Practice MCQ Questions and Answer on Decimal Fraction

11.

The value of $$\frac{{{{\left( {0.06} \right)}^2} + {{\left( {0.47} \right)}^2} + {{\left( {0.079} \right)}^2}}}{{{{\left( {0.006} \right)}^2} + {{\left( {0.047} \right)}^2} + {{\left( {0.0079} \right)}^2}}}$$       is :

• (A) 0.1
• (B) 10
• (C) 100
• (D) 1000

Show Answer

 Given expression : $$ = \frac{{{a^2} + {b^2} + {c^2}}}{{{{\left( {\frac{a}{{10}}} \right)}^2} + {{\left( {\frac{b}{{10}}} \right)}^2} + {{\left( {\frac{c}{{10}}} \right)}^2}}}$$ Where a = 0.06, b = 0.47 and c = 0.079 $$\eqalign{ & = \frac{{100\left( {{a^2} + {b^2} + {c^2}} \right)}}{{\left( {{a^2} + {b^2} + {c^2}} \right)}} \cr & = 100 \cr} $$

12.

10.0001 + 9.9999 - 8.9995 = ?

• (A) 9.0005
• (B) 10.9995
• (C) 11.0001
• (D) 11.0005

Show Answer

 10.0001 + 9.9999 - 8.9995 = 20.0000 - 8.9995 = 11.0005

13.

3889 + 12.952 - ? = 3854.002

• (A) 47.095
• (B) 47.752
• (C) 47.932
• (D) 47.95

Show Answer

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

14.

4.036 divided by 0.04 gives :

• (A) 1.009
• (B) 10.09
• (C) 100.9
• (D) None of these

Show Answer

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

15.

If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?

• (A) 0.172
• (B) 1.72
• (C) 17.2
• (D) 172

Show Answer

 = $$\frac{29.94}{1.45}$$ = $$\frac{299.4}{14.5}$$ = $$\frac{2994}{14.5}$$ × $$\frac{1}{10}$$ = $$\frac{172}{10}$$ = 17.2

16.

Solve : 1576 ÷ 45.02 + 23.99 × $$\sqrt {255} $$  = ?

• (A) 340
• (B) 420
• (C) 380
• (D) 460

Show Answer

 Given expression : = 1576 ÷ 45.02 + 23.99 × $$\sqrt {255} $$ = 1575 ÷ 45 + 24 × $$\sqrt {256} $$ = 35 + 24 × 16 = 35 + 384 = 419 $$ \approx $$ 420

17.

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 ?

• (A) 2010
• (B) 2011
• (C) 2012
• (D) 2013

Show Answer

 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

18.

555.05 + 55.5 + 5.55 + 5 + 0.55 = ?

• (A) 621.65
• (B) 634.85
• (C) 647.35
• (D) 655.75

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19.

The numerator of a fraction is decreased by 25% and the denominator is increased by 250%. If the resultant fraction is $$\frac{6}{5}$$, what is the original fraction ?

• (A) $$\frac{22}{5}$$
• (B) $$\frac{24}{5}$$
• (C) $$\frac{27}{6}$$
• (D) $$\frac{28}{5}$$

Show Answer

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

20.

The number 0.121212 ..... in the form $$\frac{p}{q}$$ is equal to :

• (A) $$\frac{2}{11}$$
• (B) $$\frac{4}{11}$$
• (C) $$\frac{2}{33}$$
• (D) $$\frac{4}{33}$$

Show Answer

 $$\eqalign{ & 0.121212...... \cr & = 0.\overline {12} \cr & = \frac{{12}}{{99}} \cr & = \frac{4}{{33}} \cr} $$