Question

x and y are two non-negative integral numbers such that $2x+y=10$. The sum of the maximum and minimum values of $\left( x+y \right)$ is
A. 6
B. 9
C. 10
D. 15

Answer 1

Hint: We first try to find the characteristics for the input $y$ in $2x+y=10$. We take even inputs for it and find the value for $x$. We find possible maximum and minimum values of $\left( x+y \right)$ and find its sum.

Complete step by step answer:
$x$ and $y$ are two non-negative integral numbers such that $2x+y=10$.
Therefore, the possible range for $x$ and $y$ will be $x,y\ge 0$.
$y$ has to be an even number as $y=10-2x$ is an even number.
Now we take even values from 0 for $y$ to find the value for $x$.
We take $y=0$. We get $2x+0=10$ which gives \[x=\dfrac{10}{2}=5\]. In this case the value of $\left( x+y \right)$ will be $\left( x+y \right)=5+0=5$.
We take $y=2$. We get $2x+2=10$ which gives \[x=\dfrac{10-2}{2}=4\]. In this case the value of $\left( x+y \right)$ will be $\left( x+y \right)=4+2=6$.
We take $y=4$. We get $2x+4=10$ which gives \[x=\dfrac{10-4}{2}=3\]. In this case the value of $\left( x+y \right)$ will be $\left( x+y \right)=3+4=7$.
We take $y=6$. We get $2x+6=10$ which gives \[x=\dfrac{10-6}{2}=2\]. In this case the value of $\left( x+y \right)$ will be $\left( x+y \right)=2+6=8$.
We take $y=8$. We get $2x+8=10$ which gives \[x=\dfrac{10-8}{2}=1\]. In this case the value of $\left( x+y \right)$ will be $\left( x+y \right)=1+8=9$.
We take $y=10$. We get $2x+10=10$ which gives \[x=\dfrac{10-10}{2}=0\]. In this case the value of $\left( x+y \right)$ will be $\left( x+y \right)=0+10=10$.
Therefore, the maximum and minimum values of $\left( x+y \right)$ is 10 and 5 respectively.
The sum will be $10+5=15$.

So, the correct answer is “Option D”.

Note: We cannot take the value of $y=10$ as in that case the value of \[x=\dfrac{10-12}{2}=-1\]. It becomes negative and creates a contradiction of $x,y\ge 0$. From the condition of $2x+10=10$, we can also tell that $x\le 5,y\le 10$ not to cross the total sum value.