Decimal Fraction — Part 2 MCQs with Explanations

$$\eqalign{ & \frac{2}{3} = 0.666 \cr & \frac{3}{5} = 0.6 \cr & \frac{2}{5} = 0.4 \cr & \frac{1}{3} = 0.333 \cr & \frac{1}{{15}} = 0.066 \cr & \frac{{31}}{{50}} = 0.62 \cr} $$ Clearly, 0.62 lies between 0.6 and 0.666 So, $$\frac{31}{50}$$ lies between $$\frac{2}{3}$$ and $$\frac{3}{5}$$

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

Given expression = $$\frac{35 × 1.4}{7}$$ = 5 × 1.4 = 7

$$\therefore \frac{{1999}}{{2111}} = 0.946$$

Given expression : $$\eqalign{ & = \frac{{{{\left( {0.943} \right)}^2} - \left( {0.943 \times 0.057} \right) + {{\left( {0.057} \right)}^2}}}{{{{\left( {0.943} \right)}^3} + {{\left( {0.057} \right)}^3}}} \cr & = \frac{{{a^2} - ab + {b^2}}}{{{a^3} + {b^3}}} \cr & = \frac{1}{{a + b}} \cr & = \frac{1}{{0.943 + 0.057}} \cr & = 1 \cr} $$

Let, $$\frac{{{x^2} + {{\left( {18} \right)}^2}}}{{125}} = 3.56$$ Then, $$\eqalign{ & {x^2} + 324 = 125 \times 3.56 = 445 \cr & \Rightarrow {x^2} = 121 \cr & \Rightarrow x = 11 \cr} $$

Let 534.596 + 61.472 - 496.708 = x + 27.271 Then, x = (534.596 + 61.472) - (496.708 + 27.271) = 596.068 - 523.979 = 72.089

$$\eqalign{ & {\left( {\frac{{18}}{4}} \right)^2} \times \left( {\frac{{455}}{{19}}} \right) \div \left( {\frac{{61}}{{799}}} \right) \cr & = \frac{{324}}{{16}} \times \frac{{455}}{{19}} \times \frac{{799}}{{61}} \cr & = 6350 \cr} $$

Let the missing number be x $$\eqalign{ & \frac{{294 \div 14 \times 5 + 11}}{x} = {8^2} \div 5 + 1.7 \cr & \Rightarrow \frac{{\frac{{294}}{{14}} \times 5 + 11}}{x} = \frac{{64}}{5} + 1.7 \cr & \Rightarrow \frac{{21 \times 5 + 11}}{x} = 12.8 + 1.7 \cr & \Rightarrow \frac{{105 + 11}}{x} = 12.8 + 1.7 \cr & \Rightarrow \frac{{116}}{x} = 14.5 \cr & \Rightarrow x = \frac{{116}}{{14.5}} \cr & \Rightarrow x = \frac{{116 \times 10}}{{145}} \cr & \Rightarrow x = 8 \cr} $$ Hence, the number is 8

324 × 115 × 85 = 3167100 Sum of decimal place = 3 ∴ 32.4 × 11.5 × 8.5 = 3167.1

$$\eqalign{ & = \frac{{3.6 \times 0.48 \times 2.50}}{{0.12 \times 0.09 \times 0.5}} \cr & = \frac{{36 \times 48 \times 250}}{{12 \times 9 \times 5}} \cr & = 800 \cr} $$

Converting each of the given fractions in to decimal form, we get $$\frac{5}{9}$$ = 0.55 $$\frac{7}{11}$$ = 0.63 $$\frac{8}{15}$$ = 0.533 $$\frac{11}{17}$$ = 0.647 Clearly, 0.647 > 0.63 > 0.55 > 0.533 So, $$\frac{11}{17}$$ > $$\frac{7}{11}$$ > $$\frac{5}{9}$$ > $$\frac{8}{15}$$

Given expression : $$\eqalign{ & = 3927 + \frac{{5526}}{{12.5}} \cr & = 3927 + \frac{{55260}}{{125}} \cr} $$ = 3927 + 442.08 = 4369.08

$$\frac{4}{5}$$ = 0.8 $$\frac{7}{13}$$ = 0.53 $$\frac{1}{2}$$ = 0.5 $$\frac{2}{3}$$ = 0.66 $$\frac{3}{4}$$ = 0.75 $$\frac{5}{7}$$ = 0.714 Clearly, 0.5 does not lie between 0.53 and 0.8 ∴ $$\frac{1}{2}$$ does not lie between $$\frac{4}{5}$$ and $$\frac{7}{13}$$

383 × 38 × 38 = 553052 Number of decimal place = 1 ∴ 383 × 38 × 3.8 = 55305.2

Converting each of the given fractions into decimal form, we get : $$\frac{2}{3}$$ = 0.66 $$\frac{3}{4}$$ = 0.75 $$\frac{4}{5}$$ = 0.8 $$\frac{5}{6}$$ = 0.833 Since 0.833 > 0.8 > 0.75 > 0.66 So, $$\frac{5}{6}$$ > $$\frac{4}{5}$$ > $$\frac{3}{4}$$ > $$\frac{2}{3}$$ ∴ Required difference = $$\left( {\frac{5}{6} - \frac{2}{3}} \right) = \frac{1}{6}$$

Given expression : $$\eqalign{ & \left[ {1 + \frac{1}{{1 \times 2}} + \frac{1}{{1 \times 2 \times 4}} + \frac{1}{{1 \times 2 \times 4 \times 8}} + \frac{1}{{1 \times 2 \times 4 \times 8 \times 16}}} \right] \cr & = \frac{{2 \times 4 \times 8 \times 16 + 4 \times 8 \times 16 + 8 \times 16 + 16 + 1}}{{2 \times 4 \times 8 \times 16}} \cr & = \frac{{1024 + 512 + 128 + 16 + 1}}{{1024}} \cr & = \frac{{1681}}{{1024}} \cr & = 1.6416 \cr} $$

Given expression : $$\eqalign{ & \frac{{{{\left( {0.013} \right)}^3} + 0.000000343}}{{{{\left( {0.013} \right)}^2} - 0.000091 + 0.000049}} \cr & a = 0.013 \text{ and } b = 0.007 \cr & = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right) \cr & = a + b \cr & = 0.013 + 0.007 \cr & = 0.020 \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & \frac{{52416}}{{312}} = 168 \cr & \Leftrightarrow \frac{{52416}}{{168}} = 312 \cr & {\text{Now,}} \cr & \frac{{52.416}}{{0.0168}} \cr & = \frac{{524160}}{{168}} \cr & = \frac{{52416}}{{168}} \times 10 \cr & = 312 \times 10 \cr & = 3120 \cr} $$

Given expression : (1.25) 3 - (0.75) 3 - 3 × (1.25) 2 × 0.75 + 3 × 1.25 × (0.75) 2 = (1.25 - 0.75) 3 = (0.5) 3 = $${\left( {\frac{1}{2}} \right)^3}$$ = $$\frac{1}{8}$$ [∵ (a - b) 3 = a 3 - b 3 - 3a 2 b + 3ab 2 ]

Let the missing number be x $$\eqalign{ & \frac{{21.5}}{5} + \frac{{21}}{6} - \frac{{13.5}}{{15}} = \frac{{{{\left( x \right)}^{\frac{1}{3}}}}}{4} + \frac{{17}}{{30}} \cr & \frac{{21.5}}{5} + \frac{{21}}{6} - \frac{{13.5}}{{15}} - \frac{{17}}{{30}} = \frac{{{{\left( x \right)}^{\frac{1}{3}}}}}{4} \cr} $$ L.C.M of 5, 6, 15 and 30 is 30 $$\eqalign{ & \frac{{129 + 105 - 27 - 17}}{{30}} = \frac{{{{\left( x \right)}^{\frac{1}{3}}}}}{4} \cr & \root 3 \of x = \frac{{190 \times 4}}{{30}} \cr & \root 3 \of x = 25.33 \approx 25 \cr & x = {25^3} \cr & x = 15625 \cr} $$ Hence, the numbers 15625

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, $$\frac{4}{11}$$ > $$\frac{9}{31}$$ > $$\frac{7}{25}$$ > $$\frac{6}{23}$$ > $$\frac{3}{17}$$

$$\eqalign{ & 0.\overline {142857} \div 0.\overline {285714} \cr & = \frac{{142857}}{{999999}} \div \frac{{285714}}{{999999}} \cr & = \left( {\frac{{142857}}{{999999}} \times \frac{{999999}}{{285714}}} \right) \cr & = \frac{1}{2} \cr} $$

Given expression : 58.621 - (13.829 + 7.302 + 1.214) = 58.621 - 22.345 = 36.276

Let original fraction be $$\frac{a}{b}$$ Now, according to the question, $$\eqalign{ & \Leftrightarrow \frac{{a - a \times \frac{{25}}{{100}}}}{{b + b \times \frac{{250}}{{100}}}} = \frac{6}{5} \cr & \Rightarrow \frac{{0.75a}}{{3.50b}} = \frac{6}{5} \cr & \Rightarrow \frac{a}{b} = \frac{6}{5} \times \frac{{3.50}}{{0.75}} \cr & \Rightarrow \frac{a}{b} = \frac{{6 \times 350 \times 100}}{{5 \times 75 \times 100}} \cr & \Rightarrow \frac{a}{b} = \frac{{28}}{5} \cr} $$

$$\frac{- 7}{10}$$ = - 0.7 $$\frac{5}{- 8}$$ = - $$\frac{5}{8}$$ = - 0.625, $$\frac{2}{- 3}$$ = - $$\frac{2}{3}$$ = - 0.66 Since, -0.7 < -0.66 < -0.625 So, $$\frac{- 7}{10}$$ < $$\frac{2}{- 3}$$ < $$\frac{5}{- 8}$$

Given expression : $$ = \frac{8 - 2.8}{1.3}$$ $$ = \frac{5.2}{1.3}$$ $$ = \frac{52}{13}$$ = 4

None of these is the correct answer to “0.3 + 3 + 3.33 + 3.3 + 3.03 + 333 =”. This question is standard in competitive examination syllabi for Decimal Fraction.

Given expression : = (11.71 + 1.78) - (0.86 + 9.20) = 13.49 - 10.06 = 3.43

9 × 7 = 63 Sum of decimal places = 5 ∴ 0.09 × 0.007 = 0.00063

Given expression : 71.808 + 95.76 = 167.568

$$\eqalign{ & = 1.\overline {27} \cr & = 1 + 0.\overline {27} \cr & = 1 + \frac{{27}}{{99}} \cr & = 1 + \frac{3}{{11}} \cr & = \frac{{11 + 3}}{{11}} \cr & = \frac{{14}}{{11}} \cr} $$

For the expressions to be equivalent, the difference between the sum of the decimal places in the numerator and that in the denominator must be equal. This difference is 1 in the given expression and 1 in (D). So, (D) is the answer.

$$\eqalign{ & = 0.4\overline {23} \cr & = \frac{{423 - 4}}{{990}} \cr & = \frac{{419}}{{990}} \cr} $$

Given expression : = 293.48 - 141.27 = 152.21

Given expression : $$\eqalign{ & = \frac{{5.32 \times \left( {56 + 44} \right)}}{{\left( {7.66 + 2.34} \right)\left( {7.66 - 2.34} \right)}} \cr & = \frac{{5.32 \times 100}}{{10 \times 5.32}} \cr & = 10 \cr} $$

$$\eqalign{ & = 0.29\overline {56} \cr & = \frac{{2956 - 29}}{{9900}} \cr & = \frac{{2927}}{{9900}} \cr} $$

Given = 1599 ÷ 39.99 + $$\frac{4}{5}$$ × 2449 - 120.05 ≈ 1600 ÷ 40 + $$\frac{4}{5}$$ × 2450 - 120 = 1600 ÷ 40 + 1960 - 120 = 40 + 1960 - 120 = 1880

None of these is the correct answer to “999.99 + 99.99 + 9.99 =”. This question is standard in competitive examination syllabi for Decimal Fraction.

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.