Number System - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & x + \left( {x + 2} \right) + \left( {x + 4} \right) = 54 \cr & 3x + 6 = 54 \cr & 3x = 48 \cr & x = 16 \cr & \cr & {\bf{Alternate: }} \cr & {\text{Middle number = }}\frac{{54}}{3} = 18 \cr & {\text{then number is 16,18,20}} \cr & {\text{Smallest number is 16}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & x \times 7 = 333333 \cr} $$
(In answer 5 digit number are given, so we take 6 digit 333333)
$$\eqalign{ & x = \frac{{333333}}{7} \cr & x = 47619 \cr} $$

$$\eqalign{ & \Rightarrow \left( {{n^3} - n} \right)\left( {n - 2} \right){\text{put n = 3}} \cr & \Rightarrow \left( {{3^3} - 3} \right)\left( {3 - 2} \right) \cr & \Rightarrow \left( {27 - 3} \right) \times 1 \cr & \Rightarrow 24 \cr & {\text{It is divisible by }}24 \cr} $$

Dividend = divisor × quotient + remainder
$$\eqalign{ & {\text{First number }} \cr & \Rightarrow {\text{ }}\left( {17 \times {\text{n}}} \right) + 13 \cr & {\text{Let n = 1}} \cr & \Rightarrow (17 \times 1) + 13 \cr & \Rightarrow 30 \cr & {\text{second number }} \cr & \Rightarrow {\text{ }}\left( {17 \times {\text{n}}} \right) + 11 \cr & \Rightarrow (17 \times 1) + 11 \cr & \Rightarrow 28 \cr & {\text{According to question}} \cr & = \frac{{30 + 28}}{{17}} \cr & = \frac{{58}}{{17}} \cr & \Rightarrow {\text{remainder = 7}} \cr} $$

Alternate :
Divisor = 1^st remainder + 2^nd remainder - 3^rd remainder
$$\eqalign{ & 17 = {\text{ }}13 + 11 - {\text{3rd remainder}} \cr & {\text{3rd remainder}} = 24 - 17 = 7 \cr} $$

$${\left( {264} \right)^{102}} + {\left( {264} \right)^{103}}$$
unit digit
$$\eqalign{ & {{\text{4}}^1} \to 4 \to 4 \cr & {4^2} \to 16 \to 6 \cr & {4^3} \to 64 \to 4 \cr} $$
Rule: When 4 has odd power, then unit digit is 4
When 4 has even power, then unit digit is 6

$$\eqalign{ & {\left( {264} \right)^{102}} + {\left( {264} \right)^{103}} \cr & \,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr & \,\,\,\,{4^{102}}\,\,\,\,\,\,\, + \,\,\,\,\,\,{4^{103}} \cr} $$
$${\text{unit digit}} = 6 + 4 = 10 \to 0$$
    (even power)   (odd power)

$$\eqalign{ & {\bf{Alternate}} \cr & \Rightarrow {\left( {264} \right)^{102}} + {\left( {264} \right)^{103}} \cr & \Rightarrow {\left( {264} \right)^{102}} + \left( {1 + 264} \right) \cr & \Rightarrow {\left( {264} \right)^{102}} + 265 \cr & {\text{Multiple of }}5\,\,{\text{and }}2 \cr & {\text{So}},{\text{unit digit is }}0 \cr} $$

$$\eqalign{ & {\text{First number = 1125}} \cr & {\text{Last number = 4950}} \cr & {\text{Number of term }} \cr & {\text{ = }}\frac{{4950 - 1125}}{{225}} + 1 \cr & = \frac{{3825}}{{225}} + 1 \cr & = 17 + 1 \cr & = 18 \cr} $$

$$\frac{{2}}{{7}}$$	$$\frac{{1}}{{3}}$$	$$\frac{{5}}{{6}}$$	$$\frac{{3}}{{4}}$$
↓	↓	↓	↓
0.2857	0.333	0.83	0.75

$$\frac{5}{6}\,{\text{is the greatest value}}$$

$$\eqalign{ & \left( x \right) + \left( {x + 6} \right) + \left( {x + 12} \right) = 63 \cr & 3x + 18 = 63 \cr & 3x = 45 \cr & x = 15 \cr & {\text{So,the largest number is}} \cr & {\text{ = }}15 + 12 \cr & = 27 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 2 \cr & {x^2} + 1 = 2x \cr & {x^2} - 2x + 1 = 0 \cr & {\left( {x - 1} \right)^2} = 0 \cr & {\text{Then, }}x{\text{ = 1}} \cr} $$

Shortcut Method :
Divisor = 1st remainder + 2nd remainder - 3rd remainder
$$\eqalign{ & = 3 + 4 - 2{\text{ }} \cr & = 7 - 2 \cr & = 5 \cr} $$