Problems On Numbers - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option D

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{number}}\,{\text{be}}\,x \cr & {\text{Then}},\,\frac{1}{3}\,{\text{of}}\,\frac{1}{{4\,}}\,{\text{of}}\,x = 15\cr & \Rightarrow \, x = 15 \times 12 = 180 \cr & {\text{So,}}\,{\text{required}}\,{\text{number}} \cr & = {\frac{3}{{10}} \times 180} \cr & = 54 \cr} $$

Answer: Option D

Let the three integers be x, x + 2 and x + 4
Then, 3x = 2(x + 4) + 3    ⇔    x = 11
∴ Third integer = x + 4 = 15

Answer: Option B

Let the ten's digit be x and unit's digit be y.
Then, (10x + y) - (10y + x) = 36
⇒ 9(x - y) = 36
⇒ x - y = 4

Answer: Option B

Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Let ten's and unit's digits be 2x and x respectively.
Then, (10 × 2x + x) - (10x + 2x) = 36
⇒ 9x = 36
⇒ x = 4
∴ Required difference
= (2x + x) - (2x - x)
= 2x
= 8

Answer: Option B

Let the ten's and unit digit be x and $$\frac{8}{x}$$ respectively
$$\eqalign{ & {\text{Then,}} \cr & \left( {10x + \frac{8}{x}} \right) + 18 = 10 \times \frac{8}{x} + x \cr & \Rightarrow 10{x^2} + 8 + 18x = 80 + {x^2} \cr & \Rightarrow 9{x^2} + 18x - 72 = 0 \cr & \Rightarrow {x^2} + 2x - 8 = 0 \cr & \Rightarrow \left( {x + 4} \right)\left( {x - 2} \right) = 0 \cr & \Rightarrow x = 2 \cr} $$
∴ first digit will be 2 and second digit will be 4.
i.e digit is 24.