Practice · One win count sets both players' scores · Sensim Math

Variant 1 easy

A box holds $20$ cards, numbered $1$ through $20$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $3$ points, and whoever draws the smaller number gets $1$ points. They play $20$ rounds. For Emma's total score to be at least $10$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Your answer

Show solution

1 · Understandwhat's really being asked

In each of $20$ rounds, Emma and Grace each draw a card; the larger number earns $3$ points and the smaller earns $1$ . Find the fewest rounds Emma must win for her total to beat Grace's by at least $10$ points.

Givens

The cards are numbered $1$ to $20$ , and both players draw at the same time from the same box.
Larger number: $3$ points. Smaller number: $1$ points.
They play $20$ rounds, and the cards go back after each round.
Emma's total must be at least $10$ points more than Grace's.

Unknowns

The smallest number of rounds in which Emma draws the larger number.

Constraints

The two cards drawn in a round are different, so every round has exactly one winner.
The number of rounds won is a whole number from $0$ to $20$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#6 Guess and Check

Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

3 · Execute4 carry out the plan

1One count decides both

#16 Count the Complement 7.EE.B.4

The two cards in a round are always different, so each round has one winner. If Emma wins

x

rounds, she loses the other

20 - x

, and Grace wins exactly those

20 - x

rounds and loses

x

.

\text{Emma: } x \text{ wins}, \; 20 - x \text{ losses} \qquad \text{Grace: } 20 - x \text{ wins}, \; x \text{ losses}

Grace's record is Emma's record turned around.

2Write both scores

#13 Convert to Algebra 7.EE.B.4

Wins earn

3

points and losses earn

1

.

\text{Emma} = 3x + 1(20 - x) = 2x + 20, \qquad \text{Grace} = 3(20 - x) + 1x = 60 - 2x

Each extra win raises Emma's score by

2

and lowers Grace's by

2

.

3Set up the gap

#13 Convert to Algebra 7.EE.B.4

Emma's score must be at least Grace's score plus

10

.

2x + 20 \ge (60 - 2x) + 10 \;\Rightarrow\; 4x \ge 50 \;\Rightarrow\; x \ge \frac{25}{2} = 12\frac{1}{2}

The gap is

4x - 40

, which grows by

4

for every extra win.

4Round up and check

#6 Guess and Check 7.EE.B.4

The number of wins is a whole number at least

12\frac{1}{2}

, so it is at least

13

. Check

13

and

12

.

x = 13: \; 46 - 34 = 12 \ge 10, \qquad x = 12: \; 44 - 36 = 8 < 10

12

wins leave the gap

2

points short;

13

wins clear it.

Answer:

13

times

4 · Reviewdoes it hold up?

With $13$ wins and $7$ losses, Emma scores $39 + 7 = 46$ and Grace scores $21 + 13 = 34$ ; the gap is $12$ . The two totals add to $80 = 20 \times 4$ , as they must, since every round hands out $3 + 1 = 4$ points.

Another way: Start from a tie in record: at $10$ wins each, the scores are equal. Each round that switches from Grace's win to Emma's win changes the gap by $2 + 2 = 4$ points. A gap of at least $10$ needs at least $3$ switches ( $4 \times 2 = 8$ is too few, $4 \times 3 = 12$ is enough), so Emma needs $10 + 3 = 13$ wins.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing both scores from Emma's number of wins, stating the gap condition, and solving $2x + 20 \ge (60 - 2x) + 10$.

💡Takeaway. When one player's win is always the other's loss, a single count tells you both scores.

Variant 2 easy

A box holds $25$ cards, numbered $1$ through $25$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $5$ points, and whoever draws the smaller number gets $3$ points. They play $22$ rounds. For Emma's total score to be at least $14$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Your answer

Show solution

1 · Understandwhat's really being asked

In each of $22$ rounds, Emma and Grace each draw a card; the larger number earns $5$ points and the smaller earns $3$ . Find the fewest rounds Emma must win for her total to beat Grace's by at least $14$ points.

Givens

The cards are numbered $1$ to $25$ , and both players draw at the same time from the same box.
Larger number: $5$ points. Smaller number: $3$ points.
They play $22$ rounds, and the cards go back after each round.
Emma's total must be at least $14$ points more than Grace's.

Unknowns

The smallest number of rounds in which Emma draws the larger number.

Constraints

The two cards drawn in a round are different, so every round has exactly one winner.
The number of rounds won is a whole number from $0$ to $22$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#6 Guess and Check

Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

3 · Execute4 carry out the plan

1One count decides both

#16 Count the Complement 7.EE.B.4

The two cards in a round are always different, so each round has one winner. If Emma wins

x

rounds, she loses the other

22 - x

, and Grace wins exactly those

22 - x

rounds and loses

x

.

\text{Emma: } x \text{ wins}, \; 22 - x \text{ losses} \qquad \text{Grace: } 22 - x \text{ wins}, \; x \text{ losses}

Grace's record is Emma's record turned around.

2Write both scores

#13 Convert to Algebra 7.EE.B.4

Wins earn

5

points and losses earn

3

.

\text{Emma} = 5x + 3(22 - x) = 2x + 66, \qquad \text{Grace} = 5(22 - x) + 3x = 110 - 2x

Each extra win raises Emma's score by

2

and lowers Grace's by

2

.

3Set up the gap

#13 Convert to Algebra 7.EE.B.4

Emma's score must be at least Grace's score plus

14

.

2x + 66 \ge (110 - 2x) + 14 \;\Rightarrow\; 4x \ge 58 \;\Rightarrow\; x \ge \frac{29}{2} = 14\frac{1}{2}

The gap is

4x - 44

, which grows by

4

for every extra win.

4Round up and check

#6 Guess and Check 7.EE.B.4

The number of wins is a whole number at least

14\frac{1}{2}

, so it is at least

15

. Check

15

and

14

.

x = 15: \; 96 - 80 = 16 \ge 14, \qquad x = 14: \; 94 - 82 = 12 < 14

14

wins leave the gap

2

points short;

15

wins clear it.

Answer:

15

times

4 · Reviewdoes it hold up?

With $15$ wins and $7$ losses, Emma scores $75 + 21 = 96$ and Grace scores $35 + 45 = 80$ ; the gap is $16$ . The two totals add to $176 = 22 \times 8$ , as they must, since every round hands out $5 + 3 = 8$ points.

Another way: Start from a tie in record: at $11$ wins each, the scores are equal. Each round that switches from Grace's win to Emma's win changes the gap by $2 + 2 = 4$ points. A gap of at least $14$ needs at least $4$ switches ( $4 \times 3 = 12$ is too few, $4 \times 4 = 16$ is enough), so Emma needs $11 + 4 = 15$ wins.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing both scores from Emma's number of wins, stating the gap condition, and solving $2x + 66 \ge (110 - 2x) + 14$.

💡Takeaway. When one player's win is always the other's loss, a single count tells you both scores.

Variant 3 easy

A box holds $30$ cards, numbered $1$ through $30$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $6$ points, and whoever draws the smaller number gets $2$ points. They play $24$ rounds. For Emma's total score to be at least $30$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Your answer

Show solution

1 · Understandwhat's really being asked

In each of $24$ rounds, Emma and Grace each draw a card; the larger number earns $6$ points and the smaller earns $2$ . Find the fewest rounds Emma must win for her total to beat Grace's by at least $30$ points.

Givens

The cards are numbered $1$ to $30$ , and both players draw at the same time from the same box.
Larger number: $6$ points. Smaller number: $2$ points.
They play $24$ rounds, and the cards go back after each round.
Emma's total must be at least $30$ points more than Grace's.

Unknowns

The smallest number of rounds in which Emma draws the larger number.

Constraints

The two cards drawn in a round are different, so every round has exactly one winner.
The number of rounds won is a whole number from $0$ to $24$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#6 Guess and Check

Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

3 · Execute4 carry out the plan

1One count decides both

#16 Count the Complement 7.EE.B.4

The two cards in a round are always different, so each round has one winner. If Emma wins

x

rounds, she loses the other

24 - x

, and Grace wins exactly those

24 - x

rounds and loses

x

.

\text{Emma: } x \text{ wins}, \; 24 - x \text{ losses} \qquad \text{Grace: } 24 - x \text{ wins}, \; x \text{ losses}

Grace's record is Emma's record turned around.

2Write both scores

#13 Convert to Algebra 7.EE.B.4

Wins earn

6

points and losses earn

2

.

\text{Emma} = 6x + 2(24 - x) = 4x + 48, \qquad \text{Grace} = 6(24 - x) + 2x = 144 - 4x

Each extra win raises Emma's score by

4

and lowers Grace's by

4

.

3Set up the gap

#13 Convert to Algebra 7.EE.B.4

Emma's score must be at least Grace's score plus

30

.

4x + 48 \ge (144 - 4x) + 30 \;\Rightarrow\; 8x \ge 126 \;\Rightarrow\; x \ge \frac{63}{4} = 15\frac{3}{4}

The gap is

8x - 96

, which grows by

8

for every extra win.

4Round up and check

#6 Guess and Check 7.EE.B.4

The number of wins is a whole number at least

15\frac{3}{4}

, so it is at least

16

. Check

16

and

15

.

x = 16: \; 112 - 80 = 32 \ge 30, \qquad x = 15: \; 108 - 84 = 24 < 30

15

wins leave the gap

6

points short;

16

wins clear it.

Answer:

16

times

4 · Reviewdoes it hold up?

With $16$ wins and $8$ losses, Emma scores $96 + 16 = 112$ and Grace scores $48 + 32 = 80$ ; the gap is $32$ . The two totals add to $192 = 24 \times 8$ , as they must, since every round hands out $6 + 2 = 8$ points.

Another way: Start from a tie in record: at $12$ wins each, the scores are equal. Each round that switches from Grace's win to Emma's win changes the gap by $4 + 4 = 8$ points. A gap of at least $30$ needs at least $4$ switches ( $8 \times 3 = 24$ is too few, $8 \times 4 = 32$ is enough), so Emma needs $12 + 4 = 16$ wins.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing both scores from Emma's number of wins, stating the gap condition, and solving $4x + 48 \ge (144 - 4x) + 30$.

💡Takeaway. When one player's win is always the other's loss, a single count tells you both scores.

Variant 4 easy

A box holds $36$ cards, numbered $1$ through $36$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $4$ points, and whoever draws the smaller number gets $2$ points. They play $28$ rounds. For Emma's total score to be at least $18$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Your answer

Show solution

1 · Understandwhat's really being asked

In each of $28$ rounds, Emma and Grace each draw a card; the larger number earns $4$ points and the smaller earns $2$ . Find the fewest rounds Emma must win for her total to beat Grace's by at least $18$ points.

Givens

The cards are numbered $1$ to $36$ , and both players draw at the same time from the same box.
Larger number: $4$ points. Smaller number: $2$ points.
They play $28$ rounds, and the cards go back after each round.
Emma's total must be at least $18$ points more than Grace's.

Unknowns

The smallest number of rounds in which Emma draws the larger number.

Constraints

The two cards drawn in a round are different, so every round has exactly one winner.
The number of rounds won is a whole number from $0$ to $28$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#6 Guess and Check

Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

3 · Execute4 carry out the plan

1One count decides both

#16 Count the Complement 7.EE.B.4

The two cards in a round are always different, so each round has one winner. If Emma wins

x

rounds, she loses the other

28 - x

, and Grace wins exactly those

28 - x

rounds and loses

x

.

\text{Emma: } x \text{ wins}, \; 28 - x \text{ losses} \qquad \text{Grace: } 28 - x \text{ wins}, \; x \text{ losses}

Grace's record is Emma's record turned around.

2Write both scores

#13 Convert to Algebra 7.EE.B.4

Wins earn

4

points and losses earn

2

.

\text{Emma} = 4x + 2(28 - x) = 2x + 56, \qquad \text{Grace} = 4(28 - x) + 2x = 112 - 2x

Each extra win raises Emma's score by

2

and lowers Grace's by

2

.

3Set up the gap

#13 Convert to Algebra 7.EE.B.4

Emma's score must be at least Grace's score plus

18

.

2x + 56 \ge (112 - 2x) + 18 \;\Rightarrow\; 4x \ge 74 \;\Rightarrow\; x \ge \frac{37}{2} = 18\frac{1}{2}

The gap is

4x - 56

, which grows by

4

for every extra win.

4Round up and check

#6 Guess and Check 7.EE.B.4

The number of wins is a whole number at least

18\frac{1}{2}

, so it is at least

19

. Check

19

and

18

.

x = 19: \; 94 - 74 = 20 \ge 18, \qquad x = 18: \; 92 - 76 = 16 < 18

18

wins leave the gap

2

points short;

19

wins clear it.

Answer:

19

times

4 · Reviewdoes it hold up?

With $19$ wins and $9$ losses, Emma scores $76 + 18 = 94$ and Grace scores $36 + 38 = 74$ ; the gap is $20$ . The two totals add to $168 = 28 \times 6$ , as they must, since every round hands out $4 + 2 = 6$ points.

Another way: Start from a tie in record: at $14$ wins each, the scores are equal. Each round that switches from Grace's win to Emma's win changes the gap by $2 + 2 = 4$ points. A gap of at least $18$ needs at least $5$ switches ( $4 \times 4 = 16$ is too few, $4 \times 5 = 20$ is enough), so Emma needs $14 + 5 = 19$ wins.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing both scores from Emma's number of wins, stating the gap condition, and solving $2x + 56 \ge (112 - 2x) + 18$.

💡Takeaway. When one player's win is always the other's loss, a single count tells you both scores.

Variant 5 medium

A box holds $40$ cards, numbered $1$ through $40$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $4$ points, and whoever draws the smaller number gets $1$ points. They play $20$ rounds. For Emma's total score to be at least $14$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Your answer

Show solution

1 · Understandwhat's really being asked

In each of $20$ rounds, Emma and Grace each draw a card; the larger number earns $4$ points and the smaller earns $1$ . Find the fewest rounds Emma must win for her total to beat Grace's by at least $14$ points.

Givens

The cards are numbered $1$ to $40$ , and both players draw at the same time from the same box.
Larger number: $4$ points. Smaller number: $1$ points.
They play $20$ rounds, and the cards go back after each round.
Emma's total must be at least $14$ points more than Grace's.

Unknowns

The smallest number of rounds in which Emma draws the larger number.

Constraints

The two cards drawn in a round are different, so every round has exactly one winner.
The number of rounds won is a whole number from $0$ to $20$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#6 Guess and Check

Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

3 · Execute4 carry out the plan

1One count decides both

#16 Count the Complement 7.EE.B.4

The two cards in a round are always different, so each round has one winner. If Emma wins

x

rounds, she loses the other

20 - x

, and Grace wins exactly those

20 - x

rounds and loses

x

.

\text{Emma: } x \text{ wins}, \; 20 - x \text{ losses} \qquad \text{Grace: } 20 - x \text{ wins}, \; x \text{ losses}

Grace's record is Emma's record turned around.

2Write both scores

#13 Convert to Algebra 7.EE.B.4

Wins earn

4

points and losses earn

1

.

\text{Emma} = 4x + 1(20 - x) = 3x + 20, \qquad \text{Grace} = 4(20 - x) + 1x = 80 - 3x

Each extra win raises Emma's score by

3

and lowers Grace's by

3

.

3Set up the gap

#13 Convert to Algebra 7.EE.B.4

Emma's score must be at least Grace's score plus

14

.

3x + 20 \ge (80 - 3x) + 14 \;\Rightarrow\; 6x \ge 74 \;\Rightarrow\; x \ge \frac{37}{3} = 12\frac{1}{3}

The gap is

6x - 60

, which grows by

6

for every extra win.

4Round up and check

#6 Guess and Check 7.EE.B.4

The number of wins is a whole number at least

12\frac{1}{3}

, so it is at least

13

. Check

13

and

12

.

x = 13: \; 59 - 41 = 18 \ge 14, \qquad x = 12: \; 56 - 44 = 12 < 14

12

wins leave the gap

2

points short;

13

wins clear it.

Answer:

13

times

4 · Reviewdoes it hold up?

With $13$ wins and $7$ losses, Emma scores $52 + 7 = 59$ and Grace scores $28 + 13 = 41$ ; the gap is $18$ . The two totals add to $100 = 20 \times 5$ , as they must, since every round hands out $4 + 1 = 5$ points.

Another way: Start from a tie in record: at $10$ wins each, the scores are equal. Each round that switches from Grace's win to Emma's win changes the gap by $3 + 3 = 6$ points. A gap of at least $14$ needs at least $3$ switches ( $6 \times 2 = 12$ is too few, $6 \times 3 = 18$ is enough), so Emma needs $10 + 3 = 13$ wins.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing both scores from Emma's number of wins, stating the gap condition, and solving $3x + 20 \ge (80 - 3x) + 14$.

💡Takeaway. When one player's win is always the other's loss, a single count tells you both scores.

Variant 6 medium

A box holds $45$ cards, numbered $1$ through $45$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $8$ points, and whoever draws the smaller number gets $3$ points. They play $26$ rounds. For Emma's total score to be at least $34$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Your answer

Show solution

1 · Understandwhat's really being asked

In each of $26$ rounds, Emma and Grace each draw a card; the larger number earns $8$ points and the smaller earns $3$ . Find the fewest rounds Emma must win for her total to beat Grace's by at least $34$ points.

Givens

The cards are numbered $1$ to $45$ , and both players draw at the same time from the same box.
Larger number: $8$ points. Smaller number: $3$ points.
They play $26$ rounds, and the cards go back after each round.
Emma's total must be at least $34$ points more than Grace's.

Unknowns

The smallest number of rounds in which Emma draws the larger number.

Constraints

The two cards drawn in a round are different, so every round has exactly one winner.
The number of rounds won is a whole number from $0$ to $26$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#6 Guess and Check

Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

3 · Execute4 carry out the plan

1One count decides both

#16 Count the Complement 7.EE.B.4

The two cards in a round are always different, so each round has one winner. If Emma wins

x

rounds, she loses the other

26 - x

, and Grace wins exactly those

26 - x

rounds and loses

x

.

\text{Emma: } x \text{ wins}, \; 26 - x \text{ losses} \qquad \text{Grace: } 26 - x \text{ wins}, \; x \text{ losses}

Grace's record is Emma's record turned around.

2Write both scores

#13 Convert to Algebra 7.EE.B.4

Wins earn

8

points and losses earn

3

.

\text{Emma} = 8x + 3(26 - x) = 5x + 78, \qquad \text{Grace} = 8(26 - x) + 3x = 208 - 5x

Each extra win raises Emma's score by

5

and lowers Grace's by

5

.

3Set up the gap

#13 Convert to Algebra 7.EE.B.4

Emma's score must be at least Grace's score plus

34

.

5x + 78 \ge (208 - 5x) + 34 \;\Rightarrow\; 10x \ge 164 \;\Rightarrow\; x \ge \frac{82}{5} = 16\frac{2}{5}

The gap is

10x - 130

, which grows by

10

for every extra win.

4Round up and check

#6 Guess and Check 7.EE.B.4

The number of wins is a whole number at least

16\frac{2}{5}

, so it is at least

17

. Check

17

and

16

.

x = 17: \; 163 - 123 = 40 \ge 34, \qquad x = 16: \; 158 - 128 = 30 < 34

16

wins leave the gap

4

points short;

17

wins clear it.

Answer:

17

times

4 · Reviewdoes it hold up?

With $17$ wins and $9$ losses, Emma scores $136 + 27 = 163$ and Grace scores $72 + 51 = 123$ ; the gap is $40$ . The two totals add to $286 = 26 \times 11$ , as they must, since every round hands out $8 + 3 = 11$ points.

Another way: Start from a tie in record: at $13$ wins each, the scores are equal. Each round that switches from Grace's win to Emma's win changes the gap by $5 + 5 = 10$ points. A gap of at least $34$ needs at least $4$ switches ( $10 \times 3 = 30$ is too few, $10 \times 4 = 40$ is enough), so Emma needs $13 + 4 = 17$ wins.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing both scores from Emma's number of wins, stating the gap condition, and solving $5x + 78 \ge (208 - 5x) + 34$.

💡Takeaway. When one player's win is always the other's loss, a single count tells you both scores.

Variant 7 medium

A box holds $50$ cards, numbered $1$ through $50$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $6$ points, and whoever draws the smaller number gets $3$ points. They play $36$ rounds. For Emma's total score to be at least $20$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Your answer

Show solution

1 · Understandwhat's really being asked

In each of $36$ rounds, Emma and Grace each draw a card; the larger number earns $6$ points and the smaller earns $3$ . Find the fewest rounds Emma must win for her total to beat Grace's by at least $20$ points.

Givens

The cards are numbered $1$ to $50$ , and both players draw at the same time from the same box.
Larger number: $6$ points. Smaller number: $3$ points.
They play $36$ rounds, and the cards go back after each round.
Emma's total must be at least $20$ points more than Grace's.

Unknowns

The smallest number of rounds in which Emma draws the larger number.

Constraints

The two cards drawn in a round are different, so every round has exactly one winner.
The number of rounds won is a whole number from $0$ to $36$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#6 Guess and Check

Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

3 · Execute4 carry out the plan

1One count decides both

#16 Count the Complement 7.EE.B.4

The two cards in a round are always different, so each round has one winner. If Emma wins

x

rounds, she loses the other

36 - x

, and Grace wins exactly those

36 - x

rounds and loses

x

.

\text{Emma: } x \text{ wins}, \; 36 - x \text{ losses} \qquad \text{Grace: } 36 - x \text{ wins}, \; x \text{ losses}

Grace's record is Emma's record turned around.

2Write both scores

#13 Convert to Algebra 7.EE.B.4

Wins earn

6

points and losses earn

3

.

\text{Emma} = 6x + 3(36 - x) = 3x + 108, \qquad \text{Grace} = 6(36 - x) + 3x = 216 - 3x

Each extra win raises Emma's score by

3

and lowers Grace's by

3

.

3Set up the gap

#13 Convert to Algebra 7.EE.B.4

Emma's score must be at least Grace's score plus

20

.

3x + 108 \ge (216 - 3x) + 20 \;\Rightarrow\; 6x \ge 128 \;\Rightarrow\; x \ge \frac{64}{3} = 21\frac{1}{3}

The gap is

6x - 108

, which grows by

6

for every extra win.

4Round up and check

#6 Guess and Check 7.EE.B.4

The number of wins is a whole number at least

21\frac{1}{3}

, so it is at least

22

. Check

22

and

21

.

x = 22: \; 174 - 150 = 24 \ge 20, \qquad x = 21: \; 171 - 153 = 18 < 20

21

wins leave the gap

2

points short;

22

wins clear it.

Answer:

22

times

4 · Reviewdoes it hold up?

With $22$ wins and $14$ losses, Emma scores $132 + 42 = 174$ and Grace scores $84 + 66 = 150$ ; the gap is $24$ . The two totals add to $324 = 36 \times 9$ , as they must, since every round hands out $6 + 3 = 9$ points.

Another way: Start from a tie in record: at $18$ wins each, the scores are equal. Each round that switches from Grace's win to Emma's win changes the gap by $3 + 3 = 6$ points. A gap of at least $20$ needs at least $4$ switches ( $6 \times 3 = 18$ is too few, $6 \times 4 = 24$ is enough), so Emma needs $18 + 4 = 22$ wins.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing both scores from Emma's number of wins, stating the gap condition, and solving $3x + 108 \ge (216 - 3x) + 20$.

💡Takeaway. When one player's win is always the other's loss, a single count tells you both scores.

Variant 8 medium

A box holds $50$ cards, numbered $1$ through $50$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $5$ points, and whoever draws the smaller number gets $2$ points. They play $30$ rounds. For Emma's total score to be at least $20$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Your answer

Show solution

1 · Understandwhat's really being asked

In each of $30$ rounds, Emma and Grace each draw a card; the larger number earns $5$ points and the smaller earns $2$ . Find the fewest rounds Emma must win for her total to beat Grace's by at least $20$ points.

Givens

The cards are numbered $1$ to $50$ , and both players draw at the same time from the same box.
Larger number: $5$ points. Smaller number: $2$ points.
They play $30$ rounds, and the cards go back after each round.
Emma's total must be at least $20$ points more than Grace's.

Unknowns

The smallest number of rounds in which Emma draws the larger number.

Constraints

The two cards drawn in a round are different, so every round has exactly one winner.
The number of rounds won is a whole number from $0$ to $30$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#6 Guess and Check

Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

3 · Execute4 carry out the plan

1One count decides both

#16 Count the Complement 7.EE.B.4

The two cards in a round are always different, so each round has one winner. If Emma wins

x

rounds, she loses the other

30 - x

, and Grace wins exactly those

30 - x

rounds and loses

x

.

\text{Emma: } x \text{ wins}, \; 30 - x \text{ losses} \qquad \text{Grace: } 30 - x \text{ wins}, \; x \text{ losses}

Grace's record is Emma's record turned around.

2Write both scores

#13 Convert to Algebra 7.EE.B.4

Wins earn

5

points and losses earn

2

.

\text{Emma} = 5x + 2(30 - x) = 3x + 60, \qquad \text{Grace} = 5(30 - x) + 2x = 150 - 3x

Each extra win raises Emma's score by

3

and lowers Grace's by

3

.

3Set up the gap

#13 Convert to Algebra 7.EE.B.4

Emma's score must be at least Grace's score plus

20

.

3x + 60 \ge (150 - 3x) + 20 \;\Rightarrow\; 6x \ge 110 \;\Rightarrow\; x \ge \frac{55}{3} = 18\frac{1}{3}

The gap is

6x - 90

, which grows by

6

for every extra win.

4Round up and check

#6 Guess and Check 7.EE.B.4

The number of wins is a whole number at least

18\frac{1}{3}

, so it is at least

19

. Check

19

and

18

.

x = 19: \; 117 - 93 = 24 \ge 20, \qquad x = 18: \; 114 - 96 = 18 < 20

18

wins leave the gap

2

points short;

19

wins clear it.

Answer:

19

times

4 · Reviewdoes it hold up?

With $19$ wins and $11$ losses, Emma scores $95 + 22 = 117$ and Grace scores $55 + 38 = 93$ ; the gap is $24$ . The two totals add to $210 = 30 \times 7$ , as they must, since every round hands out $5 + 2 = 7$ points.

Another way: Start from a tie in record: at $15$ wins each, the scores are equal. Each round that switches from Grace's win to Emma's win changes the gap by $3 + 3 = 6$ points. A gap of at least $20$ needs at least $4$ switches ( $6 \times 3 = 18$ is too few, $6 \times 4 = 24$ is enough), so Emma needs $15 + 4 = 19$ wins.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing both scores from Emma's number of wins, stating the gap condition, and solving $3x + 60 \ge (150 - 3x) + 20$.

💡Takeaway. When one player's win is always the other's loss, a single count tells you both scores.

Variant 9 hard

A box holds $60$ cards, numbered $1$ through $60$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $5$ points, and whoever draws the smaller number gets $1$ points. They play $40$ rounds. For Emma's total score to be at least $18$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Your answer

Show solution

1 · Understandwhat's really being asked

In each of $40$ rounds, Emma and Grace each draw a card; the larger number earns $5$ points and the smaller earns $1$ . Find the fewest rounds Emma must win for her total to beat Grace's by at least $18$ points.

Givens

The cards are numbered $1$ to $60$ , and both players draw at the same time from the same box.
Larger number: $5$ points. Smaller number: $1$ points.
They play $40$ rounds, and the cards go back after each round.
Emma's total must be at least $18$ points more than Grace's.

Unknowns

The smallest number of rounds in which Emma draws the larger number.

Constraints

The two cards drawn in a round are different, so every round has exactly one winner.
The number of rounds won is a whole number from $0$ to $40$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#6 Guess and Check

Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

3 · Execute4 carry out the plan

1One count decides both

#16 Count the Complement 7.EE.B.4

The two cards in a round are always different, so each round has one winner. If Emma wins

x

rounds, she loses the other

40 - x

, and Grace wins exactly those

40 - x

rounds and loses

x

.

\text{Emma: } x \text{ wins}, \; 40 - x \text{ losses} \qquad \text{Grace: } 40 - x \text{ wins}, \; x \text{ losses}

Grace's record is Emma's record turned around.

2Write both scores

#13 Convert to Algebra 7.EE.B.4

Wins earn

5

points and losses earn

1

.

\text{Emma} = 5x + 1(40 - x) = 4x + 40, \qquad \text{Grace} = 5(40 - x) + 1x = 200 - 4x

Each extra win raises Emma's score by

4

and lowers Grace's by

4

.

3Set up the gap

#13 Convert to Algebra 7.EE.B.4

Emma's score must be at least Grace's score plus

18

.

4x + 40 \ge (200 - 4x) + 18 \;\Rightarrow\; 8x \ge 178 \;\Rightarrow\; x \ge \frac{89}{4} = 22\frac{1}{4}

The gap is

8x - 160

, which grows by

8

for every extra win.

4Round up and check

#6 Guess and Check 7.EE.B.4

The number of wins is a whole number at least

22\frac{1}{4}

, so it is at least

23

. Check

23

and

22

.

x = 23: \; 132 - 108 = 24 \ge 18, \qquad x = 22: \; 128 - 112 = 16 < 18

22

wins leave the gap

2

points short;

23

wins clear it.

Answer:

23

times

4 · Reviewdoes it hold up?

With $23$ wins and $17$ losses, Emma scores $115 + 17 = 132$ and Grace scores $85 + 23 = 108$ ; the gap is $24$ . The two totals add to $240 = 40 \times 6$ , as they must, since every round hands out $5 + 1 = 6$ points.

Another way: Start from a tie in record: at $20$ wins each, the scores are equal. Each round that switches from Grace's win to Emma's win changes the gap by $4 + 4 = 8$ points. A gap of at least $18$ needs at least $3$ switches ( $8 \times 2 = 16$ is too few, $8 \times 3 = 24$ is enough), so Emma needs $20 + 3 = 23$ wins.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing both scores from Emma's number of wins, stating the gap condition, and solving $4x + 40 \ge (200 - 4x) + 18$.

💡Takeaway. When one player's win is always the other's loss, a single count tells you both scores.

Variant 10 hard

A box holds $70$ cards, numbered $1$ through $70$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $9$ points, and whoever draws the smaller number gets $4$ points. They play $32$ rounds. For Emma's total score to be at least $45$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Your answer

Show solution

1 · Understandwhat's really being asked

In each of $32$ rounds, Emma and Grace each draw a card; the larger number earns $9$ points and the smaller earns $4$ . Find the fewest rounds Emma must win for her total to beat Grace's by at least $45$ points.

Givens

The cards are numbered $1$ to $70$ , and both players draw at the same time from the same box.
Larger number: $9$ points. Smaller number: $4$ points.
They play $32$ rounds, and the cards go back after each round.
Emma's total must be at least $45$ points more than Grace's.

Unknowns

The smallest number of rounds in which Emma draws the larger number.

Constraints

The two cards drawn in a round are different, so every round has exactly one winner.
The number of rounds won is a whole number from $0$ to $32$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#6 Guess and Check

Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

3 · Execute4 carry out the plan

1One count decides both

#16 Count the Complement 7.EE.B.4

The two cards in a round are always different, so each round has one winner. If Emma wins

x

rounds, she loses the other

32 - x

, and Grace wins exactly those

32 - x

rounds and loses

x

.

\text{Emma: } x \text{ wins}, \; 32 - x \text{ losses} \qquad \text{Grace: } 32 - x \text{ wins}, \; x \text{ losses}

Grace's record is Emma's record turned around.

2Write both scores

#13 Convert to Algebra 7.EE.B.4

Wins earn

9

points and losses earn

4

.

\text{Emma} = 9x + 4(32 - x) = 5x + 128, \qquad \text{Grace} = 9(32 - x) + 4x = 288 - 5x

Each extra win raises Emma's score by

5

and lowers Grace's by

5

.

3Set up the gap

#13 Convert to Algebra 7.EE.B.4

Emma's score must be at least Grace's score plus

45

.

5x + 128 \ge (288 - 5x) + 45 \;\Rightarrow\; 10x \ge 205 \;\Rightarrow\; x \ge \frac{41}{2} = 20\frac{1}{2}

The gap is

10x - 160

, which grows by

10

for every extra win.

4Round up and check

#6 Guess and Check 7.EE.B.4

The number of wins is a whole number at least

20\frac{1}{2}

, so it is at least

21

. Check

21

and

20

.

x = 21: \; 233 - 183 = 50 \ge 45, \qquad x = 20: \; 228 - 188 = 40 < 45

20

wins leave the gap

5

points short;

21

wins clear it.

Answer:

21

times

4 · Reviewdoes it hold up?

With $21$ wins and $11$ losses, Emma scores $189 + 44 = 233$ and Grace scores $99 + 84 = 183$ ; the gap is $50$ . The two totals add to $416 = 32 \times 13$ , as they must, since every round hands out $9 + 4 = 13$ points.

Another way: Start from a tie in record: at $16$ wins each, the scores are equal. Each round that switches from Grace's win to Emma's win changes the gap by $5 + 5 = 10$ points. A gap of at least $45$ needs at least $5$ switches ( $10 \times 4 = 40$ is too few, $10 \times 5 = 50$ is enough), so Emma needs $16 + 5 = 21$ wins.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing both scores from Emma's number of wins, stating the gap condition, and solving $5x + 128 \ge (288 - 5x) + 45$.

💡Takeaway. When one player's win is always the other's loss, a single count tells you both scores.

Variant 11 hard

A box holds $80$ cards, numbered $1$ through $80$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $5$ points, and whoever draws the smaller number gets $2$ points. They play $40$ rounds. For Emma's total score to be at least $35$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Your answer

Show solution

1 · Understandwhat's really being asked

In each of $40$ rounds, Emma and Grace each draw a card; the larger number earns $5$ points and the smaller earns $2$ . Find the fewest rounds Emma must win for her total to beat Grace's by at least $35$ points.

Givens

The cards are numbered $1$ to $80$ , and both players draw at the same time from the same box.
Larger number: $5$ points. Smaller number: $2$ points.
They play $40$ rounds, and the cards go back after each round.
Emma's total must be at least $35$ points more than Grace's.

Unknowns

The smallest number of rounds in which Emma draws the larger number.

Constraints

The two cards drawn in a round are different, so every round has exactly one winner.
The number of rounds won is a whole number from $0$ to $40$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#6 Guess and Check

Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

3 · Execute4 carry out the plan

1One count decides both

#16 Count the Complement 7.EE.B.4

The two cards in a round are always different, so each round has one winner. If Emma wins

x

rounds, she loses the other

40 - x

, and Grace wins exactly those

40 - x

rounds and loses

x

.

\text{Emma: } x \text{ wins}, \; 40 - x \text{ losses} \qquad \text{Grace: } 40 - x \text{ wins}, \; x \text{ losses}

Grace's record is Emma's record turned around.

2Write both scores

#13 Convert to Algebra 7.EE.B.4

Wins earn

5

points and losses earn

2

.

\text{Emma} = 5x + 2(40 - x) = 3x + 80, \qquad \text{Grace} = 5(40 - x) + 2x = 200 - 3x

Each extra win raises Emma's score by

3

and lowers Grace's by

3

.

3Set up the gap

#13 Convert to Algebra 7.EE.B.4

Emma's score must be at least Grace's score plus

35

.

3x + 80 \ge (200 - 3x) + 35 \;\Rightarrow\; 6x \ge 155 \;\Rightarrow\; x \ge \frac{155}{6} = 25\frac{5}{6}

The gap is

6x - 120

, which grows by

6

for every extra win.

4Round up and check

#6 Guess and Check 7.EE.B.4

The number of wins is a whole number at least

25\frac{5}{6}

, so it is at least

26

. Check

26

and

25

.

x = 26: \; 158 - 122 = 36 \ge 35, \qquad x = 25: \; 155 - 125 = 30 < 35

25

wins leave the gap

5

points short;

26

wins clear it.

Answer:

26

times

4 · Reviewdoes it hold up?

With $26$ wins and $14$ losses, Emma scores $130 + 28 = 158$ and Grace scores $70 + 52 = 122$ ; the gap is $36$ . The two totals add to $280 = 40 \times 7$ , as they must, since every round hands out $5 + 2 = 7$ points.

Another way: Start from a tie in record: at $20$ wins each, the scores are equal. Each round that switches from Grace's win to Emma's win changes the gap by $3 + 3 = 6$ points. A gap of at least $35$ needs at least $6$ switches ( $6 \times 5 = 30$ is too few, $6 \times 6 = 36$ is enough), so Emma needs $20 + 6 = 26$ wins.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing both scores from Emma's number of wins, stating the gap condition, and solving $3x + 80 \ge (200 - 3x) + 35$.

💡Takeaway. When one player's win is always the other's loss, a single count tells you both scores.

Variant 12 hard

A box holds $100$ cards, numbered $1$ through $100$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $7$ points, and whoever draws the smaller number gets $3$ points. They play $30$ rounds. For Emma's total score to be at least $20$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Your answer

Show solution

1 · Understandwhat's really being asked

In each of $30$ rounds, Emma and Grace each draw a card; the larger number earns $7$ points and the smaller earns $3$ . Find the fewest rounds Emma must win for her total to beat Grace's by at least $20$ points.

Givens

The cards are numbered $1$ to $100$ , and both players draw at the same time from the same box.
Larger number: $7$ points. Smaller number: $3$ points.
They play $30$ rounds, and the cards go back after each round.
Emma's total must be at least $20$ points more than Grace's.

Unknowns

The smallest number of rounds in which Emma draws the larger number.

Constraints

The two cards drawn in a round are different, so every round has exactly one winner.
The number of rounds won is a whole number from $0$ to $30$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#6 Guess and Check

Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

3 · Execute4 carry out the plan

1One count decides both

#16 Count the Complement 7.EE.B.4

The two cards in a round are always different, so each round has one winner. If Emma wins

x

rounds, she loses the other

30 - x

, and Grace wins exactly those

30 - x

rounds and loses

x

.

\text{Emma: } x \text{ wins}, \; 30 - x \text{ losses} \qquad \text{Grace: } 30 - x \text{ wins}, \; x \text{ losses}

Grace's record is Emma's record turned around.

2Write both scores

#13 Convert to Algebra 7.EE.B.4

Wins earn

7

points and losses earn

3

.

\text{Emma} = 7x + 3(30 - x) = 4x + 90, \qquad \text{Grace} = 7(30 - x) + 3x = 210 - 4x

Each extra win raises Emma's score by

4

and lowers Grace's by

4

.

3Set up the gap

#13 Convert to Algebra 7.EE.B.4

Emma's score must be at least Grace's score plus

20

.

4x + 90 \ge (210 - 4x) + 20 \;\Rightarrow\; 8x \ge 140 \;\Rightarrow\; x \ge \frac{35}{2} = 17\frac{1}{2}

The gap is

8x - 120

, which grows by

8

for every extra win.

4Round up and check

#6 Guess and Check 7.EE.B.4

The number of wins is a whole number at least

17\frac{1}{2}

, so it is at least

18

. Check

18

and

17

.

x = 18: \; 162 - 138 = 24 \ge 20, \qquad x = 17: \; 158 - 142 = 16 < 20

17

wins leave the gap

4

points short;

18

wins clear it.

Answer:

18

times

4 · Reviewdoes it hold up?

With $18$ wins and $12$ losses, Emma scores $126 + 36 = 162$ and Grace scores $84 + 54 = 138$ ; the gap is $24$ . The two totals add to $300 = 30 \times 10$ , as they must, since every round hands out $7 + 3 = 10$ points.

Another way: Start from a tie in record: at $15$ wins each, the scores are equal. Each round that switches from Grace's win to Emma's win changes the gap by $4 + 4 = 8$ points. A gap of at least $20$ needs at least $3$ switches ( $8 \times 2 = 16$ is too few, $8 \times 3 = 24$ is enough), so Emma needs $15 + 3 = 18$ wins.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing both scores from Emma's number of wins, stating the gap condition, and solving $4x + 90 \ge (210 - 4x) + 20$.

💡Takeaway. When one player's win is always the other's loss, a single count tells you both scores.

One win count sets both players' scores · 12 practice problems

Generated variants — 12

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