Problems On Numbers - Aptitude MCQ Questions and Solutions with Explanations

Let the numbers be x and y
Then,
x + y = 40 and xy = 375
$$\eqalign{ & \therefore \frac{1}{x} + \frac{1}{y} = \frac{{x + y}}{{xy}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{40}}{{275}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{75}} \cr} $$

Let the ten's digit be x and the unit's digit be y
Then, number = 10x + y
$$\eqalign{ & \therefore 10x + y = 7\left( {x + y} \right) \cr & \Leftrightarrow 3x = 6y \cr & \Leftrightarrow x = 2y \cr} $$
Number formed by reversing the digits = 10y + x
$$\eqalign{ & \therefore \left( {10x + y} \right) - \left( {10y + x} \right) = 18 \cr & \Leftrightarrow 9x - 9y = 18 \cr & \Leftrightarrow x - y = 2 \cr & \Leftrightarrow 2y - y = 2 \cr & \Leftrightarrow y = 2 \cr & {\text{So, }}x = 2y = 4 \cr} $$
Hence,
∴ Required number
= 10x + y
= 40 + 2
= 42

Let the fraction be $$\frac{a}{1}$$
Then,
$$\eqalign{ & \Leftrightarrow \frac{1}{a} - a = \frac{9}{{20}} \cr & \Leftrightarrow \frac{{1 - {a^2}}}{a} = \frac{9}{{20}} \cr & \Leftrightarrow 20 - 20{a^2} = 9a \cr & \Leftrightarrow 20{a^2} + 9a - 20 = 0 \cr & \Leftrightarrow 20{a^2} + 25a - 16a - 20 = 0 \cr & \Leftrightarrow 5a\left( {4a + 5} \right) - 4\left( {4a + 5} \right) = 0 \cr & \Leftrightarrow \left( {4a + 5} \right)\left( {5a - 4} \right) = 0 \cr & \Leftrightarrow a = \frac{4}{5}\,\,\,\,\,\,\,\,\left[ {\because a \ne - \frac{5}{4}} \right] \cr} $$

Let the numbers be a and b
According to the question,
$$\eqalign{ & a + b = 75 \cr & a - b = 25 \cr & \because {\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} = 4ab \cr & \Rightarrow {75^2} - {25^2} = 4ab \cr & \Rightarrow 4ab = \left( {75 + 25} \right)\left( {75 - 25} \right) \cr & \left[ {\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right] \cr & \Rightarrow 4ab = 100 \times 50 \cr & \Rightarrow ab = \frac{{100 \times 50}}{4} \cr & \Rightarrow ab = 1250 \cr} $$

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow \frac{3}{4}x - \frac{1}{3}x = 60 \cr & \Leftrightarrow \frac{{5x}}{{12}} = 60 \cr & \Leftrightarrow x = \left( {\frac{{60 \times 12}}{5}} \right) \cr & \Leftrightarrow x = 144 \cr} $$

Let the numbers be a and b
Then,
ab = 17
⇒ a = 1 and b = 17
So,
$$\eqalign{ & = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr & = \frac{{{a^2} + {b^2}}}{{{a^2}{b^2}}} \cr & = \frac{{{1^2} + {{\left( {17} \right)}^2}}}{{{{\left( {1 \times 17} \right)}^2}}} \cr & = \frac{{290}}{{289}} \cr} $$

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow \frac{x}{5} + 4 = \frac{x}{4} - 10 \cr & \Leftrightarrow \frac{x}{4} - \frac{x}{5} = 14 \cr & \Leftrightarrow \frac{x}{{20}} = 14 \cr & \Leftrightarrow x = 14 \times 20 \cr & \Leftrightarrow x = 280 \cr} $$

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow \frac{{4\frac{1}{2}\left( {x + 2\frac{1}{2}} \right) + 3}}{{1\frac{1}{5}}} = 25 \cr & \Leftrightarrow \frac{{\frac{9}{2}\left( {x + \frac{5}{2}} \right) + 3}}{{\frac{6}{5}}} = 25 \cr & \Leftrightarrow \frac{{9x}}{2} + \frac{{45}}{4} + 3 = 25 \times \frac{6}{5} \cr & \Leftrightarrow \frac{{9x}}{2} + \frac{{45}}{4} + 3 = 30 \cr & \Leftrightarrow \frac{{9x}}{2} = 30 - \frac{{57}}{4} \cr & \Leftrightarrow \frac{{9x}}{2} = \frac{{63}}{4} \cr & \Leftrightarrow x = \left( {\frac{{63}}{4} \times \frac{2}{9}} \right) \cr & \Leftrightarrow x = \frac{7}{2} \cr & \Leftrightarrow x = 3\frac{1}{2} \cr} $$

Let the numbers be x and y
Then,
xy = 120 and x² + y² = 289
$$\eqalign{ & \therefore {\left( {x + y} \right)^2} \cr & = {x^2} + {y^2} + 2xy \cr & = 289 + 240 \cr & = 529 \cr & \therefore x + y \cr & = \sqrt {529} \cr & = 23 \cr} $$

Let the ten's digit be x and the unit's digit be y
Then, number = 10x + y
New number :
$$\eqalign{ & = 10 \times 2x + \frac{y}{2} \cr & = 20x + \frac{y}{2} \cr} $$
$$\eqalign{ & \therefore 20x + \frac{y}{2} = 10y + x \cr & \Leftrightarrow 40x + y = 20y + 2x \cr & \Leftrightarrow 38x = 19y \cr & \Leftrightarrow y = 2x \cr} $$
So, the unit's digit is twice the ten's digit.