Practice · Count the children two ways and equate · Sensim Math

Variant 1 easy

A kindergarten tried to put $12$ children in each class, but it was one classroom short, so $12$ children were left without a class. So some classes were given $13$ children instead. If the number of $12$ -child classes and the number of $13$ -child classes are in the ratio $3:2$ , every child is placed and no classroom is left empty. How many children are there?

Your answer

Show solution

1 · Understandwhat's really being asked

With $12$ children per room, the kindergarten needs exactly one more room than it has. Mixing $12$ -child and $13$ -child classes in the ratio $3:2$ fills every room exactly. Find the number of children.

Givens

$12$ per class leaves one full class of $12$ children without a room.
Classes of $12$ and $13$ in the ratio $3:2$ use every room and place every child.

Unknowns

The number of children.

Constraints

The number of classrooms is the same in both arrangements.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #16 Count the Complement#7 Identify Subproblems

Let the class counts be $3x$ and $2x$ , so there are $5x$ rooms. Count the children once from the first plan and once from the second plan, and set the two counts equal.

3 · Execute5 carry out the plan

1Name the class counts

#7 Identify Subproblems 6.RP.A.3

The ratio

3:2

means

3x

classes of

12

and

2x

classes of

13

, so

5x

classrooms in all.

3x + 2x = 5x

A ratio becomes actual counts once we multiply both parts by the same

x

.

2Count children from the first plan

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

With

12

per class, one extra room's worth of children is left over, so there are enough children for

5x + 1

classes of

12

.

12(5x + 1)

Being one room short means one more class of

12

than there are rooms.

3Count children from the second plan

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

Add up the

12

-child and the

13

-child classes.

12 \times 3x + 13 \times 2x = 36x + 26x = 62x

Same children, counted a second way.

4Equate and solve

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

Both expressions count the same children.

60x + 12 = 62x \;\Rightarrow\; 2x = 12 \;\Rightarrow\; x = 6

The

2x

classes of

13

each take one extra child, and together they absorb the

12

left over.

5Find the number of children

#13 Convert to Algebra 7.EE.B.3

There are

5 \times 6 = 30

classrooms.

12 \times (30 + 1) = 12 \times 31 = 372

Use either count; the first one is quickest.

Answer: 372 children

4 · Reviewdoes it hold up?

$18$ classes of $12$ and $12$ classes of $13$ hold $216 + 156 = 372$ children in $30$ rooms, in the ratio $18 : 12 = 3 : 2$ . And $372 \div 12 = 31$ , one more than $30$ rooms.

Another way: Change focus to the leftover children: turning a class of $12$ into a class of $13$ places one extra child, and $12$ children were left over, so there must be $12$ classes of $13$ . Then $2x = 12$ , $x = 6$ , and the answer follows.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the number of children two ways and setting them equal.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning the ratio $3:2$ into class counts $3x$ and $2x$.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the final number of children and checking it.

💡Takeaway. When two plans place the same children, count them both ways and set the counts equal.

Variant 2 easy

A kindergarten tried to put $14$ children in each class, but it was one classroom short, so $14$ children were left without a class. So some classes were given $15$ children instead. If the number of $14$ -child classes and the number of $15$ -child classes are in the ratio $3:2$ , every child is placed and no classroom is left empty. How many children are there?

Your answer

Show solution

1 · Understandwhat's really being asked

With $14$ children per room, the kindergarten needs exactly one more room than it has. Mixing $14$ -child and $15$ -child classes in the ratio $3:2$ fills every room exactly. Find the number of children.

Givens

$14$ per class leaves one full class of $14$ children without a room.
Classes of $14$ and $15$ in the ratio $3:2$ use every room and place every child.

Unknowns

The number of children.

Constraints

The number of classrooms is the same in both arrangements.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #16 Count the Complement#7 Identify Subproblems

Let the class counts be $3x$ and $2x$ , so there are $5x$ rooms. Count the children once from the first plan and once from the second plan, and set the two counts equal.

3 · Execute5 carry out the plan

1Name the class counts

#7 Identify Subproblems 6.RP.A.3

The ratio

3:2

means

3x

classes of

14

and

2x

classes of

15

, so

5x

classrooms in all.

3x + 2x = 5x

A ratio becomes actual counts once we multiply both parts by the same

x

.

2Count children from the first plan

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

With

14

per class, one extra room's worth of children is left over, so there are enough children for

5x + 1

classes of

14

.

14(5x + 1)

Being one room short means one more class of

14

than there are rooms.

3Count children from the second plan

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

Add up the

14

-child and the

15

-child classes.

14 \times 3x + 15 \times 2x = 42x + 30x = 72x

Same children, counted a second way.

4Equate and solve

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

Both expressions count the same children.

70x + 14 = 72x \;\Rightarrow\; 2x = 14 \;\Rightarrow\; x = 7

The

2x

classes of

15

each take one extra child, and together they absorb the

14

left over.

5Find the number of children

#13 Convert to Algebra 7.EE.B.3

There are

5 \times 7 = 35

classrooms.

14 \times (35 + 1) = 14 \times 36 = 504

Use either count; the first one is quickest.

Answer: 504 children

4 · Reviewdoes it hold up?

$21$ classes of $14$ and $14$ classes of $15$ hold $294 + 210 = 504$ children in $35$ rooms, in the ratio $21 : 14 = 3 : 2$ . And $504 \div 14 = 36$ , one more than $35$ rooms.

Another way: Change focus to the leftover children: turning a class of $14$ into a class of $15$ places one extra child, and $14$ children were left over, so there must be $14$ classes of $15$ . Then $2x = 14$ , $x = 7$ , and the answer follows.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the number of children two ways and setting them equal.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning the ratio $3:2$ into class counts $3x$ and $2x$.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the final number of children and checking it.

💡Takeaway. When two plans place the same children, count them both ways and set the counts equal.

Variant 3 easy

A kindergarten tried to put $15$ children in each class, but it was one classroom short, so $15$ children were left without a class. So some classes were given $16$ children instead. If the number of $15$ -child classes and the number of $16$ -child classes are in the ratio $2:3$ , every child is placed and no classroom is left empty. How many children are there?

Your answer

Show solution

1 · Understandwhat's really being asked

With $15$ children per room, the kindergarten needs exactly one more room than it has. Mixing $15$ -child and $16$ -child classes in the ratio $2:3$ fills every room exactly. Find the number of children.

Givens

$15$ per class leaves one full class of $15$ children without a room.
Classes of $15$ and $16$ in the ratio $2:3$ use every room and place every child.

Unknowns

The number of children.

Constraints

The number of classrooms is the same in both arrangements.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #16 Count the Complement#7 Identify Subproblems

Let the class counts be $2x$ and $3x$ , so there are $5x$ rooms. Count the children once from the first plan and once from the second plan, and set the two counts equal.

3 · Execute5 carry out the plan

1Name the class counts

#7 Identify Subproblems 6.RP.A.3

The ratio

2:3

means

2x

classes of

15

and

3x

classes of

16

, so

5x

classrooms in all.

2x + 3x = 5x

A ratio becomes actual counts once we multiply both parts by the same

x

.

2Count children from the first plan

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

With

15

per class, one extra room's worth of children is left over, so there are enough children for

5x + 1

classes of

15

.

15(5x + 1)

Being one room short means one more class of

15

than there are rooms.

3Count children from the second plan

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

Add up the

15

-child and the

16

-child classes.

15 \times 2x + 16 \times 3x = 30x + 48x = 78x

Same children, counted a second way.

4Equate and solve

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

Both expressions count the same children.

75x + 15 = 78x \;\Rightarrow\; 3x = 15 \;\Rightarrow\; x = 5

The

3x

classes of

16

each take one extra child, and together they absorb the

15

left over.

5Find the number of children

#13 Convert to Algebra 7.EE.B.3

There are

5 \times 5 = 25

classrooms.

15 \times (25 + 1) = 15 \times 26 = 390

Use either count; the first one is quickest.

Answer: 390 children

4 · Reviewdoes it hold up?

$10$ classes of $15$ and $15$ classes of $16$ hold $150 + 240 = 390$ children in $25$ rooms, in the ratio $10 : 15 = 2 : 3$ . And $390 \div 15 = 26$ , one more than $25$ rooms.

Another way: Change focus to the leftover children: turning a class of $15$ into a class of $16$ places one extra child, and $15$ children were left over, so there must be $15$ classes of $16$ . Then $3x = 15$ , $x = 5$ , and the answer follows.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the number of children two ways and setting them equal.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning the ratio $2:3$ into class counts $2x$ and $3x$.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the final number of children and checking it.

💡Takeaway. When two plans place the same children, count them both ways and set the counts equal.

Variant 4 easy

A kindergarten tried to put $16$ children in each class, but it was one classroom short, so $16$ children were left without a class. So some classes were given $17$ children instead. If the number of $16$ -child classes and the number of $17$ -child classes are in the ratio $3:4$ , every child is placed and no classroom is left empty. How many children are there?

Your answer

Show solution

1 · Understandwhat's really being asked

With $16$ children per room, the kindergarten needs exactly one more room than it has. Mixing $16$ -child and $17$ -child classes in the ratio $3:4$ fills every room exactly. Find the number of children.

Givens

$16$ per class leaves one full class of $16$ children without a room.
Classes of $16$ and $17$ in the ratio $3:4$ use every room and place every child.

Unknowns

The number of children.

Constraints

The number of classrooms is the same in both arrangements.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #16 Count the Complement#7 Identify Subproblems

Let the class counts be $3x$ and $4x$ , so there are $7x$ rooms. Count the children once from the first plan and once from the second plan, and set the two counts equal.

3 · Execute5 carry out the plan

1Name the class counts

#7 Identify Subproblems 6.RP.A.3

The ratio

3:4

means

3x

classes of

16

and

4x

classes of

17

, so

7x

classrooms in all.

3x + 4x = 7x

A ratio becomes actual counts once we multiply both parts by the same

x

.

2Count children from the first plan

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

With

16

per class, one extra room's worth of children is left over, so there are enough children for

7x + 1

classes of

16

.

16(7x + 1)

Being one room short means one more class of

16

than there are rooms.

3Count children from the second plan

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

Add up the

16

-child and the

17

-child classes.

16 \times 3x + 17 \times 4x = 48x + 68x = 116x

Same children, counted a second way.

4Equate and solve

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

Both expressions count the same children.

112x + 16 = 116x \;\Rightarrow\; 4x = 16 \;\Rightarrow\; x = 4

The

4x

classes of

17

each take one extra child, and together they absorb the

16

left over.

5Find the number of children

#13 Convert to Algebra 7.EE.B.3

There are

7 \times 4 = 28

classrooms.

16 \times (28 + 1) = 16 \times 29 = 464

Use either count; the first one is quickest.

Answer: 464 children

4 · Reviewdoes it hold up?

$12$ classes of $16$ and $16$ classes of $17$ hold $192 + 272 = 464$ children in $28$ rooms, in the ratio $12 : 16 = 3 : 4$ . And $464 \div 16 = 29$ , one more than $28$ rooms.

Another way: Change focus to the leftover children: turning a class of $16$ into a class of $17$ places one extra child, and $16$ children were left over, so there must be $16$ classes of $17$ . Then $4x = 16$ , $x = 4$ , and the answer follows.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the number of children two ways and setting them equal.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning the ratio $3:4$ into class counts $3x$ and $4x$.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the final number of children and checking it.

💡Takeaway. When two plans place the same children, count them both ways and set the counts equal.

Variant 5 medium

A kindergarten tried to put $18$ children in each class, but it was one classroom short, so $18$ children were left without a class. So some classes were given $19$ children instead. If the number of $18$ -child classes and the number of $19$ -child classes are in the ratio $4:3$ , every child is placed and no classroom is left empty. How many children are there?

Your answer

Show solution

1 · Understandwhat's really being asked

With $18$ children per room, the kindergarten needs exactly one more room than it has. Mixing $18$ -child and $19$ -child classes in the ratio $4:3$ fills every room exactly. Find the number of children.

Givens

$18$ per class leaves one full class of $18$ children without a room.
Classes of $18$ and $19$ in the ratio $4:3$ use every room and place every child.

Unknowns

The number of children.

Constraints

The number of classrooms is the same in both arrangements.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #16 Count the Complement#7 Identify Subproblems

Let the class counts be $4x$ and $3x$ , so there are $7x$ rooms. Count the children once from the first plan and once from the second plan, and set the two counts equal.

3 · Execute5 carry out the plan

1Name the class counts

#7 Identify Subproblems 6.RP.A.3

The ratio

4:3

means

4x

classes of

18

and

3x

classes of

19

, so

7x

classrooms in all.

4x + 3x = 7x

A ratio becomes actual counts once we multiply both parts by the same

x

.

2Count children from the first plan

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

With

18

per class, one extra room's worth of children is left over, so there are enough children for

7x + 1

classes of

18

.

18(7x + 1)

Being one room short means one more class of

18

than there are rooms.

3Count children from the second plan

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

Add up the

18

-child and the

19

-child classes.

18 \times 4x + 19 \times 3x = 72x + 57x = 129x

Same children, counted a second way.

4Equate and solve

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

Both expressions count the same children.

126x + 18 = 129x \;\Rightarrow\; 3x = 18 \;\Rightarrow\; x = 6

The

3x

classes of

19

each take one extra child, and together they absorb the

18

left over.

5Find the number of children

#13 Convert to Algebra 7.EE.B.3

There are

7 \times 6 = 42

classrooms.

18 \times (42 + 1) = 18 \times 43 = 774

Use either count; the first one is quickest.

Answer: 774 children

4 · Reviewdoes it hold up?

$24$ classes of $18$ and $18$ classes of $19$ hold $432 + 342 = 774$ children in $42$ rooms, in the ratio $24 : 18 = 4 : 3$ . And $774 \div 18 = 43$ , one more than $42$ rooms.

Another way: Change focus to the leftover children: turning a class of $18$ into a class of $19$ places one extra child, and $18$ children were left over, so there must be $18$ classes of $19$ . Then $3x = 18$ , $x = 6$ , and the answer follows.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the number of children two ways and setting them equal.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning the ratio $4:3$ into class counts $4x$ and $3x$.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the final number of children and checking it.

💡Takeaway. When two plans place the same children, count them both ways and set the counts equal.

Variant 6 medium

A kindergarten tried to put $20$ children in each class, but it was one classroom short, so $20$ children were left without a class. So some classes were given $21$ children instead. If the number of $20$ -child classes and the number of $21$ -child classes are in the ratio $3:4$ , every child is placed and no classroom is left empty. How many children are there?

Your answer

Show solution

1 · Understandwhat's really being asked

With $20$ children per room, the kindergarten needs exactly one more room than it has. Mixing $20$ -child and $21$ -child classes in the ratio $3:4$ fills every room exactly. Find the number of children.

Givens

$20$ per class leaves one full class of $20$ children without a room.
Classes of $20$ and $21$ in the ratio $3:4$ use every room and place every child.

Unknowns

The number of children.

Constraints

The number of classrooms is the same in both arrangements.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #16 Count the Complement#7 Identify Subproblems

Let the class counts be $3x$ and $4x$ , so there are $7x$ rooms. Count the children once from the first plan and once from the second plan, and set the two counts equal.

3 · Execute5 carry out the plan

1Name the class counts

#7 Identify Subproblems 6.RP.A.3

The ratio

3:4

means

3x

classes of

20

and

4x

classes of

21

, so

7x

classrooms in all.

3x + 4x = 7x

A ratio becomes actual counts once we multiply both parts by the same

x

.

2Count children from the first plan

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

With

20

per class, one extra room's worth of children is left over, so there are enough children for

7x + 1

classes of

20

.

20(7x + 1)

Being one room short means one more class of

20

than there are rooms.

3Count children from the second plan

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

Add up the

20

-child and the

21

-child classes.

20 \times 3x + 21 \times 4x = 60x + 84x = 144x

Same children, counted a second way.

4Equate and solve

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

Both expressions count the same children.

140x + 20 = 144x \;\Rightarrow\; 4x = 20 \;\Rightarrow\; x = 5

The

4x

classes of

21

each take one extra child, and together they absorb the

20

left over.

5Find the number of children

#13 Convert to Algebra 7.EE.B.3

There are

7 \times 5 = 35

classrooms.

20 \times (35 + 1) = 20 \times 36 = 720

Use either count; the first one is quickest.

Answer: 720 children

4 · Reviewdoes it hold up?

$15$ classes of $20$ and $20$ classes of $21$ hold $300 + 420 = 720$ children in $35$ rooms, in the ratio $15 : 20 = 3 : 4$ . And $720 \div 20 = 36$ , one more than $35$ rooms.

Another way: Change focus to the leftover children: turning a class of $20$ into a class of $21$ places one extra child, and $20$ children were left over, so there must be $20$ classes of $21$ . Then $4x = 20$ , $x = 5$ , and the answer follows.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the number of children two ways and setting them equal.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning the ratio $3:4$ into class counts $3x$ and $4x$.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the final number of children and checking it.

💡Takeaway. When two plans place the same children, count them both ways and set the counts equal.

Variant 7 medium

A kindergarten tried to put $21$ children in each class, but it was one classroom short, so $21$ children were left without a class. So some classes were given $22$ children instead. If the number of $21$ -child classes and the number of $22$ -child classes are in the ratio $4:3$ , every child is placed and no classroom is left empty. How many children are there?

Your answer

Show solution

1 · Understandwhat's really being asked

With $21$ children per room, the kindergarten needs exactly one more room than it has. Mixing $21$ -child and $22$ -child classes in the ratio $4:3$ fills every room exactly. Find the number of children.

Givens

$21$ per class leaves one full class of $21$ children without a room.
Classes of $21$ and $22$ in the ratio $4:3$ use every room and place every child.

Unknowns

The number of children.

Constraints

The number of classrooms is the same in both arrangements.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #16 Count the Complement#7 Identify Subproblems

Let the class counts be $4x$ and $3x$ , so there are $7x$ rooms. Count the children once from the first plan and once from the second plan, and set the two counts equal.

3 · Execute5 carry out the plan

1Name the class counts

#7 Identify Subproblems 6.RP.A.3

The ratio

4:3

means

4x

classes of

21

and

3x

classes of

22

, so

7x

classrooms in all.

4x + 3x = 7x

A ratio becomes actual counts once we multiply both parts by the same

x

.

2Count children from the first plan

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

With

21

per class, one extra room's worth of children is left over, so there are enough children for

7x + 1

classes of

21

.

21(7x + 1)

Being one room short means one more class of

21

than there are rooms.

3Count children from the second plan

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

Add up the

21

-child and the

22

-child classes.

21 \times 4x + 22 \times 3x = 84x + 66x = 150x

Same children, counted a second way.

4Equate and solve

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

Both expressions count the same children.

147x + 21 = 150x \;\Rightarrow\; 3x = 21 \;\Rightarrow\; x = 7

The

3x

classes of

22

each take one extra child, and together they absorb the

21

left over.

5Find the number of children

#13 Convert to Algebra 7.EE.B.3

There are

7 \times 7 = 49

classrooms.

21 \times (49 + 1) = 21 \times 50 = 1050

Use either count; the first one is quickest.

Answer: 1050 children

4 · Reviewdoes it hold up?

$28$ classes of $21$ and $21$ classes of $22$ hold $588 + 462 = 1050$ children in $49$ rooms, in the ratio $28 : 21 = 4 : 3$ . And $1050 \div 21 = 50$ , one more than $49$ rooms.

Another way: Change focus to the leftover children: turning a class of $21$ into a class of $22$ places one extra child, and $21$ children were left over, so there must be $21$ classes of $22$ . Then $3x = 21$ , $x = 7$ , and the answer follows.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the number of children two ways and setting them equal.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning the ratio $4:3$ into class counts $4x$ and $3x$.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the final number of children and checking it.

💡Takeaway. When two plans place the same children, count them both ways and set the counts equal.

Variant 8 medium

A kindergarten tried to put $22$ children in each class, but it was one classroom short, so $22$ children were left without a class. So some classes were given $23$ children instead. If the number of $22$ -child classes and the number of $23$ -child classes are in the ratio $3:2$ , every child is placed and no classroom is left empty. How many children are there?

Your answer

Show solution

1 · Understandwhat's really being asked

With $22$ children per room, the kindergarten needs exactly one more room than it has. Mixing $22$ -child and $23$ -child classes in the ratio $3:2$ fills every room exactly. Find the number of children.

Givens

$22$ per class leaves one full class of $22$ children without a room.
Classes of $22$ and $23$ in the ratio $3:2$ use every room and place every child.

Unknowns

The number of children.

Constraints

The number of classrooms is the same in both arrangements.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #16 Count the Complement#7 Identify Subproblems

Let the class counts be $3x$ and $2x$ , so there are $5x$ rooms. Count the children once from the first plan and once from the second plan, and set the two counts equal.

3 · Execute5 carry out the plan

1Name the class counts

#7 Identify Subproblems 6.RP.A.3

The ratio

3:2

means

3x

classes of

22

and

2x

classes of

23

, so

5x

classrooms in all.

3x + 2x = 5x

A ratio becomes actual counts once we multiply both parts by the same

x

.

2Count children from the first plan

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

With

22

per class, one extra room's worth of children is left over, so there are enough children for

5x + 1

classes of

22

.

22(5x + 1)

Being one room short means one more class of

22

than there are rooms.

3Count children from the second plan

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

Add up the

22

-child and the

23

-child classes.

22 \times 3x + 23 \times 2x = 66x + 46x = 112x

Same children, counted a second way.

4Equate and solve

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

Both expressions count the same children.

110x + 22 = 112x \;\Rightarrow\; 2x = 22 \;\Rightarrow\; x = 11

The

2x

classes of

23

each take one extra child, and together they absorb the

22

left over.

5Find the number of children

#13 Convert to Algebra 7.EE.B.3

There are

5 \times 11 = 55

classrooms.

22 \times (55 + 1) = 22 \times 56 = 1232

Use either count; the first one is quickest.

Answer: 1232 children

4 · Reviewdoes it hold up?

$33$ classes of $22$ and $22$ classes of $23$ hold $726 + 506 = 1232$ children in $55$ rooms, in the ratio $33 : 22 = 3 : 2$ . And $1232 \div 22 = 56$ , one more than $55$ rooms.

Another way: Change focus to the leftover children: turning a class of $22$ into a class of $23$ places one extra child, and $22$ children were left over, so there must be $22$ classes of $23$ . Then $2x = 22$ , $x = 11$ , and the answer follows.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the number of children two ways and setting them equal.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning the ratio $3:2$ into class counts $3x$ and $2x$.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the final number of children and checking it.

💡Takeaway. When two plans place the same children, count them both ways and set the counts equal.

Variant 9 hard

A kindergarten tried to put $24$ children in each class, but it was one classroom short, so $24$ children were left without a class. So some classes were given $25$ children instead. If the number of $24$ -child classes and the number of $25$ -child classes are in the ratio $5:3$ , every child is placed and no classroom is left empty. How many children are there?

Your answer

Show solution

1 · Understandwhat's really being asked

With $24$ children per room, the kindergarten needs exactly one more room than it has. Mixing $24$ -child and $25$ -child classes in the ratio $5:3$ fills every room exactly. Find the number of children.

Givens

$24$ per class leaves one full class of $24$ children without a room.
Classes of $24$ and $25$ in the ratio $5:3$ use every room and place every child.

Unknowns

The number of children.

Constraints

The number of classrooms is the same in both arrangements.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #16 Count the Complement#7 Identify Subproblems

Let the class counts be $5x$ and $3x$ , so there are $8x$ rooms. Count the children once from the first plan and once from the second plan, and set the two counts equal.

3 · Execute5 carry out the plan

1Name the class counts

#7 Identify Subproblems 6.RP.A.3

The ratio

5:3

means

5x

classes of

24

and

3x

classes of

25

, so

8x

classrooms in all.

5x + 3x = 8x

A ratio becomes actual counts once we multiply both parts by the same

x

.

2Count children from the first plan

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

With

24

per class, one extra room's worth of children is left over, so there are enough children for

8x + 1

classes of

24

.

24(8x + 1)

Being one room short means one more class of

24

than there are rooms.

3Count children from the second plan

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

Add up the

24

-child and the

25

-child classes.

24 \times 5x + 25 \times 3x = 120x + 75x = 195x

Same children, counted a second way.

4Equate and solve

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

Both expressions count the same children.

192x + 24 = 195x \;\Rightarrow\; 3x = 24 \;\Rightarrow\; x = 8

The

3x

classes of

25

each take one extra child, and together they absorb the

24

left over.

5Find the number of children

#13 Convert to Algebra 7.EE.B.3

There are

8 \times 8 = 64

classrooms.

24 \times (64 + 1) = 24 \times 65 = 1560

Use either count; the first one is quickest.

Answer: 1560 children

4 · Reviewdoes it hold up?

$40$ classes of $24$ and $24$ classes of $25$ hold $960 + 600 = 1560$ children in $64$ rooms, in the ratio $40 : 24 = 5 : 3$ . And $1560 \div 24 = 65$ , one more than $64$ rooms.

Another way: Change focus to the leftover children: turning a class of $24$ into a class of $25$ places one extra child, and $24$ children were left over, so there must be $24$ classes of $25$ . Then $3x = 24$ , $x = 8$ , and the answer follows.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the number of children two ways and setting them equal.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning the ratio $5:3$ into class counts $5x$ and $3x$.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the final number of children and checking it.

💡Takeaway. When two plans place the same children, count them both ways and set the counts equal.

Variant 10 hard

A kindergarten tried to put $25$ children in each class, but it was one classroom short, so $25$ children were left without a class. So some classes were given $26$ children instead. If the number of $25$ -child classes and the number of $26$ -child classes are in the ratio $4:5$ , every child is placed and no classroom is left empty. How many children are there?

Your answer

Show solution

1 · Understandwhat's really being asked

With $25$ children per room, the kindergarten needs exactly one more room than it has. Mixing $25$ -child and $26$ -child classes in the ratio $4:5$ fills every room exactly. Find the number of children.

Givens

$25$ per class leaves one full class of $25$ children without a room.
Classes of $25$ and $26$ in the ratio $4:5$ use every room and place every child.

Unknowns

The number of children.

Constraints

The number of classrooms is the same in both arrangements.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #16 Count the Complement#7 Identify Subproblems

Let the class counts be $4x$ and $5x$ , so there are $9x$ rooms. Count the children once from the first plan and once from the second plan, and set the two counts equal.

3 · Execute5 carry out the plan

1Name the class counts

#7 Identify Subproblems 6.RP.A.3

The ratio

4:5

means

4x

classes of

25

and

5x

classes of

26

, so

9x

classrooms in all.

4x + 5x = 9x

A ratio becomes actual counts once we multiply both parts by the same

x

.

2Count children from the first plan

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

With

25

per class, one extra room's worth of children is left over, so there are enough children for

9x + 1

classes of

25

.

25(9x + 1)

Being one room short means one more class of

25

than there are rooms.

3Count children from the second plan

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

Add up the

25

-child and the

26

-child classes.

25 \times 4x + 26 \times 5x = 100x + 130x = 230x

Same children, counted a second way.

4Equate and solve

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

Both expressions count the same children.

225x + 25 = 230x \;\Rightarrow\; 5x = 25 \;\Rightarrow\; x = 5

The

5x

classes of

26

each take one extra child, and together they absorb the

25

left over.

5Find the number of children

#13 Convert to Algebra 7.EE.B.3

There are

9 \times 5 = 45

classrooms.

25 \times (45 + 1) = 25 \times 46 = 1150

Use either count; the first one is quickest.

Answer: 1150 children

4 · Reviewdoes it hold up?

$20$ classes of $25$ and $25$ classes of $26$ hold $500 + 650 = 1150$ children in $45$ rooms, in the ratio $20 : 25 = 4 : 5$ . And $1150 \div 25 = 46$ , one more than $45$ rooms.

Another way: Change focus to the leftover children: turning a class of $25$ into a class of $26$ places one extra child, and $25$ children were left over, so there must be $25$ classes of $26$ . Then $5x = 25$ , $x = 5$ , and the answer follows.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the number of children two ways and setting them equal.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning the ratio $4:5$ into class counts $4x$ and $5x$.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the final number of children and checking it.

💡Takeaway. When two plans place the same children, count them both ways and set the counts equal.

Variant 11 hard

A kindergarten tried to put $28$ children in each class, but it was one classroom short, so $28$ children were left without a class. So some classes were given $29$ children instead. If the number of $28$ -child classes and the number of $29$ -child classes are in the ratio $5:4$ , every child is placed and no classroom is left empty. How many children are there?

Your answer

Show solution

1 · Understandwhat's really being asked

With $28$ children per room, the kindergarten needs exactly one more room than it has. Mixing $28$ -child and $29$ -child classes in the ratio $5:4$ fills every room exactly. Find the number of children.

Givens

$28$ per class leaves one full class of $28$ children without a room.
Classes of $28$ and $29$ in the ratio $5:4$ use every room and place every child.

Unknowns

The number of children.

Constraints

The number of classrooms is the same in both arrangements.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #16 Count the Complement#7 Identify Subproblems

Let the class counts be $5x$ and $4x$ , so there are $9x$ rooms. Count the children once from the first plan and once from the second plan, and set the two counts equal.

3 · Execute5 carry out the plan

1Name the class counts

#7 Identify Subproblems 6.RP.A.3

The ratio

5:4

means

5x

classes of

28

and

4x

classes of

29

, so

9x

classrooms in all.

5x + 4x = 9x

A ratio becomes actual counts once we multiply both parts by the same

x

.

2Count children from the first plan

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

With

28

per class, one extra room's worth of children is left over, so there are enough children for

9x + 1

classes of

28

.

28(9x + 1)

Being one room short means one more class of

28

than there are rooms.

3Count children from the second plan

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

Add up the

28

-child and the

29

-child classes.

28 \times 5x + 29 \times 4x = 140x + 116x = 256x

Same children, counted a second way.

4Equate and solve

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

Both expressions count the same children.

252x + 28 = 256x \;\Rightarrow\; 4x = 28 \;\Rightarrow\; x = 7

The

4x

classes of

29

each take one extra child, and together they absorb the

28

left over.

5Find the number of children

#13 Convert to Algebra 7.EE.B.3

There are

9 \times 7 = 63

classrooms.

28 \times (63 + 1) = 28 \times 64 = 1792

Use either count; the first one is quickest.

Answer: 1792 children

4 · Reviewdoes it hold up?

$35$ classes of $28$ and $28$ classes of $29$ hold $980 + 812 = 1792$ children in $63$ rooms, in the ratio $35 : 28 = 5 : 4$ . And $1792 \div 28 = 64$ , one more than $63$ rooms.

Another way: Change focus to the leftover children: turning a class of $28$ into a class of $29$ places one extra child, and $28$ children were left over, so there must be $28$ classes of $29$ . Then $4x = 28$ , $x = 7$ , and the answer follows.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the number of children two ways and setting them equal.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning the ratio $5:4$ into class counts $5x$ and $4x$.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the final number of children and checking it.

💡Takeaway. When two plans place the same children, count them both ways and set the counts equal.

Variant 12 hard

A kindergarten tried to put $30$ children in each class, but it was one classroom short, so $30$ children were left without a class. So some classes were given $31$ children instead. If the number of $30$ -child classes and the number of $31$ -child classes are in the ratio $5:6$ , every child is placed and no classroom is left empty. How many children are there?

Your answer

Show solution

1 · Understandwhat's really being asked

With $30$ children per room, the kindergarten needs exactly one more room than it has. Mixing $30$ -child and $31$ -child classes in the ratio $5:6$ fills every room exactly. Find the number of children.

Givens

$30$ per class leaves one full class of $30$ children without a room.
Classes of $30$ and $31$ in the ratio $5:6$ use every room and place every child.

Unknowns

The number of children.

Constraints

The number of classrooms is the same in both arrangements.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #16 Count the Complement#7 Identify Subproblems

Let the class counts be $5x$ and $6x$ , so there are $11x$ rooms. Count the children once from the first plan and once from the second plan, and set the two counts equal.

3 · Execute5 carry out the plan

1Name the class counts

#7 Identify Subproblems 6.RP.A.3

The ratio

5:6

means

5x

classes of

30

and

6x

classes of

31

, so

11x

classrooms in all.

5x + 6x = 11x

A ratio becomes actual counts once we multiply both parts by the same

x

.

2Count children from the first plan

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

With

30

per class, one extra room's worth of children is left over, so there are enough children for

11x + 1

classes of

30

.

30(11x + 1)

Being one room short means one more class of

30

than there are rooms.

3Count children from the second plan

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

Add up the

30

-child and the

31

-child classes.

30 \times 5x + 31 \times 6x = 150x + 186x = 336x

Same children, counted a second way.

4Equate and solve

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

Both expressions count the same children.

330x + 30 = 336x \;\Rightarrow\; 6x = 30 \;\Rightarrow\; x = 5

The

6x

classes of

31

each take one extra child, and together they absorb the

30

left over.

5Find the number of children

#13 Convert to Algebra 7.EE.B.3

There are

11 \times 5 = 55

classrooms.

30 \times (55 + 1) = 30 \times 56 = 1680

Use either count; the first one is quickest.

Answer: 1680 children

4 · Reviewdoes it hold up?

$25$ classes of $30$ and $30$ classes of $31$ hold $750 + 930 = 1680$ children in $55$ rooms, in the ratio $25 : 30 = 5 : 6$ . And $1680 \div 30 = 56$ , one more than $55$ rooms.

Another way: Change focus to the leftover children: turning a class of $30$ into a class of $31$ places one extra child, and $30$ children were left over, so there must be $30$ classes of $31$ . Then $6x = 30$ , $x = 5$ , and the answer follows.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the number of children two ways and setting them equal.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Turning the ratio $5:6$ into class counts $5x$ and $6x$.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the final number of children and checking it.

💡Takeaway. When two plans place the same children, count them both ways and set the counts equal.

Count the children two ways and equate · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?