The product of two numbers is 120 and the sum of their squares is 289. The sum of the 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:

Let the ten digit be x, unit digit is y.
Then (10x + y) - (10y + x) = 36
=> 9x - 9y = 36
=> x - y = 4.

Answer: Option A

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 C

Explanation:

Friends, this sort of question is quite important in competitive exams, whenever any question come which have relation between sum, product and difference, this formula do the magic:
$$\begin{aligned} => (x+y)^2 = (x-y)^2 + 4xy \end{aligned}$$

$$\begin{aligned} <=> (25)^2 = (13)^2 + 4xy \end{aligned}$$

$$\begin{aligned} <=> 4xy = (25)^2 - (13)^2 \end{aligned}$$
$$\begin{aligned} <=> xy = \frac{456}{4} = 114 \end{aligned}$$