Decimal Fraction — Part 1 MCQs with Explanations

$$\eqalign{ & \text{Let } a = 2.39 \text{ and } b = 1.61 \cr & {\text{Given Expression}} \cr & = \frac{{{a^2} - {b^2}}}{{a - b}} \cr & = \frac{{\left( {a + b} \right)\left( {a - b} \right)}}{{ {a - b} }} \cr & = {a + b} \cr & = {2.39 + 1.61} \cr & = 4 \cr} $$

$$\eqalign{ & {\text{Required decimal}} \cr & = \frac{1}{{60 \times 60}} \cr & = \frac{1}{{3600}} \cr & = .00027 \cr} $$

$$\eqalign{ & {\text{Given}}\,{\text{expression}} \cr & = \frac{{{{\left( {0.96} \right)}^3} - {{\left( {0.1} \right)}^3}}}{{{{\left( {0.96} \right)}^2} + \left( {0.96 \times 0.1} \right) + {{\left( {0.1} \right)}^2}}} \cr & = {\frac{{{a^3} - {b^3}}}{{{a^2} + ab + {b^2}}}} \cr & = \frac{\left(a-b\right) \left(a^2 + ab + b^2 \right)}{\left(a^2 + ab + b^2 \right)} \cr & = {a - b} \cr & = {0.96 - 0.1} \cr & = 0.86 \cr} $$

$$\eqalign{ & {\text{Give}}\,{\text{expression}} \cr & = \frac{{{{\left( {0.1} \right)}^3} + {{\left( {0.02} \right)}^3}}}{{{2^3}\left[ {{{\left( {0.1} \right)}^3} + {{\left( {0.02} \right)}^3}} \right]}} \cr & = \frac{1}{8} \cr & = 0.125 \cr} $$

$$\eqalign{ & \frac{{29.94}}{{1.45}} \cr & = \frac{{299.4}}{{14.5}} \cr} $$ $$ = {\frac{{2994}}{{14.5}} \times \frac{1}{{10}}} $$ [Here, Substitute 172 in the place of $${\frac{{2994}}{{14.5}}}$$ ] $$\eqalign{ & = \frac{{172}}{{10}} \cr & = 17.2 \cr} $$

$$0.232323... = 0.\overline {23} = \frac{{23}}{{99}}$$

$$\eqalign{ & {\text{Let}}\,\frac{{.009}}{x} = .01; \cr & {\text{Then}}\,x = \frac{{.009}}{{.01}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{.9}}{1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = .9 \cr} $$

Given expression = (11.98) 2 + (0.02) 2 + 11.98 × X For the given expression to be a perfect square, (a + b) 2 = a 2 + b 2 + 2ab we must have 11.98 × X = 2 × 11.98 × 0.02 ∴ X = 0.04

$$\eqalign{ & {\text{Given}}\,{\text{expression}} \cr & = \frac{{ {0.3333} }}{{ {0.2222} }} \times \frac{{\left( {0.1667} \right)\left( {0.8333} \right)}}{{\left( {0.6667} \right)\left( {0.1250} \right)}} \cr & = \frac{{3333}}{{2222}} \times \frac{{\frac{1}{6} \times \frac{5}{6}}}{{\frac{2}{3} \times \frac{{125}}{{1000}}}} {\text{ [where }} {0.3333} = {\frac{1}{6}}, {0.8333} = {\frac{5}{6}} {\text{ and }} {0.6667} = {\frac{2}{3}} {\text{ ] }} \cr & = {\frac{3}{2} \times \frac{1}{6} \times \frac{5}{6} \times \frac{3}{2} \times 8} \cr & = \frac{5}{2} \cr & = 2.50 \cr} $$

Let 3889 + 12.952 - x = 3854.002. Then x = (3889 + 12.952) - 3854.002 = 3901.952 - 3854.002 = 47.95

4 x 162 = 648. Sum of decimal places = 6. So, 0.04 x 0.0162 = 0.000648 = 6.48 x 10 -4

$$\eqalign{ & \text{Let } a = 4.2 \text{ and } b = 1.9 \cr & {\text{Given Expression}} \cr & = \frac{{ {{a^2} - {b^2}} }}{{\left( {a + b} \right)\left( {a - b} \right)}} \cr & = \frac{{ {{a^2} - {b^2}} }}{{ {{a^2} - {b^2}} }} \cr & = 1 \cr} $$

$$\eqalign{ & \frac{{144}}{{0.144}} = \frac{{14.4}}{x} \cr & \Rightarrow \frac{{144 \times 1000}}{{144}} = \frac{{14.4}}{x} \cr & \Rightarrow x = \frac{{14.4}}{{1000}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.0144 \cr} $$

Suppose commodity X will cost 40 paise more than Y after z years Then,(4.20 + 0.40z) − (6.30 + 0.15z) = 0.40 ⇒ 0.25z = 0.40 + 2.10 $$\eqalign{ & \Rightarrow z = \frac{{2.50}}{{0.25}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{250}}{{25}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 10 \cr} $$ ∴ X will cost 40 paise more than Y 10 years After 2001 i.e., 2011

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, $$\frac{6}{7}$$ > $$\frac{5}{6}$$ > $$\frac{4}{5}$$ > $$\frac{3}{7}$$ > $$\frac{2}{5}$$ > $$\frac{1}{3}$$

$$\eqalign{ & \frac{3}{4} = 0.75,{\kern 1pt} \cr & \frac{5}{6} = 0.833,{\kern 1pt} \cr & \frac{1}{2} = 0.5, \cr & \frac{2}{3} = 0.66, \cr & {\kern 1pt} \frac{4}{5} = 0.8, \cr & \frac{9}{{10}} = 0.9 \cr} $$ Clearly, 0.8 lies between 0.75 and 0.833 $$\therefore \frac{4}{5}{\text{lies between}}\frac{3}{4}{\text{and}}\frac{5}{6}$$

$$0.125125... = 0.\overline {125} = \frac{{125}}{{999}}$$

$$\eqalign{ & 617.00 \cr & \,\,\,\,\,\,\,6.017 \cr & \,\,\,\,\,\,\,0.617 \cr & + \,\,\,6.0017 \cr & - - - - - - \cr & \,\,\,629.6357 \cr & - - - - - - \cr} $$

$$\eqalign{ & \frac{{489.1375 \times 0.0483 \times 1.956}}{{0.0873 \times 92.581 \times 99.749}} \approx \frac{{489 \times 0.05 \times 2}}{{0.09 \times 93 \times 100}} \cr & = \frac{{489}}{{9 \times 93 \times 10}} \cr & = \frac{{163}}{{279}} \times \frac{1}{{10}} \cr & = \frac{{0.58}}{{10}} \cr & = 0.058 \approx 0.06 \cr} $$

2 x 5 = 10. Sum of decimal places = 4 ∴ 0.002 x 0.5 = 0.001

$$\eqalign{ & \,\,\,\,34.95 \cr & \,240.016 \cr & + 23.98 \cr & - - - - - \cr & 298.946 \cr & - - - - - \cr} $$

3.14 x 10 6 = 3.14 x 1000000 = 3140000

$$\eqalign{ & 1 \div 0.2 = \frac{1}{{0.2}} = \frac{{10}}{2} = 5; \cr & 0.\overline 2 = 0.222...; \cr & {\left( {0.2} \right)^2} = 0.04 \cr & 0.04 < 0.2 < 0.22.... < 5 \cr & {\text{Since}}\,0.04\,{\text{is}}\,{\text{the}}\,{\text{least,}}\,{\text{so}}\,{\left( {0.2} \right)^2}\,{\text{is}}\,{\text{the}}\,{\text{least}}. \cr} $$

$$\eqalign{ & {\text{Given Expression}} \cr & = \frac{{8 - 2.8}}{{1.3}} \cr & = \frac{{5.2}}{{1.3}} \cr & = \frac{{52}}{{13}} \cr & = 4 \cr} $$

Sum of decimal places = 7. Since the last digit to the extreme right will be zero (since 5 x 4 = 20), so there will be 6 significant digits to the right of the decimal point.

$$\eqalign{ & 6.\overline {46} \cr & = 6 + 0.\overline {46} \cr & = 6 + \frac{{46}}{{99}} \cr & = \frac{{594 + 46}}{{99}} \cr & = \frac{{640}}{{99}} \cr} $$

$$\eqalign{ & 101\frac{{27}}{{100000}} \cr & = 101 + \frac{{27}}{{100000}} \cr & = 101 + .00027 \cr & = 101.00027 \cr} $$

$$\eqalign{ & \frac{{0.0203 \times 2.92}}{{0.0073 \times 14.5 \times 0.7}} \cr & = \frac{{203 \times 292}}{{73 \times 145 \times 7}} \cr & = \frac{4}{5} \cr & = 0.8 \cr} $$

$$\eqalign{ & \frac{{4.036}}{{0.04}} \cr & = \frac{{403.6}}{4} \cr & = 100.9 \cr} $$

$$\eqalign{ & 3.\overline {87} - 2.\overline {59} \cr & = \left( {3 + 0.\overline {87} } \right) - \left( {2 + 0.\overline {59} } \right) \cr & = \left( {3 + \frac{{87}}{{99}}} \right) - \left( {2 + \frac{{59}}{{99}}} \right) \cr & = 1 + \left( {\frac{{87}}{{99}} - \frac{{59}}{{99}}} \right) \cr & = 1 + \frac{{28}}{{99}} \cr & = 1.\overline {28} \cr} $$

$$ = 7\frac{1}{2} - $$ $$\left[ {2\frac{1}{4} \div \left\{ {1\frac{1}{4} - \frac{1}{2}\left( {1\frac{1}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right]$$ $$ = \frac{{15}}{2} - $$ $$\left[ {\frac{9}{4} \div \left\{ {\frac{5}{4} - \frac{1}{2}\left( {\frac{3}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right]$$ $$ = \frac{{15}}{2} - $$ $$\left[ {\frac{9}{4} \div \left\{ {\frac{5}{4} - \frac{1}{2}\left( {\frac{{9 - 2 - 1}}{6}} \right)} \right\}} \right]$$ $$\eqalign{ & = \frac{{15}}{2} - \left[ {\frac{9}{4} \div \left\{ {\frac{5}{4} - \frac{1}{2}} \right\}} \right] \cr & = \frac{{15}}{2} - \left[ {\frac{9}{4} \div \left\{ {\frac{{5 - 2}}{4}} \right\}} \right] \cr & = \frac{{15}}{2} - \left[ {\frac{9}{4} \div \frac{3}{4}} \right] \cr & = \frac{{15}}{2} - \left[ {\frac{9}{4} \times \frac{4}{3}} \right] \cr & = \frac{{15}}{2} - 3 \cr & = \frac{{15 - 6}}{2} \cr & = \frac{9}{2} \cr & = 4\frac{1}{2} \cr} $$

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, $$\frac{11}{14}$$ < $$\frac{16}{19}$$ < $$\frac{19}{21}$$

-The first digit after the decimal point represents tenths (1/10). -The second digit represents hundredths (1/100). -The third digit represents thousandths (1/1000), and so on. In the number 0.06945: - 0 is in the tenths place. - 6 is in the hundredths place. - 9 is in the thousandths place. Therefore, the place value of 9 in 0.06945 is 9/1000 (nine thousandths). So, the correct answer is Option D: 9/1000 .

$$\eqalign{ & \frac{1}{3} = 0.333 \cr & \frac{3}{4} = 0.75 \cr & \frac{{117}}{{300}} = 0.39 \cr & \frac{{287}}{{400}} = 0.7175 \cr & \frac{{95}}{{300}} = 0.316 \cr & \frac{{301}}{{400}} = 0.7525 \cr & \frac{{99}}{{300}} = 0.33 \cr & \frac{{97}}{{300}} = 0.323 \cr & \frac{{299}}{{500}} = 0.598 \cr} $$ Clearly, each one of 0.39 and 0.7175 lies between 0.333 and 0.75 So, $$\frac{117}{300}$$ and $$\frac{287}{400}$$ lie between $$\frac{1}{3}$$ and $$\frac{3}{4}$$

$$\eqalign{ & = \frac{{3.157 \times 4126 \times 3.198}}{{63.972 \times 2835.121}} \cr & \approx \frac{{3.2 \times 4126 \times 3.2}}{{64 \times 2835}} \cr & = \frac{{32 \times 4126 \times 32}}{{64 \times 2835}} \times \frac{1}{{100}} \cr & = \frac{{66016}}{{2835}} \times \frac{1}{{100}} \cr & = \frac{{23.28}}{{100}} \cr & = 0.23 \cr & \approx 0.2 \cr} $$

477 × 124 × 86 = 5086728 Sum of decimal place = 3 ∴ 47.7 × 12.4 × 8.6 = 5086.728

Given, 168 × 32 = 5376 or 5376 ÷ 168 = 32 Now, $$\eqalign{ & = \frac{{5.376}}{{16.8}} \cr & = \frac{{53.76}}{{168}} \cr & = \left( {\frac{{5376}}{{168}} \times \frac{1}{{100}}} \right) \cr & = \frac{{32}}{{100}} \cr & = 0.32 \cr} $$

Converting each of the given fractions into decimal form, we get : $$\frac{2}{3}$$ = 0.66 $$\frac{3}{5}$$ = 0.6 $$\frac{7}{9}$$ = 0.77 $$\frac{9}{11}$$ = 0.81 $$\frac{8}{9}$$ = 0.88 Clearly, 0.6 < 0.66 < 0.77 < 0.81 < 0.88 So, $$\frac{3}{5}$$ < $$\frac{2}{3}$$ < $$\frac{7}{9}$$ < $$\frac{9}{11}$$ < $$\frac{8}{9}$$

Given expression : $$\eqalign{ & = \frac{{{{\left( {0.1} \right)}^3} + {{\left( {0.02} \right)}^3}}}{{{2^3}\left[ {{{\left( {0.1} \right)}^3} + {{\left( {0.02} \right)}^3}} \right]}} \cr & = \frac{1}{8} \cr & = 0.125 \cr} $$

$$\eqalign{ & 0.125125.... \cr & = 0.\overline {125} \cr & = \frac{{125}}{{999}} \cr} $$