Number System - Aptitude MCQ Questions and Solutions with Explanations

$${\text{0}}{\text{.1 = }}\frac{1}{{10}}\left( {{\text{mud}}} \right),\,\,\,\,\frac{5}{8}\left( {{\text{water}}} \right)$$
Let total length of bamboo = 40x ( LCM of 8, 10 )
$$\eqalign{ & {\text{Mud part :}} \cr & {\text{ = }}\frac{1}{{10}} \times 40x = 4x \cr & {\text{Water part :}} \cr & {\text{ = }}\frac{5}{8} \times 40x = 25x \cr & {\text{Remaining part :}} \cr & = (40x - 25x - 4x) \cr & = 11x \cr & 11x{\text{ }} \to {\text{ }}2.75m \cr & {\text{ }}x{\text{ }} \to {\text{ }}0.25m \cr & 40x{\text{ }} \to {\text{ }}10m \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{(1001)^{2008}} + (1002) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr & {\text{unit digit}} \to {\text{ }}{{\text{1}}^{2008}} + 1002 \cr} $$
Unit digit will be 1 in case of 1 respective of power
⇒ 1 + 1002 = 1003
⇒ 3 unit digit (last digit)

$$\eqalign{ & {\text{0}}{\text{.121212}}........ \cr & \Rightarrow 0.\overline {12} = \frac{{12}}{{99}} = \frac{4}{{33}} \cr} $$

$$\eqalign{ & \frac{6}{{\frac{7}{8}}} = \frac{{6 \times 8}}{7} = \frac{{48}}{7} \cr & \frac{{\frac{6}{7}}}{8} = \frac{6}{{7 \times 8}} = \frac{6}{{56}} = \frac{3}{{28}} \cr & {\text{Difference - }} \cr & = \frac{{48}}{7} - \frac{3}{{28}} \cr & = \frac{{192 - 3}}{{28}} \cr & = \frac{{189}}{{28}} \cr & = \frac{{27}}{4} \cr & = 6\frac{3}{4} \cr} $$

$$\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right)\left( {1 - \frac{1}{5}} \right)$$     . . . . .$$\left( {1 - \frac{1}{{99}}} \right)$$ $$\left( {1 - \frac{1}{{100}}} \right)$$
$$\eqalign{ & = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times ..... \times \frac{{98}}{{99}} \times \frac{{99}}{{100}} \cr & = \frac{2}{{100}} \cr & = \frac{1}{{50}} \cr} $$

(xyz , xyz) is number
⇒ 100000x + 10000y + 1000z + 100x + 10y + z
⇒ 1001 (100x + 10y + z)
Since 1001 has factors i.e., 7, 11 and 13
So, the answer is 1001

$$\eqalign{ & {\text{By option }}\frac{8}{9},{\text{ }} \cr & {\text{Proper fraction = }}\frac{8}{9} \cr & {\text{Its reciprocal = }}\frac{9}{8} \cr & {\text{According to question}} \cr & = \frac{9}{8} - \frac{8}{9} \cr & = \frac{{81 - 64}}{{72}} \cr & = \frac{{17}}{{72}}{\text{ satisfied}} \cr} $$

Let the 3 consecutive number is
x, x + 1, x + 2
According to question,
$$ \Rightarrow {x^2} + {\left( {x + 1} \right)^2} + {\left( {x + 2} \right)^2}$$     = 365
$$ \Rightarrow {x^2} + {x^2} + 1 + 2x + {x^2} + 4 + 4x$$       = 365
$$\eqalign{ & \Rightarrow 3{x^2} + 6x = 360 \cr & \Rightarrow {x^2} + 2x - 120 = 0 \cr & \Rightarrow \left( {x - 10} \right)\left( {x + 12} \right) = 0 \cr & \Rightarrow x = 10 \cr & {\text{Sum of numbers :}} \cr & = 10 + 11 + 12 \cr & = 33 \cr} $$

= 142² - 1
= (142 - 1) (142 + 1) [a² - b² = (a - b) (a + b)]
= 141 × 143
= 141 × 11 × 13

Let the number is a, b, c
a = 3b       $$\frac{a}{b}$$ = $$\frac{3}{1}$$
c = $$\frac{2}{3}$$a       $$\frac{c}{a}$$ = $$\frac{2}{3}$$

c	a	b
2	3
	3	1

c : a : b = 2 : 3 : 1
So, (2 + 3 + 1) = 252 units
6 units → 252
1 unit → 42
3 units → 42 × 3
So, second number is = 42