Number System - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \left( {{n^3} - n} \right){\text{ and n is any integer}} \cr & {\text{put n = 2 so, }}{{\text{2}}^3} - 2 = 6 \cr & {\text{It will be always divisible by 6}} \cr & {\text{(put n = 2, 3, 4}}....{\text{)}} \cr & {\text{(n = 2, 3, 4}}.....{\text{)}} \cr} $$

Take LCM of 10 and 13 = 130
Any number divisible by 130 will be divisible by 13 and 10
i.e., 130, 260, 390, 520, 650, 780 & 910
⇒ total numbers are 7

$$\frac{8}{{25}},\frac{7}{{23}},\frac{{11}}{{23}},\frac{{14}}{{53}}$$

(184)	,	(175)
↑		↑
$$\frac{{8}}{{25}}$$	⤩	$$\frac{{7}}{{23}}$$	= $$\frac{{8}}{{25}}$$ > $$\frac{{7}}{{23}}$$
(371)	,	(322)
↑		↑
$$\frac{{7}}{{23}}$$	⤩	$$\frac{{14}}{{53}}$$	= $$\frac{{7}}{{23}}$$ > $$\frac{{14}}{{53}}$$

$${\text{So}}\frac{{14}}{{53}}\,{\text{is smallest}}$$

$$\eqalign{ & {\text{No}}{\text{. of digits}} \cr & {\text{0 - 9 }} \to {\text{ 1}} \times {\text{9 = 9}} \cr & {\text{10 - 19 }} \to {\text{ 2}} \times {\text{10 = 20}} \cr & {\text{20 - 29 }} \to {\text{ 2}} \times {\text{10 = 20}} \cr & {\text{30 - 39 }} \to {\text{ 2}} \times {\text{10 = 20}} \cr & 40{\text{ - 49 }} \to {\text{ 2}} \times {\text{10 = 20}} \cr & 50\, \to \,2 \times {\text{1 = 2 = 91}} \cr} $$

$$\eqalign{ & \Leftrightarrow \left( {{2^{71}} + {2^{72}} + {2^{73}} + {2^{74}}} \right)\, \cr & = {2^{71}}\left( {{2^0} + {2^1} + {2^2} + {2^3}} \right)\, \cr & = {2^{71}}(1 + 2 + 4 + 8) \cr & = {2^{71}} \times 15 \cr & = {2^{70}} \times 30 \cr & {\text{It is divisible by 10}} \cr} $$

$$\eqalign{ & = 999\frac{{98}}{{99}} \times 99 \cr & = (1000 - 1)\frac{{98}}{{99}} \times 99 \cr & = \left[ {\left( {1000 - 1} \right) + \frac{{98}}{{99}}} \right] \times 99 \cr & = 99000 - 99 + 98 \cr & = 99000 - 1 \cr & = 98999 \cr} $$

Let 'a' be the smallest even number
According to question
$$\eqalign{ & (a) + (a + 2) + (a + 4) + (a + 6) = 748 \cr & 4a + 12 = 748 \cr & 4a = 736 \cr & a = 184 \cr & \cr & {\bf{Alternate :}} \cr & {\text{Middle term }} = \frac{{748}}{4} = 187 \cr & \underline {184} \,\,\,\,\underline {186} \,\,\,\underline {188} \,\,\,\underline {190} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {187}\limits^ \downarrow \cr & {\text{Smallest number = 184}} \cr} $$

$$\eqalign{ & 1.\overline {27} \, = \frac{{127 - 1}}{{99}}{\text{ = }}\frac{{126}}{{99}}{\text{ = }}\frac{{14}}{{11}} \cr & {\text{ }}\therefore {\text{ }}\frac{{\text{p}}}{{\text{q}}} = \frac{{14}}{{11}} \cr} $$

$$\eqalign{ & {\text{Let the 3 numbers be }}x,y,\& \,\,z \cr & {\text{According to question}} \cr & x + y = 55 \to y = 55 - x \cr & y + z = 65 \to 55 - x + z = 65 \cr & z - x = 10.....(i) \cr & z + 3x = 110.....(ii) \cr & {\text{Solve (i)}}\,{\text{and (ii)}} \cr & {\bf{z = 35,}} \cr & x = 25, \cr & y = 30 \cr} $$

$$\eqalign{ & {\text{Let the two digit number be }} \cr & 10y + x\,\,\,\,\,{\text{where}}\,\,\,y > x \cr & 10x + y - 10y - x = 63 \cr & 9x - 9y = 63 \cr & x - y = 7 \cr & {\bf{x = 7,8,9}}{\text{ and}} \cr & y = 0,1,2 \cr} $$