if the sum of $$\begin{aligned} \frac{1}{2} \end{aligned}$$ and $$\begin{aligned} \frac{1}{5} \end{aligned}$$ of a number exceeds $$\begin{aligned} \frac{1}{3} \end{aligned}$$ of the number by $$\begin{aligned} 7\frac {1}{3} \end{aligned}$$ , then number is

Answer: Option A

Explanation:

If the numbers are a, b, then ab = 17,
as 17 is a prime number, so a = 1, b = 17.

$$\begin{aligned} \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{1^2} + \frac{1}{17^2} \end{aligned}$$
$$\begin{aligned} = \frac{290}{289} \end{aligned}$$

Answer: Option B

Explanation:

$$\begin{aligned} => x + \frac{1}{x} = \frac{13}{6} \end{aligned}$$
$$\begin{aligned} => \frac{x^2+1}{x} = \frac{13}{6} \end{aligned}$$
$$\begin{aligned} => 6x^2-13x+6 = 0 \end{aligned}$$

$$\begin{aligned} => 6x^2-9x-4x+6 = 0 \end{aligned}$$
$$\begin{aligned} => (3x-2)(2x-3) \end{aligned}$$
$$\begin{aligned} => x = 2/3 or 3/2 \end{aligned}$$

Answer: Option C

Explanation:

$$\begin{aligned} => \frac{(x+y)}{(x-y)} = \frac{40}{4} \end{aligned}$$

$$\begin{aligned} => (x+y)= 10(x-y) \end{aligned}$$

$$\begin{aligned} => 9x = 11y => \frac{x}{y} = \frac{11}{9} \end{aligned}$$

Answer: Option C

Explanation:

Let the three integers be x, x+2 and x+4.
Then, 3x = 2(x+4)+3,
x= 11
Therefore, third integer x+4 = 15

Answer: Option A

Explanation:

Let the number be x.
7x -15 = 2x + 10 => 5x = 25 => x = 5