Sum of two numbers is 25 and their difference is 13. Find their product.

Answer: Option A

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 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 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 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}