Question 14 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, $x y$ must be a constant.
Therefore, $8 \times y _1= x _1 \times 4=16 \times 5= x _2 \times 2=80 \times y _2$
Now, $16 \times 5=8 \times y_1$
$\Rightarrow\frac{80}{4}=\text{x}_1$
$\therefore \ \text{x}_1=20$
$16\times5=\text{x}_2\times2$
$\Rightarrow\frac{80}{2}=\text{x}_2$
$\therefore \ \text{x}_2=40$
$16\times5=80\times\text{y}_2$
$\Rightarrow\frac{80}{80}=\text{y}_2$
$\therefore \ \text{y}_2=1$
Hence, $\text{y}_1=10,\text{x}_1=20,\text{x}_2=40$ and $\text{y}_2=1$
Therefore, $8 \times y _1= x _1 \times 4=16 \times 5= x _2 \times 2=80 \times y _2$
Now, $16 \times 5=8 \times y_1$
$\Rightarrow\frac{80}{4}=\text{x}_1$
$\therefore \ \text{x}_1=20$
$16\times5=\text{x}_2\times2$
$\Rightarrow\frac{80}{2}=\text{x}_2$
$\therefore \ \text{x}_2=40$
$16\times5=80\times\text{y}_2$
$\Rightarrow\frac{80}{80}=\text{y}_2$
$\therefore \ \text{y}_2=1$
Hence, $\text{y}_1=10,\text{x}_1=20,\text{x}_2=40$ and $\text{y}_2=1$