Decimal Fraction - Aptitude MCQ Questions and Solutions with Explanations

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

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

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

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