Math worksheet featuring word problems on dividing fractions, requiring students to solve practical scenarios and reduce answers to simplest form.

Let's solve each of these word problems involving division of fractions, step by step, and reduce the answers to their simplest form.

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Problem 1:

Jack needs to make sandwiches for a class picnic with $ \frac{2}{9} $ of a kilogram of sugar. If each sandwich needs $ \frac{3}{18} $ of a kilogram of sugar, then how many total sandwiches can be made?

#### Step 1: Simplify the fraction for each sandwich
$$
\frac{3}{18} = \frac{1}{6}
$$

So, each sandwich uses $ \frac{1}{6} $ kg of sugar.

#### Step 2: Divide total sugar by sugar per sandwich
We want to know: How many $ \frac{1}{6} $-kg portions are in $ \frac{2}{9} $ kg?

$$
\frac{2}{9} \div \frac{1}{6} = \frac{2}{9} \times \frac{6}{1} = \frac{12}{9} = \frac{4}{3}
$$

#### Step 3: Interpret the result
$ \frac{4}{3} = 1 \frac{1}{3} $

But since Jack can't make a third of a sandwich (assuming full sandwiches only), we interpret this as 1 full sandwich can be made, but he doesn't have enough sugar for a second one.

Wait — let’s double-check the logic.

Actually, the question asks "how many total sandwiches can be made?" — so if it’s not a whole number, we should consider whether partial sandwiches are allowed. But typically, you can’t make a fraction of a sandwich in this context.

But wait — is the answer supposed to be exact or rounded down?

Let’s recheck:

$$
\frac{2}{9} \div \frac{1}{6} = \frac{2}{9} \times 6 = \frac{12}{9} = \frac{4}{3} \approx 1.33
$$

So, only 1 full sandwich can be made.

But let’s confirm: Does the problem allow fractional sandwiches? It says “how many total sandwiches can be made?” — usually, that implies whole sandwiches.

However, sometimes word problems like this expect the exact mathematical answer, even if it's a fraction.

But here’s a catch: Let's verify the amount of sugar used.

Each sandwich uses $ \frac{1}{6} $ kg → $ \frac{1}{6} \approx 0.1667 $ kg
Total sugar: $ \frac{2}{9} \approx 0.2222 $ kg

So:
- One sandwich: $ 0.1667 $ kg → remaining: $ 0.2222 - 0.1667 = 0.0555 $ kg
- Not enough for a second sandwich (needs 0.1667)

So only 1 sandwich can be made.

But let’s see what the math gives: $ \frac{4}{3} $, which is not a whole number.

Wait — maybe I made a mistake in simplifying?

Wait: $ \frac{3}{18} = \frac{1}{6} $ — correct.

Then $ \frac{2}{9} \div \frac{1}{6} = \frac{2}{9} \times 6 = \frac{12}{9} = \frac{4}{3} $

So the mathematical answer is $ \frac{4}{3} $, but in real life, only 1 sandwich can be made.

But since the problem says "how many total sandwiches can be made?" and doesn’t specify partial ones, we must assume whole sandwiches only, so the answer is 1.

But let’s check if the problem expects the fractional answer.

Looking at other problems, they may expect exact division.

Wait — perhaps I misread.

Let’s re-read: “each sandwich needs $ \frac{3}{18} $” — that’s $ \frac{1}{6} $. So yes.

But $ \frac{2}{9} \div \frac{1}{6} = \frac{4}{3} $ — so $ \frac{4}{3} $ sandwiches.

But unless partial sandwiches are allowed, this isn't practical.

But the problem might be testing division of fractions, so likely wants $ \frac{4}{3} $ as the answer.

But let's check: Is there a possibility that $ \frac{3}{18} $ was meant to be simplified?

Yes — $ \frac{3}{18} = \frac{1}{6} $, so correct.

Alternatively, maybe keep it as $ \frac{3}{18} $ and divide:

$$
\frac{2}{9} \div \frac{3}{18} = \frac{2}{9} \times \frac{18}{3} = \frac{2}{9} \times 6 = \frac{12}{9} = \frac{4}{3}
$$

Same result.

So, answer: $ \frac{4}{3} $ — but reduced to simplest form: $ \boxed{\frac{4}{3}} $ or $ 1\frac{1}{3} $

But again, can you make $ \frac{4}{3} $ sandwiches? Probably not.

But since the problem is about dividing fractions, and no restriction is mentioned, we’ll go with the exact mathematical answer: $ \boxed{\frac{4}{3}} $

But let’s hold off and see if others are similar.

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Problem 2:

John has a piece of metal rod that is $ \frac{3}{4} $ of a meter long. He needs to cut pieces from the rod that are $ \frac{5}{16} $ of a meter long. How many pieces can John cut?

We divide the total length by the length of each piece:

$$
\frac{3}{4} \div \frac{5}{16} = \frac{3}{4} \times \frac{16}{5} = \frac{48}{20} = \frac{12}{5} = 2.4
$$

So, 2.4 pieces — but he can only cut whole pieces.

So, he can cut 2 full pieces (since 2 × $ \frac{5}{16} $ = $ \frac{10}{16} = \frac{5}{8} $), and leftover: $ \frac{3}{4} - \frac{5}{8} = \frac{6}{8} - \frac{5}{8} = \frac{1}{8} $, which is less than $ \frac{5}{16} $, so not enough for a third.

So, 2 pieces can be cut.

But again, the mathematical answer is $ \frac{12}{5} $, but practical answer is 2.

But since the problem says “how many pieces can John cut?” — and cutting is physical — we need whole pieces.

So answer: 2

But again, let’s see what the expected format is.

Wait — perhaps the problem wants the exact quotient, not rounded down.

But in most such problems, especially when asking “how many pieces”, they expect the maximum whole number that fits.

So: $ \frac{3}{4} \div \frac{5}{16} = \frac{12}{5} = 2.4 $ → 2 whole pieces

So answer: $ \boxed{2} $

But let’s confirm:

- $ \frac{5}{16} \times 2 = \frac{10}{16} = \frac{5}{8} $
- $ \frac{3}{4} = \frac{6}{8} $, so $ \frac{6}{8} - \frac{5}{8} = \frac{1}{8} $ left → not enough

So yes, 2 pieces

Answer: $ \boxed{2} $

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Problem 3:

$ \frac{3}{7} $ of a 1 liter container is filled with water. If a mug can contain $ \frac{9}{84} $ of a liter, then how many mugs of water are needed to fill up the bucket?

First, simplify $ \frac{9}{84} $:

$$
\frac{9}{84} = \frac{3}{28}
$$

So each mug holds $ \frac{3}{28} $ liters.

The container has $ \frac{3}{7} $ liters of water already, but the question says: "how many mugs of water are needed to fill up the bucket?"

Wait — it says: "$ \frac{3}{7} $ of a 1 liter container is filled" — so currently $ \frac{3}{7} $ L is in it.

To fill up the bucket, we need to add $ 1 - \frac{3}{7} = \frac{4}{7} $ liters.

Now, each mug holds $ \frac{3}{28} $ L.

So number of mugs needed to add $ \frac{4}{7} $ L:

$$
\frac{4}{7} \div \frac{3}{28} = \frac{4}{7} \times \frac{28}{3} = \frac{112}{21} = \frac{16}{3} \approx 5.333
$$

So, $ \frac{16}{3} $ mugs.

But we need to fill the bucket, so we need enough mugs to cover $ \frac{4}{7} $ L.

Since $ \frac{16}{3} = 5\frac{1}{3} $, we need 6 mugs to fill it completely.

But the question says: “how many mugs of water are needed to fill up the bucket?”

This could mean: how many full mugs are required to fill it.

So, we need to round up to the next whole number: $ \lceil \frac{16}{3} \rceil = 6 $

But again, if it's asking for the exact amount, it might be $ \frac{16}{3} $

But let’s read carefully: “how many mugs of water are needed to fill up the bucket.”

If we pour $ \frac{16}{3} $ mugs, that’s exactly $ \frac{4}{7} $ L added.

But since mugs are discrete, we can't pour a third of a mug — unless we're measuring precisely.

But in real-world terms, you’d need 6 mugs to ensure the bucket is full.

But again, since this is a fraction division problem, likely wants the exact quotient.

So let’s compute:

$$
\frac{4}{7} \div \frac{3}{28} = \frac{4}{7} \times \frac{28}{3} = \frac{112}{21} = \frac{16}{3}
$$

Simplify: $ \frac{16}{3} $ is already in simplest form.

So answer: $ \boxed{\frac{16}{3}} $ or $ 5\frac{1}{3} $

But the question is: “how many mugs... are needed to fill up the bucket?”

This implies minimum number of mugs required, so we need to round up.

So 6 mugs.

But let’s see: is the bucket already partially filled? Yes — $ \frac{3}{7} $ is already filled.

We need to add $ \frac{4}{7} $ L.

Each mug adds $ \frac{3}{28} $ L.

So number of mugs: $ \frac{4}{7} \div \frac{3}{28} = \frac{16}{3} \approx 5.33 $

So 5 mugs add: $ 5 \times \frac{3}{28} = \frac{15}{28} $

But $ \frac{4}{7} = \frac{16}{28} $, so 5 mugs give $ \frac{15}{28} $, still short by $ \frac{1}{28} $

So need 6 mugs to get $ \frac{18}{28} = \frac{9}{14} $, which is more than $ \frac{4}{7} = \frac{8}{14} $, so overfilled.

So 6 mugs are needed.

Thus, answer: $ \boxed{6} $

But the problem says “how many mugs of water are needed” — meaning how many mugs must be poured — so 6

Even though $ \frac{16}{3} $ is the mathematical division, the practical answer is 6.

But let’s compare with other problems.

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Problem 4:

A box of table tennis balls weighs $ \frac{5}{9} $ of a kg. If each ball weighs $ \frac{15}{81} $ of a kg, then how many balls are there in the box?

First, simplify $ \frac{15}{81} $:

$$
\frac{15}{81} = \frac{5}{27}
$$

So each ball weighs $ \frac{5}{27} $ kg.

Total weight: $ \frac{5}{9} $ kg

Number of balls:

$$
\frac{5}{9} \div \frac{5}{27} = \frac{5}{9} \times \frac{27}{5} = \frac{135}{45} = 3
$$

So, 3 balls

Perfectly divisible.

Answer: $ \boxed{3} $

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Now, Re-evaluating Problem 1 and 2

In Problem 1: $ \frac{2}{9} \div \frac{1}{6} = \frac{4}{3} $ — but can you make $ \frac{4}{3} $ sandwiches? No.

But maybe the problem allows fractional sandwiches? Unlikely.

But let’s check: Is $ \frac{3}{18} $ really $ \frac{1}{6} $? Yes.

But $ \frac{2}{9} \div \frac{1}{6} = \frac{4}{3} $ — so only 1 full sandwich can be made.

But the exact number of sandwiches that can be made is $ \frac{4}{3} $, but physically, only 1.

But in math problems, especially in early education, they often accept the exact division result, even if it's a fraction.

For example, “how many sandwiches can be made?” might be interpreted as “how many times does the amount per sandwich fit into the total?”

So even if it’s not a whole number, they want the quotient.

Similarly, in Problem 2, $ \frac{3}{4} \div \frac{5}{16} = \frac{12}{5} = 2.4 $, but you can only cut 2 full pieces.

But again, some problems ask for how many pieces can be cut, meaning maximum whole pieces, so floor function.

But in Problem 4, it worked out to an integer.

Let’s go back to Problem 1.

Wait — is it possible that $ \frac{3}{18} $ is not simplified? But we did.

But let’s calculate:

Total sugar: $ \frac{2}{9} \approx 0.2222 $ kg
Per sandwich: $ \frac{3}{18} = 0.1667 $ kg

So $ 0.2222 \div 0.1667 \approx 1.333 $ → $ \frac{4}{3} $

So $ \frac{4}{3} $ sandwiches.

But if the problem allows partial sandwiches, then answer is $ \frac{4}{3} $

But if not, then 1.

But looking at the context: making sandwiches for a picnic — likely wants whole sandwiches, so 1

But in math, the operation is division of fractions.

Perhaps the intended answer is $ \frac{4}{3} $

Let’s look at Problem 3 again.

It says: “how many mugs of water are needed to fill up the bucket”

We calculated $ \frac{4}{7} \div \frac{3}{28} = \frac{16}{3} $, but we need mugs to fill, so must round up to 6.

But if the question were: “how many mugs can be filled from the remaining water?” then it would be $ \frac{16}{3} $, but it says “needed to fill up”, so minimum number required, so 6

Similarly, in Problem 2: “how many pieces can John cut?” — maximum whole pieces, so 2

In Problem 1: “how many total sandwiches can be made?” — likely 1

But let’s check if the sugar per sandwich is $ \frac{3}{18} $, which is $ \frac{1}{6} $, and $ \frac{2}{9} \div \frac{1}{6} = \frac{4}{3} $, but that’s not a whole number.

But perhaps the problem expects us to leave it as a fraction.

Let’s see: maybe I made a mistake.

Wait — is it possible that $ \frac{3}{18} $ is meant to be $ \frac{3}{18} $ of a kg, and $ \frac{2}{9} $ is total?

Yes.

But $ \frac{2}{9} \div \frac{3}{18} = \frac{2}{9} \times \frac{18}{3} = \frac{2}{9} \times 6 = \frac{12}{9} = \frac{4}{3} $

So $ \frac{4}{3} $

Now, if the problem allows fractional sandwiches, then answer is $ \frac{4}{3} $

But if not, then 1.

But since the other problems involve physical objects (pieces, mugs, balls), and they expect whole numbers, likely Problem 1 also expects a whole number.

But Problem 4 gave an integer, Problem 2 gives 2, Problem 3 gives 6, Problem 1 gives 1.

But let’s try to see if there’s a different interpretation.

Wait — perhaps the sugar is shared among sandwiches, and we’re to find how many sandwiches can be made, so maximum number of full sandwiches.

So for Problem 1: $ \frac{2}{9} \div \frac{3}{18} = \frac{4}{3} $, so only 1 full sandwich.

Similarly, Problem 2: $ \frac{3}{4} \div \frac{5}{16} = \frac{12}{5} = 2.4 $, so 2 pieces.

Problem 3: $ \frac{4}{7} \div \frac{3}{28} = \frac{16}{3} \approx 5.33 $, so 6 mugs needed.

Problem 4: $ \frac{5}{9} \div \frac{5}{27} = 3 $, so 3 balls.

So likely, the answers are:

1. 1
2. 2
3. 6
4. 3

But the problem says: “Solve each problem and reduce the answer to the simplest form.”

“Reduce the answer” suggests that the answer might be a fraction.

But in cases where it’s not a whole number, and it’s a count of items, it must be rounded down or up.

But let’s see Problem 3 again: “how many mugs of water are needed to fill up the bucket?”

If we use $ \frac{16}{3} $ mugs, that’s $ 5\frac{1}{3} $, but you can’t pour a third of a mug unless you measure.

But the question is: “how many mugs are needed” — so you need 6 mugs to have enough.

Similarly, in Problem 1: you can only make 1 sandwich.

But let’s see what happens if we don’t round.

Maybe the problem wants the exact quotient, even if it’s not a whole number.

Let’s assume that the problems expect the exact division result, not rounded.

Then:

1. $ \frac{2}{9} \div \frac{3}{18} = \frac{2}{9} \times \frac{18}{3} = \frac{36}{27} = \frac{4}{3} $ → $ \boxed{\frac{4}{3}} $

2. $ \frac{3}{4} \div \frac{5}{16} = \frac{3}{4} \times \frac{16}{5} = \frac{48}{20} = \frac{12}{5} $ → $ \boxed{\frac{12}{5}} $

3. $ \frac{4}{7} \div \frac{3}{28} = \frac{4}{7} \times \frac{28}{3} = \frac{112}{21} = \frac{16}{3} $ → $ \boxed{\frac{16}{3}} $

4. $ \frac{5}{9} \div \frac{15}{81} = \frac{5}{9} \times \frac{81}{15} = \frac{405}{135} = 3 $ → $ \boxed{3} $

And reduce to simplest form.

So:

1. $ \frac{4}{3} $
2. $ \frac{12}{5} $
3. $ \frac{16}{3} $
4. $ 3 $

But for problems 1, 2, 3, these are not whole numbers, but represent the number of items.

But in reality, you can't have a fraction of a sandwich, etc.

But since the instruction is “solve the problem and reduce the answer to the simplest form”, and the operations are divisions of fractions, likely they want the exact mathematical result.

Moreover, in many textbooks, such problems expect the quotient, even if it’s a fraction.

For example, “how many $ \frac{1}{6} $-kg portions are in $ \frac{2}{9} $ kg?” — answer is $ \frac{4}{3} $, regardless of practicality.

So we'll go with the exact division results, reduced.

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Final Answers:

#### 1.
$$
\frac{2}{9} \div \frac{3}{18} = \frac{2}{9} \times \frac{18}{3} = \frac{2}{9} \times 6 = \frac{12}{9} = \boxed{\frac{4}{3}}
$$

#### 2.
$$
\frac{3}{4} \div \frac{5}{16} = \frac{3}{4} \times \frac{16}{5} = \frac{48}{20} = \boxed{\frac{12}{5}}
$$

#### 3.
Water needed to fill bucket: $ 1 - \frac{3}{7} = \frac{4}{7} $ L
Mug capacity: $ \frac{9}{84} = \frac{3}{28} $ L
$$
\frac{4}{7} \div \frac{3}{28} = \frac{4}{7} \times \frac{28}{3} = \frac{112}{21} = \boxed{\frac{16}{3}}
$$

#### 4.
$$
\frac{5}{9} \div \frac{15}{81} = \frac{5}{9} \times \frac{81}{15} = \frac{405}{135} = \boxed{3}
$$

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✔ Final Answers (Reduced to Simplest Form):

1. $ \boxed{\frac{4}{3}} $
2. $ \boxed{\frac{12}{5}} $
3. $ \boxed{\frac{16}{3}} $
4. $ \boxed{3} $

These are the exact mathematical solutions to the division problems, reduced to simplest form.

Even though some answers are not whole numbers, they represent the number of times the smaller quantity fits into the larger one, which is what division of fractions represents.

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✔ Note: If the context requires whole items only, then answers would be rounded down (for making) or up (for filling). But based on the instruction “solve the problem and reduce the answer,” we provide the exact quotients.