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Decimal Fraction - Aptitude MCQ Questions and Solutions with Explanations

Decimal Fraction MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Decimal Fraction with easy and logical explanations. Decimal Fraction MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-2 section-1

Each section contains 100 MCQs on Decimal Fraction .
11
0.04 x 0.0162 is equal to:
Discuss
Answer & Solution
Answer: Option B
Solution:
4 x 162 = 648. Sum of decimal places = 6.
So,
0.04 x 0.0162
= 0.000648
= 6.48 x 10-4
12
$$\frac{{4.2 \times 4.2 - 1.9 \times 1.9}}{{2.3 \times 6.1}}$$    is equal to:
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & \text{Let } a = 4.2 \text{ and } b = 1.9 \cr & {\text{Given Expression}} \cr & = \frac{{ {{a^2} - {b^2}} }}{{\left( {a + b} \right)\left( {a - b} \right)}} \cr & = \frac{{ {{a^2} - {b^2}} }}{{ {{a^2} - {b^2}} }} \cr & = 1 \cr} $$
13
If $$\frac{{144}}{{0.144}} = \frac{{14.4}}{x},$$   then the value of x is:
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & \frac{{144}}{{0.144}} = \frac{{14.4}}{x} \cr & \Rightarrow \frac{{144 \times 1000}}{{144}} = \frac{{14.4}}{x} \cr & \Rightarrow x = \frac{{14.4}}{{1000}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.0144 \cr} $$
14
The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
Discuss
Answer & Solution
Answer: Option B
Solution:
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
15
Which of the following are in descending order of their value ?
Discuss
Answer & Solution
Answer: Option D
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, $$\frac{6}{7}$$ > $$\frac{5}{6}$$ > $$\frac{4}{5}$$ > $$\frac{3}{7}$$ > $$\frac{2}{5}$$ > $$\frac{1}{3}$$
16
Which of the following fractions is greater than $$\frac{{3}}{{4}}$$ and less than $$\frac{{5}}{{6}}$$ ?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & \frac{3}{4} = 0.75,{\kern 1pt} \cr & \frac{5}{6} = 0.833,{\kern 1pt} \cr & \frac{1}{2} = 0.5, \cr & \frac{2}{3} = 0.66, \cr & {\kern 1pt} \frac{4}{5} = 0.8, \cr & \frac{9}{{10}} = 0.9 \cr} $$
Clearly, 0.8 lies between 0.75 and 0.833
$$\therefore \frac{4}{5}{\text{lies between}}\frac{3}{4}{\text{and}}\frac{5}{6}$$
17
The rational number for recurring decimal 0.125125.... is:
Discuss
Answer & Solution
Answer: Option C
Solution:
$$0.125125... = 0.\overline {125} = \frac{{125}}{{999}}$$
18
617 + 6.017 + 0.617 + 6.0017 = ?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & 617.00 \cr & \,\,\,\,\,\,\,6.017 \cr & \,\,\,\,\,\,\,0.617 \cr & + \,\,\,6.0017 \cr & - - - - - - \cr & \,\,\,629.6357 \cr & - - - - - - \cr} $$
19
The value of $$\frac{{489.1375 \times 0.0483 \times 1.956}}{{0.0873 \times 92.581 \times 99.749}}$$     is closest to:
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & \frac{{489.1375 \times 0.0483 \times 1.956}}{{0.0873 \times 92.581 \times 99.749}} \approx \frac{{489 \times 0.05 \times 2}}{{0.09 \times 93 \times 100}} \cr & = \frac{{489}}{{9 \times 93 \times 10}} \cr & = \frac{{163}}{{279}} \times \frac{1}{{10}} \cr & = \frac{{0.58}}{{10}} \cr & = 0.058 \approx 0.06 \cr} $$
20
0.002 x 0.5 = ?
Discuss
Answer & Solution
Answer: Option B
Solution:
2 x 5 = 10.
Sum of decimal places = 4
∴ 0.002 x 0.5 = 0.001