21.
Out of the fractions $$\frac{9}{31}$$, $$\frac{3}{17}$$, $$\frac{6}{23}$$, $$\frac{4}{11}$$ and $$\frac{7}{25}$$ which is the largest ?
(A) $$\frac{9}{31}$$
(B) $$\frac{3}{17}$$
(C) $$\frac{6}{23}$$
(D) $$\frac{4}{11}$$
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Solution:
Converting each of the given fractions into decimal form, we get: $$\frac{9}{31}$$ = 0.29 $$\frac{3}{17}$$ = 0.176 $$\frac{6}{23}$$ = 0.26 $$\frac{4}{11}$$ = 0.363 $$\frac{7}{25}$$ = 0.28 Clearly, 0.363 > 0.29 > 0.28 > 0.26 > 0.176 So, $$\frac{4}{11}$$ > $$\frac{9}{31}$$ > $$\frac{7}{25}$$ > $$\frac{6}{23}$$ > $$\frac{3}{17}$$
22.
The value of $$\left[ {35.7 - \left( {3 + \frac{1}{{3 + \frac{1}{3}}}} \right) - \left( {2 + \frac{1}{{2 + \frac{1}{2}}}} \right)} \right]$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ is :
(A) 30
(B) 34.8
(C) 36.6
(D) 41.4
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Solution:
Given expression : $$\eqalign{ & = 35.7 - \left( {3 + \frac{1}{{\frac{{10}}{3}}}} \right) - \left( {2 + \frac{1}{{\frac{5}{2}}}} \right) \cr & = 35.7 - \left( {3 + \frac{3}{{10}}} \right) - \left( {2 + \frac{2}{5}} \right) \cr & = 35.7 - \frac{{33}}{{10}} - \frac{{12}}{5} \cr & = 35.7 - \left( {\frac{{33}}{{10}} + \frac{{12}}{5}} \right) \cr & = 35.7 - \frac{{57}}{{10}} \cr & = 35.7 - 5.7 \cr & = 30 \cr} $$
23.
The value of $$\frac{{{{\left( {0.06} \right)}^2} + {{\left( {0.47} \right)}^2} + {{\left( {0.079} \right)}^2}}}{{{{\left( {0.006} \right)}^2} + {{\left( {0.047} \right)}^2} + {{\left( {0.0079} \right)}^2}}}$$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ is :
(A) 0.1
(B) 10
(C) 100
(D) 1000
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Solution:
Given expression : $$ = \frac{{{a^2} + {b^2} + {c^2}}}{{{{\left( {\frac{a}{{10}}} \right)}^2} + {{\left( {\frac{b}{{10}}} \right)}^2} + {{\left( {\frac{c}{{10}}} \right)}^2}}}$$ Where a = 0.06, b = 0.47 and c = 0.079 $$\eqalign{ & = \frac{{100\left( {{a^2} + {b^2} + {c^2}} \right)}}{{\left( {{a^2} + {b^2} + {c^2}} \right)}} \cr & = 100 \cr} $$
24.
0.3 + 3 + 3.33 + 3.3 + 3.03 + 333 = ?
(A) 345.99
(B) 355.96
(C) 375.66
(D) 375.93
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Solution:
Answer & Solution Answer: Option E Solution:
25.
Solve this, $$\frac{3.5 ÃÂÃÂÃÂÃÂÃÂÃÂ 1.4}{0.7}$$ÃÂÃÂÃÂÃÂ = ?
(A) 0.7
(B) 2.4
(C) 3.5
(D) 7.1
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Solution:
Answer & Solution Answer: Option E Solution: Given expression = $$\frac{35 × 1.4}{7}$$ = 5 × 1.4 = 7
26.
Solve this : $$\frac{0.0203 ÃÂÃÂÃÂÃÂÃÂÃÂ 2.92}{0.0073 ÃÂÃÂÃÂÃÂÃÂÃÂ 14.5 ÃÂÃÂÃÂÃÂÃÂÃÂ 0.7}$$ ÃÂÃÂÃÂÃÂ ÃÂÃÂÃÂÃÂ = ?
(A) 0.8
(B) 1.45
(C) 2.40
(D) 3.25
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Solution:
= $$\frac{0.0203 × 2.92}{0.0073 × 14.5 × 0.7}$$ = $$\frac{203 × 292}{73 × 145 × 7}$$ = $$\frac{4}{5}$$ = 0.8
27.
If 13 + 23 + 33 + .... + 93 = 2025, then the value of (0.11)3 + (0.22)3 + .... + (0.99)3 is close to :
(A) 0.2695
(B) 0.3695
(C) 2.695
(D) 3.695
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Solution:
$$\eqalign{ & {\left( {0.11} \right)^3} + {\left( {0.22} \right)^3} + .... + {\left( {0.99} \right)^3} \cr & = {\left( {0.11} \right)^3}\left( {{1^3} + {2^3} + .... + {9^3}} \right) \cr & = 0.001331 \times 2025 \cr & = 2.695275 \approx 2.695 \cr} $$
28.
$$\frac{{{{\left( {0.013} \right)}^3} + 0.000000343}}{{{{\left( {0.013} \right)}^2} - 0.000091 + 0.000049}} = \,?$$
(A) 0.002
(B) 0.020
(C) 0.021
(D) 0.023
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Solution:
Given expression : $$\eqalign{ & \frac{{{{\left( {0.013} \right)}^3} + 0.000000343}}{{{{\left( {0.013} \right)}^2} - 0.000091 + 0.000049}} \cr & a = 0.013 \text{ and } b = 0.007 \cr & = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right) \cr & = a + b \cr & = 0.013 + 0.007 \cr & = 0.020 \cr} $$
29.
Given expression : (11.6 ÃÂÃÂÃÂ÷ 0.8) (13.5 ÃÂÃÂÃÂ÷ 2) = ?
(A) 98
(B) 99
(C) 100
(D) None of these
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Solution:
Given expression : = $$\frac{116}{8}$$ × $$\frac{13.5}{2}$$ = 14.5 × 6.75 = 97.875
30.
Solve : $$\frac{{{{\left( {36.54} \right)}^2} - {{\left( {3.46} \right)}^2}}}{?} = 40$$
(A) 3.308
(B) 4
(C) 33.08
(D) 330.8
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Solution:
Let, the missing number be x, Then, $$\frac{{{{\left( {36.54} \right)}^2} - {{\left( {3.46} \right)}^2}}}{x} = 40$$ x = $$\frac{{{{\left( {36.54} \right)}^2} - {{\left( {3.46} \right)}^2}}}{40} $$ x = $$\frac{{{{\left( {36.54} \right)}^2} - {{\left( {3.46} \right)}^2}}}{{36.54 + 3.46}}$$ Let 36.54 = $$a$$ and 3.46 = $$b$$ x = $$\frac{{{a^2} - {b^2}}}{{a + b}}$$ x = (a - b) x = (36.54 - 3.46) x = 33.08