Practice MCQ Questions and Answer on Decimal Fraction

91.

$$\frac{96.54-89.63}{96.54+89.63}$$   ÷ $$\frac{965.4-896.3}{9.654+8.963}$$   = ?

• (A) 10- 2
• (B) 10- 1
• (C) 10
• (D) None of these

Show Answer

 Given expression : = $$\frac{(96.54 - 89.63)}{(96.54 + 89.63)}$$   × $$\frac{(9.654 + 8.963)}{(965.4 - 896.3)}$$ = $$\frac{(96.54 - 89.63)}{(965.4 - 896.3)}$$   × $$\frac{(9.654 + 8.963)}{(96.54 + 89.63)}$$ = $$\frac{(96.54 - 89.63)}{10 (96.54 - 89.63)}$$   × $$\frac{(96.54 + 89.63)}{10 (96.54 + 89.63)}$$ = $$\frac{1}{10}$$ × $$\frac{1}{10}$$ = $$\frac{1}{100}$$ = $$\frac{1}{{{{10}^2}}}$$ = 10 -2

92.

The rational numbers lying between $$\frac{1}{3}$$ and $$\frac{3}{4}$$ are :

• (A) $$\frac{117}{300},\frac{287}{400}$$
• (B) $$\frac{95}{300},\frac{301}{400}$$
• (C) $$\frac{99}{300},\frac{301}{400}$$
• (D) $$\frac{97}{300},\frac{299}{500}$$

Show Answer

 $$\eqalign{ & \frac{1}{3} = 0.333 \cr & \frac{3}{4} = 0.75 \cr & \frac{{117}}{{300}} = 0.39 \cr & \frac{{287}}{{400}} = 0.7175 \cr & \frac{{95}}{{300}} = 0.316 \cr & \frac{{301}}{{400}} = 0.7525 \cr & \frac{{99}}{{300}} = 0.33 \cr & \frac{{97}}{{300}} = 0.323 \cr & \frac{{299}}{{500}} = 0.598 \cr} $$ 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 $$\frac{3}{4}$$

93.

Which of the following fractions is greater than $$\frac{3}{4}$$ and less than $$\frac{5}{6}$$ ?

• (A) $$\frac{1}{2}$$
• (B) $$\frac{2}{3}$$
• (C) $$\frac{4}{5}$$
• (D) $$\frac{9}{10}$$

Show Answer

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

94.

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

95.

11.71 - 0.86 + 1.78 - 9.20 = ?

• (A) 2.43
• (B) 3.13
• (C) 3.43
• (D) 4.13

Show Answer

 Given expression : = (11.71 + 1.78) - (0.86 + 9.20) = 13.49 - 10.06 = 3.43

96.

$$\begin{array}{rl}& \text{The}\text{value}\text{of}\\ & \frac{0.1\times 0.1\times 0.1+0.02\times 0.02\times 0.02}{0.2\times 0.2\times 0.2+0.04\times 0.04\times 0.04}\text{is}:\end{array}$$

• (A) 0.0125
• (B) 0.125
• (C) 0.25
• (D) 0.5

Show Answer

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

97.

When 52416 is divided by 312, the quotient is 168. What will be the quotient when 52.416 is divided by 0.0168 ?

• (A) 3.12
• (B) 312
• (C) 3120
• (D) None of these

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 $$\eqalign{ & {\text{Given,}} \cr & \frac{{52416}}{{312}} = 168 \cr & \Leftrightarrow \frac{{52416}}{{168}} = 312 \cr & {\text{Now,}} \cr & \frac{{52.416}}{{0.0168}} \cr & = \frac{{524160}}{{168}} \cr & = \frac{{52416}}{{168}} \times 10 \cr & = 312 \times 10 \cr & = 3120 \cr} $$

98.

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

99.

Solve $${\left(\frac{18}{4}\right)}^2\times$$ $$\left(\frac{455}{19}\right)÷$$  $$\left(\frac{61}{799}\right)=?$$

• (A) 6320
• (B) 6400
• (C) 6350
• (D) 6430

Show Answer

 $$\eqalign{ & {\left( {\frac{{18}}{4}} \right)^2} \times \left( {\frac{{455}}{{19}}} \right) \div \left( {\frac{{61}}{{799}}} \right) \cr & = \frac{{324}}{{16}} \times \frac{{455}}{{19}} \times \frac{{799}}{{61}} \cr & = 6350 \cr} $$

100.

$$\left(8.3\bar{1}+0.\bar{6}+0.00\bar{2}\right)$$    is equal to :

• (A) $$8.9\overline{12}$$
• (B) $$8.\overline{912}$$
• (C) $$8.9\overline{79}$$
• (D) $$8.97\bar{9}$$

Show Answer

 $$\eqalign{ & \left( {8.3\bar 1 + 0.\bar 6 + 0.00\bar 2} \right) \cr & = 8 + \frac{{31 - 3}}{{90}} + \frac{6}{9} + \frac{2}{{900}} \cr & = \frac{{7200 + 280 + 600 + 2}}{{900}} \cr & = \frac{{8082}}{{900}} \cr & = 8\frac{{882}}{{900}} \cr & = 8 + \frac{{979 - 97}}{{900}} \cr & = 8.97\bar 9 \cr} $$