1.
The value of $$\frac{241.6 ÃÂÃÂÃÂÃÂÃÂÃÂ 0.3814 ÃÂÃÂÃÂÃÂÃÂÃÂ 6.842}{0.4618 ÃÂÃÂÃÂÃÂÃÂÃÂ 38.25 ÃÂÃÂÃÂÃÂÃÂÃÂ 73.65}$$ ÃÂÃÂÃÂÃÂ ÃÂÃÂÃÂÃÂ is close to :
(A) 0.2
(B) 0.4
(C) 0.6
(D) 1
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Solution:
$$\frac{241.6 × 0.3814 × 6.842}{0.4618 × 38.25 × 73.65}$$ ≈ $$\frac{240 × 0.38 × 6.9}{0.46 × 38 × 75}$$ = $$\frac{240 × 38 × 69}{46 × 38 × 75}$$ × $$\frac{1}{10}$$ = $$\frac{24}{5}$$ × $$\frac{1}{10}$$ = $$\frac{4.8}{10}$$ = 0.48 So, the value is close to 0.4
2.
Which of the following fractions is the largest ?
(A) $$\frac{7}{3}$$
(B) $$\frac{5}{4}$$
(C) $$\frac{7}{2}$$
(D) $$\frac{3}{2}$$
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Solution:
Each fraction is equivalent to decimal $$\frac{3}{2}$$ = 1.5 $$\frac{7}{3}$$ = 2.3 $$\frac{5}{4}$$ = 1.25 $$\frac{7}{2}$$ = 3.5 Hence, $$\frac{7}{2}$$ is largest fraction.
3.
$$\frac{{4.2 \times 4.2 - 1.9 \times 1.9}}{{2.3 \times 6.1}}$$ ÃÂÃÂ ÃÂÃÂ is equal to:
(A) 0.5
(B) 1
(C) 20
(D) 22
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Solution:
$$\eqalign{ & \text{Let } a = 4.2 \text{ and } b = 1.9 \cr & {\text{Given Expression}} \cr & = \frac{{ {{a^2} - {b^2}} }}{{\left( {a + b} \right)\left( {a - b} \right)}} \cr & = \frac{{ {{a^2} - {b^2}} }}{{ {{a^2} - {b^2}} }} \cr & = 1 \cr} $$
4.
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
5.
$$2\frac{1.5}{5}$$ + 2$$\frac{1}{6}$$ - $$1\frac{3.5}{15}$$ = $$\left( {\frac{{{{(?)}^{\frac{1}{3}}}}}{4}} \right)$$ÃÂÃÂ + $$1\frac{7}{30}$$
(A) 2
(B) 8
(C) 512
(D) 324
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Solution:
$$\eqalign{ & 2\frac{{1.5}}{5} + 2\frac{1}{6} - 1\frac{{3.5}}{{15}} = \frac{{{x^{\frac{1}{3}}}}}{4} + 1\frac{7}{{30}} \cr & \Rightarrow \frac{{11.5}}{5} + \frac{{13}}{6} - \frac{{18.5}}{{15}} = \frac{{{x^{\frac{1}{3}}}}}{4} + \frac{{37}}{{30}} \cr} $$ ⇒ L.C.M. of 5, 6 and 15 is 30 $$\eqalign{ & \Rightarrow \frac{{69 + 65 - 37}}{{30}} = \frac{{{x^{\frac{1}{3}}}}}{4} + \frac{{37}}{{30}} \cr & \Rightarrow \frac{{97}}{{30}} = \frac{{{x^{\frac{1}{3}}}}}{4} + \frac{{37}}{{30}} \cr & \Rightarrow \frac{{{x^{\frac{1}{3}}}}}{4} = \frac{{97}}{{30}} - \frac{{37}}{{30}} \cr & \Rightarrow {x^{\frac{1}{3}}} = \frac{{60}}{{30}} \times 4 \cr & \Rightarrow {x^{\frac{1}{3}}} = 8 \cr & \Rightarrow x = {\left( 8 \right)^3} \cr & \Rightarrow x = 512 \cr} $$ Hence, the number is 512
6.
0.5 ÃÂÃÂÃÂÃÂÃÂà0.5 + 0.5 ÃÂÃÂÃÂÃÂÃÂÃÂÃÂ÷ 5 is equal to :
(A) 0.15
(B) 0.25
(C) 0.35
(D) 0.45
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Solution:
Given expression : 0.5 × 0.5 + $$\frac{0.5}{5}$$ = 0.25 + 0.1 = 0.35
7.
Which of the following are in descending order of their value ?
(A) $$\frac{1}{3},\frac{2}{5},\frac{3}{7},\frac{4}{5},\frac{5}{6},\frac{6}{7}$$
(B) $$\frac{1}{3},\frac{2}{5},\frac{3}{5},\frac{4}{7},\frac{5}{6},\frac{6}{7}$$
(C) $$\frac{1}{3},\frac{2}{5},\frac{3}{5},\frac{4}{6},\frac{5}{7},\frac{6}{7}$$
(D) $$\frac{6}{7},\frac{5}{6},\frac{4}{5},\frac{3}{7},\frac{2}{5},\frac{1}{3}$$
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Solution:
Converting each of the given fractions in to decimal form, we get $$\eqalign{ & \frac{1}{3} = 0.33 \cr & \frac{2}{5} = 0.4 \cr & \frac{3}{7} = 0.42 \cr & \frac{4}{5} = 0.8 \cr & \frac{5}{6} = 0.83 \cr & \frac{6}{7} = 0.85 \cr} $$ Clearly, 0.85 > 0.83 > 0.8 > 0.42 > 0.4 > 0.33 So, $$\frac{6}{7}$$ > $$\frac{5}{6}$$ > $$\frac{4}{5}$$ > $$\frac{3}{7}$$ > $$\frac{2}{5}$$ > $$\frac{1}{3}$$
8.
The product of 0.09 and 0.007 is :
(A) 0.6300
(B) 0.00063
(C) 0.00630
(D) 0.000063
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Solution:
9 × 7 = 63 Sum of decimal places = 5 ∴ 0.09 × 0.007 = 0.00063
9.
$$0.\overline {142857} \div 0.\overline {285714} $$ ÃÂÃÂ ÃÂÃÂ is equal to :
(A) $$\frac{1}{2}$$
(B) $$\frac{1}{3}$$
(C) 2
(D) 10
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Solution:
$$\eqalign{ & 0.\overline {142857} \div 0.\overline {285714} \cr & = \frac{{142857}}{{999999}} \div \frac{{285714}}{{999999}} \cr & = \left( {\frac{{142857}}{{999999}} \times \frac{{999999}}{{285714}}} \right) \cr & = \frac{1}{2} \cr} $$
10.
34.95 + 240.016 + 23.98 = ?
(A) 298.0946
(B) 298.111
(C) 298.946
(D) 299.09
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Solution:
$$\eqalign{ & \,\,\,\,34.95 \cr & \,240.016 \cr & + 23.98 \cr & - - - - - \cr & 298.946 \cr & - - - - - \cr} $$