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Fill a table and complete a line graph from clues · 8 practice problems

4.OA.A.35.MD.B.2

Generated variants — 8

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 answer: June = 200 books, July = 300 books

The line graph shows the books sold recorded each month. In June the count was $40$ more than in May, and the number in June and July together is $500$ . Complete the line graph.

(Figure) A line graph titled "Books Sold." The horizontal axis shows the month - $3$ (March), $4$ (April), $5$ (May), $6$ (June), $7$ (July). The vertical axis shows the number of books, with major gridlines at $0$ , $100$ , $200$ , $300$ , $400$ and each small grid square worth $20$ books. Points are plotted and joined for March at $120$ ; April at $220$ ; May at $160$ ; the points for June and July have not yet been plotted.

Show solution

Understand

A line graph of books sold each month shows March = 120, April = 220, May = 160; June and July are blank. June is 40 more than May, and June + July together is 500. I must find the June and July counts to complete the graph.

Givens

March = 120 books, April = 220 books, May = 160 books
June = May + 40
June + July = 500
Each small grid square = 20 books

Unknowns

Number of books made in June
Number of books made in July

Constraints

Plotted June and July values must land on grid squares worth 20 each

Plan

#7 Identify Subproblems · also uses: #11 Work Backwards

I solve in order from what I can already know: first June (May plus 40), then July (the 500 total minus June). Each is a one-step subproblem, and finding July uses the total backwards.

Execute

#7 Identify Subproblems 4.OA.A.3

June is 40 more than May, and May is 160, so June = 160 + 40 = 200.

160 + 40 = 200

Start from the value you already know and apply the given step.

#11 Work Backwards 4.OA.A.3

June and July together make 500, so July = 500 - June = 500 - 200 = 300.

500 - 200 = 300

Knowing the total and one part, subtract to recover the other part.

#7 Identify Subproblems 5.MD.B.2

On the graph (each square = 20), June 200 sits 0 squares above 200, and July 300 sits 0 squares above 300; join the points with straight segments.

200 = 200 + 0\times 20,\quad 300 = 300 + 0\times 20

Converting the numbers into squares lets you place each point exactly.

Answer: June = 200 books, July = 300 books

Review

Check the clues: June 200 is indeed 40 more than May 160, and June 200 + July 300 = 500 as required. Both values land neatly on grid lines (multiples of 20), so the plotted points are valid.

Convert to algebra (tool 13): let June = J = 160 + 40 and July = 500 - J; solving gives the same J = 200 and July = 300.

Standards · min grade 5

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Computing June from May and July from the total
5.MD.B.2 Make a line plot to display a data set and solve problems using the data — Placing the June and July values on the scaled graph

💡 Fill in what you can figure out first, then use the total to get the rest - then just plot!

Variant 2 answer: July = 260 frames, August = 380 frames

The line graph shows the picture frames made recorded each month. In July the count was $80$ more than in June, and the number in July and August together is $640$ . Complete the line graph.

(Figure) A line graph titled "Picture Frames Made." The horizontal axis shows the month - $4$ (April), $5$ (May), $6$ (June), $7$ (July), $8$ (August). The vertical axis shows the number of frames, with major gridlines at $0$ , $100$ , $200$ , $300$ , $400$ and each small grid square worth $20$ frames. Points are plotted and joined for April at $280$ ; May at $380$ ; June at $180$ ; the points for July and August have not yet been plotted.

Show solution

Understand

A line graph of picture frames made each month shows April = 280, May = 380, June = 180; July and August are blank. July is 80 more than June, and July + August together is 640. I must find the July and August counts to complete the graph.

Givens

April = 280 frames, May = 380 frames, June = 180 frames
July = June + 80
July + August = 640
Each small grid square = 20 frames

Unknowns

Number of frames made in July
Number of frames made in August

Constraints

Plotted July and August values must land on grid squares worth 20 each

Plan

#7 Identify Subproblems · also uses: #11 Work Backwards

I solve in order from what I can already know: first July (June plus 80), then August (the 640 total minus July). Each is a one-step subproblem, and finding August uses the total backwards.

Execute

#7 Identify Subproblems 4.OA.A.3

July is 80 more than June, and June is 180, so July = 180 + 80 = 260.

180 + 80 = 260

Start from the value you already know and apply the given step.

#11 Work Backwards 4.OA.A.3

July and August together make 640, so August = 640 - July = 640 - 260 = 380.

640 - 260 = 380

Knowing the total and one part, subtract to recover the other part.

#7 Identify Subproblems 5.MD.B.2

On the graph (each square = 20), July 260 sits 3 squares above 200, and August 380 sits 4 squares above 300; join the points with straight segments.

260 = 200 + 3\times 20,\quad 380 = 300 + 4\times 20

Converting the numbers into squares lets you place each point exactly.

Answer: July = 260 frames, August = 380 frames

Review

Check the clues: July 260 is indeed 80 more than June 180, and July 260 + August 380 = 640 as required. Both values land neatly on grid lines (multiples of 20), so the plotted points are valid.

Convert to algebra (tool 13): let July = J = 180 + 80 and August = 640 - J; solving gives the same J = 260 and August = 380.

Standards · min grade 5

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Computing July from June and August from the total
5.MD.B.2 Make a line plot to display a data set and solve problems using the data — Placing the July and August values on the scaled graph

💡 Fill in what you can figure out first, then use the total to get the rest - then just plot!

Variant 3 answer: Week 4 = 300 cars, Week 5 = 400 cars

The line graph shows the toy cars made recorded each week. In Week 4 the count was $60$ more than in Week 3, and the number in Week 4 and Week 5 together is $700$ . Complete the line graph.

(Figure) A line graph titled "Toy Cars Made." The horizontal axis shows the week - $1$ (Week 1), $2$ (Week 2), $3$ (Week 3), $4$ (Week 4), $5$ (Week 5). The vertical axis shows the number of cars, with major gridlines at $0$ , $100$ , $200$ , $300$ , $400$ and each small grid square worth $20$ cars. Points are plotted and joined for Week 1 at $200$ ; Week 2 at $300$ ; Week 3 at $240$ ; the points for Week 4 and Week 5 have not yet been plotted.

Show solution

Understand

A line graph of toy cars made each week shows Week 1 = 200, Week 2 = 300, Week 3 = 240; Week 4 and Week 5 are blank. Week 4 is 60 more than Week 3, and Week 4 + Week 5 together is 700. I must find the Week 4 and Week 5 counts to complete the graph.

Givens

Week 1 = 200 cars, Week 2 = 300 cars, Week 3 = 240 cars
Week 4 = Week 3 + 60
Week 4 + Week 5 = 700
Each small grid square = 20 cars

Unknowns

Number of cars made in Week 4
Number of cars made in Week 5

Constraints

Plotted Week 4 and Week 5 values must land on grid squares worth 20 each

Plan

#7 Identify Subproblems · also uses: #11 Work Backwards

I solve in order from what I can already know: first Week 4 (Week 3 plus 60), then Week 5 (the 700 total minus Week 4). Each is a one-step subproblem, and finding Week 5 uses the total backwards.

Execute

#7 Identify Subproblems 4.OA.A.3

Week 4 is 60 more than Week 3, and Week 3 is 240, so Week 4 = 240 + 60 = 300.

240 + 60 = 300

Start from the value you already know and apply the given step.

#11 Work Backwards 4.OA.A.3

Week 4 and Week 5 together make 700, so Week 5 = 700 - Week 4 = 700 - 300 = 400.

700 - 300 = 400

Knowing the total and one part, subtract to recover the other part.

#7 Identify Subproblems 5.MD.B.2

On the graph (each square = 20), Week 4 300 sits 0 squares above 300, and Week 5 400 sits 0 squares above 400; join the points with straight segments.

300 = 300 + 0\times 20,\quad 400 = 400 + 0\times 20

Converting the numbers into squares lets you place each point exactly.

Answer: Week 4 = 300 cars, Week 5 = 400 cars

Review

Check the clues: Week 4 300 is indeed 60 more than Week 3 240, and Week 4 300 + Week 5 400 = 700 as required. Both values land neatly on grid lines (multiples of 20), so the plotted points are valid.

Convert to algebra (tool 13): let Week 4 = J = 240 + 60 and Week 5 = 700 - J; solving gives the same J = 300 and Week 5 = 400.

Standards · min grade 5

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Computing Week 4 from Week 3 and Week 5 from the total
5.MD.B.2 Make a line plot to display a data set and solve problems using the data — Placing the Week 4 and Week 5 values on the scaled graph

💡 Fill in what you can figure out first, then use the total to get the rest - then just plot!

Variant 4 answer: July = 300 tickets, August = 300 tickets

The line graph shows the tickets sold recorded each month. In July the count was $80$ more than in June, and the number in July and August together is $600$ . Complete the line graph.

(Figure) A line graph titled "Tickets Sold." The horizontal axis shows the month - $4$ (April), $5$ (May), $6$ (June), $7$ (July), $8$ (August). The vertical axis shows the number of tickets, with major gridlines at $0$ , $100$ , $200$ , $300$ , $400$ and each small grid square worth $20$ tickets. Points are plotted and joined for April at $240$ ; May at $340$ ; June at $220$ ; the points for July and August have not yet been plotted.

Show solution

Understand

A line graph of tickets sold each month shows April = 240, May = 340, June = 220; July and August are blank. July is 80 more than June, and July + August together is 600. I must find the July and August counts to complete the graph.

Givens

April = 240 tickets, May = 340 tickets, June = 220 tickets
July = June + 80
July + August = 600
Each small grid square = 20 tickets

Unknowns

Number of tickets made in July
Number of tickets made in August

Constraints

Plotted July and August values must land on grid squares worth 20 each

Plan

#7 Identify Subproblems · also uses: #11 Work Backwards

I solve in order from what I can already know: first July (June plus 80), then August (the 600 total minus July). Each is a one-step subproblem, and finding August uses the total backwards.

Execute

#7 Identify Subproblems 4.OA.A.3

July is 80 more than June, and June is 220, so July = 220 + 80 = 300.

220 + 80 = 300

Start from the value you already know and apply the given step.

#11 Work Backwards 4.OA.A.3

July and August together make 600, so August = 600 - July = 600 - 300 = 300.

600 - 300 = 300

Knowing the total and one part, subtract to recover the other part.

#7 Identify Subproblems 5.MD.B.2

On the graph (each square = 20), July 300 sits 0 squares above 300, and August 300 sits 0 squares above 300; join the points with straight segments.

300 = 300 + 0\times 20,\quad 300 = 300 + 0\times 20

Converting the numbers into squares lets you place each point exactly.

Answer: July = 300 tickets, August = 300 tickets

Review

Check the clues: July 300 is indeed 80 more than June 220, and July 300 + August 300 = 600 as required. Both values land neatly on grid lines (multiples of 20), so the plotted points are valid.

Convert to algebra (tool 13): let July = J = 220 + 80 and August = 600 - J; solving gives the same J = 300 and August = 300.

Standards · min grade 5

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Computing July from June and August from the total
5.MD.B.2 Make a line plot to display a data set and solve problems using the data — Placing the July and August values on the scaled graph

💡 Fill in what you can figure out first, then use the total to get the rest - then just plot!

Variant 5 answer: Day 4 = 200 loaves, Day 5 = 260 loaves

The line graph shows the loaves baked recorded each day. In Day 4 the count was $60$ more than in Day 3, and the number in Day 4 and Day 5 together is $460$ . Complete the line graph.

(Figure) A line graph titled "Loaves Baked." The horizontal axis shows the day - $1$ (Day 1), $2$ (Day 2), $3$ (Day 3), $4$ (Day 4), $5$ (Day 5). The vertical axis shows the number of loaves, with major gridlines at $0$ , $100$ , $200$ , $300$ , $400$ and each small grid square worth $20$ loaves. Points are plotted and joined for Day 1 at $180$ ; Day 2 at $280$ ; Day 3 at $140$ ; the points for Day 4 and Day 5 have not yet been plotted.

Show solution

Understand

A line graph of loaves baked each day shows Day 1 = 180, Day 2 = 280, Day 3 = 140; Day 4 and Day 5 are blank. Day 4 is 60 more than Day 3, and Day 4 + Day 5 together is 460. I must find the Day 4 and Day 5 counts to complete the graph.

Givens

Day 1 = 180 loaves, Day 2 = 280 loaves, Day 3 = 140 loaves
Day 4 = Day 3 + 60
Day 4 + Day 5 = 460
Each small grid square = 20 loaves

Unknowns

Number of loaves made in Day 4
Number of loaves made in Day 5

Constraints

Plotted Day 4 and Day 5 values must land on grid squares worth 20 each

Plan

#7 Identify Subproblems · also uses: #11 Work Backwards

I solve in order from what I can already know: first Day 4 (Day 3 plus 60), then Day 5 (the 460 total minus Day 4). Each is a one-step subproblem, and finding Day 5 uses the total backwards.

Execute

#7 Identify Subproblems 4.OA.A.3

Day 4 is 60 more than Day 3, and Day 3 is 140, so Day 4 = 140 + 60 = 200.

140 + 60 = 200

Start from the value you already know and apply the given step.

#11 Work Backwards 4.OA.A.3

Day 4 and Day 5 together make 460, so Day 5 = 460 - Day 4 = 460 - 200 = 260.

460 - 200 = 260

Knowing the total and one part, subtract to recover the other part.

#7 Identify Subproblems 5.MD.B.2

On the graph (each square = 20), Day 4 200 sits 0 squares above 200, and Day 5 260 sits 3 squares above 200; join the points with straight segments.

200 = 200 + 0\times 20,\quad 260 = 200 + 3\times 20

Converting the numbers into squares lets you place each point exactly.

Answer: Day 4 = 200 loaves, Day 5 = 260 loaves

Review

Check the clues: Day 4 200 is indeed 60 more than Day 3 140, and Day 4 200 + Day 5 260 = 460 as required. Both values land neatly on grid lines (multiples of 20), so the plotted points are valid.

Convert to algebra (tool 13): let Day 4 = J = 140 + 60 and Day 5 = 460 - J; solving gives the same J = 200 and Day 5 = 260.

Standards · min grade 5

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Computing Day 4 from Day 3 and Day 5 from the total
5.MD.B.2 Make a line plot to display a data set and solve problems using the data — Placing the Day 4 and Day 5 values on the scaled graph

💡 Fill in what you can figure out first, then use the total to get the rest - then just plot!

Variant 6 answer: Week 4 = 260 kites, Week 5 = 320 kites

The line graph shows the kites made recorded each week. In Week 4 the count was $80$ more than in Week 3, and the number in Week 4 and Week 5 together is $580$ . Complete the line graph.

(Figure) A line graph titled "Kites Made." The horizontal axis shows the week - $1$ (Week 1), $2$ (Week 2), $3$ (Week 3), $4$ (Week 4), $5$ (Week 5). The vertical axis shows the number of kites, with major gridlines at $0$ , $100$ , $200$ , $300$ , $400$ and each small grid square worth $20$ kites. Points are plotted and joined for Week 1 at $140$ ; Week 2 at $240$ ; Week 3 at $180$ ; the points for Week 4 and Week 5 have not yet been plotted.

Show solution

Understand

A line graph of kites made each week shows Week 1 = 140, Week 2 = 240, Week 3 = 180; Week 4 and Week 5 are blank. Week 4 is 80 more than Week 3, and Week 4 + Week 5 together is 580. I must find the Week 4 and Week 5 counts to complete the graph.

Givens

Week 1 = 140 kites, Week 2 = 240 kites, Week 3 = 180 kites
Week 4 = Week 3 + 80
Week 4 + Week 5 = 580
Each small grid square = 20 kites

Unknowns

Number of kites made in Week 4
Number of kites made in Week 5

Constraints

Plotted Week 4 and Week 5 values must land on grid squares worth 20 each

Plan

#7 Identify Subproblems · also uses: #11 Work Backwards

I solve in order from what I can already know: first Week 4 (Week 3 plus 80), then Week 5 (the 580 total minus Week 4). Each is a one-step subproblem, and finding Week 5 uses the total backwards.

Execute

#7 Identify Subproblems 4.OA.A.3

Week 4 is 80 more than Week 3, and Week 3 is 180, so Week 4 = 180 + 80 = 260.

180 + 80 = 260

Start from the value you already know and apply the given step.

#11 Work Backwards 4.OA.A.3

Week 4 and Week 5 together make 580, so Week 5 = 580 - Week 4 = 580 - 260 = 320.

580 - 260 = 320

Knowing the total and one part, subtract to recover the other part.

#7 Identify Subproblems 5.MD.B.2

On the graph (each square = 20), Week 4 260 sits 3 squares above 200, and Week 5 320 sits 1 squares above 300; join the points with straight segments.

260 = 200 + 3\times 20,\quad 320 = 300 + 1\times 20

Converting the numbers into squares lets you place each point exactly.

Answer: Week 4 = 260 kites, Week 5 = 320 kites

Review

Check the clues: Week 4 260 is indeed 80 more than Week 3 180, and Week 4 260 + Week 5 320 = 580 as required. Both values land neatly on grid lines (multiples of 20), so the plotted points are valid.

Convert to algebra (tool 13): let Week 4 = J = 180 + 80 and Week 5 = 580 - J; solving gives the same J = 260 and Week 5 = 320.

Standards · min grade 5

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Computing Week 4 from Week 3 and Week 5 from the total
5.MD.B.2 Make a line plot to display a data set and solve problems using the data — Placing the Week 4 and Week 5 values on the scaled graph

💡 Fill in what you can figure out first, then use the total to get the rest - then just plot!

Variant 7 answer: August = 300 cups, September = 380 cups

The line graph shows the cups sold recorded each month. In August the count was $40$ more than in July, and the number in August and September together is $680$ . Complete the line graph.

(Figure) A line graph titled "Cups Sold." The horizontal axis shows the month - $5$ (May), $6$ (June), $7$ (July), $8$ (August), $9$ (September). The vertical axis shows the number of cups, with major gridlines at $0$ , $100$ , $200$ , $300$ , $400$ and each small grid square worth $20$ cups. Points are plotted and joined for May at $200$ ; June at $320$ ; July at $260$ ; the points for August and September have not yet been plotted.

Show solution

Understand

A line graph of cups sold each month shows May = 200, June = 320, July = 260; August and September are blank. August is 40 more than July, and August + September together is 680. I must find the August and September counts to complete the graph.

Givens

May = 200 cups, June = 320 cups, July = 260 cups
August = July + 40
August + September = 680
Each small grid square = 20 cups

Unknowns

Number of cups made in August
Number of cups made in September

Constraints

Plotted August and September values must land on grid squares worth 20 each

Plan

#7 Identify Subproblems · also uses: #11 Work Backwards

I solve in order from what I can already know: first August (July plus 40), then September (the 680 total minus August). Each is a one-step subproblem, and finding September uses the total backwards.

Execute

#7 Identify Subproblems 4.OA.A.3

August is 40 more than July, and July is 260, so August = 260 + 40 = 300.

260 + 40 = 300

Start from the value you already know and apply the given step.

#11 Work Backwards 4.OA.A.3

August and September together make 680, so September = 680 - August = 680 - 300 = 380.

680 - 300 = 380

Knowing the total and one part, subtract to recover the other part.

#7 Identify Subproblems 5.MD.B.2

On the graph (each square = 20), August 300 sits 0 squares above 300, and September 380 sits 4 squares above 300; join the points with straight segments.

300 = 300 + 0\times 20,\quad 380 = 300 + 4\times 20

Converting the numbers into squares lets you place each point exactly.

Answer: August = 300 cups, September = 380 cups

Review

Check the clues: August 300 is indeed 40 more than July 260, and August 300 + September 380 = 680 as required. Both values land neatly on grid lines (multiples of 20), so the plotted points are valid.

Convert to algebra (tool 13): let August = J = 260 + 40 and September = 680 - J; solving gives the same J = 300 and September = 380.

Standards · min grade 5

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Computing August from July and September from the total
5.MD.B.2 Make a line plot to display a data set and solve problems using the data — Placing the August and September values on the scaled graph

💡 Fill in what you can figure out first, then use the total to get the rest - then just plot!

Variant 8 answer: Day 4 = 300 bottles, Day 5 = 340 bottles

The line graph shows the bottles filled recorded each day. In Day 4 the count was $100$ more than in Day 3, and the number in Day 4 and Day 5 together is $640$ . Complete the line graph.

(Figure) A line graph titled "Bottles Filled." The horizontal axis shows the day - $1$ (Day 1), $2$ (Day 2), $3$ (Day 3), $4$ (Day 4), $5$ (Day 5). The vertical axis shows the number of bottles, with major gridlines at $0$ , $100$ , $200$ , $300$ , $400$ and each small grid square worth $20$ bottles. Points are plotted and joined for Day 1 at $160$ ; Day 2 at $260$ ; Day 3 at $200$ ; the points for Day 4 and Day 5 have not yet been plotted.

Show solution

Understand

A line graph of bottles filled each day shows Day 1 = 160, Day 2 = 260, Day 3 = 200; Day 4 and Day 5 are blank. Day 4 is 100 more than Day 3, and Day 4 + Day 5 together is 640. I must find the Day 4 and Day 5 counts to complete the graph.

Givens

Day 1 = 160 bottles, Day 2 = 260 bottles, Day 3 = 200 bottles
Day 4 = Day 3 + 100
Day 4 + Day 5 = 640
Each small grid square = 20 bottles

Unknowns

Number of bottles made in Day 4
Number of bottles made in Day 5

Constraints

Plotted Day 4 and Day 5 values must land on grid squares worth 20 each

Plan

#7 Identify Subproblems · also uses: #11 Work Backwards

I solve in order from what I can already know: first Day 4 (Day 3 plus 100), then Day 5 (the 640 total minus Day 4). Each is a one-step subproblem, and finding Day 5 uses the total backwards.

Execute

#7 Identify Subproblems 4.OA.A.3

Day 4 is 100 more than Day 3, and Day 3 is 200, so Day 4 = 200 + 100 = 300.

200 + 100 = 300

Start from the value you already know and apply the given step.

#11 Work Backwards 4.OA.A.3

Day 4 and Day 5 together make 640, so Day 5 = 640 - Day 4 = 640 - 300 = 340.

640 - 300 = 340

Knowing the total and one part, subtract to recover the other part.

#7 Identify Subproblems 5.MD.B.2

On the graph (each square = 20), Day 4 300 sits 0 squares above 300, and Day 5 340 sits 2 squares above 300; join the points with straight segments.

300 = 300 + 0\times 20,\quad 340 = 300 + 2\times 20

Converting the numbers into squares lets you place each point exactly.

Answer: Day 4 = 300 bottles, Day 5 = 340 bottles

Review

Check the clues: Day 4 300 is indeed 100 more than Day 3 200, and Day 4 300 + Day 5 340 = 640 as required. Both values land neatly on grid lines (multiples of 20), so the plotted points are valid.

Convert to algebra (tool 13): let Day 4 = J = 200 + 100 and Day 5 = 640 - J; solving gives the same J = 300 and Day 5 = 340.

Standards · min grade 5

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Computing Day 4 from Day 3 and Day 5 from the total
5.MD.B.2 Make a line plot to display a data set and solve problems using the data — Placing the Day 4 and Day 5 values on the scaled graph

💡 Fill in what you can figure out first, then use the total to get the rest - then just plot!