Number System - Aptitude MCQ Questions and Solutions with Explanations

For divisibility by 8, last 3 number should be divisible by 8.
Now, put the value 0, 1 ... 2 ... and check.
When use * = $$\frac{632}{8}$$ = 79
So, hence it is divisible by 8 answer is 3

→( 1, 3, 5, 7. . . . . 99 ) × 128

5	→	51
15	→	51
25	→	52
35	→	51
45	→	51
55	→	51
65	→	51
75	→	52
85	→	51
95	→	51
×
128	→	27

5¹² + 2⁷ will make zero but since 2 comes 7 times, so only 7 zero will come.

(0.1)2	→	.01
$$\sqrt {0.0121} $$	→	0.11
0.12	→	0.12	→ Largest
$$\sqrt {.0004} $$	→	0.02

Let the two number are a and b
According to question,
$$\eqalign{ & {a^2} + {b^2} = 289 \cr & ab{\text{ }} = {\text{ }}120 \cr & {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \cr & {\left( {a + b} \right)^2} = {\left( {289} \right)^2} + \left( {2 \times 120} \right) \cr & {\left( {a + b} \right)^2} = 529 \cr & \left( {a + b} \right) = 23 \cr} $$

$$\eqalign{ & {\left( {2153} \right)^{167}} \cr & {\text{unit digit = }}{{\text{3}}^{167}} \cr & {\text{unit digit }} \cr & {{\text{3}}^1} \to {\text{3}} \to {\text{3}} \cr & {{\text{3}}^2} \to 9 \to 9 \cr & {{\text{3}}^3} \to 27 \to 7 \cr & {{\text{3}}^4} \to 81 \to 1 \cr} $$
This cycle will continue
⇒ divide the power of 3 by 4
$$\eqalign{ & \frac{{167}}{4} \Rightarrow {\text{remainder is 3}} \cr & {{\text{3}}^3} \Rightarrow 7 \cr & {\text{unit digit = }}1 \times 7 = {\text{ }}7 \cr} $$

$$\eqalign{ & \sqrt {a + b\,} \,{\text{and }}\sqrt a + \sqrt b \cr & {\text{Squaring both sides}} \cr & {(\sqrt {a + b\,} )^2}{\text{and (}}\sqrt a + \sqrt b {)^2} \cr & \Rightarrow a + b\,\,{\text{and}}\,a + b + 2\sqrt a \sqrt b \cr & So\,\sqrt {a + b\,} \, < {\text{ }}\sqrt a + \sqrt b \cr} $$

$$\frac{7}{6},\,\frac{7}{9},\,\frac{4}{5},\,\frac{5}{7}$$
Step 1 : Compare two fractions
$$\frac{{7}}{{6}}$$ ⤩ $$\frac{{7}}{{9}}$$
Cross multiply
63 , 42
 ↑      ↑
$$\frac{7}{6}$$ ⤩ $$\frac{7}{9}$$
42 is smaller than 63
So, $$\frac{7}{9}$$ is smaller than $$\frac{7}{6}$$
Step 2 : Compare
(i) $$\frac{7}{9},\frac{4}{5}$$
(ii) $$\frac{7}{9}$$ ⤩ $$\frac{4}{5}$$
Cross multiply it
35 , 36
 ↑      ↑
$$\frac{7}{9}$$ ⤩ $$\frac{4}{5}$$
35 is smaller than 36
So, $$\frac{7}{9}$$ is smaller than $$\frac{4}{5}$$
Step 3 : Compare
(i) $$\frac{7}{9},\frac{5}{7}$$
(ii) $$\frac{7}{9}$$ ⤩ $$\frac{5}{7}$$
Cross multiply it
49 , 45
 ↑      ↑
$$\frac{7}{9}$$ ⤩ $$\frac{5}{7}$$
45 is smaller than 49
S0, $$\frac{5}{7}$$ is the smallest

Alternate :
$$\eqalign{ & \frac{7}{6} = 1.166, \cr & \frac{7}{6} = 0.777 \cr & \frac{4}{5} = 0.8, \cr & \frac{5}{{7\,}} = 0.714 \cr & {\text{Thus }}\frac{5}{7}{\text{ is the smallest}}{\text{.}} \cr} $$

Let, the number be 'x'
According to the question
$$\eqalign{ & \frac{1}{5}x - \frac{1}{7}x = 10 \cr & \frac{{7x - 5x}}{{35}} = 10 \cr & 2x = 350 \cr & x = \frac{{350}}{2} \cr & x = 175 \cr} $$

$$\eqalign{ & {\text{Let the one number be }}x \cr & {\text{And the other number be }}\frac{1}{5}x \cr & {\text{According to question}} \cr & x \times \frac{1}{5}x = 0.008 \cr & {x^2} = 0.040 \cr & {\text{ }}x = 0.2 \cr & {\text{Other number}} \cr & = \frac{1}{5} \times 0.2\, \cr & = 0.04 \cr & {\text{Smaller number is 0}}{\text{.04}} \cr} $$

Always do these types of question by option to save time
Pick up the option and follow the the question instruction
Take option 3
64 ⇒ Divided 3 it gives remainder 1
Now add 1 to 64
$$\frac{{65}}{5}$$ ⇒ remainder '0' is satisfies
So K = 64 this is answer