Question

$5$ pencils and $7$ pens together cost Rs. $50$, whereas $7$ pencils and $5$ pens together cost Rs. $46$. Find the cost of one pencil and that of one pen by graphical method.

• A. Cost of one pencil $=$ Rs. $2$ and that of one pen $=$ Rs. $4$
• B. Cost of one pencil $=$ Rs. $3$ and that of one pen $=$ Rs. $5$
• C. Cost of one pencil $=$ Rs. $5$ and that of one pen $=$ Rs. $6$
• D. Cost of one pencil $=$ Rs. $2$ and that of one pen $=$ Rs. $6$

Answer 1

The correct option is B Cost of one pencil $=$ Rs. $3$ and that of one pen $=$ Rs. $5$

Let cost of pencil be Rs. $x$

Cost of pens be Rs. $y$

$5$ pencils and $7$ pens together cost Rs. $50$,

So we get

$5x+7y=50$

Subtract $7y$ both side we get

$5x=50–7y$

Divide by $5$ we get

$x=10-\frac{7}{5}y$

Plug value of $y$ which is factor of $5$ to get whole number so plug $y=5,10,15$ we get

for$y=5$

$x=10-\frac{7}{5}y=10-7=3$

for $y=10$

$x=10-\frac{7}{5}y=10-14=-4$

for$y=15$

$x=10-\frac{7}{5}y=10-21=-11$

Therefore, the required points are $(3,5),(-4,10),(-11,15)$.

Given that $7$ pencils and $5$ pens together cost Rs. $46$

$7x+5y=46$

Subtract $7x$ both side we get

$5y=46–7x$

Divide by $5$ we get

$y=9.2–1.4x$

Plug $x=0,2,4$ we get

for $x=0$

$y=9.2–0=9.2$

for $x=2$

$y=9.2–2.8=6.4$

for$x=4$

$y=9.2–5.6=3.6$

Therefore, the required points are $(0,9.2),(2,6.4),(4,3.6)$.

The graph is as shown below:

Since the point of intersection is $(3,5)$,

Hence, the cost of one pencil is Rs. $3$ and the cost of one pen is Rs. $5$