Practice MCQ Questions and Answer on Decimal Fraction

81.

$$\frac{.009}{?}=.01$$

• (A) .0009
• (B) .09
• (C) .9
• (D) 9

Show Answer

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

82.

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 $$\eqalign{ & = \frac{{100\left( {{a^2} + {b^2} + {c^2}} \right)}}{{\left( {{a^2} + {b^2} + {c^2}} \right)}} \cr & = 100 \cr} $$

83.

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

84.

32.4 × 11.5 × 8.5 = ?

• (A) 3149.5
• (B) 3129.1
• (C) 3167.1
• (D) 3162.5

Show Answer

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

85.

Which of the following is equal to 3.14 x 10⁶ ?

• (A) 314
• (B) 3140
• (C) 3140000
• (D) None of these

Show Answer

 3.14 x 106 = 3.14 x 1000000 = 3140000

86.

If $$\frac{144}{0.144}$$ = $$\frac{14.4}{x},$$  then the value of $$x$$ is :

• (A) 0.0144
• (B) 1.44
• (C) 14.4
• (D) 144

Show Answer

 $$\frac{144}{0.144}$$ = $$\frac{14.4}{x}$$ ⇒ $$\frac{144 × 1000}{144}$$   = $$\frac{14.4}{x}$$ ⇒ x = $$\frac{14.4}{1000}$$ ⇒ x = 0.0144

87.

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

88.

4.036 divided by 0.04 gives :

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

Show Answer

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

89.

$$5.5-\left[6.5-\left\{3.5÷\left(6.5-\overline{5.5-2.5}\right)\right\}\right]$$       is equal to :

• (A) - 1
• (B) 0
• (C) 0.1
• (D) 1

Show Answer

 Given expression : = 5.5 - [6.5 - {3.5 ÷ (6.5 - 3)}] = 5.5 - [6.5 - {3.5 ÷ 3.5}] = 5.5 - [6.5 - 1] = 5.5 - 5.5 = 0

90.

The least among the following is:

• (A) 0.2
• (B) 1 ÷ 0.2
• (C) $$0.\bar{2}$$
• (D) (0.2)2

Show Answer

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