Question

What will be the cost of painting a wall of area $(m-4y)$ sq m. If cost of painting is $\$(3m-5y)$ per sq m.

• A. $\$(3m^2-17my+20y^2)$
• B. $\$(3m^2+17my+20y^2)$
• C. $\$(3m^2-17my-20y^2)$
• D. $\$(3m^2+17my-20y^2)$

Answer 1

The correct option is A $\$(3m^2-17my+20y^2)$

$\underset{–}{\text{Given}:}$
$\bullet$ Area of the wall $=(m-4y)$ sq m.
$\bullet$ Cost of painting $=(3m-5y)$ per sq m.

$\underset{–}{\text{Need to Find}:}$
Cost of painting of the wall

Cost of painting of the wall
$=$ Area of the wall $\times$ Cost of painting one sq. m

$=(m-4y)\times \$(3m-5y)$

Applying the $FOIL$ rule, we get


$m\times \$3m=\$3m^{(1+1)}=\$3m^2$ $(\because a^m\times a^n=a^{m+n})$

The value of first term in the product $(m-4y)\times \$(3m-5y)$ is $\$3m^2$.


Now, we would multiply the "OUTER" terms in the product $(m-4y)\times \$(3m-5y)$. $($i.e., $m$ and $\$(-5y))$

$m\times \$(-5y)=\$(-5my)$

The value of second term in the product $(m-4y)\times \$(3m-5y)$ is $\$(-5my)$.


Now, we would multiply the "INNER" terms in the product $(m-4y)\times \$(3m-5y)$. $($i.e., $-4y$ and $\$3m$)

$-4y\times \$3m=\$(-12my)$

The value of third term in the product $(m-4y)\times \$(3m-5y)$ is $\$(-12my)$.

Now, we would multiply the "LAST" terms in the product $(m-4y)\times \$(3m-5y)$. $($i.e., $-4y$ and $\$(-5y)$)


$-4y\times \$(-5y)=\$20y^{(1+1)}=\$20y^2$ $(\because a^m\times a^n=a^{m+n})$

The value of fourth term in the product $(m-4y)\times \$(3m-5y)$ is $\$20y^2$.


$\therefore$ Cost of painting of the wall

$=(m-4y)\times \$(3m-5y)$

$=\$[3m^2+(-5my)+(-12my)+20y^2]$

$=\$[3m^2+\underset{–}{(-5my)+(-12my)}+20y^2]$

Combining the like terms, we get

$=\$(3m^2-17my+20y^2)$

The value of area of rectangle is $\$(3m^2-17my+20y^2)$. Therefore, option (a.) is the correct answer.