Number System - Aptitude MCQ Questions and Solutions with Explanations

\[\begin{gathered} {\text{ unit place}} \hfill \\ \left[ \begin{gathered} {7^1} \to 7 \to 7 \hfill \\ {7^2} \to 49 \to 9 \hfill \\ {7^3} \to 343 \to 3 \hfill \\ {7^4} \to 2401 \to 1 \hfill \\ \end{gathered} \right]{\text{cycle }} \hfill \\ \end{gathered} \]
$$\eqalign{ & {\left( {4387} \right)^{245}}\,\,\,\,\, \times \,\,\,\,\,{\left( {621} \right)^{72}} \cr & \,\,\,\,\mathop {{{\left( 7 \right)}^{245}}\,\,\,\,\,\,\,\,}\limits^ \downarrow \times \,\,\,\,\,\,\,\,\,{\left( 1 \right)^{72}} \cr & \,\,\,\,\mathop {{{\left( 7 \right)}^{4 \times 61 + 1}}\,\,\,}\limits^ \downarrow \times \,\,\,\,\,\,\,\,\,\,\,1 \cr & \,\mathop {{{\left( 7 \right)}^{4 \times 61}} \times 7\,\,}\limits^ \downarrow \times \,\,\,\,\,\,\,\,\,\,\,1 \cr & \,\,\,\,\,\,\mathop {{{\left( 1 \right)}^{61 }\times 7}}\limits^ \downarrow \,\,\,\,\, \times \,\,\,\,\,\,\,\,\,\,\,1 \cr} $$
unit digit = 7

$$\eqalign{ & \frac{6}{7},\frac{5}{6},\frac{7}{8},\frac{4}{5} \cr & \frac{6}{7} = 0.85, \cr & \frac{5}{6} = 0.83, \cr & \frac{7}{8} = 0.87, \cr & \frac{4}{5} = 0.80 \cr & \frac{7}{8}\,\,{\text{is}}\,{\text{largest}} \cr} $$

Explanation :
When the fractions are given in this order when the difference between the numerator and denominator is same so largest fraction will be largest numerator fraction and the smallest will be smallest numerator fraction.
Note: - Fraction should be in the order where denominator will be larger than numerator.
In this question : largest is $$\frac{7}{8}$$and smallest is $$\frac{4}{5}$$

$$\eqalign{ & {\text{ }}\frac{2}{5},\frac{5}{6},\frac{{11}}{{15}}{\text{ & }}\frac{7}{8} \cr & \frac{7}{8}\,{\text{is largest among }}\frac{2}{5},\frac{5}{6},\frac{7}{8} \cr & {\text{now compare}} \cr } $$

(105)		(88)
$$\frac{7}{8}$$	⤧	$$\frac{{11}}{{15}}$$

So $$\frac{{7}}{{8}}$$ is largest

4x² - 12x + k
for perfect square must be
b² = 4ac
12 × 12 = 4 × 4 × k
9 = k
k = 9

$$\eqalign{ & {\text{Let the unit digit }} \cr & {\text{Ten digit }} = x - 2 \cr & \therefore {\text{ Number :}} \cr & = {\text{ }}10(x - 2){\text{ }} + {\text{ }}x \cr & = {\text{ }}10x{\text{ }} - {\text{ }}20{\text{ }} + {\text{ }}x \cr & = {\text{ }}11x{\text{ }} - {\text{ }}20 \cr} $$
New number obtained after reversing the digits
$$\eqalign{ & = 10x + x - 2{\text{ }} \cr & = {\text{ }}11x - 2 \cr & {\text{According to the question,}} \cr & {\text{3}}\left( {11x - 20} \right) + \frac{6}{7}\left( {11x - 2} \right) = 108 \cr & 7\left( {11x - 20} \right) + 2\left( {11x - 2} \right) = 36 \times 7 \cr & 77x - 140 + 22x - 4 = 252 \cr & 99x = 252 + 144 \cr & x = \frac{{396}}{{99}} = 4 \cr & {\text{Number :}} \cr & = {\text{ }}11x - 20{\text{ }} \cr & = {\text{ }}11 \times 4 - 20 \cr & = {\text{ }}24 \cr & {\text{Sum of digit }} = 2 + 4 = 6 \cr} $$

Let the numerator and denominator br x and 11 - x
According to question,
$$\eqalign{ & {\text{Fraction }} = \frac{x}{{11 - x}} \cr & \Rightarrow \frac{{x + 2}}{{11 - x + 2}} = \frac{x}{{11 - x}} + \frac{1}{{24}} \cr & \Rightarrow \frac{{x + 2}}{{13 - x}} - \frac{x}{{11 - x}} = \frac{1}{{24}} \cr} $$
$$ \Rightarrow \frac{{11x + 22 - {x^2} - 2x - 13x + {x^2}}}{{\left( {13 - x} \right)\left( {11 - x} \right)}}$$       $$ = \frac{1}{{24}}$$
$$\eqalign{ & \Rightarrow \frac{{22 - 4x}}{{143 - 24x + {x^2}}} = \frac{1}{{24}} \cr & \Rightarrow 528 - 96x = 143 - 24x + {x^2} \cr & \Rightarrow {x^2} + 72x - 385 = 0 \cr & \Rightarrow \left( {x + 77} \right)\left( {x - 5} \right) = 0 \cr & \Rightarrow x = 5 \cr & {\text{Numerator }}(x) = 5 \cr & {\text{Denominator }} = 11 - 5 = 6 \cr & {\text{Difference }} = 6 - 5 = 1 \cr} $$

	(1570)2	+	(1571)2	+	(1572)2	+	(1573)2
unit digit →	↓		↓		↓		↓
	02	+	12	+	22	+	32
	↓		↓		↓		↓
	0	+	1	+	4	+	9 = 14

unit digit = 4

$$\eqalign{ & \frac{{{\text{remainder of number}}}}{{16}} = \frac{{54}}{{16}} \cr & \Rightarrow {\text{Remainder = 6}} \cr} $$

$$\eqalign{ & {{\text{7}}^{105}} \Rightarrow \cr & {{\text{7}}^1} \to 7 \to 7 \cr & {{\text{7}}^2} \to 49 \to 9 \cr & {{\text{7}}^3} \to 343 \to 3 \cr & {{\text{7}}^4} \to 2401 \to 1 \cr & {\text{Divided power of 7 by 4}} \cr & \frac{{105}}{4} \to {\text{ remainder = 1}} \cr & \Rightarrow {{\text{7}}^1}\,{\text{is left unit digit = 1}} \times 7 = 7 \cr} $$

Let us assume that he answered x question correctly. Marks scored by him in x question = 4x
Wrong answer would be 200 - x

According to question,
4x - (200 - x) = 200
or 4x - 200 + x = 200
or 5x = 200 + 200 = 400
or x = 80