141.
Which of the following is closest to zero ?
- (A) $${\left( {0.09} \right)^2}$$
- (B) $$0.09$$
- (C) $${\left( {1 - 0.9} \right)^2}$$
- (D) $$1 - {\left( {0.9} \right)^2}$$
Solution:
(a) (0.09)2 = 0.0081 (b) 0.09 (c) (1 - 0.9)2 = (0.1)2 = 0.01 (d) 1 - (0.9)2 = 1 - 0.81 = 0.19 Clearly, 0.0081 0.01 0.09 0.19 So, 0.0081 is closest to zero.
142.
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
Solution:
Given expression : $$\eqalign{ & = 35.7 - \left( {3 + \frac{1}{{\frac{{10}}{3}}}} \right) - \left( {2 + \frac{1}{{\frac{5}{2}}}} \right) \cr & = 35.7 - \left( {3 + \frac{3}{{10}}} \right) - \left( {2 + \frac{2}{5}} \right) \cr & = 35.7 - \frac{{33}}{{10}} - \frac{{12}}{5} \cr & = 35.7 - \left( {\frac{{33}}{{10}} + \frac{{12}}{5}} \right) \cr & = 35.7 - \frac{{57}}{{10}} \cr & = 35.7 - 5.7 \cr & = 30 \cr} $$
143.
$$1.\overline {27} $$ ÃÂÃÂ in the from $$\frac{p}{q}$$ is equal to :
- (A) $$\frac{127}{100}$$
- (B) $$\frac{14}{11}$$
- (C) $$\frac{73}{100}$$
- (D) $$\frac{11}{14}$$
Solution:
$$\eqalign{ & = 1.\overline {27} \cr & = 1 + 0.\overline {27} \cr & = 1 + \frac{{27}}{{99}} \cr & = 1 + \frac{3}{{11}} \cr & = \frac{{11 + 3}}{{11}} \cr & = \frac{{14}}{{11}} \cr} $$
144.
$$\frac{{\left( {0.1667} \right)\left( {0.8333} \right)\left( {0.3333} \right)}}{{\left( {0.2222} \right)\left( {0.6667} \right)\left( {0.1250} \right)}}$$ ÃÂÃÂ ÃÂÃÂ is approximately equal to:
- (A) 2
- (B) 2.40
- (C) 2.43
- (D) 2.50
Solution:
$$\eqalign{ & {\text{Given}}\,{\text{expression}} \cr & = \frac{{ {0.3333} }}{{ {0.2222} }} \times \frac{{\left( {0.1667} \right)\left( {0.8333} \right)}}{{\left( {0.6667} \right)\left( {0.1250} \right)}} \cr & = \frac{{3333}}{{2222}} \times \frac{{\frac{1}{6} \times \frac{5}{6}}}{{\frac{2}{3} \times \frac{{125}}{{1000}}}} {\text{ [where }} {0.3333} = {\frac{1}{6}}, {0.8333} = {\frac{5}{6}} {\text{ and }} {0.6667} = {\frac{2}{3}} {\text{ ] }} \cr & = {\frac{3}{2} \times \frac{1}{6} \times \frac{5}{6} \times \frac{3}{2} \times 8} \cr & = \frac{5}{2} \cr & = 2.50 \cr} $$
145.
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) $$\frac{4}{11}$$
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, $$\frac{4}{11}$$ > $$\frac{9}{31}$$ > $$\frac{7}{25}$$ > $$\frac{6}{23}$$ > $$\frac{3}{17}$$
146.
Which of the following is equal to 3.14 x 106 ?
- (A) 314
- (B) 3140
- (C) 3140000
- (D) None of these
Solution:
3.14 x 106 = 3.14 x 1000000 = 3140000
147.
The value of :
$$\left( {\frac{{0.051 \times 0.051 \times 0.051 + 0.041 \times 0.041 \times 0.041}}{{0.051 \times 0.051 - 0.051 \times 0.041 + 0.041 \times 0.041}}} \right)$$
- (A) 0.00092
- (B) 0.0092
- (C) 0.092
- (D) 0.92
Solution:
Given expression : $$ = \frac{{{{\left( {0.051} \right)}^3} + {{\left( {0.041} \right)}^3}}}{{{{\left( {0.051} \right)}^2} - \left( {0.051 \times 0.041} \right) + {{\left( {0.041} \right)}^2}}}$$ Let 0.051 = $$a$$ and 0.041 = $$b$$ $$\eqalign{ & = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right) \cr & = (a + b) \cr & = \left( {0.051 + 0.041} \right) \cr & = 0.092 \cr} $$