Problems On Numbers - Aptitude MCQ Questions and Solutions with Explanations

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

Let the numbers be x and y
Then, xy = 120 and x² + y² = 289
∴ (x + y)² = x² + y² + 2xy = 289 + (2 x 120) = 529
∴ x + y = $$\sqrt {529} $$  = 23

Let the middle digit be x
Then, 2x = 10 or x = 5. So, the number is either 253 or 352
Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit.
Hence, required number = 253

Let the numbers be x and y
Then, x + y = 25 and x - y = 13
4xy = (x + y)² - (x- y)²
   = (25)² - (13)²
   = (625 - 169)
   = 456
∴ xy = 114

Let the numbers be x and x + 2.
Then, (x + 2)² - x² = 84
⇒ 4x + 4 = 84
⇒ 4x = 80
⇒ x = 20.
∴ The required sum
= x + (x + 2)
= 2x + 2
= 42

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow 2x + 3 \times 42 = 238 \cr & \Leftrightarrow 2x + 126 = 238 \cr & \Leftrightarrow 2x = 112 \cr & \Leftrightarrow x = 56 \cr} $$
∴ Required sum :
$$\eqalign{ & = 3x + 2 \times 42 \cr & = 3 \times 56 + 2 \times 42 \cr & = 168 + 84 \cr & = 252 \cr} $$

$$\eqalign{ & = \frac{7}{8}{\text{ of }}1008 - \frac{3}{4}{\text{ of }}568 \cr & = \left( {1008 \times \frac{7}{8}} \right) - \left( {568 \times \frac{3}{4}} \right) \cr & = 882 - 426 \cr & = 456 \cr} $$

Let the number be x
Then,
$$\eqalign{ & \Leftrightarrow 3{x^2} - 4x = x + 50 \cr & \Leftrightarrow 3{x^2} - 5x - 50 = 0 \cr & \Leftrightarrow \left( {3x + 10} \right)\left( {x - 5} \right) = 0 \cr & \Leftrightarrow x = 5 \cr} $$

Let the numbers be x and y
Then,
$$xy = 1092.....(i)$$
And,
$$\eqalign{ & \Leftrightarrow \left( {x + y} \right) - \left( {x - y} \right) = 42 \cr & \Leftrightarrow 2y = 42 \cr & \Leftrightarrow y = 21 \cr} $$
Putting y = 21 in (i), we get :
$$x = \frac{{1092}}{{21}} = 52$$
Hence, greater number = 52

Let the numbers be x, x + 2 and x + 4
Then,
$$\eqalign{ & \Leftrightarrow \frac{{x\left( {x + 2} \right)\left( {x + 4} \right)}}{8} = 720 \cr & \Leftrightarrow x\left( {x + 2} \right)\left( {x + 4} \right) = 5760 \cr & \therefore \sqrt x \times \sqrt {\left( {x + 2} \right)} \times \sqrt {\left( {x + 4} \right)} \cr & = \sqrt {x\left( {x + 2} \right)\left( {x + 4} \right)} \cr & = \sqrt {5760} \cr & = 24\sqrt {10} \cr} $$