Question

The area of the triangular region in the first quadrant bounded on the left by the y-axis, bounded above by the line $7x+4y=168$ and bounded below by the line $5x+3y=121$, is?

• A. $\frac{50}{3}$
• B. $\frac{52}{7}$
• C. $\frac{53}{10}$
• D. $\frac{7}{10}$

Answer 1

The correct option is A $\frac{50}{3}$

The intersection point of $7x+4y=168$ and $5x+3y=121$ is found as

$x+\frac{4}{7}y-x-\frac{3}{5}y=\frac{168}{7}-\frac{121}{5}⟹y=7⟹x=20$

The triangle is has vertices $(0,42)$, $(20,7)$ and $(0,40.33)$.

Hence, area of triangle = $\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ 0& 20& 0\\ 42& 7& 40.33\end{array}\right|$

$=\frac{1}{2}|\frac{2420}{3}-840|$

$=\frac{1}{2}\times \frac{2520-2420}{3}$

$=\frac{50}{3}$

This is the required answer.