81.
The number, whose square is equal to the difference of the squares of 75.15 and 60.12, is = ?
(A) 46.09
(B) 48.09
(C) 45.09
(D) 47.09
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Solution:
$$\eqalign{ & {\text{Let the number be = }}x \cr & {\text{According to question,}} \cr & {x^2} = {\left( {75.15} \right)^2} - {\left( {60.12} \right)^2} \cr & \Rightarrow {x^2} = {\left( {75.15 + 60.12} \right)}\,{\left( {75.15 - 60.12} \right)} \cr & \Rightarrow {x^2} = 135.27 \times 15.03 \cr & \Rightarrow {x^2} = 2033.1081 \cr & \Rightarrow x = 45.09 \cr} $$
82.
The denominator of a fraction is 4 more than twice the numerator. When the numerator is increased by 3 and the denominator is decreased by 3, the fraction becomes ÃÂÃÂ What is the difference between the denominator and numerator of the original fraction?
(A) 13
(B) 10
(C) 12
(D) 11
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Solution:
$$\eqalign{ & \frac{{x + 3}}{{2x + 4 - 3}} = \frac{2}{3} \cr & 3x + 9 = 4x + 2 \cr & x = 7 \cr & {\text{Original fraction}} \cr & \frac{x}{{2x + 4}} = \frac{7}{{18}} > 11 \cr} $$ Differentiate between numerator and denumerator = 11
83.
Simplification of
(A) 6.91
(B) 7
(C) 6.81
(D) 7.1
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Solution:
$$\eqalign{ & {\text{According to the question}} \cr & {\text{ }}\frac{{{{\left( {3.4567} \right)}^2} - {{\left( {3.4533} \right)}^2}}}{{0.0034}} \cr & \left[ {\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right] \cr & \Rightarrow \frac{{\left( {3.4567 + 3.4533} \right)\left( {3.4567 - 3.4533} \right)}}{{0.0034}} \cr & \Rightarrow \frac{{6.91 \times 0.0034}}{{0.0034}} \cr & \Rightarrow 6.91 \cr} $$
84.
The simplified value of ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ = ?
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Solution:
$$\eqalign{ & {\text{According to question,}} \cr & \,\,\,\, \sqrt {5 + \sqrt {11 + \sqrt {19 + \sqrt {29 + \sqrt {49} } } } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + \sqrt {19 + \sqrt {29 + 7} } } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + \sqrt {19 + \sqrt {36} } } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + \sqrt {19 + 6} } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + \sqrt {25} } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + 5} } \cr & \Rightarrow \sqrt {5 + \sqrt {16} } \cr & \Rightarrow \sqrt {5 + 4} \cr & \Rightarrow \sqrt 9 \cr & \Rightarrow 3 \cr} $$
85.
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
86.
Find the value of
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Solution:
$$\eqalign{ & x = \sqrt {30 + \sqrt {30 + \sqrt {30 + \sqrt {30\,........\,\infty } } } } \cr & 5 \times 6 = 30 \cr & {\text{Hence }}x = 6 \cr} $$
87.
The value of
(A) 5
(B) 23.25
(C) 23.75
(D) 25
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Solution:
According to the question, $$\eqalign{ & \frac{{25 - 5\left[ {2 + 3\left\{ {2 - 2\left( {5 - 3} \right) + 5} \right\} - 10} \right]}}{4} \cr & = \frac{{25 - 5\left[ {2 + 3\left\{ {2 - 2 * 2 + 5} \right\} - 10} \right]}}{4} \cr & = \frac{{25 - 5\left[ {2 + 3\left\{ {2 - 4 + 5} \right\} - 10} \right]}}{4} \cr & = \frac{{25 - 5\left[ {2 + 3 * 3 - 10} \right]}}{4} \cr & = \frac{{25 - 5\left[ {11 - 10} \right]}}{4} \cr & = 25 - \frac{5}{4} \cr & = \frac{{100 - 5}}{4} \cr & = \frac{{95}}{4} \cr & = 23.75 \cr} $$
88.
ÃÂÃÂ ÃÂÃÂ is equal to ?
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Solution:
Given expression, $$\left[ {\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - x.\frac{1}{x}} \right)} \right]$$ $$\left[ {\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + x.\frac{1}{x}} \right)} \right]$$ $$\eqalign{ & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} - \frac{1}{{{x^3}}}} \right) \cr & = {x^6} - \frac{1}{{{x^6}}} \cr} $$
89.
If 3.352 - (9.759 - x) - 19.64 = 7.052, then what is the value of x?
(A) -6.181
(B) 13.581
(C) 33.099
(D) 39.803
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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
90.
ÃÂÃÂ is equal to = ?
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Solution:
$$\eqalign{ & {\text{Given expression ,}} \cr & \left( {1000 - \frac{1}{{1000}}} \right) \times 7 \cr & = \left( {7000 - \frac{7}{{1000}}} \right) \cr & = 6999\frac{{993}}{{1000}} \cr} $$