Question

Consider the inequality $40x+20y\le 120$, where $x$ and $y$ are whole numbers. Then,

• A. $(0,0),(0,5),(1,1),(2,2),(3,0)$ are some solutions of given inequality
• B. The graphical representation of given inequality is
• C. Both (a) and (b)
• D. None of these

Answer 1

Answer: (A)

Given, $40x+20y\le 120,x$ and $y$ are whole numbers.
To start with, let $x=0$. Then, LHS of given inequality is
$40x+20y=40(0)+20y=20y$
Thus, we have
$20y\le 120$
or $y\le 6$
For $x=0$, the corresponding values of $y$ can be $0,1,2,3,4$, 5,6 only. Inthis case, the solutions of given inequality are $(0,0),(0,1),(0,2),(0,3),(0,4),(0,5)$ and $(0,6)$

Similarly, other solutions of given inequality when $x=1,2$ and 3 are $(1,0),(1,1),(1,2),(1,3),(1,4),(2,0),(2,1),(2,2)$, $(3,0)$.
This is shown in above figure.
Statement (b) is incorrect, since $x$ and $y$ are whole numbers. $x$ and $y$ can only take values $\{0,1,2,3\dots \infty \}$. Given shaded region shows negative as well as fractional values of $x$ and $y$.