71.
$$\frac{3.25 ÃÂÃÂÃÂÃÂÃÂÃÂ 3.20 - 3.20 ÃÂÃÂÃÂÃÂÃÂÃÂ 3.05}{0.064}$$ ÃÂÃÂÃÂÃÂ ÃÂÃÂÃÂÃÂ ÃÂÃÂÃÂÃÂ is equal to :
(A) 1
(B) $$\frac{1}{2}$$
(C) $$\frac{1}{10}$$
(D) 10
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Solution:
Given expression : = $$\frac{3.20 (3.25 - 3.05)}{0.064}$$ = $$\frac{3.20 × 0.2}{0.064}$$ = $$\frac{0.64}{0.064}$$ = $$\frac{64}{6.4}$$ = 10
72.
Solve : 48.2 ÃÂÃÂÃÂÃÂÃÂÃÂ 2.5 ÃÂÃÂÃÂÃÂÃÂÃÂ 2.2 + ? = 270
(A) 6.5
(B) 2.8
(C) 4.9
(D) 3.4
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Solution:
Let the missing number be x Given, 48.2 × 2.5 × 2.2 + x = 270 ⇒ x = 270 - 48.2 × 2.5 × 2.2 ⇒ x = 270 - 265.1 ⇒ x = 4.9 Hence, the number is 4.9
73.
$$\left( {0.34\overline {67} + 0.13\overline {33} } \right)$$ ÃÂÃÂ is equal to :
(A) $${0.4\overline 8 }$$
(B) $${0.\overline 48 }$$
(C) $${0.48\overline {01} }$$
(D) $${0.48}$$
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Solution:
$${0.34\overline {67} + 0.13\overline {33} }$$ = $$\frac{3467 - 34}{9900}$$ + $$\frac{1333 - 13}{9900}$$ = $$\frac{3433 + 1320}{9900}$$ = $$\frac{4753}{9900}$$ = $$\frac{4801 - 48}{9900}$$ = $${0.48\overline {01} }$$
74.
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
Show Answer
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 $$\eqalign{ & = \frac{{100\left( {{a^2} + {b^2} + {c^2}} \right)}}{{\left( {{a^2} + {b^2} + {c^2}} \right)}} \cr & = 100 \cr} $$
75.
$$\frac{5 ÃÂÃÂÃÂÃÂÃÂÃÂ 1.6 - 2 ÃÂÃÂÃÂÃÂÃÂÃÂ 1.4}{1.3}$$ ÃÂÃÂÃÂÃÂ = ?
(A) 0.4
(B) 1.2
(C) 1.4
(D) 4
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Solution:
Given expression : $$ = \frac{8 - 2.8}{1.3}$$ $$ = \frac{5.2}{1.3}$$ $$ = \frac{52}{13}$$ = 4
76.
Which part contains the fractions in ascending order ?
(A) $$\frac{{11}}{{14}},\frac{{16}}{{19}},\frac{{19}}{{21}}$$
(B) $$\frac{{16}}{{19}},\frac{{11}}{{14}},\frac{{19}}{{21}}$$
(C) $$\frac{{16}}{{19}},\frac{{19}}{{21}},\frac{{11}}{{14}}$$
(D) $$\frac{{19}}{{21}},\frac{{11}}{{14}},\frac{{16}}{{19}}$$
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Solution:
Clearly, $$\eqalign{ & \frac{{11}}{{14}} = 0.785 \cr & \frac{{16}}{{19}} = 0.842 \cr & \frac{{19}}{{21}} = 0.904 \cr} $$ Now, 0.785 0.842 0.904 So, $$\frac{11}{14}$$ $$\frac{16}{19}$$ $$\frac{19}{21}$$
77.
Which of the following fractions lies between $$\frac{2}{3}$$ and $$\frac{3}{5}$$ = ?
(A) $$\frac{2}{5}$$
(B) $$\frac{1}{3}$$
(C) $$\frac{1}{15}$$
(D) $$\frac{31}{50}$$
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Solution:
$$\eqalign{ & \frac{2}{3} = 0.666 \cr & \frac{3}{5} = 0.6 \cr & \frac{2}{5} = 0.4 \cr & \frac{1}{3} = 0.333 \cr & \frac{1}{{15}} = 0.066 \cr & \frac{{31}}{{50}} = 0.62 \cr} $$ Clearly, 0.62 lies between 0.6 and 0.666 So, $$\frac{31}{50}$$ lies between $$\frac{2}{3}$$ and $$\frac{3}{5}$$
78.
$$\frac{5}{9}$$ of a number is equal to twenty five percent of second number. Second number is equal to $$\frac{1}{4}$$ of third number. The value of third number is 2960. What is 30% of first number ?
(A) 99.9
(B) 88.8
(C) 77.7
(D) None of these
Show Answer
Solution:
Let the third number be 2960 ∵ Second number = $$\frac{1}{4}$$ of the third number = $$\frac{1}{4}$$ × 2960 = 740 $$\frac{5}{9}$$ of first number = 25% of second number $$\frac{5}{9}$$ first number = $$\frac{25 × 740}{100}$$ = 185 ⇒ First number = $$\frac{185 × 9}{5}$$ = 333 ∴ 30% of 333 = $$\frac{30}{100}$$ × 333 = 99.9
79.
6425 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ 125 ÃÂÃÂÃÂÃÂÃÂà8 = ?
(A) 41.12
(B) 64.25
(C) 411.2
(D) 421.25
Show Answer
Solution:
Given expression : 51.4 × 8 = 411.2
80.
$$3.\overline {87} - 2.\overline {59} = ?$$
(A) 1.20
(B) $$1.\overline 2 $$
(C) $$1.\overline {27} $$
(D) $$1.\overline {28} $$
Show Answer
Solution:
$$\eqalign{ & 3.\overline {87} - 2.\overline {59} \cr & = \left( {3 + 0.\overline {87} } \right) - \left( {2 + 0.\overline {59} } \right) \cr & = \left( {3 + \frac{{87}}{{99}}} \right) - \left( {2 + \frac{{59}}{{99}}} \right) \cr & = 1 + \left( {\frac{{87}}{{99}} - \frac{{59}}{{99}}} \right) \cr & = 1 + \frac{{28}}{{99}} \cr & = 1.\overline {28} \cr} $$