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

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}

Answer: Option B

Explanation:

We know
\begin{aligned}
(x + y)^2 = x^2 + y^2 + 2xy
\end{aligned}

\begin{aligned}
=> (x + y)^2 = 289 + 2(120)
\end{aligned}

\begin{aligned}
=> (x + y) = \sqrt{529} = 23
\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 sons age = x. Then, fathers age = 2x.
12(x—2O) = (2x - 20) so x = 22
Fathers present age = 44 years

Answer: Option D

Explanation:

Let the number = x,
Then, x-(3/5)x = 50,
=> (2/5)x = 50 => 2x = 50*5,
=> x = 125