Question

The sum of two numbers is $80$ and the greater number exceed twice the smaller by $11$. Find the larger number.

Answer 1

Solution:-

Let the two numbers be $x$ and $y$ such that $x>y$..

Now according to the question-

Condition I:-

$x+y=80.....\left(1\right)$

Condition II:-

$x=2y+11.....\left(2\right)$

Substituting the value of $x$ in ${eq}^n\left(1\right)$ we have

$(2y+11)+y=80$

$3y+11=80$

$3y=80-11$

$y=\frac{69}{3}=23$

Substituting the value of $y$ in ${eq}^n\left(2\right)$, we have

$x=2\times 23+11$

$\Rightarrow x=46+11=57$

Hence the required numbers are $57$ and $23$.