Decomposing Fractions Worksheets - Math Monks - Free Printable

Let’s solve each number bond problem step by step. Remember: the big circle (the whole) is equal to the sum of all the smaller circles connected to it. So we add up the known parts and subtract from the whole to find the missing part(s).

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Problem 1:
Whole = 6/7
Known parts: 1/7, 1/7 → Sum = 2/7
Missing parts: two empty circles → Let’s call them A and B
So: A + B = 6/7 - 2/7 = 4/7
We can split 4/7 into any two fractions that add to 4/7 — but since this is decomposing with same denominators, likely they want simple splits like 2/7 and 2/7, or 3/7 and 1/7. But looking at the example, they used different numerators. Actually, in Problem 1, there are TWO missing circles — so we need to fill both. Since no other info, we can choose any pair that adds to 4/7. But let’s check if maybe one is meant to be filled first? Wait — actually, looking again: the diagram has three small circles under the whole. Two are given (1/7 and 1/7), one is blank on left, one on right? Wait — no, looking at the image description: for Problem 1, it shows:

Top: 6/7
Bottom-left: 1/7
Bottom-right: 1/7
Left side: empty circle
Right side: empty circle

Wait — that would be FOUR parts? That doesn’t match the example. Let me re-express based on standard number bonds.

Actually, looking back at the Example:
Example: Whole = 4/5, broken into 1/5, 1/5, 2/5 → three parts.

In Problem 1: Whole = 6/7, with two known parts (1/7 and 1/7), and two empty circles → total four parts? That seems odd.

Wait — perhaps I misread. Let me think differently.

Looking at the structure: In the example, the top circle connects to three bottom circles. Similarly, in Problem 1, the top circle (6/7) connects to three bottom circles: left, middle, right. Middle is 1/7, right is 1/7, left is empty? And then also a circle on the far left? No — probably the diagram has three branches.

Actually, upon closer inspection of typical worksheets: each “number bond” has one top circle (whole) and usually 2 or 3 bottom circles (parts). In Problem 1, it appears there are three parts: left, middle, right. Middle and right are given as 1/7 each, left is blank. But then why is there another blank on the far left? Maybe it's a typo in my interpretation.

Wait — let’s look at Problem 2: Whole = 6/11, with left part = 3/11, bottom-right = 1/11, and two blanks: one bottom-left, one right-side. Again, possibly three parts? Or four?

This is confusing without seeing the exact layout. But based on common design, and the example having three parts, I’ll assume each problem has exactly three parts unless shown otherwise.

But in Problem 1, if whole is 6/7, and two parts are 1/7 and 1/7, then third part must be 6/7 - 1/7 - 1/7 = 4/7. So only one missing part? But the diagram shows two empty circles? Hmm.

Alternatively, perhaps some problems have more than three parts. Let’s count the connections.

To avoid confusion, let’s go problem by problem using the rule: sum of all parts = whole.

I’ll list each problem with what’s given and calculate missing values.

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Problem 1:
Whole: 6/7
Given parts: 1/7, 1/7
Assume there are three parts total → missing part = 6/7 - 1/7 - 1/7 = 4/7
But the diagram might show two missing circles? If so, we need to split 4/7 into two parts. For simplicity, let’s say 2/7 and 2/7, or 3/7 and 1/7. But since the worksheet likely expects specific answers, and looking at other problems, perhaps it’s designed to have unique solutions.

Wait — let’s look at Problem 3: Whole = 8/9, parts: 2/9, 3/9, and two blanks. So 8/9 - 2/9 - 3/9 = 3/9 left for two blanks → could be 1/9 and 2/9, etc.

Perhaps for consistency, we should assume that when multiple blanks exist, we can choose any combination that sums correctly, but ideally use simplest forms or match pattern.

But to be precise, let’s do exact calculation for each.

I think I need to proceed systematically.

Let me define for each problem:

- Identify the whole (top circle)
- List all given parts
- Subtract sum of given parts from whole to get sum of missing parts
- If one missing part, that’s the answer
- If multiple missing parts, we may need to infer from context or provide possible values — but since this is a worksheet, likely each blank has a unique intended value.

Looking back at the user’s image description, for Problem 1: "two empty circles" — but in the text, it says for Problem 1: left circle empty, right circle empty, and bottom has 1/7 and 1/7 — that would be four parts? That doesn't make sense.

Perhaps the diagram is: top circle connected to three circles below: left, center, right. In Problem 1, center and right are 1/7, left is empty — so only one missing part. But the user said "two empty circles" — maybe including the sides?

I think there's a miscommunication. To resolve, I'll use the following approach:

For each problem, I'll calculate the missing value(s) based on the whole minus the sum of known parts, and if there are multiple blanks, I'll distribute the remainder evenly or in a logical way, but since this is math, the sum must be correct.

Let’s start over with clear assumptions:

Assumption: Each number bond has the whole at the top, and several parts below. The sum of all parts equals the whole. We need to find the missing part(s).

Now, let's tackle each problem one by one.

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Problem 1:
Whole: 6/7
Known parts: 1/7, 1/7
If there are three parts total, missing part = 6/7 - 1/7 - 1/7 = 4/7
But if there are four parts, and two are given, then sum of two missing = 4/7, so each could be 2/7.
Given that in the example, they had three parts, and here likely three, I'll go with one missing part of 4/7. But the user mentioned "two empty circles" for Problem 1 — let's check the original request.

In the initial problem list:
1. Has two empty circles? From the text: "for 1: left circle empty, right circle empty, and bottom has 1/7 and 1/7" — that suggests four parts: left, right, bottom-left, bottom-right? This is ambiguous.

To move forward, I'll use the most logical interpretation based on standard worksheets: typically, number bonds for fractions have 2 or 3 parts. Given the example has 3 parts, I'll assume each problem has 3 parts unless specified otherwise.

For Problem 1: if two parts are given (1/7 and 1/7), then the third part is 4/7. So the missing value is 4/7.

But the user said "two empty circles" — perhaps for Problem 1, it's structured as: whole connected to three circles: one on left (empty), one on right (empty), and one at bottom which is split into two? No.

Another idea: perhaps the "number bond" allows for multiple levels, but that's complicated.

Let's look at Problem 4: Whole = 10/13, parts: 4/13, 2/13, and one empty — so clearly three parts, missing one.

Similarly, Problem 5: Whole = 8/13, part = 5/13, one empty — so two parts? Or three? It shows one given and one empty, so likely two parts.

In Problem 5, it's probably two parts: 5/13 and ? = 8/13, so ? = 3/13.

For consistency, I'll assume that the number of parts varies, and we just need to ensure sum of all parts = whole.

So let's do each problem carefully.

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Problem 1:
Whole: 6/7
Given: two parts of 1/7 each → sum = 2/7
Remaining for other parts: 6/7 - 2/7 = 4/7
If there is one missing part, it's 4/7.
If there are two missing parts, their sum is 4/7, so possible pairs: (1/7,3/7), (2/7,2/7), etc.
Since the worksheet likely intends specific answers, and looking at other problems, I think for Problem 1, it's meant to have three parts: so missing part is 4/7.
But the user's description says "two empty circles" — perhaps it's a mistake, or perhaps in the diagram, there are two blanks to fill, meaning two missing parts.

To resolve, let's calculate the required sum and see.

Perhaps for Problem 1, the two 1/7 are not both parts; maybe one is part of a sub-bond? Unlikely.

I recall that in some number bonds, the whole is decomposed into parts, and sometimes parts are further decomposed, but in this case, the example shows direct decomposition.

Let's skip to Problem 2 for clarity.

Problem 2:
Whole: 6/11
Given: 3/11 (left), 1/11 (bottom-right)
Sum of given = 4/11
Remaining = 6/11 - 4/11 = 2/11
If there is one missing part, it's 2/11.
If two missing parts, sum to 2/11, e.g., 1/11 and 1/11.
Again, ambiguity.

Perhaps the key is to notice that in the example, they used different numerators, and for these problems, the missing parts are to be filled such that all parts are positive fractions with the same denominator.

Moreover, in Problem 6, there are multiple parts given, so we can calculate.

Let's do Problem 6 first as it has more data.

Problem 6:
Whole: 15/17
Given parts: 3/17, 7/17, 3/17
Sum = 3+7+3 = 13/17
Remaining = 15/17 - 13/17 = 2/17
And there is one empty circle, so it must be 2/17.

Similarly, Problem 4:
Whole: 10/13
Given: 4/13, 2/13
Sum = 6/13
Remaining = 4/13
One empty circle, so 4/13.

Problem 5:
Whole: 8/13
Given: 5/13
One empty circle, so 8/13 - 5/13 = 3/13.

Problem 7:
Whole: 5/7
Given: 2/7, 1/7
Sum = 3/7
Remaining = 2/7
One empty circle, so 2/7.

Now for Problem 3:
Whole: 8/9
Given: 2/9, 3/9
Sum = 5/9
Remaining = 3/9 = 1/3, but keep as 3/9
Two empty circles? User said for Problem 3: "left circle empty, right circle has 3/9, bottom has 2/9, and another empty" — so perhaps three parts: left, bottom, right. Given bottom=2/9, right=3/9, so left = 8/9 - 2/9 - 3/9 = 3/9. But then why two empties? Unless one is already filled.

In the user's text for Problem 3: "for 3: left circle empty, right circle has 3/9, bottom has 2/9, and another empty" — this suggests four parts? That can't be.

Perhaps "another empty" refers to the same as left circle. I think there's a language issue.

To simplify, I'll assume that for each problem, the number of missing parts is such that the sum works out, and for cases with multiple blanks, we can fill them with values that sum to the remainder, but since the worksheet likely has unique answers, I'll use the following strategy:

- If only one blank, calculate directly.
- If two blanks, and no other info, split the remainder equally or use smallest integers.

But let's look at Problem 1 again. Perhaps in Problem 1, the two 1/7 are not both separate; maybe one is part of a group. Another idea: in some number bonds, the whole is split into two parts, and one of those is split further. But in the example, it's direct.

I think I need to accept that for problems with multiple blanks, the intended answer is to have the sum correct, and perhaps the blanks are to be filled with specific values based on context.

Let's list all problems with calculations:

Problem 1:
Whole: 6/7
Given: 1/7, 1/7
Sum given: 2/7
Remainder: 4/7
If one missing part: 4/7
If two missing parts: e.g., 2/7 and 2/7, or 1/7 and 3/7
Since the example used 1/5,1/5,2/5 for 4/5, perhaps here they want 1/7,1/7,4/7, but 4/7 is large. Or maybe 2/7,2/7,2/7 for 6/7, but given are 1/7,1/7, so not.

Perhaps the two 1/7 are not both parts; maybe the diagram has the whole connected to three circles: one is 1/7, another is 1/7, and the third is to be found, so 4/7. I think that's it. The "two empty circles" might be a misdescription.

Similarly for others.

I'll proceed with the calculation for each, assuming the number of parts is consistent with the given data.

Final decision: for each problem, calculate the missing value(s) as whole minus sum of given parts, and if multiple blanks, assign values that sum to the remainder, preferring equal split or simple fractions.

But to provide definite answers, I'll do the following:

- For problems with one blank, give the exact value.
- For problems with two blanks, give two values that sum to the remainder, and since the worksheet might expect specific ones, I'll use the most straightforward.

Let's begin:

Problem 1:
Whole: 6/7
Given parts: 1/7, 1/7
Sum given: 2/7
Remainder: 4/7
Number of missing parts: let's say 1 (as per standard), so missing part = 4/7
But if two missing parts, then 2/7 and 2/7 (since 2/7 + 2/7 = 4/7)
Given that in the example, they had three parts, and here likely three, but with two given, one missing, I'll go with 4/7 for the single missing part. However, the user said "two empty circles", so perhaps for Problem 1, there are two missing parts.

Upon second thought, in the initial problem list, for Problem 1, it's described as having two empty circles, while for Problem 4, it has one empty circle. So for Problem 1, we need to fill two circles whose sum is 4/7.

What values to choose? Perhaps 2/7 and 2/7, as it's symmetric.

Similarly for Problem 2: whole 6/11, given 3/11 and 1/11, sum 4/11, remainder 2/11, and if two missing parts, then 1/11 and 1/11.

For Problem 3: whole 8/9, given 2/9 and 3/9, sum 5/9, remainder 3/9, and if two missing parts, then 1.5/9 each, but fractions should be integer numerators, so 1/9 and 2/9 or 3/9 and 0, but 0 not allowed. So 1/9 and 2/9.

But 3/9 can be split as 1/9 and 2/9.

Let's systematize.

I think the best way is to calculate the required sum for missing parts and provide reasonable values.

So here goes:

Problem 1:
Missing sum: 4/7
Two missing parts: let's say 2/7 and 2/7

Problem 2:
Whole: 6/11
Given: 3/11, 1/11
Sum given: 4/11
Remainder: 2/11
Two missing parts: 1/11 and 1/11

Problem 3:
Whole: 8/9
Given: 2/9, 3/9
Sum given: 5/9
Remainder: 3/9 = 1/3
Two missing parts: 1/9 and 2/9 (since 1/9 + 2/9 = 3/9)

Problem 4:
Whole: 10/13
Given: 4/13, 2/13
Sum given: 6/13
Remainder: 4/13
One missing part: 4/13

Problem 5:
Whole: 8/13
Given: 5/13
One missing part: 3/13

Problem 6:
Whole: 15/17
Given: 3/17, 7/17, 3/17
Sum given: 13/17
Remainder: 2/17
One missing part: 2/17

Problem 7:
Whole: 5/7
Given: 2/7, 1/7
Sum given: 3/7
Remainder: 2/7
One missing part: 2/7

Now, for problems with two missing parts, I've chosen values that sum correctly and are simple.

So final answers:

1. 2/7 and 2/7 (or other pairs, but 2/7 and 2/7 is fine)
2. 1/11 and 1/11
3. 1/9 and 2/9
4. 4/13
5. 3/13
6. 2/17
7. 2/7

But to confirm, let's verify Problem 3: 2/9 + 3/9 + 1/9 + 2/9 = 8/9? 2+3+1+2=8, yes, but that's four parts. If the diagram has only three parts, then it should be 2/9, 3/9, and 3/9, but 2+3+3=8, so 3/9 for the missing part. But the user said two empty circles for Problem 3, so perhaps four parts.

This is inconsistent.

Perhaps for each problem, the number of parts is indicated by the diagram, and in the absence of that, we go by the number of given parts plus missing.

To resolve, I'll output the answers as per the calculation above, as it's the most reasonable.

So for the final answer, I'll list the missing values for each problem.

Final Answer:
1. \frac{2}{7}, \frac{2}{7}
2. \frac{1}{11}, \frac{1}{11}
3. \frac{1}{9}, \frac{2}{9}
4. \frac{4}{13}
5. \frac{3}{13}
6. \frac{2}{17}
7. \frac{2}{7}

Note: For problems with two missing parts, the values are chosen to sum to the required remainder and are in simplest form where possible.