Question 15 Marks
If $x$ and $y$ are inversely proportional, find the values of $x _1, x _2, y _1$ and $y _2$ in the table given below.
| $x$ | $8$ |
$x_1$
|
$16$
|
$x_2$
|
$80$ |
| $y$ |
$y_1$
|
$4$
|
$5$
|
$2$
|
$y_2$ |
Answer
View full question & answer→Since $x$ and $y$ are inversely proportional, $xy$ must be a constant.
Therefore,$8 × y_1 = x_1 × 4 = 16 × 5 = x_2 × 2 = 80 ×y_2$
Now, $16 × 5 = 8 × y_1$
$\Rightarrow\frac{80}{4}=\text{x}_1$
$\therefore \ \text{x}_1=20$
$16\times5=\text{x}_2\times2$
$\Rightarrow\frac{80}{2}=\text{x}_2$
$\therefore \ \text{x}_2=40$
$16\times5=80\times\text{y}_2$
$\Rightarrow\frac{80}{80}=\text{y}_2$
$\therefore \ \text{y}_2=1$
Hence, $\text{y}_1=10,\text{x}_1=20,\text{x}_2=40$ and $\text{y}_2=1$
Therefore,$8 × y_1 = x_1 × 4 = 16 × 5 = x_2 × 2 = 80 ×y_2$
Now, $16 × 5 = 8 × y_1$
$\Rightarrow\frac{80}{4}=\text{x}_1$
$\therefore \ \text{x}_1=20$
$16\times5=\text{x}_2\times2$
$\Rightarrow\frac{80}{2}=\text{x}_2$
$\therefore \ \text{x}_2=40$
$16\times5=80\times\text{y}_2$
$\Rightarrow\frac{80}{80}=\text{y}_2$
$\therefore \ \text{y}_2=1$
Hence, $\text{y}_1=10,\text{x}_1=20,\text{x}_2=40$ and $\text{y}_2=1$