Practice · Split a whole into proper-fraction portions · Sensim Math

Variant 1 easy

Jisoo weighed the strawberries she picked at a weekend farm, and they weighed $\dfrac{53}{5}\ \text{kg}$ . She wants to pack the strawberries into boxes holding $2\ \text{kg}$ each and sell them. How many boxes in all can she sell?

Your answer

Show solution

1 · Understandwhat's really being asked

Jisoo has $\frac{53}{5}$ kg of strawberries. She packs them into 2 kg boxes. We need to count how many complete 2 kg boxes she can fill and sell.

Givens

Total weight of strawberries is $\frac{53}{5}$ kg.
Each box holds 2 kg of strawberries.

Unknowns

The number of complete 2 kg boxes she can sell.

Constraints

Only full boxes (each containing 2 kg) can be sold; leftover less than 2 kg cannot make a box.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #5 Look for a Pattern

Turn the improper fraction $\frac{53}{5}$ into an easier decimal weight, then repeatedly take away 2 kg (one box at a time) and count how many full boxes come out before less than 2 kg remains.

3 · Execute3 carry out the plan

1Write the total weight in an easier form

#9 Solve an Easier Related Problem 3.NF.A.1

Change the improper fraction

\frac{53}{5}

into a decimal so the weight is easy to picture: 53 divided by 5 is 10.6.

\dfrac{53}{5} = 10.6\ \text{kg}

Seeing

\frac{53}{5}

as 10.6 kg lets a young learner reason with a familiar weight instead of a scary fraction.

2Take away 2 kg for each box

#5 Look for a Pattern 3.OA.A.2

Each box uses up 2 kg. Counting boxes is the same as seeing how many whole groups of 2 kg fit inside 10.6 kg: 10 kg fills 5 boxes, and 12 kg would be too much.

2 + 2 + 2 + 2 + 2 = 10\ \text{kg} \le 10.6\ \text{kg} < 12\ \text{kg}

Repeatedly subtracting 2 kg is just sharing the strawberries into equal 2 kg boxes.

3Count full boxes and check the leftover

#9 Solve an Easier Related Problem 3.OA.A.2

After 5 boxes she has used 10 kg, leaving 10.6 - 10 = 0.6 kg. That is less than 2 kg, so it cannot fill another full box. She can sell 5 boxes.

10.6 - 10 = 0.6\ \text{kg} < 2\ \text{kg}

A box must be full to be sold, so the leftover that is not enough for a whole box does not count.

Answer: 5 boxes

4 · Reviewdoes it hold up?

10.6 kg of strawberries in 2 kg boxes should be about $\frac{10.6}{2}$ = about 5.30 boxes, and only whole full boxes count, so 5 boxes is the right size answer.

Another way: Guess and check (tool 6): test 5 boxes = 10 kg (fits) and 6 boxes = 12 kg (too much, more than 10.6 kg), confirming 5 is the largest number of full boxes.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading the improper fraction $\frac{53}{5}$ kg as a real weight (10.6 kg).
3.OA.A.2 Interpret whole-number quotients of whole numbers — Finding how many full groups of 2 kg fit in the total weight.

💡Takeaway. This only needs Grade 3 sharing-into-equal-groups sense: keep taking out 2 kg until less than 2 kg is left!

Variant 2 easy

Jisoo weighed the strawberries she picked at a weekend farm, and they weighed $\dfrac{73}{5}\ \text{kg}$ . She wants to pack the strawberries into boxes holding $3\ \text{kg}$ each and sell them. How many boxes in all can she sell?

Your answer

Show solution

1 · Understandwhat's really being asked

Jisoo has $\frac{73}{5}$ kg of strawberries. She packs them into 3 kg boxes. We need to count how many complete 3 kg boxes she can fill and sell.

Givens

Total weight of strawberries is $\frac{73}{5}$ kg.
Each box holds 3 kg of strawberries.

Unknowns

The number of complete 3 kg boxes she can sell.

Constraints

Only full boxes (each containing 3 kg) can be sold; leftover less than 3 kg cannot make a box.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #5 Look for a Pattern

Turn the improper fraction $\frac{73}{5}$ into an easier decimal weight, then repeatedly take away 3 kg (one box at a time) and count how many full boxes come out before less than 3 kg remains.

3 · Execute3 carry out the plan

1Write the total weight in an easier form

#9 Solve an Easier Related Problem 3.NF.A.1

Change the improper fraction

\frac{73}{5}

into a decimal so the weight is easy to picture: 73 divided by 5 is 14.6.

\dfrac{73}{5} = 14.6\ \text{kg}

Seeing

\frac{73}{5}

as 14.6 kg lets a young learner reason with a familiar weight instead of a scary fraction.

2Take away 3 kg for each box

#5 Look for a Pattern 3.OA.A.2

Each box uses up 3 kg. Counting boxes is the same as seeing how many whole groups of 3 kg fit inside 14.6 kg: 12 kg fills 4 boxes, and 15 kg would be too much.

3 + 3 + 3 + 3 = 12\ \text{kg} \le 14.6\ \text{kg} < 15\ \text{kg}

Repeatedly subtracting 3 kg is just sharing the strawberries into equal 3 kg boxes.

3Count full boxes and check the leftover

#9 Solve an Easier Related Problem 3.OA.A.2

After 4 boxes she has used 12 kg, leaving 14.6 - 12 = 2.6 kg. That is less than 3 kg, so it cannot fill another full box. She can sell 4 boxes.

14.6 - 12 = 2.6\ \text{kg} < 3\ \text{kg}

A box must be full to be sold, so the leftover that is not enough for a whole box does not count.

Answer: 4 boxes

4 · Reviewdoes it hold up?

14.6 kg of strawberries in 3 kg boxes should be about $\frac{14.6}{3}$ = about 4.87 boxes, and only whole full boxes count, so 4 boxes is the right size answer.

Another way: Guess and check (tool 6): test 4 boxes = 12 kg (fits) and 5 boxes = 15 kg (too much, more than 14.6 kg), confirming 4 is the largest number of full boxes.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading the improper fraction $\frac{73}{5}$ kg as a real weight (14.6 kg).
3.OA.A.2 Interpret whole-number quotients of whole numbers — Finding how many full groups of 3 kg fit in the total weight.

💡Takeaway. This only needs Grade 3 sharing-into-equal-groups sense: keep taking out 3 kg until less than 3 kg is left!

Variant 3 easy

Jisoo weighed the strawberries she picked at a weekend farm, and they weighed $\dfrac{89}{10}\ \text{kg}$ . She wants to pack the strawberries into boxes holding $4\ \text{kg}$ each and sell them. How many boxes in all can she sell?

Your answer

Show solution

1 · Understandwhat's really being asked

Jisoo has $\frac{89}{10}$ kg of strawberries. She packs them into 4 kg boxes. We need to count how many complete 4 kg boxes she can fill and sell.

Givens

Total weight of strawberries is $\frac{89}{10}$ kg.
Each box holds 4 kg of strawberries.

Unknowns

The number of complete 4 kg boxes she can sell.

Constraints

Only full boxes (each containing 4 kg) can be sold; leftover less than 4 kg cannot make a box.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #5 Look for a Pattern

Turn the improper fraction $\frac{89}{10}$ into an easier decimal weight, then repeatedly take away 4 kg (one box at a time) and count how many full boxes come out before less than 4 kg remains.

3 · Execute3 carry out the plan

1Write the total weight in an easier form

#9 Solve an Easier Related Problem 3.NF.A.1

Change the improper fraction

\frac{89}{10}

into a decimal so the weight is easy to picture: 89 divided by 10 is 8.9.

\dfrac{89}{10} = 8.9\ \text{kg}

Seeing

\frac{89}{10}

as 8.9 kg lets a young learner reason with a familiar weight instead of a scary fraction.

2Take away 4 kg for each box

#5 Look for a Pattern 3.OA.A.2

Each box uses up 4 kg. Counting boxes is the same as seeing how many whole groups of 4 kg fit inside 8.9 kg: 8 kg fills 2 boxes, and 12 kg would be too much.

4 + 4 = 8\ \text{kg} \le 8.9\ \text{kg} < 12\ \text{kg}

Repeatedly subtracting 4 kg is just sharing the strawberries into equal 4 kg boxes.

3Count full boxes and check the leftover

#9 Solve an Easier Related Problem 3.OA.A.2

After 2 boxes she has used 8 kg, leaving 8.9 - 8 = 0.9 kg. That is less than 4 kg, so it cannot fill another full box. She can sell 2 boxes.

8.9 - 8 = 0.9\ \text{kg} < 4\ \text{kg}

A box must be full to be sold, so the leftover that is not enough for a whole box does not count.

Answer: 2 boxes

4 · Reviewdoes it hold up?

8.9 kg of strawberries in 4 kg boxes should be about 8.9 / 4 = about 2.23 boxes, and only whole full boxes count, so 2 boxes is the right size answer.

Another way: Guess and check (tool 6): test 2 boxes = 8 kg (fits) and 3 boxes = 12 kg (too much, more than 8.9 kg), confirming 2 is the largest number of full boxes.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading the improper fraction $\frac{89}{10}$ kg as a real weight (8.9 kg).
3.OA.A.2 Interpret whole-number quotients of whole numbers — Finding how many full groups of 4 kg fit in the total weight.

💡Takeaway. This only needs Grade 3 sharing-into-equal-groups sense: keep taking out 4 kg until less than 4 kg is left!

Variant 4 easy

Jisoo weighed the strawberries she picked at a weekend farm, and they weighed $\dfrac{97}{4}\ \text{kg}$ . She wants to pack the strawberries into boxes holding $5\ \text{kg}$ each and sell them. How many boxes in all can she sell?

Your answer

Show solution

1 · Understandwhat's really being asked

Jisoo has $\frac{97}{4}$ kg of strawberries. She packs them into 5 kg boxes. We need to count how many complete 5 kg boxes she can fill and sell.

Givens

Total weight of strawberries is $\frac{97}{4}$ kg.
Each box holds 5 kg of strawberries.

Unknowns

The number of complete 5 kg boxes she can sell.

Constraints

Only full boxes (each containing 5 kg) can be sold; leftover less than 5 kg cannot make a box.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #5 Look for a Pattern

Turn the improper fraction $\frac{97}{4}$ into an easier decimal weight, then repeatedly take away 5 kg (one box at a time) and count how many full boxes come out before less than 5 kg remains.

3 · Execute3 carry out the plan

1Write the total weight in an easier form

#9 Solve an Easier Related Problem 3.NF.A.1

Change the improper fraction

\frac{97}{4}

into a decimal so the weight is easy to picture: 97 divided by 4 is 24.25.

\dfrac{97}{4} = 24.25\ \text{kg}

Seeing

\frac{97}{4}

as 24.25 kg lets a young learner reason with a familiar weight instead of a scary fraction.

2Take away 5 kg for each box

#5 Look for a Pattern 3.OA.A.2

Each box uses up 5 kg. Counting boxes is the same as seeing how many whole groups of 5 kg fit inside 24.25 kg: 20 kg fills 4 boxes, and 25 kg would be too much.

5 + 5 + 5 + 5 = 20\ \text{kg} \le 24.25\ \text{kg} < 25\ \text{kg}

Repeatedly subtracting 5 kg is just sharing the strawberries into equal 5 kg boxes.

3Count full boxes and check the leftover

#9 Solve an Easier Related Problem 3.OA.A.2

After 4 boxes she has used 20 kg, leaving 24.25 - 20 = 4.25 kg. That is less than 5 kg, so it cannot fill another full box. She can sell 4 boxes.

24.25 - 20 = 4.25\ \text{kg} < 5\ \text{kg}

A box must be full to be sold, so the leftover that is not enough for a whole box does not count.

Answer: 4 boxes

4 · Reviewdoes it hold up?

24.25 kg of strawberries in 5 kg boxes should be about $\frac{24.25}{5}$ = about 4.85 boxes, and only whole full boxes count, so 4 boxes is the right size answer.

Another way: Guess and check (tool 6): test 4 boxes = 20 kg (fits) and 5 boxes = 25 kg (too much, more than 24.25 kg), confirming 4 is the largest number of full boxes.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading the improper fraction $\frac{97}{4}$ kg as a real weight (24.25 kg).
3.OA.A.2 Interpret whole-number quotients of whole numbers — Finding how many full groups of 5 kg fit in the total weight.

💡Takeaway. This only needs Grade 3 sharing-into-equal-groups sense: keep taking out 5 kg until less than 5 kg is left!

Variant 5 medium

Jisoo weighed the strawberries she picked at a weekend farm, and they weighed $\dfrac{137}{20}\ \text{kg}$ . She wants to pack the strawberries into boxes holding $2\ \text{kg}$ each and sell them. How many boxes in all can she sell?

Your answer

Show solution

1 · Understandwhat's really being asked

Jisoo has $\frac{137}{20}$ kg of strawberries. She packs them into 2 kg boxes. We need to count how many complete 2 kg boxes she can fill and sell.

Givens

Total weight of strawberries is $\frac{137}{20}$ kg.
Each box holds 2 kg of strawberries.

Unknowns

The number of complete 2 kg boxes she can sell.

Constraints

Only full boxes (each containing 2 kg) can be sold; leftover less than 2 kg cannot make a box.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #5 Look for a Pattern

Turn the improper fraction $\frac{137}{20}$ into an easier decimal weight, then repeatedly take away 2 kg (one box at a time) and count how many full boxes come out before less than 2 kg remains.

3 · Execute3 carry out the plan

1Write the total weight in an easier form

#9 Solve an Easier Related Problem 3.NF.A.1

Change the improper fraction

\frac{137}{20}

into a decimal so the weight is easy to picture: 137 divided by 20 is 6.85.

\dfrac{137}{20} = 6.85\ \text{kg}

Seeing

\frac{137}{20}

as 6.85 kg lets a young learner reason with a familiar weight instead of a scary fraction.

2Take away 2 kg for each box

#5 Look for a Pattern 3.OA.A.2

Each box uses up 2 kg. Counting boxes is the same as seeing how many whole groups of 2 kg fit inside 6.85 kg: 6 kg fills 3 boxes, and 8 kg would be too much.

2 + 2 + 2 = 6\ \text{kg} \le 6.85\ \text{kg} < 8\ \text{kg}

Repeatedly subtracting 2 kg is just sharing the strawberries into equal 2 kg boxes.

3Count full boxes and check the leftover

#9 Solve an Easier Related Problem 3.OA.A.2

After 3 boxes she has used 6 kg, leaving 6.85 - 6 = 0.85 kg. That is less than 2 kg, so it cannot fill another full box. She can sell 3 boxes.

6.85 - 6 = 0.85\ \text{kg} < 2\ \text{kg}

A box must be full to be sold, so the leftover that is not enough for a whole box does not count.

Answer: 3 boxes

4 · Reviewdoes it hold up?

6.85 kg of strawberries in 2 kg boxes should be about $\frac{6.85}{2}$ = about 3.42 boxes, and only whole full boxes count, so 3 boxes is the right size answer.

Another way: Guess and check (tool 6): test 3 boxes = 6 kg (fits) and 4 boxes = 8 kg (too much, more than 6.85 kg), confirming 3 is the largest number of full boxes.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading the improper fraction $\frac{137}{20}$ kg as a real weight (6.85 kg).
3.OA.A.2 Interpret whole-number quotients of whole numbers — Finding how many full groups of 2 kg fit in the total weight.

💡Takeaway. This only needs Grade 3 sharing-into-equal-groups sense: keep taking out 2 kg until less than 2 kg is left!

Variant 6 medium

Jisoo weighed the strawberries she picked at a weekend farm, and they weighed $\dfrac{143}{10}\ \text{kg}$ . She wants to pack the strawberries into boxes holding $3\ \text{kg}$ each and sell them. How many boxes in all can she sell?

Your answer

Show solution

1 · Understandwhat's really being asked

Jisoo has $\frac{143}{10}$ kg of strawberries. She packs them into 3 kg boxes. We need to count how many complete 3 kg boxes she can fill and sell.

Givens

Total weight of strawberries is $\frac{143}{10}$ kg.
Each box holds 3 kg of strawberries.

Unknowns

The number of complete 3 kg boxes she can sell.

Constraints

Only full boxes (each containing 3 kg) can be sold; leftover less than 3 kg cannot make a box.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #5 Look for a Pattern

Turn the improper fraction $\frac{143}{10}$ into an easier decimal weight, then repeatedly take away 3 kg (one box at a time) and count how many full boxes come out before less than 3 kg remains.

3 · Execute3 carry out the plan

1Write the total weight in an easier form

#9 Solve an Easier Related Problem 3.NF.A.1

Change the improper fraction

\frac{143}{10}

into a decimal so the weight is easy to picture: 143 divided by 10 is 14.3.

\dfrac{143}{10} = 14.3\ \text{kg}

Seeing

\frac{143}{10}

as 14.3 kg lets a young learner reason with a familiar weight instead of a scary fraction.

2Take away 3 kg for each box

#5 Look for a Pattern 3.OA.A.2

Each box uses up 3 kg. Counting boxes is the same as seeing how many whole groups of 3 kg fit inside 14.3 kg: 12 kg fills 4 boxes, and 15 kg would be too much.

3 + 3 + 3 + 3 = 12\ \text{kg} \le 14.3\ \text{kg} < 15\ \text{kg}

Repeatedly subtracting 3 kg is just sharing the strawberries into equal 3 kg boxes.

3Count full boxes and check the leftover

#9 Solve an Easier Related Problem 3.OA.A.2

After 4 boxes she has used 12 kg, leaving 14.3 - 12 = 2.3 kg. That is less than 3 kg, so it cannot fill another full box. She can sell 4 boxes.

14.3 - 12 = 2.3\ \text{kg} < 3\ \text{kg}

A box must be full to be sold, so the leftover that is not enough for a whole box does not count.

Answer: 4 boxes

4 · Reviewdoes it hold up?

14.3 kg of strawberries in 3 kg boxes should be about $\frac{14.3}{3}$ = about 4.77 boxes, and only whole full boxes count, so 4 boxes is the right size answer.

Another way: Guess and check (tool 6): test 4 boxes = 12 kg (fits) and 5 boxes = 15 kg (too much, more than 14.3 kg), confirming 4 is the largest number of full boxes.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading the improper fraction $\frac{143}{10}$ kg as a real weight (14.3 kg).
3.OA.A.2 Interpret whole-number quotients of whole numbers — Finding how many full groups of 3 kg fit in the total weight.

💡Takeaway. This only needs Grade 3 sharing-into-equal-groups sense: keep taking out 3 kg until less than 3 kg is left!

Variant 7 medium

Jisoo weighed the strawberries she picked at a weekend farm, and they weighed $\dfrac{179}{8}\ \text{kg}$ . She wants to pack the strawberries into boxes holding $6\ \text{kg}$ each and sell them. How many boxes in all can she sell?

Your answer

Show solution

1 · Understandwhat's really being asked

Jisoo has $\frac{179}{8}$ kg of strawberries. She packs them into 6 kg boxes. We need to count how many complete 6 kg boxes she can fill and sell.

Givens

Total weight of strawberries is $\frac{179}{8}$ kg.
Each box holds 6 kg of strawberries.

Unknowns

The number of complete 6 kg boxes she can sell.

Constraints

Only full boxes (each containing 6 kg) can be sold; leftover less than 6 kg cannot make a box.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #5 Look for a Pattern

Turn the improper fraction $\frac{179}{8}$ into an easier decimal weight, then repeatedly take away 6 kg (one box at a time) and count how many full boxes come out before less than 6 kg remains.

3 · Execute3 carry out the plan

1Write the total weight in an easier form

#9 Solve an Easier Related Problem 3.NF.A.1

Change the improper fraction

\frac{179}{8}

into a decimal so the weight is easy to picture: 179 divided by 8 is 22.375.

\dfrac{179}{8} = 22.375\ \text{kg}

Seeing

\frac{179}{8}

as 22.375 kg lets a young learner reason with a familiar weight instead of a scary fraction.

2Take away 6 kg for each box

#5 Look for a Pattern 3.OA.A.2

Each box uses up 6 kg. Counting boxes is the same as seeing how many whole groups of 6 kg fit inside 22.375 kg: 18 kg fills 3 boxes, and 24 kg would be too much.

6 + 6 + 6 = 18\ \text{kg} \le 22.375\ \text{kg} < 24\ \text{kg}

Repeatedly subtracting 6 kg is just sharing the strawberries into equal 6 kg boxes.

3Count full boxes and check the leftover

#9 Solve an Easier Related Problem 3.OA.A.2

After 3 boxes she has used 18 kg, leaving 22.375 - 18 = 4.375 kg. That is less than 6 kg, so it cannot fill another full box. She can sell 3 boxes.

22.375 - 18 = 4.375\ \text{kg} < 6\ \text{kg}

A box must be full to be sold, so the leftover that is not enough for a whole box does not count.

Answer: 3 boxes

4 · Reviewdoes it hold up?

22.375 kg of strawberries in 6 kg boxes should be about $\frac{22.375}{6}$ = about 3.73 boxes, and only whole full boxes count, so 3 boxes is the right size answer.

Another way: Guess and check (tool 6): test 3 boxes = 18 kg (fits) and 4 boxes = 24 kg (too much, more than 22.375 kg), confirming 3 is the largest number of full boxes.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading the improper fraction $\frac{179}{8}$ kg as a real weight (22.375 kg).
3.OA.A.2 Interpret whole-number quotients of whole numbers — Finding how many full groups of 6 kg fit in the total weight.

💡Takeaway. This only needs Grade 3 sharing-into-equal-groups sense: keep taking out 6 kg until less than 6 kg is left!

Variant 8 medium

Jisoo weighed the strawberries she picked at a weekend farm, and they weighed $\dfrac{187}{20}\ \text{kg}$ . She wants to pack the strawberries into boxes holding $2\ \text{kg}$ each and sell them. How many boxes in all can she sell?

Your answer

Show solution

1 · Understandwhat's really being asked

Jisoo has $\frac{187}{20}$ kg of strawberries. She packs them into 2 kg boxes. We need to count how many complete 2 kg boxes she can fill and sell.

Givens

Total weight of strawberries is $\frac{187}{20}$ kg.
Each box holds 2 kg of strawberries.

Unknowns

The number of complete 2 kg boxes she can sell.

Constraints

Only full boxes (each containing 2 kg) can be sold; leftover less than 2 kg cannot make a box.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #5 Look for a Pattern

Turn the improper fraction $\frac{187}{20}$ into an easier decimal weight, then repeatedly take away 2 kg (one box at a time) and count how many full boxes come out before less than 2 kg remains.

3 · Execute3 carry out the plan

1Write the total weight in an easier form

#9 Solve an Easier Related Problem 3.NF.A.1

Change the improper fraction

\frac{187}{20}

into a decimal so the weight is easy to picture: 187 divided by 20 is 9.35.

\dfrac{187}{20} = 9.35\ \text{kg}

Seeing

\frac{187}{20}

as 9.35 kg lets a young learner reason with a familiar weight instead of a scary fraction.

2Take away 2 kg for each box

#5 Look for a Pattern 3.OA.A.2

Each box uses up 2 kg. Counting boxes is the same as seeing how many whole groups of 2 kg fit inside 9.35 kg: 8 kg fills 4 boxes, and 10 kg would be too much.

2 + 2 + 2 + 2 = 8\ \text{kg} \le 9.35\ \text{kg} < 10\ \text{kg}

Repeatedly subtracting 2 kg is just sharing the strawberries into equal 2 kg boxes.

3Count full boxes and check the leftover

#9 Solve an Easier Related Problem 3.OA.A.2

After 4 boxes she has used 8 kg, leaving 9.35 - 8 = 1.35 kg. That is less than 2 kg, so it cannot fill another full box. She can sell 4 boxes.

9.35 - 8 = 1.35\ \text{kg} < 2\ \text{kg}

A box must be full to be sold, so the leftover that is not enough for a whole box does not count.

Answer: 4 boxes

4 · Reviewdoes it hold up?

9.35 kg of strawberries in 2 kg boxes should be about $\frac{9.35}{2}$ = about 4.67 boxes, and only whole full boxes count, so 4 boxes is the right size answer.

Another way: Guess and check (tool 6): test 4 boxes = 8 kg (fits) and 5 boxes = 10 kg (too much, more than 9.35 kg), confirming 4 is the largest number of full boxes.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading the improper fraction $\frac{187}{20}$ kg as a real weight (9.35 kg).
3.OA.A.2 Interpret whole-number quotients of whole numbers — Finding how many full groups of 2 kg fit in the total weight.

💡Takeaway. This only needs Grade 3 sharing-into-equal-groups sense: keep taking out 2 kg until less than 2 kg is left!

Variant 9 hard

Jisoo weighed the strawberries she picked at a weekend farm, and they weighed $\dfrac{211}{25}\ \text{kg}$ . She wants to pack the strawberries into boxes holding $4\ \text{kg}$ each and sell them. How many boxes in all can she sell?

Your answer

Show solution

1 · Understandwhat's really being asked

Jisoo has $\frac{211}{25}$ kg of strawberries. She packs them into 4 kg boxes. We need to count how many complete 4 kg boxes she can fill and sell.

Givens

Total weight of strawberries is $\frac{211}{25}$ kg.
Each box holds 4 kg of strawberries.

Unknowns

The number of complete 4 kg boxes she can sell.

Constraints

Only full boxes (each containing 4 kg) can be sold; leftover less than 4 kg cannot make a box.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #5 Look for a Pattern

Turn the improper fraction $\frac{211}{25}$ into an easier decimal weight, then repeatedly take away 4 kg (one box at a time) and count how many full boxes come out before less than 4 kg remains.

3 · Execute3 carry out the plan

1Write the total weight in an easier form

#9 Solve an Easier Related Problem 3.NF.A.1

Change the improper fraction

\frac{211}{25}

into a decimal so the weight is easy to picture: 211 divided by 25 is 8.44.

\dfrac{211}{25} = 8.44\ \text{kg}

Seeing

\frac{211}{25}

as 8.44 kg lets a young learner reason with a familiar weight instead of a scary fraction.

2Take away 4 kg for each box

#5 Look for a Pattern 3.OA.A.2

Each box uses up 4 kg. Counting boxes is the same as seeing how many whole groups of 4 kg fit inside 8.44 kg: 8 kg fills 2 boxes, and 12 kg would be too much.

4 + 4 = 8\ \text{kg} \le 8.44\ \text{kg} < 12\ \text{kg}

Repeatedly subtracting 4 kg is just sharing the strawberries into equal 4 kg boxes.

3Count full boxes and check the leftover

#9 Solve an Easier Related Problem 3.OA.A.2

After 2 boxes she has used 8 kg, leaving 8.44 - 8 = 0.44 kg. That is less than 4 kg, so it cannot fill another full box. She can sell 2 boxes.

8.44 - 8 = 0.44\ \text{kg} < 4\ \text{kg}

A box must be full to be sold, so the leftover that is not enough for a whole box does not count.

Answer: 2 boxes

4 · Reviewdoes it hold up?

8.44 kg of strawberries in 4 kg boxes should be about $\frac{8.44}{4}$ = about 2.11 boxes, and only whole full boxes count, so 2 boxes is the right size answer.

Another way: Guess and check (tool 6): test 2 boxes = 8 kg (fits) and 3 boxes = 12 kg (too much, more than 8.44 kg), confirming 2 is the largest number of full boxes.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading the improper fraction $\frac{211}{25}$ kg as a real weight (8.44 kg).
3.OA.A.2 Interpret whole-number quotients of whole numbers — Finding how many full groups of 4 kg fit in the total weight.

💡Takeaway. This only needs Grade 3 sharing-into-equal-groups sense: keep taking out 4 kg until less than 4 kg is left!

Variant 10 hard

Jisoo weighed the strawberries she picked at a weekend farm, and they weighed $\dfrac{251}{100}\ \text{kg}$ . She wants to pack the strawberries into boxes holding $1\ \text{kg}$ each and sell them. How many boxes in all can she sell?

Your answer

Show solution

1 · Understandwhat's really being asked

Jisoo has $\frac{251}{100}$ kg of strawberries. She packs them into 1 kg boxes. We need to count how many complete 1 kg boxes she can fill and sell.

Givens

Total weight of strawberries is $\frac{251}{100}$ kg.
Each box holds 1 kg of strawberries.

Unknowns

The number of complete 1 kg boxes she can sell.

Constraints

Only full boxes (each containing 1 kg) can be sold; leftover less than 1 kg cannot make a box.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #5 Look for a Pattern

Turn the improper fraction $\frac{251}{100}$ into an easier decimal weight, then repeatedly take away 1 kg (one box at a time) and count how many full boxes come out before less than 1 kg remains.

3 · Execute3 carry out the plan

1Write the total weight in an easier form

#9 Solve an Easier Related Problem 3.NF.A.1

Change the improper fraction

\frac{251}{100}

into a decimal so the weight is easy to picture: 251 divided by 100 is 2.51.

\dfrac{251}{100} = 2.51\ \text{kg}

Seeing

\frac{251}{100}

as 2.51 kg lets a young learner reason with a familiar weight instead of a scary fraction.

2Take away 1 kg for each box

#5 Look for a Pattern 3.OA.A.2

Each box uses up 1 kg. Counting boxes is the same as seeing how many whole groups of 1 kg fit inside 2.51 kg: 2 kg fills 2 boxes, and 3 kg would be too much.

1 + 1 = 2\ \text{kg} \le 2.51\ \text{kg} < 3\ \text{kg}

Repeatedly subtracting 1 kg is just sharing the strawberries into equal 1 kg boxes.

3Count full boxes and check the leftover

#9 Solve an Easier Related Problem 3.OA.A.2

After 2 boxes she has used 2 kg, leaving 2.51 - 2 = 0.51 kg. That is less than 1 kg, so it cannot fill another full box. She can sell 2 boxes.

2.51 - 2 = 0.51\ \text{kg} < 1\ \text{kg}

A box must be full to be sold, so the leftover that is not enough for a whole box does not count.

Answer: 2 boxes

4 · Reviewdoes it hold up?

2.51 kg of strawberries in 1 kg boxes should be about $\frac{2.51}{1}$ = about 2.51 boxes, and only whole full boxes count, so 2 boxes is the right size answer.

Another way: Guess and check (tool 6): test 2 boxes = 2 kg (fits) and 3 boxes = 3 kg (too much, more than 2.51 kg), confirming 2 is the largest number of full boxes.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading the improper fraction $\frac{251}{100}$ kg as a real weight (2.51 kg).
3.OA.A.2 Interpret whole-number quotients of whole numbers — Finding how many full groups of 1 kg fit in the total weight.

💡Takeaway. This only needs Grade 3 sharing-into-equal-groups sense: keep taking out 1 kg until less than 1 kg is left!

Variant 11 hard

Jisoo weighed the strawberries she picked at a weekend farm, and they weighed $\dfrac{321}{50}\ \text{kg}$ . She wants to pack the strawberries into boxes holding $3\ \text{kg}$ each and sell them. How many boxes in all can she sell?

Your answer

Show solution

1 · Understandwhat's really being asked

Jisoo has $\frac{321}{50}$ kg of strawberries. She packs them into 3 kg boxes. We need to count how many complete 3 kg boxes she can fill and sell.

Givens

Total weight of strawberries is $\frac{321}{50}$ kg.
Each box holds 3 kg of strawberries.

Unknowns

The number of complete 3 kg boxes she can sell.

Constraints

Only full boxes (each containing 3 kg) can be sold; leftover less than 3 kg cannot make a box.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #5 Look for a Pattern

Turn the improper fraction $\frac{321}{50}$ into an easier decimal weight, then repeatedly take away 3 kg (one box at a time) and count how many full boxes come out before less than 3 kg remains.

3 · Execute3 carry out the plan

1Write the total weight in an easier form

#9 Solve an Easier Related Problem 3.NF.A.1

Change the improper fraction

\frac{321}{50}

into a decimal so the weight is easy to picture: 321 divided by 50 is 6.42.

\dfrac{321}{50} = 6.42\ \text{kg}

Seeing

\frac{321}{50}

as 6.42 kg lets a young learner reason with a familiar weight instead of a scary fraction.

2Take away 3 kg for each box

#5 Look for a Pattern 3.OA.A.2

Each box uses up 3 kg. Counting boxes is the same as seeing how many whole groups of 3 kg fit inside 6.42 kg: 6 kg fills 2 boxes, and 9 kg would be too much.

3 + 3 = 6\ \text{kg} \le 6.42\ \text{kg} < 9\ \text{kg}

Repeatedly subtracting 3 kg is just sharing the strawberries into equal 3 kg boxes.

3Count full boxes and check the leftover

#9 Solve an Easier Related Problem 3.OA.A.2

After 2 boxes she has used 6 kg, leaving 6.42 - 6 = 0.42 kg. That is less than 3 kg, so it cannot fill another full box. She can sell 2 boxes.

6.42 - 6 = 0.42\ \text{kg} < 3\ \text{kg}

A box must be full to be sold, so the leftover that is not enough for a whole box does not count.

Answer: 2 boxes

4 · Reviewdoes it hold up?

6.42 kg of strawberries in 3 kg boxes should be about $\frac{6.42}{3}$ = about 2.14 boxes, and only whole full boxes count, so 2 boxes is the right size answer.

Another way: Guess and check (tool 6): test 2 boxes = 6 kg (fits) and 3 boxes = 9 kg (too much, more than 6.42 kg), confirming 2 is the largest number of full boxes.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading the improper fraction $\frac{321}{50}$ kg as a real weight (6.42 kg).
3.OA.A.2 Interpret whole-number quotients of whole numbers — Finding how many full groups of 3 kg fit in the total weight.

💡Takeaway. This only needs Grade 3 sharing-into-equal-groups sense: keep taking out 3 kg until less than 3 kg is left!

Variant 12 hard

Jisoo weighed the strawberries she picked at a weekend farm, and they weighed $\dfrac{467}{40}\ \text{kg}$ . She wants to pack the strawberries into boxes holding $5\ \text{kg}$ each and sell them. How many boxes in all can she sell?

Your answer

Show solution

1 · Understandwhat's really being asked

Jisoo has $\frac{467}{40}$ kg of strawberries. She packs them into 5 kg boxes. We need to count how many complete 5 kg boxes she can fill and sell.

Givens

Total weight of strawberries is $\frac{467}{40}$ kg.
Each box holds 5 kg of strawberries.

Unknowns

The number of complete 5 kg boxes she can sell.

Constraints

Only full boxes (each containing 5 kg) can be sold; leftover less than 5 kg cannot make a box.

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #5 Look for a Pattern

Turn the improper fraction $\frac{467}{40}$ into an easier decimal weight, then repeatedly take away 5 kg (one box at a time) and count how many full boxes come out before less than 5 kg remains.

3 · Execute3 carry out the plan

1Write the total weight in an easier form

#9 Solve an Easier Related Problem 3.NF.A.1

Change the improper fraction

\frac{467}{40}

into a decimal so the weight is easy to picture: 467 divided by 40 is 11.675.

\dfrac{467}{40} = 11.675\ \text{kg}

Seeing

\frac{467}{40}

as 11.675 kg lets a young learner reason with a familiar weight instead of a scary fraction.

2Take away 5 kg for each box

#5 Look for a Pattern 3.OA.A.2

Each box uses up 5 kg. Counting boxes is the same as seeing how many whole groups of 5 kg fit inside 11.675 kg: 10 kg fills 2 boxes, and 15 kg would be too much.

5 + 5 = 10\ \text{kg} \le 11.675\ \text{kg} < 15\ \text{kg}

Repeatedly subtracting 5 kg is just sharing the strawberries into equal 5 kg boxes.

3Count full boxes and check the leftover

#9 Solve an Easier Related Problem 3.OA.A.2

After 2 boxes she has used 10 kg, leaving 11.675 - 10 = 1.675 kg. That is less than 5 kg, so it cannot fill another full box. She can sell 2 boxes.

11.675 - 10 = 1.675\ \text{kg} < 5\ \text{kg}

A box must be full to be sold, so the leftover that is not enough for a whole box does not count.

Answer: 2 boxes

4 · Reviewdoes it hold up?

11.675 kg of strawberries in 5 kg boxes should be about $\frac{11.675}{5}$ = about 2.33 boxes, and only whole full boxes count, so 2 boxes is the right size answer.

Another way: Guess and check (tool 6): test 2 boxes = 10 kg (fits) and 3 boxes = 15 kg (too much, more than 11.675 kg), confirming 2 is the largest number of full boxes.

Standardsmin grade 3

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading the improper fraction $\frac{467}{40}$ kg as a real weight (11.675 kg).
3.OA.A.2 Interpret whole-number quotients of whole numbers — Finding how many full groups of 5 kg fit in the total weight.

💡Takeaway. This only needs Grade 3 sharing-into-equal-groups sense: keep taking out 5 kg until less than 5 kg is left!

Split a whole into proper-fraction portions · 12 practice problems

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