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

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 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:

Let the number be x.
Then, 15x = x + 196
=› 14 x= 196
=› x = 14.

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 B

Explanation:

Let the numbers be a, b and c.
Then,
\begin{aligned}
a^2 + b^2 + c^2 = 138
\end{aligned}
and (ab + bc + ca) = 131
\begin{aligned}
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
\end{aligned}
= 138 + 2 x 131 = 400

\begin{aligned}
=> (a + b + c) = \sqrt{400} = 20.
\end{aligned}