Number System - Aptitude MCQ Questions and Solutions with Explanations

Let x and y are two numbers (x > y)
According to the question,
$$\eqalign{ & xy = {\text{ }}11520 \cr & \frac{x}{y} = \frac{9}{5} \Rightarrow \boxed{x = \frac{{9y}}{5}} \cr & {\text{put the value of x in eq}}{\text{. (i)}} \cr & \frac{{9{\text{y}}}}{5} \times {\text{y = 11520 }} \cr & \Rightarrow {\text{ }}{{\text{y}}^2} = 6400 \cr & \Rightarrow \boxed{{\text{y = 80}}} \cr & 80{\text{x = 11520 }} \cr & \Rightarrow \boxed{{\text{x = 144}}} \cr & {\text{x - y = 144 - 80 = 64 Ans}}{\text{.}} \cr} $$

a + b + 2ab is even
Explanation : Sum of two odd number is even number and multiplication of two odd number with 2 always given an even number.
Alternate :
Put , a = 3,   b = 5, option (D) will satisfy

Let the numerator be n and the denominator be n + 3
$$\eqalign{ & {\text{According to question}} \cr & \frac{{n + 7}}{{\left( {n + 3} \right) - 2}} = \frac{2}{1} \cr & \frac{{n + 7}}{{n + 1}} = \frac{2}{1} \cr & 2n + 2 = n + 7 \cr & n = 5 \cr & {\text{Fraction = }}\frac{5}{{5 + 3}} = \frac{5}{8} \cr & {\text{Sum = 5 + 8 = 13}} \cr} $$

$$\eqalign{ & \Rightarrow \left( {99\frac{{95}}{{99}}} \right) \times 99 \cr & = \left( {99 + \frac{{95}}{{99}}} \right) \times 99 \cr & = \left\{ {\left( {100 - 1} \right) + \frac{{95}}{{99}}} \right\} \times 99 \cr & = 9900 - 99 + 95 \cr & = 9896 \cr} $$

$$\eqalign{ & \frac{{{\text{remainder of number}}}}{{13}} = \frac{{17}}{{13}} \cr & \Rightarrow {\text{Remainder = 4}} \cr} $$

$$\eqalign{ & {\text{Let the total number of picnic party = x}} \cr & {\text{Each member contributed = 2x}} \cr & {\text{Accroding to question, }} \cr & (2x){\text{ }}(x) = 3042 \cr & 2{x^2} = 3042 \cr & {x^2} = 1521 \cr & x = 39 \cr} $$

$$\eqalign{ & \left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{3}} \right)\left( {1 + \frac{1}{4}} \right).....\left( {1 + \frac{1}{{120}}} \right) \cr & = \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4}.........\frac{{121}}{{120}} \cr & = \frac{{121}}{2} \cr & = 60.5{\text{ Ans}}. \cr} $$

$$\eqalign{ & {{\text{a}}^3} + {b^3} = (a + b)({{\text{a}}^2} + {b^2} - ab) \cr & {\text{Take option (A)}} \cr & {\text{ = }}{{\text{2}}^{96}} + 1{\text{ }} \cr & = {({2^{32}})^3} + {(1)^3} \cr & = ({{\text{2}}^{32}} + 1)[{({2^{32}})^2}{\text{ + }}{{\text{1}}^2}{\text{ - }}{{\text{2}}^{32}}] \cr & = ({{\text{2}}^{32}} + 1)[{2^{64}} + 1 - {2^{32}}] \cr & {\text{Clearly, }} \cr & {{\text{2}}^{32}} + 1{\text{ is a factor of }}{{\text{2}}^{96}} + 1 \cr} $$

25⁶²⁵¹ = 5⁶²⁵¹ = 5 (because unit digit of 5 is always 5)
36⁵²⁸ = 6⁵²⁸ = 6 (because unit digit of 6 is always 6)
73⁵⁴ = 3⁵⁴ = 9(unit digit of 3⁵⁴ = 3² ($$\frac{{54}}{4}$$ = 2 remainder) = 9)
now add all the unit digit 5 + 6 + 9 = 20 = 0

$$\eqalign{ & {\text{Let thr number be x}} \cr & {\text{According to question}} \cr & \frac{3}{4}x - \frac{3}{{14}}x = 150 \cr & \frac{{21x - 6x}}{{28}} = 150 \cr & 15x = 150 \times 28 \cr & x = 280 \cr} $$