Problems On Numbers - Aptitude MCQ Questions and Solutions with Explanations

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow x \times \frac{2}{3}x = 864 \cr & \Leftrightarrow \frac{2}{3}{x^2} = 864 \cr & \Leftrightarrow {x^2} = \left( {\frac{{864 \times 3}}{2}} \right) \cr & \Leftrightarrow {x^2} = 1296 \cr & \Leftrightarrow x = \sqrt {1296} \cr & \Leftrightarrow x = 36 \cr} $$

Let the numbers be x and y
Then,
$$xy = 9375{\text{ and }}\frac{x}{y} = 15$$
$$\eqalign{ & \Leftrightarrow \frac{{xy}}{{\left( {\frac{x}{y}} \right)}} = \frac{{9375}}{{15}} \cr & \Leftrightarrow {y^2} = 625 \cr & \Leftrightarrow y = 25 \cr & \therefore x = 15y \cr & \Rightarrow x = 15 \times 25 \cr & \Rightarrow x = 375 \cr} $$
∴ Sum of the numbers :
= 375 + 25
= 400

$$\eqalign{ & 120: \cr & = 2 \times 2 \times 2 \times 3 \times 5 \cr & = \left( {2 \times 2} \right) \times 5 \times \left( {2 \times 3} \right) \cr & = 4 \times 5 \times 6 \cr} $$
Clearly, the three consecutive integers whose product is 120 are 4, 5 and 6
Required sum :
= 4 + 5 + 6
= 15

Let the ten's and unit's digits be $$x$$ and $$\frac{8}{x}$$ respectively
Then,
$$\eqalign{ & \Leftrightarrow \left( {10x + \frac{8}{x}} \right) + 18 = 10 \times \frac{8}{x} + x \cr & \Leftrightarrow 10{x^2} + 8 + 18x = 80 + {x^2} \cr & \Leftrightarrow 9{x^2} + 18x - 72 = 0 \cr & \Leftrightarrow {x^2} + 2x - 8 = 0 \cr & \Leftrightarrow \left( {x + 4} \right)\left( {x - 2} \right) = 0 \cr & \Leftrightarrow x = 2 \cr} $$
So, ten's digit = 2 and unit's digit = 4
Hence, required number = 24

Let the numbers be a, b and c.
Then, a² + b² + c² = 138
And (ab + bc + ca) = 131
∴ (a + b + c)²
= a² + b² + c² + 2(ab + bc + ca)
= 138 + 2 × 131
= 400
⇒ (a + b + c) = $$\sqrt {400} $$  = 20

Let the positive integers be a and b where a > b
According to the question,
$$\eqalign{ & {a^2} + {b^2} = 100.....(\text{i}) \cr & {a^2} - {b^2} = 28.....(\text{ii}) \cr} $$
By adding (i) and (ii), we get :
$$\eqalign{ & \therefore {a^2} + {b^2} + {a^2} - {b^2} = 100 + 28 \cr & \Rightarrow 2{a^2} = 128 \cr & \Rightarrow {a^2} = \frac{{128}}{2} \cr & \Rightarrow a = \sqrt {64} \cr & \Rightarrow a = 8 \cr} $$
From equation (i)
$$\eqalign{ & \Rightarrow {8^2} + {b^2} = 100 \cr & \Rightarrow {b^2} = 100 - 64 \cr & \Rightarrow b = \sqrt {36} \cr & \Rightarrow b = 6 \cr & \therefore a + b = 8 + 6 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \cr} $$

Let the number be x
Then,
$$\eqalign{ & \Rightarrow 13x = x + 180 \cr & \Rightarrow 12x = 180 \cr & \Rightarrow x = \frac{{180}}{{12}} \cr & \Rightarrow x = 15 \cr} $$

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow x - 4 = \frac{{21}}{x} \cr & \Leftrightarrow {x^2} - 4x - 21 = 0 \cr & \Leftrightarrow \left( {x - 7} \right)\left( {x + 3} \right) = 0 \cr & \Leftrightarrow x = 7 \cr} $$

Let the numbers be x and (x + 1365)
Then,
⇒ x + 1365 = 6x + 15
⇒ 5x = 1350
⇒ x = 270

Let the ten's digit be x
Then, number = 10x + 3 and
Sum of digits = (x + 3)
So, (x + 3) = $$\frac{1}{7}$$(10x + 3)
⇔ 7x + 21 = 10x + 3
⇔ 3x = 18
⇔ x = 6
Hence, the number is :
= (10x + 3)
= (10 × 6 + 3)
= 63