Sum of three numbers 264, If the first number be twice then second and third number be one third of the first, then the second number is

Answer: Option B

Explanation:

Let the number is x,
\begin{aligned}
=> \frac{2}{3}x-50 = \frac{1}{4}x + 40
\end{aligned}

\begin{aligned}
<=> \frac{2}{3}x-\frac{1}{4}x = 90
\end{aligned}
\begin{aligned}
<=> \frac{5x}{12} = 90
\end{aligned}
\begin{aligned}
<=> x = 216
\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 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 D

Explanation:

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

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}