191.
What will come at place of x, (x 10) for $$\frac{{\left( {132 \div 12 \times x - 3 \times 3} \right)}}{{\left( {{5^2} - 6 \times 4 + {x^2}} \right)}} = 1?$$
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Solution:
$$\eqalign{ & \frac{{132 \div 12 \times x - 3 \times 3}}{{{5^2} - 6 \times 4 + {x^2}}} = 1 \cr & 11 \times x - 9 = 25 - 24 + {x^2} \cr & 11x - 9 = 1 + {x^2} \cr & {x^2} + 1 + 9 - 11x = 0 \cr & {x^2} - 11x + 10 = 0 \cr & {x^2} - 10x - x + 10 = 0 \cr & x\left( {x - 10} \right) - 1\left( {x - 10} \right) = 0 \cr & \left( {x - 10} \right)\left( {x - 1} \right) = 0 \cr & x = 10,\,1 \cr & {\text{For the condition}}\left( {x 10} \right) \cr & x = 1 \cr} $$
192.
$$\frac{1}{{10}}$$ of a pole is coloured red, $$\frac{1}{{20}}$$ white, $$\frac{1}{{30}}$$ blue, $$\frac{1}{{40}}$$ black, $$\frac{1}{{50}}$$ violet, $$\frac{1}{{60}}$$ yellow and the rest is green. If the length of the green portion of the pole is 12.08 metres, then the length of the pole is = ?
(A) 16m
(B) 18m
(C) 20m
(D) 30m
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Solution:
$$\eqalign{ & {\text{Green portion}} \cr & = \left[ {1 - \left( {\frac{1}{{10}} + \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{40}} + \frac{1}{{50}} + \frac{1}{{60}}} \right)} \right] \cr & = \left[ {1 - \frac{1}{{10}}\left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}} \right)} \right] \cr & = 1 - \frac{1}{{10}} \times \frac{{147}}{{60}} \cr & = 1 - \frac{{147}}{{600}} \cr & = \frac{{453}}{{600}} \cr & {\text{let the length of the pole be }}x{\text{ metres}} \cr & {\text{Then, }}\frac{{453}}{{600}}x = 12.08 \cr & \Leftrightarrow x = \left( {\frac{{12.08 \times 600}}{{453}}} \right) = 16 \cr} $$
193.
The value of $$0.4\overline 6 + 0.7\overline {23} - 0.3\overline 9 \times 0.\overline 7 $$ ÃÂÃÂ ÃÂÃÂ is:
(A) $$0.\overline {57} $$
(B) $$0.\overline {87} $$
(C) $$0.\overline {97} $$
(D) $$0.\overline {77} $$
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Solution:
$$\eqalign{ & 0.4\overline 6 + 0.7\overline {23} - 0.3\overline 9 \times 0.\overline 7 \cr & = \frac{{46 - 4}}{{90}} + \frac{{723 - 7}}{{990}} - \frac{{39 - 3}}{{90}} \times \frac{7}{9} \cr & = \frac{{42}}{{90}} + \frac{{716}}{{990}} - \frac{{36}}{{90}} \times \frac{7}{9} \cr & = \frac{{42}}{{90}} + \frac{{716}}{{990}} - \frac{{28}}{{90}} \cr & = \frac{{14}}{{90}} + \frac{{716}}{{990}} \cr & = \frac{{154 + 716}}{{990}} \cr & = \frac{{870}}{{990}} \cr & = 0.\overline {87} \cr} $$
194.
The value of 3 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ 18 of 3 ÃÂÃÂÃÂÃÂÃÂà6 + 21 ÃÂÃÂÃÂÃÂÃÂà6 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ 18 - 3 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ 2 + 3 - 3 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ 9 of 3 ÃÂÃÂÃÂÃÂÃÂà9 is:
(A) $$\frac{{29}}{6}$$
(B) $$\frac{{41}}{6}$$
(C) $$\frac{{35}}{9}$$
(D) $$\frac{{47}}{6}$$
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Solution:
$$\eqalign{ & 3 \div 18{\text{ of }}3 \times 6 + 21 \times 6 \div 18 - 3 \div 2 + 3 - 3 \div 9{\text{ of }}3 \times 9 \cr & = \frac{3}{{54}} \times 6 + \frac{{21 \times 6}}{{18}} - \frac{3}{2} + 3 - \frac{3}{{27}} \times 9 \cr & = \frac{1}{3} + 7 - \frac{3}{2} + 3 - 1 \cr & = 9 + \frac{1}{3} - \frac{3}{2} \cr & = \frac{{54 + 2 - 9}}{6} \cr & = \frac{{47}}{6} \cr} $$
195.
If $$\sqrt {{\text{4096}}} $$ÃÂÃÂ = 64, then the value of $$\sqrt {{\text{40}}{\text{.96}}} $$ ÃÂÃÂ $$ + $$ $$\sqrt {{\text{0}}{\text{.4096}}} $$ ÃÂÃÂ $$ + $$ $$\sqrt {{\text{0}}{\text{.004096}}} $$ ÃÂÃÂ ÃÂÃÂ $$ + $$ $$\sqrt {{\text{0}}{\text{.00004096}}} $$ ÃÂÃÂ ÃÂÃÂ up to two place of decimals is = ?
(A) 7.09
(B) 7.10
(C) 7.11
(D) 7.12
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Solution:
$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {40.96} {\text{ + }}\sqrt {0.4096} {\text{ + }}\sqrt {0.004096} {\text{ + }}\sqrt {0.00004096} \cr & \Rightarrow \sqrt {\frac{{4096}}{{100}}} + \sqrt {\frac{{4096}}{{10000}}} + \sqrt {\frac{{4096}}{{1000000}}} + \sqrt {\frac{{4096}}{{100000000}}} \cr & \Rightarrow \frac{{64}}{{10}} + \frac{{64}}{{100}} + \frac{{64}}{{1000}} + \frac{{64}}{{10000}} \cr & \Rightarrow 6.4 + 0.64 + 0.064 + 0.0064 \cr & \Rightarrow 7.11 \cr} $$
196.
If $$\frac{1}{{4.263}} = 0.2346,$$ ÃÂÃÂ find the value of $$\frac{1}{{0.0004263}}.$$
(A) 2346
(B) 4.263
(C) 2.346
(D) 4263
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Solution:
$$\eqalign{ & \frac{1}{{4.263}} = 0.2346 \cr & \frac{{1000}}{{4263}} = 0.2346 \cr & \frac{1}{{0.0004263}} \cr & = \frac{{10000000}}{{4263}} \cr & = 10000 \times 0.2346 \cr & = 2346 \cr} $$
197.
Find the value of 2.1 + 2.25 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ [63 - {7.5 ÃÂÃÂÃÂÃÂÃÂà8 + (13 - 2.5 ÃÂÃÂÃÂÃÂÃÂà5)}].
(A) 2.8
(B) 2.9
(C) 3.0
(D) 3.1
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Solution:
2.1 + 2.25 ÷ [63 - {7.5 × 8 + (13 - 2.5 × 5)}] = 2.1 + 2.25 ÷ [63 - {60 + 0.5}] = 2.1 + 2.25 ÷ [63 - 60.5] = 2.1 + $$\frac{{2.25}}{{2.50}}$$ = 2.1 + 0.9 = 3.0
198.
The simplified value of $$\sqrt {5 + \sqrt {11 + \sqrt {19 + \sqrt {29 + \sqrt {49} } } } } $$ ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ = ?
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Solution:
$$\eqalign{ & {\text{According to question,}} \cr & \,\,\,\, \sqrt {5 + \sqrt {11 + \sqrt {19 + \sqrt {29 + \sqrt {49} } } } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + \sqrt {19 + \sqrt {29 + 7} } } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + \sqrt {19 + \sqrt {36} } } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + \sqrt {19 + 6} } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + \sqrt {25} } } \cr & \Rightarrow \sqrt {5 + \sqrt {11 + 5} } \cr & \Rightarrow \sqrt {5 + \sqrt {16} } \cr & \Rightarrow \sqrt {5 + 4} \cr & \Rightarrow \sqrt 9 \cr & \Rightarrow 3 \cr} $$
199.
Given that $$\sqrt {574.6} $$ÃÂÃÂ = 23.97, $$\sqrt {5746} $$ÃÂÃÂ = 75.8 then $$\sqrt {0.00005746} $$ ÃÂÃÂ = ?
(A) 0.002394
(B) 0.0002397
(C) 0.0007580
(D) 0.00758
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Solution:
$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {0.00005746} \cr & \Rightarrow \sqrt {\frac{{5746}}{{100000000}}} \cr & \Rightarrow \frac{{75.8}}{{10000}} \cr & \Rightarrow 0.00758 \cr} $$
200.
Let 0 x 1, then the correct inequality is = ?
(A) $$x \sqrt x {x^2}$$
(B) $$\sqrt x x {x^2}$$
(C) $${x^2} x \sqrt x $$
(D) $$\sqrt x {x^2} x$$
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Solution:
$$\eqalign{ & 0 x 1, \cr & {\text{Let }}x = \frac{4}{{10}} \cr & {\text{So}},\sqrt x = \frac{2}{{\sqrt {10} }}\,\& \, \cr & \,\,{x^2} = \frac{{16}}{{100}} = 0.16 \cr & {\text{Now}}, \cr & \because 0.16 \frac{4}{{10}} \frac{2}{{\sqrt {10} }} \cr & \therefore {x^2} x \sqrt x \cr} $$