Inverse Proportion Worksheet | Printable Maths Worksheets - Free Printable

Explanation:
We are given three sections, each about inverse proportion. In inverse proportion, when one quantity increases, the other decreases so that their product stays constant.

Let’s go through each section step by step.

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Section A: $ y $ is inversely proportional to $ x $

1) Proportional relationship:
If $ y $ is inversely proportional to $ x $, then
$$
y \propto \frac{1}{x} \quad \text{or} \quad y = \frac{k}{x}
$$
where $ k $ is the constant of proportionality.

2) Equation using $ k $:
$$
y = \frac{k}{x}
\quad \text{or} \quad xy = k
$$

3) Use given values to find $ k $, then fill missing cells.

First row:
Given $ y = 3 $, $ x = 4 $.
So $ k = xy = 3 \times 4 = 12 $

→ Equation: $ y = \frac{12}{x} $ or $ xy = 12 $

Now find missing values:

- $ k = 12 $
- Equation in $ y $ and $ x $: $ xy = 12 $ (or $ y = \frac{12}{x} $)
- Given $ x = 2 $, find $ y $:
$ y = \frac{12}{2} = 6 $

So first row:
$ y = 3 $, $ x = 4 $, $ k = 12 $, equation: $ xy = 12 $, $ y = 6 $ when $ x = 2 $

Wait — but in the table, the last column says “x = 2” for the first row, and they want us to find the corresponding $ y $. So yes, $ y = 6 $.

But hold on — the table layout is:

| y | x | k | Equation | y = | x = |
|---|---|----|-----------|------|------|
| 3 | 4 | ? | ? | ? | 2 |
| 1/2 | 10 | ? | ? | 15 | ? |

So in row 1:
- Given: y = 3, x = 4 → find k = 3×4 = 12
- Equation: $ y = \frac{12}{x} $ or $ xy = 12 $
- Then, when x = 2, y = 12 / 2 = 6

So fill:
k = 12
Equation: $ xy = 12 $ (or $ y = \frac{12}{x} $)
y = 6
x = 2 (given)

Row 2:
Given: $ y = \frac{1}{2} $, $ x = 10 $
So $ k = xy = \frac{1}{2} \times 10 = 5 $

Equation: $ y = \frac{5}{x} $ or $ xy = 5 $

Given $ y = 15 $, find $ x $:
From $ y = \frac{5}{x} $,
$ 15 = \frac{5}{x} \Rightarrow x = \frac{5}{15} = \frac{1}{3} $

So row 2:
k = 5
Equation: $ xy = 5 $
y = 15 (given), so x = 1/3

✔ Section A done:

Row 1: k = 12, equation = $ xy = 12 $, y = 6
Row 2: k = 5, equation = $ xy = 5 $, x = 1/3

---

Section B: $ p $ is inversely proportional to $ h^2 $

1) Proportional relationship:
$ p \propto \frac{1}{h^2} $

2) Equation:
$$
p = \frac{k}{h^2} \quad \text{or} \quad ph^2 = k
$$

3) Fill table:

Row 1:
Given $ p = 1.5 $, $ h = 10 $
So $ k = p h^2 = 1.5 \times 10^2 = 1.5 \times 100 = 150 $

Equation: $ p = \frac{150}{h^2} $ or $ ph^2 = 150 $

Given $ h = 0.1 $, find $ p $:
$ p = \frac{150}{(0.1)^2} = \frac{150}{0.01} = 15000 $

So row 1:
k = 150
Equation: $ ph^2 = 150 $
p = 15000
h = 0.1 (given)

Row 2:
Given $ p = 1 $, $ h = 6 $
So $ k = 1 \times 6^2 = 1 \times 36 = 36 $

Equation: $ p = \frac{36}{h^2} $

Given $ p = 4 $, find $ h $:
$ 4 = \frac{36}{h^2} \Rightarrow h^2 = \frac{36}{4} = 9 \Rightarrow h = \sqrt{9} = 3 $
(Since h is likely positive in context, we take positive root.)

So row 2:
k = 36
Equation: $ ph^2 = 36 $
p = 4 (given)
h = 3

✔ Section B done:

Row 1: k = 150, equation = $ ph^2 = 150 $, p = 15000
Row 2: k = 36, equation = $ ph^2 = 36 $, h = 3

---

Section C: $ Z $ is inversely proportional to $ \sqrt[3]{w} $

That is, $ Z \propto \frac{1}{\sqrt[3]{w}} $

1) Proportional relationship:
$ Z = \frac{k}{\sqrt[3]{w}} $

2) Equation:
$$
Z = \frac{k}{\sqrt[3]{w}} \quad \text{or} \quad Z \cdot \sqrt[3]{w} = k
$$

Note: $ \sqrt[3]{w} = w^{1/3} $, so $ Z w^{1/3} = k $

3) Fill table:

Row 1:
Given $ Z = 2 $, $ w = 64 $
Find $ \sqrt[3]{64} = 4 $, since $ 4^3 = 64 $

So $ k = Z \cdot \sqrt[3]{w} = 2 \times 4 = 8 $

Equation: $ Z = \frac{8}{\sqrt[3]{w}} $ or $ Z \cdot \sqrt[3]{w} = 8 $

Given $ w = 8 $, find $ Z $:
$ \sqrt[3]{8} = 2 $, so $ Z = \frac{8}{2} = 4 $

So row 1:
k = 8
Equation: $ Z \cdot \sqrt[3]{w} = 8 $
Z = 4
w = 8 (given)

Row 2:
Given $ Z = 12 $, $ w = 125 $
$ \sqrt[3]{125} = 5 $, since $ 5^3 = 125 $

So $ k = 12 \times 5 = 60 $

Equation: $ Z = \frac{60}{\sqrt[3]{w}} $

Given $ Z = 10 $, find $ w $:
$ 10 = \frac{60}{\sqrt[3]{w}} \Rightarrow \sqrt[3]{w} = \frac{60}{10} = 6 $

Then $ w = 6^3 = 216 $

So row 2:
k = 60
Equation: $ Z \cdot \sqrt[3]{w} = 60 $
Z = 10 (given)
w = 216

✔ Section C done:

Row 1: k = 8, equation = $ Z \cdot \sqrt[3]{w} = 8 $, Z = 4
Row 2: k = 60, equation = $ Z \cdot \sqrt[3]{w} = 60 $, w = 216

Now compile all answers clearly.

Final Answer:
Section A:
- Row 1: k = 12, equation = $ xy = 12 $, y = 6
- Row 2: k = 5, equation = $ xy = 5 $, x = $ \frac{1}{3} $

Section B:
- Row 1: k = 150, equation = $ ph^2 = 150 $, p = 15000
- Row 2: k = 36, equation = $ ph^2 = 36 $, h = 3

Section C:
- Row 1: k = 8, equation = $ Z\sqrt[3]{w} = 8 $, Z = 4
- Row 2: k = 60, equation = $ Z\sqrt[3]{w} = 60 $, w = 216

But the question asks to fill the table — so final answer should be the filled values in order as per the table columns.

Let me reconstruct the tables with only the missing entries (since some are given):

Section A table missing cells:
Row 1: k = 12, Equation = $ xy = 12 $, y = 6
Row 2: k = 5, Equation = $ xy = 5 $, x = $ \frac{1}{3} $

Section B table missing cells:
Row 1: k = 150, Equation = $ ph^2 = 150 $, p = 15000
Row 2: k = 36, Equation = $ ph^2 = 36 $, h = 3

Section C table missing cells:
Row 1: k = 8, Equation = $ Z\sqrt[3]{w} = 8 $, Z = 4
Row 2: k = 60, Equation = $ Z\sqrt[3]{w} = 60 $, w = 216

Since the output format expects only the final answer (and no extra text), I’ll list all missing values in order as they appear in the tables, row by row, column by column (k, Equation, y or x, etc.), but the instruction says: *Provide only the final correct answer* — likely meaning the completed table values.

However, to be safe and precise, and since the user expects a single “Final Answer” block, I will format it as the filled-in missing cells in the order of the tables:

Section A:
- Row 1: k = 12, Equation: $ xy = 12 $, y = 6
- Row 2: k = 5, Equation: $ xy = 5 $, x = $ \frac{1}{3} $

Section B:
- Row 1: k = 150, Equation: $ ph^2 = 150 $, p = 15000
- Row 2: k = 36, Equation: $ ph^2 = 36 $, h = 3

Section C:
- Row 1: k = 8, Equation: $ Z\sqrt[3]{w} = 8 $, Z = 4
- Row 2: k = 60, Equation: $ Z\sqrt[3]{w} = 60 $, w = 216

But since the instruction says “only the final correct answer”, and this is a multi-part fill-in, the cleanest is to give each missing value in order as per the blank cells in the original layout.

Let me count blanks per row:

Section A has 6 blanks total:
Row 1: k, Equation, y (since x=2 given) → 3 blanks
Row 2: k, Equation, x (since y=1/2, x=10 given, and y=15 given in y column? Wait careful.)

Actually, looking again at the table structure:

Columns: y | x | k | Equation | y = | x =

Row 1: y=3, x=4, [k], [Equation], [y= ?], x=2
→ So they give y and x for first two columns, and x=2 in last column, asking for corresponding y. So blanks: k, Equation, y (when x=2)

Row 2: y=1/2, x=10, [k], [Equation], y=15, [x=?]
→ blanks: k, Equation, x (when y=15)

So 6 blanks in Section A:
1. k₁ = 12
2. Eq₁: xy = 12
3. y₁ = 6
4. k₂ = 5
5. Eq₂: xy = 5
6. x₂ = 1/3

Section B: columns p | h | k | Equation | p = | h =

Row 1: p=1.5, h=10, [k], [Eq], [p=?], h=0.1
→ blanks: k, Eq, p

Row 2: p=1, h=6, [k], [Eq], p=4, [h=?]
→ blanks: k, Eq, h

So 6 blanks:
7. k₁ = 150
8. Eq₁: ph² = 150
9. p₁ = 15000
10. k₂ = 36
11. Eq₂: ph² = 36
12. h₂ = 3

Section C: Z | w | k | Equation | Z = | w =

Row 1: Z=2, w=64, [k], [Eq], [Z=?], w=8
→ blanks: k, Eq, Z

Row 2: Z=12, w=125, [k], [Eq], Z=10, [w=?]
→ blanks: k, Eq, w

13. k₁ = 8
14. Eq₁: Z∛w = 8
15. Z₁ = 4
16. k₂ = 60
17. Eq₂: Z∛w = 60
18. w₂ = 216

But the problem likely expects just the numeric and expression answers in the table positions.

Given the format, I'll provide the final answer as a structured list matching the table rows.

However, per instructions: Only the final answer, after reasoning.

I will write the completed missing values clearly:

Final Answer:
Section A:
- First row: k = 12, equation = $ xy = 12 $, y = 6
- Second row: k = 5, equation = $ xy = 5 $, x = $ \frac{1}{3} $

Section B:
- First row: k = 150, equation = $ ph^2 = 150 $, p = 15000
- Second row: k = 36, equation = $ ph^2 = 36 $, h = 3

Section C:
- First row: k = 8, equation = $ Z\sqrt[3]{w} = 8 $, Z = 4
- Second row: k = 60, equation = $ Z\sqrt[3]{w} = 60 $, w = 216