Problems On Numbers - Aptitude MCQ Questions and Solutions with Explanations

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

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

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

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

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.

Let the ten's digit be x and unit's digit be y
Then, x + y = 15 and x - y = 3   or   y - x = 3
Solving x + y = 15   and   x - y = 3, we get: x = 9, y = 6
Solving x + y = 15   and   y - x = 3, we get: x = 6, y = 9
So, the number is either 96 or 69
Hence, the number cannot be determined.

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 x 131 = 400
⇒ (a + b + c) = $$\sqrt {400} $$  = 20

Let the ten's digit be x and unit's digit be y.
Then, number = 10x + y
Number obtained by interchanging the digits = 10y + x
∴ (10x + y) + (10y + x) = 11(x + y), which is divisible by 11

Let the ten's digit be x
Then, unit's digit = x + 2
Number = 10x + (x + 2) = 11x + 2
Sum of digits = x + (x + 2) = 2x + 2
∴ (11x + 2)(2x + 2) = 144
⇒ 22x² + 26x - 140 = 0
⇒ 11x² + 13x - 70 = 0
⇒ 11x² + (35 - 22)x - 70 = 0
⇒ 11x² + 35x - 22x - 70 = 0
⇒ (x - 2)(11x + 35) = 0
⇒ x = 2
Hence, required number = 11x + 2 = 24

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{number}}\,{\text{be}}\,x \cr & {\text{Then}},\,x + 17 = \frac{{60}}{x} \cr & \Rightarrow {x^2} + 17x - 60 = 0 \cr & \Rightarrow (x + 20)(x - 3) = 0 \cr & \Rightarrow x = 3 \cr} $$