Question

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

Answer 1

Hint: Let the greater number be x and the smaller number be y.
Then, form linear equations using the information provided in the question.
Finally, solve the pair of linear equations to get the values of x and y i.e. greater number and smaller number respectively.

Complete step-by-step answer:
Let the greater number be x and the smaller number be y.
It is given that the sum of the numbers is 80.
 $\Rightarrow x + y = 80$ … (1)
Also, it is given that the greater number exceeds twice the smaller number by 11.
 $\therefore x = 2y + 11$ … (2)
Now, substituting the equation (2) in equation (1) to get the value of y.
 $
  \Rightarrow 2y + 11 + y = 80 \\
  \Rightarrow 3y = 80 - 11 \\
  \Rightarrow 3y = 69 \\
  \Rightarrow y = \dfrac{{69}}{3} \\
  \Rightarrow y = 23 \\
 $
Substituting y = 23, in equation (1) gives the value of x.
 $
  \Rightarrow x + 23 = 80 \\
  \Rightarrow x = 80 - 23 \\
  \Rightarrow x = 57 \\
 $
Thus, the greater number is 57 and the smaller number is 23.

Note: Alternate method:
Let the greater number be x and the smaller number be y.
It is given that the sum of the numbers is 80.
 $\Rightarrow x + y = 80$ … (1)
Also, it is given that the greater number exceeds twice the smaller number by 11.
 $\Rightarrow x = 2y + 11$
 $\Rightarrow x - 2y = 11$ … (2)
Now, we will find the value of y, by multiplying the equation (1) by -1.
 $
  \Rightarrow - 1\left( {x + y} \right) = - 80 \\
  \Rightarrow - x - y = - 80 \\
 $
So,
 $
   - x - y = - 80 \\
  x - 2y = 11 \\
 $

 $ - 3y = - 69$
 $
  \Rightarrow y = \dfrac{{ - 69}}{{ - 3}} \\
  \Rightarrow y = 23 \\
 $
Substituting y = 23, in equation (1) gives the value of x.
 $
  \Rightarrow x + 23 = 80 \\
  \Rightarrow x = 80 - 23 \\
  \Rightarrow x = 57 \\
 $
Thus, the greater number is 57 and the smaller number is 23.