Practice MCQ Questions and Answer on Decimal Fraction

141.

The value of (1.25)³ - 2.25 (1.25)² + 3.75 (0.75)² - (0.75)³ is :

• (A) 1
• (B) $$\frac{1}{2}$$
• (C) $$\frac{1}{4}$$
• (D) $$\frac{1}{8}$$

Show Answer

 Given expression : (1.25)3 - (0.75)3 - 3 × (1.25)2 × 0.75 + 3 × 1.25 × (0.75)2 = (1.25 - 0.75)3 = (0.5)3 = $${\left( {\frac{1}{2}} \right)^3}$$ = $$\frac{1}{8}$$ [∵ (a - b)3 = a3 - b3 - 3a2 b + 3ab2]

142.

4.036 divided by 0.04 gives :

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

Show Answer

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

143.

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

144.

The vulgar fraction of $$\text{0}\text{.39}\overline{\text{39}}$$   is ?

• (A) $$\frac{15}{33}$$
• (B) $$\frac{11}{39}$$
• (C) $$\frac{17}{39}$$
• (D) $$\frac{13}{33}$$

Show Answer

 The given expression can be written in this form also N = $${\text{ 0}}{\text{.39}}\overline {{\text{39}}} $$ ..... (i) Multiply equation (i) with 100 on both sides. 100N = $${\text{39}}{\text{.}}\overline {{\text{39}}} $$ ..... (ii) Subtracting equation (i) from (ii) we get, ⇒ 100N - N = $${\text{39}}\overline {{\text{.39}}} $$  - $${\text{0}}\overline {{\text{.39}}} $$ ⇒ 99N = 39 ⇒ N = $$\frac{39}{99}$$ = $$\frac{13}{33}$$

145.

Given, 168 × 32 = 5376 , then 5.376 ÷ 16.8 is equal to :

• (A) 0.032
• (B) 0.32
• (C) 3.2
• (D) 32

Show Answer

 Given, 168 × 32 = 5376 or 5376 ÷ 168 = 32 Now, $$\eqalign{ & = \frac{{5.376}}{{16.8}} \cr & = \frac{{53.76}}{{168}} \cr & = \left( {\frac{{5376}}{{168}} \times \frac{1}{{100}}} \right) \cr & = \frac{{32}}{{100}} \cr & = 0.32 \cr} $$

146.

Solve $$\frac{3}{5}$$ of $$\frac{4}{7}$$ of $$\frac{5}{12}$$ of 1015 = ?

• (A) 220
• (B) 340
• (C) 240
• (D) 145

Show Answer

 Given $$\frac{3}{5}$$ of $$\frac{4}{7}$$ of $$\frac{5}{12}$$ of 1015 ⇒ x = $$\frac{3}{5}$$ × $$\frac{4}{7}$$ × $$\frac{5}{12}$$ × 1015 ⇒ x = 145

147.

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