Problems On Numbers - Aptitude MCQ Questions and Solutions with Explanations

Let the hundred's digit = x
Then, ten's digit = (x + 1)
Unit's digit :
$$\eqalign{ & = 75\% {\text{ of }}\left( {x + 1} \right) \cr & = \frac{3}{4}\left( {x + 1} \right) \cr} $$
$$\eqalign{ & \therefore \left( {x + 1} \right) + x = 15 \cr & \Leftrightarrow 2x = 14 \cr & \Leftrightarrow x = 7 \cr} $$
So, hundred's digit = 7
Ten's digit = 8
Unit's digit :
$$\eqalign{ & = \frac{3}{4}\left( {x + 1} \right) \cr & = \frac{3}{4}\left( {7 + 1} \right) \cr & = \frac{3}{4}\left( 8 \right) \cr & = 6 \cr} $$
Hence, required number = 786

Let the numbers be a and b, where a > b
According to the question,
$$\eqalign{ & a + b = 37\& {a^2} - {b^2} = 185 \cr & \Rightarrow \left( {a + b} \right)\left( {a - b} \right) = 185 \cr & \Rightarrow 37\left( {a - b} \right) = 185 \cr & \Rightarrow a - b = \frac{{185}}{{37}} \cr & \Rightarrow a - b = 5 \cr} $$

Let the number be x
Then,
$$\eqalign{ & \frac{3}{5}{\text{of }}\frac{2}{3}{\text{of }}x - \frac{2}{5}{\text{of }}\frac{1}{4}{\text{of }}x = 288 \cr & \Leftrightarrow \left( {x \times \frac{3}{5} \times \frac{2}{3}} \right) - \left( {x \times \frac{2}{5} \times \frac{1}{4}} \right) = 288 \cr & \Leftrightarrow \frac{2}{5}x - \frac{1}{{10}}x = 288 \cr & \Leftrightarrow \frac{{3x}}{{10}} = 288 \cr & \Leftrightarrow x = \left( {\frac{{288 \times 10}}{3}} \right) \cr & \Leftrightarrow x = 960 \cr} $$

Let the numbers be x and y
Then,
$$\eqalign{ & x + y = 12\,\, \& \,\, xy = 35 \cr & \therefore \frac{1}{x} + \frac{1}{y} = \frac{{x + y}}{{xy}} = \frac{{12}}{{35}} \cr} $$

Let the numbers be x and (x + 16)
Then,
$$\eqalign{ & \Leftrightarrow \frac{x}{3} - \frac{{\left( {x + 16} \right)}}{7} = 4 \cr & \Leftrightarrow 7x - 3\left( {x + 16} \right) = 84 \cr & \Leftrightarrow 4x = 84 + 48 \cr & \Leftrightarrow 4x = 132 \cr & \Leftrightarrow x = 33 \cr} $$
Hence, the numbers are 33 and 49

Let the ten's digit be x and the unit's digit be y
Then, number = 10x + y
$$\eqalign{ & \therefore 10x + y = k\left( {x + y} \right) \cr & \Rightarrow k = \frac{{10x + y}}{{x + y}} \cr} $$
Number formed by interchanging the digits = 10y + x
Let, 10y + x = h(x + y)
Then,
$$\eqalign{ & h = \frac{{10y + x}}{{x + y}} \cr & \,\,\,\,\, = \frac{{11\left( {x + y} \right) - \left( {10x + y} \right)}}{{x + y}} \cr & \,\,\,\,\, = 11 - \frac{{10x + y}}{{x + y}} \cr & \,\,\,\,\, = 11 - k \cr} $$

Let the two fractions be a and b
Then,
$$\eqalign{ & ab = \frac{{14}}{{15}}\, \& \, \frac{a}{b} = \frac{{35}}{{24}} \cr & \Leftrightarrow \frac{{ab}}{{\left( {\frac{a}{b}} \right)}} = \left( {\frac{{14}}{{15}} \times \frac{{24}}{{35}}} \right) \cr & \Leftrightarrow {b^2} = \frac{{16}}{{25}} \cr & \Leftrightarrow b = \frac{4}{5} \cr & {\text{So, }} \cr & \Leftrightarrow ab = \frac{{14}}{{15}} \cr & \Leftrightarrow a = \left( {\frac{{14}}{{15}} \times \frac{5}{4}} \right) \cr & \Leftrightarrow a = \frac{7}{6} \cr} $$
Since a > b,
So, greater fraction is $$\frac{7}{6}$$

Let the total number of eggs bought be a
10% of eggs are rotten
∴ Remaining eggs :
$$\eqalign{ & = a - 10\% {\text{ of }}a \cr & = a - \frac{{10a}}{{100}} \cr & = \frac{{100a - 10a}}{{100}} \cr & = \frac{{90a}}{{100}} \cr & = \frac{{9a}}{{10}} \cr} $$
Man gives 80% of $$\frac{{9a}}{{100}}$$ eggs to his neighbour
$$\eqalign{ & = \frac{{80}}{{100}} \times \frac{{9a}}{{10}} \cr & = \frac{{72a}}{{100}} \cr} $$
Remaining eggs :
$$\eqalign{ & = \frac{{9a}}{{10}} - \frac{{72a}}{{100}} \cr & = \frac{{90a - 72a}}{{100}} \cr & = \frac{{18a}}{{100}} \cr & = \frac{{9a}}{{50}} \cr} $$
According the question,
$$\eqalign{ & \Rightarrow \frac{{9a}}{{50}} = 36 \cr & \Rightarrow 9a = 36 \times 50 \cr & \Rightarrow a = \frac{{36 \times 50}}{9} \cr & \Rightarrow a = 200 \cr} $$
Hence, the total number of eggs bought be = 200

Let the number be x
Then,
⇔ 3(2x + 9) = 75
⇔ 2x + 9 = 25
⇔ 2x = 16
⇔ x = 8

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow x + \frac{1}{x} = 3\left( {x - \frac{1}{x}} \right) \cr & \Leftrightarrow \frac{{{x^2} + 1}}{x} = 3\left( {\frac{{{x^2} - 1}}{x}} \right) \cr & \Leftrightarrow {x^2} + 1 = 3{x^2} - 3 \cr & \Leftrightarrow 2{x^2} = 4 \cr & \Leftrightarrow {x^2} = 2 \cr & \Leftrightarrow x = \sqrt 2 \cr} $$