1.
A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be:
(A) 22
(B) 23
(C) 24
(D) 26
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Solution:
Let the number of hens be x and the number of cows be y. Then, x + y = 48 . . . . . (i) and 2x + 4y = 140 ⇒ x + 2y = 70 . . . . . (ii) Solving (i) and (ii) we get: x = 26, y = 22 ∴ The required answer = 26
2.
If $$x = \sqrt 3 {\text{ + }}\sqrt 2 {\text{,}}$$ ÃÂÃÂ ÃÂÃÂ then the value of $${x^3} - \frac{1}{{{x^3}}}$$ ÃÂÃÂ is?
(A) $$10\sqrt 2 $$
(B) $$14\sqrt 2 $$
(C) $$22\sqrt 2 $$
(D) $$8\sqrt 2 $$
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Solution:
$$\eqalign{ & x = \sqrt 3 + \sqrt 2 \cr & \frac{1}{x} = \frac{1}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \frac{1}{x} = \sqrt 3 - \sqrt 2 \cr & {x^3} - \frac{1}{{{x^3}}} \cr & = {\left[ {x - \frac{1}{x}} \right]^3} + 3 \times x \times \frac{1}{x}\left( {x - \frac{1}{x}} \right) \cr} $$ $$ = {\left( {\sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 } \right)^3} + $$ $$3\left( {\sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 } \right)$$ $$\eqalign{ & = {\left( {2\sqrt 2 } \right)^3} + 3\left( {2\sqrt 2 } \right) \cr & = 16\sqrt 2 + 6\sqrt 2 \cr & = 22\sqrt 2 \cr} $$
3.
The simplification of $$\frac{1}{8} + $$ $$\frac{1}{{{8^2}}} + $$ $$\frac{1}{{{8^3}}} + $$ $$\frac{1}{{{8^4}}} + $$ $$\frac{1}{{{8^5}}}$$ upto three place of decimals yields = ?
(A) 0.143
(B) 0.163
(C) 0.215
(D) 0.715
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Solution:
$$\eqalign{ & {\text{According to question,}} \cr & \frac{1}{8} + \frac{1}{{{8^2}}} + \frac{1}{{{8^3}}} + \frac{1}{{{8^4}}} + \frac{1}{{{8^5}}} \cr & \Rightarrow \frac{1}{8} + \frac{1}{{64}} + \frac{1}{{512}} + \frac{1}{{4096}} + \frac{1}{{32768}} \cr} $$ ⇒ 0.125 + 0.015625 + 0.00195313 + 0.00024414 + 0.0000305175 ⇒ 0.143
4.
The value of $$\frac{{1.6 \times 1.6 \times 1.6 - 0.6 \times 0.6 \times 0.6}}{{1.6 \times 1.6 + 1.6 \times 0.6 + 0.6 \times 0.6}}$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ is:
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Solution:
$$\eqalign{ & \frac{{1.6 \times 1.6 \times 1.6 - 0.6 \times 0.6 \times 0.6}}{{1.6 \times 1.6 + 1.6 \times 0.6 + 0.6 \times 0.6}} \cr & = \frac{{{{\left( {1.6} \right)}^3} - {{\left( {0.6} \right)}^3}}}{{{{\left( {1.6} \right)}^2} + 1.6 \times 0.6 + {{\left( {0.6} \right)}^2}}} \cr & = \frac{{\left( {1.6 - 0.6} \right)\left\{ {{{\left( {1.6} \right)}^2} + 1.6 \times 0.6 + {{\left( {0.6} \right)}^2}} \right\}}}{{{{\left( {1.6} \right)}^2} + 1.6 \times 0.6 + {{\left( {0.6} \right)}^2}}} \cr & = 1.6 - 0.6 \cr & = 1 \cr} $$
5.
A students was asked to find the value of $$9\frac{4}{9} \div 11\frac{1}{3}{\text{ of }}\frac{1}{6} + \left( {1\frac{1}{3} \times 1\frac{4}{5} \div \frac{3}{5}} \right) \times 2\frac{1}{6}{\text{ of }}\frac{2}{3} \div \frac{4}{3}{\text{ of }}\frac{2}{3}.$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ His answer was $$19\frac{1}{4}.$$ ÃÂÃÂ What is the difference between his answer and the correct answer?
(A) $$7\frac{1}{2}$$
(B) $$6\frac{1}{3}$$
(C) $$7\frac{3}{4}$$
(D) $$6\frac{2}{3}$$
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Solution:
$$\eqalign{ & 9\frac{4}{9} \div 11\frac{1}{3}{\text{ of }}\frac{1}{6} + \left( {1\frac{1}{3} \times 1\frac{4}{5} \div \frac{3}{5}} \right) \times 2\frac{1}{6}{\text{ of }}\frac{2}{3} \div \frac{4}{3}{\text{ of }}\frac{2}{3} \cr & = \frac{{85}}{9} \div \frac{{34}}{3}{\text{ of }}\frac{1}{6} + \left( {\frac{4}{3} \times \frac{9}{5} \div \frac{3}{5}} \right) \times \frac{{13}}{6}{\text{ of }}\frac{2}{3} \div \frac{4}{3}{\text{ of }}\frac{2}{3} \cr & = \frac{{85}}{9} \times \frac{{18}}{{34}} + \left( 4 \right) \times \frac{{13}}{9} \times \frac{9}{8} \cr & = 5 + \frac{{13}}{2} \cr & = \frac{{23}}{2} \cr & {\text{Difference of answer}} = \frac{{77}}{4} - \frac{{46}}{4} \cr & = \frac{{31}}{4} \cr & = 7\frac{3}{4} \cr} $$
6.
The difference of $${\text{1}}\frac{3}{{16}}$$ ÃÂÃÂ and its reciprocal is equal to = ?
(A) $$1\frac{1}{8}$$
(B) $$\frac{4}{3}$$
(C) $$\frac{{15}}{{16}}$$
(D) None of these
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Solution:
$$\eqalign{ & {\text{Required differnce}} \cr & {\text{ = }}\frac{{19}}{{16}} - \frac{{16}}{{19}} = \frac{{{{19}^2} - {{16}^2}}}{{304}} \cr & = \frac{{\left( {19 + 16} \right)\left( {19 - 16} \right)}}{{304}} \cr & = \frac{{35 \times 3}}{{304}} \cr & = \frac{{105}}{{304}} \cr} $$
7.
A canteen requires 798 bananas for a week. Total how many bananas did it require for the months of January, February and March 2008?
(A) 10277
(B) 10374
(C) 10480
(D) 10586
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Solution:
$$\eqalign{ & {\text{Number of bananas required daily}} \cr & {\text{ = }}\frac{{798}}{7} \cr & = 144 \cr & \therefore \,\,\,{\text{Required number}} \cr & {\text{ = }}\left[ {114 \times \left( {\mathop {31}\limits^{{\text{Jan}}} + \mathop {29}\limits^{{\text{Feb}}} + \mathop {31}\limits^{{\text{Mar}}} } \right)} \right] \cr & = \left( {114 \times 91} \right) \cr & = 10374 \cr} $$
8.
The value of 1 + 3 + 5 + 7 + . . . . . . (2n - 1) is:
(A) (2n - 1) × (2n - 1)
(B) $$\frac{n}{2}$$
(C) n × n
(D) $$\frac{{n\left( {n + 1} \right)}}{2}$$
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Solution:
1 + 3 + 5 + 7 + . . . . . . (2n - 1) We know that such of N odd number = n2 = n × n
9.
$$\frac{{{{\left( {0.73} \right)}^3} + {{\left( {0.27} \right)}^3}}}{{{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27}}$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ = ?
(A) 0.27
(B) 0.4087
(C) 0.73
(D) 1
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Solution:
$$\eqalign{ & {\text{Given,}} \cr & \frac{{{{\left( {0.73} \right)}^3} + {{\left( {0.27} \right)}^3}}}{{{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27}} \cr & {\text{let }}a = 0.73{\text{ and }}b = 0.27 \cr} $$ $$ = \frac{{\left( {0.73 + 0.27} \right)\left[ {{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27} \right]}}{{{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - 0.73 \times 0.27}}$$ $$\eqalign{ & \left[ {\because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = 0.73 + 0.27 \cr & = 1 \cr} $$
10.
The simplest value of $$\left( {\frac{1}{{\sqrt 9 - \sqrt 8 }} - \frac{1}{{\sqrt 8 - \sqrt 7 }} + \frac{1}{{\sqrt 7 - \sqrt 6 }} - \frac{1}{{\sqrt 6 - \sqrt 5 }}} \right)$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ is = ?
(A) $$3 - \sqrt 5 $$
(B) 3
(C) $$\sqrt 5 $$
(D) $$\sqrt 5 - 3$$
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Solution:
$$\left( {\frac{1}{{\sqrt 9 - \sqrt 8 }} - \frac{1}{{\sqrt 8 - \sqrt 7 }} + \frac{1}{{\sqrt 7 - \sqrt 6 }} - \frac{1}{{\sqrt 6 - \sqrt 5 }}} \right)$$ $$\eqalign{ & = \sqrt 9 + \sqrt 8 - \sqrt 8 - \sqrt 7 + \sqrt 7 + \sqrt 6 - \sqrt 6 - \sqrt 5 \cr & = \sqrt 9 - \sqrt 5 \cr & = 3 - \sqrt 5 \cr} $$