Question 14 Marks
A rectangular floor area can be completely tiled with 200 square tiles. If the side leng th of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor
(i) Assuming the original length of each side of a tile be $x$ units, make a quadratic equation from the above information.
(ii) Write the corresponding quadratic equation in standard form.
(iii) (a) Find the value of $x$, the length of side of tile by factorisation.
OR
(b) Solve the quadratic equation for $x$, using quadratic formula.
(i) Assuming the original length of each side of a tile be $x$ units, make a quadratic equation from the above information.
(ii) Write the corresponding quadratic equation in standard form.
(iii) (a) Find the value of $x$, the length of side of tile by factorisation.
OR
(b) Solve the quadratic equation for $x$, using quadratic formula.
Answer
View full question & answer→(i) Area of rectangular floor when 200 square tiles of side $x$ units cover the floor completely is $200 x^2$.
When 128 tiles of side $(x+1)$ units cover the floor completely, the area of floor is $128(x+1)^2$, Therefore, $200 x^2=128(x+1)^2$.
(ii) We have,
$
200 x^2=128(x+1)^2 \Rightarrow 72 x^2-256 x-128=0 \Rightarrow 9 x^2-32 x-16=0
$
(iii) (a) $9 x^2-32 x-16=0$
$
\begin{array}{ll}
\Rightarrow & 9 x^2-36 x+4 x-16=0 \\
\Rightarrow & (x-4)(9 x+4)=0 \Rightarrow x-4=0 \Rightarrow x=4
\end{array}
$
$
[\because 9 x+4 \neq 0]
$
OR
(b) $9 x^2-32 x-16=0$
$
x=\frac{32 \pm \sqrt{(-32)^2+4 \times 9 \times 16}}{18}=\frac{32 \pm \sqrt{1600}}{18}=\frac{32 \pm 40}{18}=4,-\frac{4}{9} \Rightarrow x=4 \qquad [\because x > y]
$
When 128 tiles of side $(x+1)$ units cover the floor completely, the area of floor is $128(x+1)^2$, Therefore, $200 x^2=128(x+1)^2$.
(ii) We have,
$
200 x^2=128(x+1)^2 \Rightarrow 72 x^2-256 x-128=0 \Rightarrow 9 x^2-32 x-16=0
$
(iii) (a) $9 x^2-32 x-16=0$
$
\begin{array}{ll}
\Rightarrow & 9 x^2-36 x+4 x-16=0 \\
\Rightarrow & (x-4)(9 x+4)=0 \Rightarrow x-4=0 \Rightarrow x=4
\end{array}
$
$
[\because 9 x+4 \neq 0]
$
OR
(b) $9 x^2-32 x-16=0$
$
x=\frac{32 \pm \sqrt{(-32)^2+4 \times 9 \times 16}}{18}=\frac{32 \pm \sqrt{1600}}{18}=\frac{32 \pm 40}{18}=4,-\frac{4}{9} \Rightarrow x=4 \qquad [\because x > y]
$
