therefore \,\frac{y}{{y - z}} + \frac{x}{{x - z}} \Rightarrow \frac{1}{{ - 1}} + \frac{3}{1} = 2$$