Sum of two numbers is 40 and their difference is 4. The ratio of the numbers is

Answer: Option D

Explanation:

If the number is x,

Then, x + 17 = 60/x

x2 + 17x - 60 = 0

(x + 20)(x - 3) = 0

x = 3, -20, so x = 3 (as 3 is positive)

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}

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 D

Explanation:

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

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}