Math worksheet on constant of proportionality with five problems involving graphs, tables, and word problems.

Let's solve each row step by step to find the constant of proportionality for each box and identify the "odd man out" — the one that doesn't belong.

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🔷 What is the Constant of Proportionality?

In a proportional relationship, $ y = kx $, where:
- $ k $ is the constant of proportionality.
- It’s found by dividing $ y $ by $ x $: $ k = \frac{y}{x} $

We’ll compute $ k $ for each box and see which one has a different value. That one is the "odd man out."

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## ✔ Row 1

| Box | Content |
|-----|--------|
| A | Graph: Points (1,12), (2,24), (3,36) → slope = $ \frac{24-12}{2-1} = 12 $ → $ k = 12 $ |
| B | Jaylin earned $36 in 3 hours → $ \frac{36}{3} = 12 $. $60 in 5 hours → $ \frac{60}{5} = 12 $ → $ k = 12 $ |
| C | Table: $ x=12, y=1 $ → $ \frac{1}{12} = 0.083... $; $ x=24, y=2 $ → $ \frac{2}{24} = 0.083... $; $ x=60, y=5 $ → $ \frac{5}{60} = 0.083... $ → $ k = \frac{1}{12} $ |
| D | Equation: $ y = 12x $ → $ k = 12 $ |

✔ All except C have $ k = 12 $
✘ C has $ k = \frac{1}{12} $ → This is the ODD MAN OUT

👉 Shade Box C

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## ✔ Row 2

| Box | Content |
|-----|--------|
| A | Sophie read 4 books in 2 months → $ \frac{4}{2} = 2 $ books/month. Then 10 books in 5 months → $ \frac{10}{5} = 2 $ → $ k = 2 $ |
| B | Table: $ x=0.1, y=0.1 $ → $ \frac{0.1}{0.1} = 1 $; $ x=2.6, y=2.6 $ → $ \frac{2.6}{2.6} = 1 $; $ x=15, y=15 $ → $ \frac{15}{15} = 1 $ → $ k = 1 $ |
| C | Equation: $ y = x $ → $ k = 1 $ |
| D | Graph: Points (1,1), (2,2), (3,3), (4,4) → $ \frac{y}{x} = 1 $ → $ k = 1 $ |

✔ All except A have $ k = 1 $
✘ A has $ k = 2 $ → This is the ODD MAN OUT

👉 Shade Box A

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## ✔ Row 3

| Box | Content |
|-----|--------|
| A | Equation: $ y = \frac{1}{3}x $ → $ k = \frac{1}{3} \approx 0.333 $ |
| B | Graph: Points (2,6), (4,12), (6,18), (8,24) → $ \frac{6}{2} = 3 $, $ \frac{12}{4} = 3 $, etc. → $ k = 3 $ |
| C | Table: $ x=4, y=12 $ → $ \frac{12}{4} = 3 $; $ x=6, y=18 $ → $ \frac{18}{6} = 3 $; $ x=9, y=27 $ → $ \frac{27}{9} = 3 $ → $ k = 3 $ |
| D | Alexis made 24 cookies with 8 tsp sugar → $ \frac{8}{24} = \frac{1}{3} $; 45 cookies with 15 tsp → $ \frac{15}{45} = \frac{1}{3} $ → $ k = \frac{1}{3} $ |

✔ All except B and C have $ k = \frac{1}{3} $
Wait! Let's check:

- A: $ k = \frac{1}{3} $
- B: $ k = 3 $
- C: $ k = 3 $
- D: $ k = \frac{1}{3} $

So both B and C have $ k = 3 $, while A and D have $ k = \frac{1}{3} $

But we're told there's one odd man out, so let's recheck.

Wait — actually, B and C are consistent with each other ($ k = 3 $), and A and D are consistent with each other ($ k = \frac{1}{3} $). But since the problem says *one* doesn't belong, maybe we misread.

Wait — Box B: The graph shows $ y $ increasing faster than $ x $. From point (2,6): $ \frac{6}{2} = 3 $, yes. So $ k = 3 $

But Box A: $ y = \frac{1}{3}x $ → $ k = \frac{1}{3} $

Box D: Sugar per cookie: $ \frac{8}{24} = \frac{1}{3} $, same as $ \frac{15}{45} = \frac{1}{3} $ → $ k = \frac{1}{3} $

So:
- A: $ k = \frac{1}{3} $
- B: $ k = 3 $
- C: $ k = 3 $
- D: $ k = \frac{1}{3} $

So B and C have $ k = 3 $, A and D have $ k = \frac{1}{3} $

But again, two boxes don’t match.

Wait — is the constant of proportionality defined as $ y/x $ or $ x/y $?

In this case:
- For A: $ y = \frac{1}{3}x $ → $ k = \frac{1}{3} $
- For D: sugar (y) vs cookies (x): $ y = \frac{1}{3}x $ → $ k = \frac{1}{3} $
- For B: y-axis is paint output, x-axis is input → $ \frac{y}{x} = 3 $
- For C: $ \frac{y}{x} = 3 $

So A and D are $ k = \frac{1}{3} $, B and C are $ k = 3 $

But we need only one odd man out.

Wait — maybe the question asks for the constant of proportionality in terms of how much y changes per x, but perhaps in D, it's sugar per cookie, so $ \frac{\text{sugar}}{\text{cookies}} = \frac{1}{3} $, which matches A.

But in B, the graph shows $ y = 3x $, so $ k = 3 $

So only two values: $ \frac{1}{3} $ and $ 3 $

But unless one is flipped...

Wait — perhaps the odd man out is B, because it's a graph, but let's check the value.

Wait — look at box B: the graph has points (2,6), (4,12), (6,18), (8,24)

So $ y = 3x $ → $ k = 3 $

Box C: $ y = 3x $ → $ k = 3 $

Box A: $ y = \frac{1}{3}x $ → $ k = \frac{1}{3} $

Box D: $ \frac{8}{24} = \frac{1}{3} $, $ \frac{15}{45} = \frac{1}{3} $ → $ k = \frac{1}{3} $

So A and D have $ k = \frac{1}{3} $, B and C have $ k = 3 $

But the problem says “there is an ODD MAN OUT” — only one.

Ah! Wait — maybe I misread box D.

Box D says: "Alexis made 24 cookies with 8 teaspoons of sugar and 45 cookies with 15 teaspoons of sugar."

So sugar used per cookie: $ \frac{8}{24} = \frac{1}{3} $ tsp per cookie

So if $ y $ = sugar, $ x $ = cookies → $ y = \frac{1}{3}x $

So $ k = \frac{1}{3} $

Now, box B graph: $ y = 3x $

But box C table: $ x=4, y=12 $ → $ y = 3x $

So all except A have $ k = 3 $? No — A has $ k = \frac{1}{3} $, others have $ k = 3 $?

Wait no — A says $ y = \frac{1}{3}x $ → $ k = \frac{1}{3} $

B: graph shows $ y = 3x $ → $ k = 3 $

C: table shows $ y = 3x $ → $ k = 3 $

D: sugar per cookie = $ \frac{1}{3} $ → $ k = \frac{1}{3} $

So A and D have $ k = \frac{1}{3} $, B and C have $ k = 3 $

But then both A and D have same $ k $, B and C same.

But the problem says one odd man out.

Wait — perhaps the constant of proportionality is defined differently.

Wait — in box D, is the constant $ \frac{\text{cookies}}{\text{sugar}} $ or $ \frac{\text{sugar}}{\text{cookies}} $?

The question says: "What is the constant of proportionality?"

Typically, constant of proportionality is $ \frac{y}{x} $, where $ y $ is dependent on $ x $.

Here, sugar depends on number of cookies → $ \text{sugar} = k \times \text{cookies} $

So $ k = \frac{8}{24} = \frac{1}{3} $

So $ k = \frac{1}{3} $

But box A also has $ k = \frac{1}{3} $

Box B: $ y = 3x $ → $ k = 3 $

Box C: $ y = 3x $ → $ k = 3 $

So two groups: $ \frac{1}{3} $ and $ 3 $

But only one odd man out — so maybe box B is the odd one?

No — B and C are both $ k = 3 $

Wait — look at the graph in B — does it go through origin?

Yes: (0,0), (2,6), etc. → yes.

But wait — the equation in A is $ y = \frac{1}{3}x $, but in B, the graph shows $ y = 3x $

So they’re inverses.

But no, the issue is: which one has a different constant?

But A and D have $ k = \frac{1}{3} $, B and C have $ k = 3 $

So unless one is wrong.

Wait — box C: $ x=4, y=12 $ → $ \frac{12}{4} = 3 $

$ x=6, y=18 $ → $ \frac{18}{6} = 3 $

$ x=9, y=27 $ → $ \frac{27}{9} = 3 $ → $ k = 3 $

Box B: graph — (2,6): $ 6/2 = 3 $, (4,12): $ 12/4 = 3 $, (6,18): $ 18/6 = 3 $, (8,24): $ 24/8 = 3 $ → $ k = 3 $

Box D: $ \frac{8}{24} = \frac{1}{3} $, $ \frac{15}{45} = \frac{1}{3} $ → $ k = \frac{1}{3} $

Box A: $ y = \frac{1}{3}x $ → $ k = \frac{1}{3} $

So A and D have $ k = \frac{1}{3} $, B and C have $ k = 3 $

But the problem says one odd man out, so likely I made a mistake in interpreting D.

Wait — what is the constant of proportionality?

In D, the ratio is sugar to cookies = $ \frac{8}{24} = \frac{1}{3} $, so $ k = \frac{1}{3} $

But if the constant is cookies per teaspoon of sugar, then $ \frac{24}{8} = 3 $

But typically, we say sugar needed per cookie → $ \frac{1}{3} $ tsp per cookie

So $ k = \frac{1}{3} $

So A and D both have $ k = \frac{1}{3} $

B and C have $ k = 3 $

So no single odd man out?

Wait — but look at the equation in A: $ y = \frac{1}{3}x $

And graph in B: $ y = 3x $

But table in C: $ y = 3x $

D: $ y = \frac{1}{3}x $

So A and D are $ k = \frac{1}{3} $, B and C are $ k = 3 $

But unless one of them is not proportional, but all are.

Wait — perhaps the odd man out is the one with a different form?

No — the instruction is to find the constant of proportionality.

Wait — let’s recheck box D: "Alexis made 24 cookies with 8 teaspoons of sugar..."

So sugar = $ \frac{1}{3} \times $ cookies → $ k = \frac{1}{3} $

Same as A.

But box B: graph — is it really $ y = 3x $? Yes.

But box C: $ y = 3x $, yes.

So why would one be odd?

Wait — look at the graph in B: the axes — x from 2 to 10, y from 0 to 30

Points: (2,6), (4,12), (6,18), (8,24)

Yes, $ y = 3x $

But box A: $ y = \frac{1}{3}x $

So A and D have $ k = \frac{1}{3} $, B and C have $ k = 3 $

But unless the constant is defined as $ \frac{x}{y} $?

No — standard is $ \frac{y}{x} $

Wait — perhaps the odd man out is box A, because it's an equation, and others are data?

No — the problem says to find the constant.

Wait — maybe I misread box D.

Wait — box D says: "Alexis made 24 cookies with 8 teaspoons of sugar and 45 cookies with 15 teaspoons of sugar"

So sugar / cookies = 8/24 = 1/3, 15/45 = 1/3 → k = 1/3

So $ y = \frac{1}{3}x $ if y=sugar, x=cookies

So same as A.

But box B has $ y = 3x $, box C has $ y = 3x $

So A and D have $ k = 1/3 $, B and C have $ k = 3 $

But the problem says "there is an ODD MAN OUT", so likely one of these is incorrect.

Wait — look at box B: the graph — is it possible that the axis labels are switched?

No — x-axis is horizontal, y-axis vertical.

Point (2,6): when x=2, y=6 → $ y = 3x $

Yes.

But wait — box A says $ y = \frac{1}{3}x $, which is different.

But box D also has $ k = \frac{1}{3} $

So unless box D is the odd one?

No — it matches A.

Wait — perhaps the constant of proportionality is meant to be the rate, but in D, it's sugar per cookie, but in A, it's something else?

No — both have $ k = \frac{1}{3} $

Wait — maybe the odd man out is box C, because it's a table with q instead of a number?

Look at box C: $ x=4,6,q $, $ y=12,18,27 $

So $ \frac{27}{q} = 3 $ → $ q = 9 $

So it's consistent.

So $ k = 3 $

So still, B and C have $ k = 3 $, A and D have $ k = \frac{1}{3} $

But the problem says "one odd man out", so perhaps box A is the odd one, because it's the only one with $ k = \frac{1}{3} $, but D also has it.

Wait — unless in D, the constant is not $ \frac{1}{3} $, but $ 3 $?

Let’s think: if you want to know how many cookies per teaspoon of sugar: 24 cookies / 8 tsp = 3 cookies per tsp

So $ \text{cookies} = 3 \times \text{sugar} $

So if $ x $ = sugar, $ y $ = cookies → $ y = 3x $ → $ k = 3 $

Ah! Here’s the key!

What is the independent variable?

In the problem, it says: "Alexis made 24 cookies with 8 teaspoons of sugar"

So cookies depend on sugar, so $ \text{cookies} = k \times \text{sugar} $

So $ k = \frac{24}{8} = 3 $, or $ \frac{45}{15} = 3 $

So $ k = 3 $

Similarly, in box A: $ y = \frac{1}{3}x $ → $ k = \frac{1}{3} $

But now D has $ k = 3 $, same as B and C

Then A is the only one with $ k = \frac{1}{3} $

Yes!

So:

- A: $ y = \frac{1}{3}x $ → $ k = \frac{1}{3} $
- B: graph: $ y = 3x $ → $ k = 3 $
- C: table: $ y = 3x $ → $ k = 3 $
- D: cookies = 3 × sugar → $ k = 3 $

So A is the odd man out — it has $ k = \frac{1}{3} $, others have $ k = 3 $

👉 Shade Box A

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## ✔ Row 4

| Box | Content |
|-----|--------|
| A | Jazmyn uses 2 gallons for 5 walls, 4 gallons for 10 walls → $ \frac{2}{5} = 0.4 $ gallons per wall → $ k = 0.4 = \frac{2}{5} $ |
| B | Table: $ x=0.5, y=0.2 $ → $ \frac{0.2}{0.5} = 0.4 $; $ x=1.5, y=0.6 $ → $ \frac{0.6}{1.5} = 0.4 $; $ x=3.5, y=1.4 $ → $ \frac{1.4}{3.5} = 0.4 $ → $ k = 0.4 = \frac{2}{5} $ |
| C | Graph: Points (2,5), (4,10), (6,15), (8,20) → $ \frac{5}{2} = 2.5 $, $ \frac{10}{4} = 2.5 $, etc. → $ k = 2.5 $ |
| D | Equation: $ y = \frac{2}{5}x $ → $ k = \frac{2}{5} = 0.4 $ |

So:
- A: $ k = 0.4 $
- B: $ k = 0.4 $
- C: $ k = 2.5 $
- D: $ k = 0.4 $

Only C has $ k = 2.5 $, others have $ k = 0.4 $

👉 Shade Box C

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## ✔ Row 5

| Box | Content |
|-----|--------|
| A | Table: $ x=1, y=1.2 $ → $ \frac{1.2}{1} = 1.2 $; $ x=2, y=2.4 $ → $ \frac{2.4}{2} = 1.2 $; $ x=3, y=3.6 $ → $ \frac{3.6}{3} = 1.2 $ → $ k = 1.2 $ |
| B | Equation: $ y = 1.25x $ → $ k = 1.25 $ |
| C | 4 lbs cost $5 → $ \frac{5}{4} = 1.25 $ per lb; 7 lbs cost $8.75 → $ \frac{8.75}{7} = 1.25 $ → $ k = 1.25 $ |
| D | Graph: Points (4,5), (8,10), (12,15), (16,20) → $ \frac{5}{4} = 1.25 $, $ \frac{10}{8} = 1.25 $, etc. → $ k = 1.25 $ |

So:
- A: $ k = 1.2 $
- B: $ k = 1.25 $
- C: $ k = 1.25 $
- D: $ k = 1.25 $

Only A has $ k = 1.2 $, others have $ k = 1.25 $

👉 Shade Box A

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✔ Final Answers:

| Row | Odd Man Out (to shade) |
|-----|-------------------------|
| 1 | C (table with x=12,y=1, etc.) |
| 2 | A (Sophie's reading) |
| 3 | A (equation $ y = \frac{1}{3}x $) |
| 4 | C (graph with points like (2,5)) |
| 5 | A (table with x=1,y=1.2) |

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📝 Summary of Constants:

| Row | Constant of Proportionality |
|-----|-------------------------------|
| 1 | Most: 12, One: 1/12 → Odd: C |
| 2 | Most: 1, One: 2 → Odd: A |
| 3 | Most: 3, One: 1/3 → Odd: A |
| 4 | Most: 0.4, One: 2.5 → Odd: C |
| 5 | Most: 1.25, One: 1.2 → Odd: A |

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✔ Final Answer: Shade the following boxes:

- Row 1: Box C
- Row 2: Box A
- Row 3: Box A
- Row 4: Box C
- Row 5: Box A

Let me know if you'd like a visual guide!