Find the number, when 15 is subtracted from 7 times the number, the result is 10 more than twice of the number

Answer: Option A

Explanation:

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

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 D

Explanation:

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

Answer: Option D

Explanation:

Friends you remember,
\begin{aligned}
=> (x+y)^2 = (x-y)^2 + 4xy
\end{aligned}

\begin{aligned}
=> (x+y)^2 = (5)^2 + 4(336)
\end{aligned}

\begin{aligned}
=> (x+y) = \sqrt{1369} = 37
\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}