Constant of Proportionality Worksheets - Free Printable

Let's solve each problem on the "Constant of Proportionality - Table" worksheet step by step.

We are given tables where $ x $ and $ y $ are proportional. That means:

$$
y = kx \quad \text{or} \quad k = \frac{y}{x}
$$

The constant of proportionality, $ k $, is the ratio $ \frac{y}{x} $, and it should be the same for all pairs in a proportional relationship.

We’ll find $ k $ using known values and then use it to fill in missing values.

---

1)

Given:
| x | 9 | 18 | 27 | ? |
|---|---|----|----|---|
| y | 36 | 72 | 108 | 8 |

Find $ k $:
Use first pair: $ \frac{y}{x} = \frac{36}{9} = 4 $

So $ k = 4 $

Now find missing $ x $ when $ y = 8 $:
$$
y = kx \Rightarrow 8 = 4x \Rightarrow x = \frac{8}{4} = 2
$$

✔ Answer: Missing $ x = 2 $

---

2)

Given:
| x | 4 | 8 | 15 | ? |
|---|---|---|----|---|
| y | 4 | ? | ? | 20 |

Find $ k $:
Use first pair: $ \frac{4}{4} = 1 $

So $ k = 1 $

Now:
- When $ x = 8 $, $ y = 1 \times 8 = 8 $
- When $ x = 15 $, $ y = 1 \times 15 = 15 $
- When $ y = 20 $, $ x = \frac{20}{1} = 20 $

✔ Answers:
$ y = 8 $, $ y = 15 $, $ x = 20 $

---

3)

Given:
| x | 12 | 30 | 48 |
|---|----|----|----|
| y | ? | 3 | 8 |

Find $ k $ using second pair: $ \frac{y}{x} = \frac{3}{30} = 0.1 $

Check with third pair: $ \frac{8}{48} = \frac{1}{6} \approx 0.166... $ → Not matching!

Wait — this is inconsistent. Let’s check if we made a mistake.

Wait — maybe we need to find consistent $ k $. Try using $ x = 12 $, $ y = ? $, but we don’t have $ y $. Use other pairs.

But we only know two values: $ (30, 3) $ and $ (48, 8) $

Try $ k $ from $ (30, 3) $: $ \frac{3}{30} = 0.1 $

From $ (48, 8) $: $ \frac{8}{48} = \frac{1}{6} \approx 0.1667 $ → Not equal.

Hmm — not proportional? But the problem says they are proportional.

Wait — perhaps I misread.

Look again:

| x | 12 | 30 | 48 |
|---|----|----|----|
| y | ? | 3 | 8 |

Is $ \frac{3}{30} = \frac{8}{48} $?
Left: $ 0.1 $, Right: $ \frac{8}{48} = \frac{1}{6} \approx 0.1667 $ → No.

Not equal.

But wait — maybe the table is misaligned?

Wait — let's double-check the image.

Actually, looking at the original image, it seems like:

For problem 3:
- $ x = 12 $, $ y = ? $
- $ x = 30 $, $ y = 3 $
- $ x = 48 $, $ y = 8 $

But $ \frac{3}{30} = 0.1 $, $ \frac{8}{48} \approx 0.1667 $ — not proportional.

Wait — could there be a typo? Or maybe I'm reading wrong.

Wait — perhaps the $ y $ values are not corresponding to those $ x $'s?

No — the table shows them aligned.

Alternatively, maybe the constant is not consistent — but the problem states that $ x $ and $ y $ are proportional.

So perhaps one of the values is wrong?

Wait — let’s suppose the relationship is proportional, so $ y = kx $. Then $ k $ must be constant.

Let’s try to find $ k $ from the two known pairs:

From $ (30, 3) $: $ k = \frac{3}{30} = 0.1 $

Then for $ x = 48 $, $ y = 0.1 \times 48 = 4.8 $, but actual $ y = 8 $ → not match.

From $ (48, 8) $: $ k = \frac{8}{48} = \frac{1}{6} \approx 0.1667 $

Then for $ x = 30 $, $ y = \frac{1}{6} \times 30 = 5 $, but actual $ y = 3 $ → no.

So contradiction.

Wait — unless the $ y $ value for $ x = 30 $ is not 3? Let's recheck the image.

Looking back at the user's image:

> 3)
> x: 12, 30, 48
> y: ?, 3, 8

Yes — that's what it says.

But this cannot be proportional because $ \frac{3}{30} = 0.1 $, $ \frac{8}{48} = \frac{1}{6} \approx 0.1667 $

So unless the data is incorrect, or I'm misreading.

Wait — maybe the $ y $ values are not for those $ x $'s? No — they are aligned.

Wait — perhaps the table is asking us to assume proportionality and find the missing $ y $, but the other values are not consistent?

That can't be — the problem says "the values of $ x $ and $ y $ are proportional".

So likely, the values are intended to be consistent.

Wait — perhaps I misread the numbers.

Let me assume the problem is correct and re-express.

Wait — maybe the $ y $ value for $ x = 30 $ is not 3?

But the image says:
```
x 12 30 48
y ? 3 8
```

Wait — could it be that the $ y $ values are in order, but not necessarily matching?

No — in a table, columns correspond.

Alternatively, maybe the missing value is the key.

Suppose $ k $ is constant.

Let’s suppose $ k = \frac{y}{x} $ is constant.

Let’s denote the missing $ y $ as $ y_1 $ for $ x = 12 $

Then:
- $ \frac{y_1}{12} = \frac{3}{30} = \frac{8}{48} $ — but $ \frac{3}{30} = 0.1 $, $ \frac{8}{48} = 0.1667 $ — not equal.

So unless both are meant to give same $ k $, it’s impossible.

Wait — unless the $ y $ value for $ x = 48 $ is not 8?

But it says 8.

Wait — maybe it's a typo in the problem? Or I’m misreading.

Wait — perhaps the $ y $ values are not for the same row?

No — standard table format.

Wait — let's look at the image again.

Actually, upon closer inspection of the original image (as described), problem 3 is:

```
x 12 30 48
y ? 3 8
```

But $ \frac{3}{30} = 0.1 $, $ \frac{8}{48} = \frac{1}{6} \approx 0.1667 $ — not equal.

So unless the constant is different, it’s not proportional.

But the problem says they are proportional.

So perhaps the $ y $ value for $ x = 30 $ is not 3?

Wait — maybe it's 30? But it says 3.

Wait — maybe the number is misprinted.

Alternatively, perhaps the $ y $ values are in reverse?

Wait — let's try assuming that $ y = kx $, and use $ (30, 3) $ to get $ k = 0.1 $

Then:
- For $ x = 12 $, $ y = 0.1 \times 12 = 1.2 $
- For $ x = 48 $, $ y = 0.1 \times 48 = 4.8 $ — but actual $ y = 8 $ — doesn't match.

If we use $ (48, 8) $, $ k = 8/48 = 1/6 \approx 0.1667 $

Then $ y $ for $ x = 30 $: $ 0.1667 \times 30 = 5 $, but given $ y = 3 $ — no.

So inconsistency.

Unless the $ y $ value for $ x = 30 $ is actually 5, not 3.

But the image says 3.

Wait — perhaps the $ x $ values are not in order?

Wait — maybe the table has a typo.

Alternatively, perhaps I should skip and come back.

Let’s move to problem 4 — maybe it's clearer.

---

4)

Given:
| x | ? | 4 | 8 | 12 |
|---|---|---|---|---|
| y | 25 | 40 | 20 | 60 |

So $ x $: ?, 4, 8, 12
$ y $: 25, 40, 20, 60

Find $ k $ using known pairs.

Use $ x = 4 $, $ y = 40 $: $ k = \frac{40}{4} = 10 $

Check $ x = 8 $, $ y = 20 $: $ \frac{20}{8} = 2.5 $ → not 10 → not proportional?

Wait — $ 20/8 = 2.5 $, $ 40/4 = 10 $ — not same.

But problem says they are proportional.

Wait — maybe $ y $ values are not matching?

Wait — let's see:

| x | ? | 4 | 8 | 12 |
|---|---|---|---|---|
| y | 25 | 40 | 20 | 60 |

So:
- $ x=4 $, $ y=40 $ → $ k = 10 $
- $ x=8 $, $ y=20 $ → $ k = 2.5 $
- $ x=12 $, $ y=60 $ → $ k = 5 $

All different — not proportional.

But problem says they are.

This suggests either:
- The values are misread, or
- The table is not correctly transcribed.

Wait — perhaps the $ y $ values are not in the same order?

Or maybe the $ x $ values are not aligned?

Wait — let’s go back to the original image description.

User said:

> 4)
> x: ? 4 8 12
> y: 25 40 20 60

But $ y/x $:
- 40/4 = 10
- 20/8 = 2.5
- 60/12 = 5

Not constant.

But if we assume $ k $ is constant, then something is wrong.

Wait — unless the $ y $ value for $ x = 4 $ is not 40?

But it says 40.

Wait — perhaps the $ y $ values are not proportional to $ x $, but we are to find the missing $ x $ such that $ y = 25 $, and $ k $ is consistent?

But without a consistent $ k $, we can’t.

Unless we use one pair to find $ k $, and ignore others.

But that defeats the purpose.

Wait — perhaps the table is:

| x | ? | 4 | 8 | 12 |
|---|---|---|---|---|
| y | 25 | 40 | 20 | 60 |

But maybe the $ y $ values are not for the same rows?

No — standard convention.

Wait — let's try to see if any two pairs give same $ k $.

- $ \frac{40}{4} = 10 $
- $ \frac{20}{8} = 2.5 $
- $ \frac{60}{12} = 5 $
- $ \frac{25}{?} = k $

No common $ k $.

So unless the problem has a typo, this is problematic.

But perhaps I misread the numbers.

Wait — maybe the $ y $ values are:
- for $ x = ? $: $ y = 25 $
- $ x = 4 $: $ y = 40 $
- $ x = 8 $: $ y = 20 $
- $ x = 12 $: $ y = 60 $

But $ 40/4 = 10 $, $ 60/12 = 5 $ — not same.

Wait — unless the $ y $ value for $ x = 4 $ is not 40?

But it is.

Wait — perhaps the $ x $ values are not in order?

No.

Wait — let’s look at problem 5.

---

5)

Given:
| x | 7 | 11 | 4 |
|---|---|----|---|
| y | ? | 32 | 24 |

So:
- $ x = 7 $, $ y = ? $
- $ x = 11 $, $ y = 32 $
- $ x = 4 $, $ y = 24 $

Find $ k $ from $ (11, 32) $: $ k = \frac{32}{11} \approx 2.909 $

From $ (4, 24) $: $ k = \frac{24}{4} = 6 $ — not equal.

Again, not proportional.

This is concerning.

Wait — perhaps the $ y $ value for $ x = 4 $ is not 24?

But it says 24.

Wait — maybe the table is:

| x | 7 | 11 | 4 |
|---|---|----|---|
| y | ? | 32 | 24 |

But $ 32/11 \approx 2.909 $, $ 24/4 = 6 $ — not same.

So unless there's a typo, these aren't proportional.

But the problem says they are.

Perhaps the values are switched?

Wait — let’s try to assume that $ y = kx $, and use the pair $ (4, 24) $: $ k = 6 $

Then:
- $ y = 6 \times 7 = 42 $
- $ y = 6 \times 11 = 66 $, but given $ y = 32 $ — no.

Use $ (11, 32) $: $ k = 32/11 \approx 2.909 $

Then $ y $ for $ x = 4 $: $ 2.909 \times 4 \approx 11.636 $, but given 24 — no.

So not working.

Wait — perhaps the $ y $ values are for different $ x $?

No.

Maybe the table is meant to be read differently.

Wait — perhaps the first column is $ x = 7 $, $ y = ? $; second: $ x = 11 $, $ y = 32 $; third: $ x = 4 $, $ y = 24 $

But still, not proportional.

Unless the constant is not the same — but the problem says it is.

I think there might be a mistake in how the tables are presented.

Let’s go back to the original image description.

User wrote:

> 1) x: 9, 18, 27, ? ; y: 36, 72, 108, 8
> 2) x: 4, 8, 15, ? ; y: 4, ?, ?, 20
> 3) x: 12, 30, 48 ; y: ?, 3, 8
> 4) x: ?, 4, 8, 12 ; y: 25, 40, 20, 60
> 5) x: 7, 11, 4 ; y: ?, 32, 24
> 6) x: 14, ?, 15, 3 ; y: 21, 6, ?, ?
> 7) x: 4, 5, 36 ; y: 14, ?, 12
> 8) x: 3, 4, 7 ; y: 18, ?, ?

Now, let’s try to fix the issues.

Wait — for problem 3: $ x = 12, 30, 48 $; $ y = ?, 3, 8 $

Let’s assume the relationship is proportional.

Let’s suppose $ k = \frac{y}{x} $ is constant.

Let’s use $ (30, 3) $: $ k = 3/30 = 0.1 $

Then for $ x = 12 $, $ y = 0.1 \times 12 = 1.2 $

For $ x = 48 $, $ y = 0.1 \times 48 = 4.8 $, but given $ y = 8 $ — not match.

But if we use $ (48, 8) $: $ k = 8/48 = 1/6 \approx 0.1667 $

Then $ y $ for $ x = 30 $: $ 0.1667 \times 30 = 5 $, but given $ y = 3 $ — no.

So unless the $ y $ value for $ x = 30 $ is 5, or for $ x = 48 $ is 4.8, it’s not proportional.

But it’s not.

Wait — perhaps the $ y $ value for $ x = 30 $ is not 3, but 30?

But it says 3.

Wait — maybe the number is 30, but written as 3?

Unlikely.

Alternatively, perhaps the $ x $ values are not in order.

But even then, the ratios must be constant.

Let’s try problem 6.

---

6)

| x | 14 | ? | 15 | 3 |
|---|----|---|----|---|
| y | 21 | 6 | ? | ? |

So:
- $ x = 14 $, $ y = 21 $ → $ k = 21/14 = 1.5 $
- $ x = ? $, $ y = 6 $ → $ x = 6 / 1.5 = 4 $
- $ x = 15 $, $ y = ? $ → $ y = 1.5 \times 15 = 22.5 $
- $ x = 3 $, $ y = ? $ → $ y = 1.5 \times 3 = 4.5 $

So:
- $ x = 4 $
- $ y = 22.5 $
- $ y = 4.5 $

✔ So answers: $ x = 4 $, $ y = 22.5 $, $ y = 4.5 $

And $ k = 1.5 $

---

7)

| x | 4 | 5 | 36 |
|---|---|---|----|
| y | 14 | ? | 12 |

Use $ (4, 14) $: $ k = 14/4 = 3.5 $

Check $ (36, 12) $: $ k = 12/36 = 1/3 \approx 0.333 $ — not 3.5

Not proportional.

But problem says they are.

Wait — unless the $ y $ value for $ x = 36 $ is not 12?

But it says 12.

Wait — if $ k = 3.5 $, then $ y = 3.5 \times 36 = 126 $, not 12.

If $ k = 12/36 = 1/3 $, then $ y $ for $ x = 4 $: $ (1/3) \times 4 = 1.333 $, not 14.

So not proportional.

But the problem says they are.

Wait — perhaps the $ y $ value for $ x = 4 $ is 14, and for $ x = 36 $ is 12, but that would require $ k = 14/4 = 3.5 $, $ k = 12/36 = 1/3 $ — not same.

So not possible.

Unless the missing $ y $ for $ x = 5 $ is to be found, but we need consistent $ k $.

But no consistent $ k $.

So perhaps there is a typo in the problem.

Wait — maybe the $ y $ value for $ x = 36 $ is 126, not 12?

But it says 12.

Alternatively, maybe the $ x $ value for $ y = 12 $ is not 36?

But it is.

I think there may be errors in the provided table.

But let’s look at problem 8.

---

8)

| x | 3 | 4 | 7 |
|---|---|---|---|
| y | 18 | ? | ? |

Use $ (3, 18) $: $ k = 18/3 = 6 $

Then:
- $ x = 4 $, $ y = 6 \times 4 = 24 $
- $ x = 7 $, $ y = 6 \times 7 = 42 $

✔ So $ y = 24 $, $ y = 42 $

---

Now, going back, only problems 1, 2, 6, 8 seem consistent.

Let’s re-examine problem 3.

Wait — perhaps the $ y $ value for $ x = 30 $ is not 3, but 30?

But it says 3.

Wait — maybe it's 30, and the digit is small.

But in the text, it says "3".

Similarly, in problem 4, if $ y = 40 $ for $ x = 4 $, $ k = 10 $, then $ y = 25 $ implies $ x = 25/10 = 2.5 $

But then $ x = 8 $, $ y = 20 $: $ 20/8 = 2.5 $ — not 10

So not consistent.

But if we use $ (8, 20) $: $ k = 2.5 $, then $ x = 25 / 2.5 = 10 $

Then $ x = 4 $, $ y = 2.5 \times 4 = 10 $, but given $ y = 40 $ — no.

So not.

Wait — unless the $ y $ value for $ x = 4 $ is 10, not 40.

But it says 40.

I think there might be typos in the problem.

But let’s assume that in problem 3, the $ y $ value for $ x = 30 $ is 3, and for $ x = 48 $ is 8, and we are to find the missing $ y $ for $ x = 12 $, and assume proportionality.

But since $ 3/30 = 0.1 $, $ 8/48 = 1/6 $, not equal, it’s impossible.

Unless the constant is not the same — but the problem says it is.

Perhaps the table is not meant to be read as proportional across all rows, but only some.

But the instruction says: "Determine the constant of proportionality and find the missing values."

So it must be proportional.

Given the inconsistencies, perhaps the user made a mistake in transcription.

But based on what's given, let’s assume that in each table, the relationship is proportional, and the values are correct, and try to find a consistent $ k $.

Let’s try problem 3 again.

Suppose the $ y $ value for $ x = 30 $ is 3, and for $ x = 48 $ is 8.

Then $ k = 3/30 = 0.1 $, $ k = 8/48 = 1/6 $

Not equal.

But if we take the average or something — no.

Alternatively, perhaps the $ y $ value for $ x = 30 $ is 30, not 3.

Let’s assume that.

Then $ k = 30/30 = 1 $

Then for $ x = 48 $, $ y = 48 $, but given 8 — no.

If $ y = 8 $ for $ x = 48 $, $ k = 8/48 = 1/6 $

Then for $ x = 30 $, $ y = 30 * 1/6 = 5 $, not 3.

So not.

Wait — perhaps the $ x $ values are not 12, 30, 48, but something else.

But the user said: "x: 12, 30, 48"

I think there might be a typo in the problem.

But for now, let’s focus on the ones that work.

---

Final Answers (for the ones that make sense):

#### 1)
- $ k = 36/9 = 4 $
- When $ y = 8 $, $ x = 8/4 = 2 $
- ✔ Missing $ x = 2 $

#### 2)
- $ k = 4/4 = 1 $
- $ y = 1 \times 8 = 8 $
- $ y = 1 \times 15 = 15 $
- When $ y = 20 $, $ x = 20/1 = 20 $
- ✔ $ y = 8 $, $ y = 15 $, $ x = 20 $

#### 3) Inconsistent — cannot determine.

#### 4) Inconsistent — cannot determine.

#### 5) Inconsistent — cannot determine.

#### 6)
- $ k = 21/14 = 1.5 $
- $ x = 6 / 1.5 = 4 $
- $ y = 1.5 \times 15 = 22.5 $
- $ y = 1.5 \times 3 = 4.5 $
- ✔ $ x = 4 $, $ y = 22.5 $, $ y = 4.5 $

#### 7) Inconsistent — cannot determine.

#### 8)
- $ k = 18/3 = 6 $
- $ y = 6 \times 4 = 24 $
- $ y = 6 \times 7 = 42 $
- ✔ $ y = 24 $, $ y = 42 $

---

Conclusion:

Only problems 1, 2, 6, and 8 have consistent proportional relationships.

For the others, the given values are not proportional, suggesting possible typos in the worksheet.

But based on the assumption that the relationship is proportional and using the available data:

---

✔ Final Answers:

1) $ x = 2 $
2) $ y = 8 $, $ y = 15 $, $ x = 20 $
3) Cannot determine (inconsistent)
4) Cannot determine (inconsistent)
5) Cannot determine (inconsistent)
6) $ x = 4 $, $ y = 22.5 $, $ y = 4.5 $
7) Cannot determine (inconsistent)
8) $ y = 24 $, $ y = 42 $

---

If you can provide a clearer image or confirm the values, I can refine the solution.