Practice · Model sum as rectangle area for means · Sensim Math

Variant 1 easy

At one school, $30$ fifth-grade students took part in a math competition. The overall mean score was $57$ points. If the boys' mean score was $55$ points and the girls' mean score was $60$ points, find how many girls took part in the math competition.

Your answer

Show solution

1 · Understandwhat's really being asked

30 students averaged 57 points. Split by sex, the boys averaged 55 and the girls 60. We need the number of girls.

Givens

30 students in all, mean 57.
Boys' mean: 55.
Girls' mean: 60.

Unknowns

How many girls took part.

Constraints

The overall mean lies between the two group means, since it is built from both.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

Draw each group as a rectangle: how many students wide, how many points tall, so its area is that group's total score. The overall mean is a rectangle of the same total area over the whole width, which turns the question into a balance.

3 · Execute5 carry out the plan

1Picture each mean as a rectangle's area

#1 Draw a Diagram 6.SP.A.3

A group of n students averaging m points has a total of n times m -- the area of an n-by-m rectangle.

\text{total score} = \text{students} \times \text{mean}

Means become heights and counts become widths.

2Find the area of the whole class rectangle

#7 Identify Subproblems 6.NS.B.3

All 30 students at a mean of 57.

30 \times 57 = 1710

1710 points between everyone.

3Compare the overall mean to each group's mean

#1 Draw a Diagram 6.SP.B.5

The overall mean sits 2 above the boys' and 3 below the girls'. Every boy is short of the overall mean by 2, and every girl is over it by 3.

57 - 55 = 2,\quad 60 - 57 = 3

The girls' surplus has to cover the boys' shortfall exactly.

4Balance the two pieces to split the 30 students

#6 Guess and Check 6.RP.A.3

If there are g girls then 30 minus g boys, and the surplus must equal the shortfall.

g \times 3 = (30 - g) \times 2 \Rightarrow g = 12

12 girls and 18 boys.

5Check the totals rebuild 1710

#7 Identify Subproblems 6.NS.B.3

Add the two group totals back up.

18 \times 55 + 12 \times 60 = 1710

It matches, so 12 girls is right.

Answer: 12 girls

4 · Reviewdoes it hold up?

The overall mean 57 is nearer the boys' mean, so that group should be the larger -- and it is, 18 against 12.

Another way: Work in totals instead: the boys alone would give 1650, which is 60 short, and each girl swapped in adds 5 -- again 12.

Standardsmin grade 6

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Reading each mean as a total shared over a count.
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Measuring how far the overall mean sits from each group's.
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems — Balancing surplus against shortfall to split the group.
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals — The decimal multiplications on both sides of the check.

💡Takeaway. An overall mean sits closer to the bigger group. How far it leans tells you how the group splits.

Variant 2 easy

At one school, $50$ fifth-grade students took part in a math competition. The overall mean score was $61.8$ points. If the boys' mean score was $61$ points and the girls' mean score was $63$ points, find how many girls took part in the math competition.

Your answer

Show solution

1 · Understandwhat's really being asked

50 students averaged 61.8 points. Split by sex, the boys averaged 61 and the girls 63. We need the number of girls.

Givens

50 students in all, mean 61.8.
Boys' mean: 61.
Girls' mean: 63.

Unknowns

How many girls took part.

Constraints

The overall mean lies between the two group means, since it is built from both.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

Draw each group as a rectangle: how many students wide, how many points tall, so its area is that group's total score. The overall mean is a rectangle of the same total area over the whole width, which turns the question into a balance.

3 · Execute5 carry out the plan

1Picture each mean as a rectangle's area

#1 Draw a Diagram 6.SP.A.3

A group of n students averaging m points has a total of n times m -- the area of an n-by-m rectangle.

\text{total score} = \text{students} \times \text{mean}

Means become heights and counts become widths.

2Find the area of the whole class rectangle

#7 Identify Subproblems 6.NS.B.3

All 50 students at a mean of 61.8.

50 \times 61.8 = 3090

3090 points between everyone.

3Compare the overall mean to each group's mean

#1 Draw a Diagram 6.SP.B.5

The overall mean sits 0.8 above the boys' and 1.2 below the girls'. Every boy is short of the overall mean by 0.8, and every girl is over it by 1.2.

61.8 - 61 = 0.8,\quad 63 - 61.8 = 1.2

The girls' surplus has to cover the boys' shortfall exactly.

4Balance the two pieces to split the 50 students

#6 Guess and Check 6.RP.A.3

If there are g girls then 50 minus g boys, and the surplus must equal the shortfall.

g \times 1.2 = (50 - g) \times 0.8 \Rightarrow g = 20

20 girls and 30 boys.

5Check the totals rebuild 3090

#7 Identify Subproblems 6.NS.B.3

Add the two group totals back up.

30 \times 61 + 20 \times 63 = 3090

It matches, so 20 girls is right.

Answer: 20 girls

4 · Reviewdoes it hold up?

The overall mean 61.8 is nearer the boys' mean, so that group should be the larger -- and it is, 30 against 20.

Another way: Work in totals instead: the boys alone would give 3050, which is 40 short, and each girl swapped in adds 2 -- again 20.

Standardsmin grade 6

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Reading each mean as a total shared over a count.
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Measuring how far the overall mean sits from each group's.
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems — Balancing surplus against shortfall to split the group.
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals — The decimal multiplications on both sides of the check.

💡Takeaway. An overall mean sits closer to the bigger group. How far it leans tells you how the group splits.

Variant 3 easy

At one school, $36$ fifth-grade students took part in a math competition. The overall mean score was $61$ points. If the boys' mean score was $59$ points and the girls' mean score was $65$ points, find how many girls took part in the math competition.

Your answer

Show solution

1 · Understandwhat's really being asked

36 students averaged 61 points. Split by sex, the boys averaged 59 and the girls 65. We need the number of girls.

Givens

36 students in all, mean 61.
Boys' mean: 59.
Girls' mean: 65.

Unknowns

How many girls took part.

Constraints

The overall mean lies between the two group means, since it is built from both.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

Draw each group as a rectangle: how many students wide, how many points tall, so its area is that group's total score. The overall mean is a rectangle of the same total area over the whole width, which turns the question into a balance.

3 · Execute5 carry out the plan

1Picture each mean as a rectangle's area

#1 Draw a Diagram 6.SP.A.3

A group of n students averaging m points has a total of n times m -- the area of an n-by-m rectangle.

\text{total score} = \text{students} \times \text{mean}

Means become heights and counts become widths.

2Find the area of the whole class rectangle

#7 Identify Subproblems 6.NS.B.3

All 36 students at a mean of 61.

36 \times 61 = 2196

2196 points between everyone.

3Compare the overall mean to each group's mean

#1 Draw a Diagram 6.SP.B.5

The overall mean sits 2 above the boys' and 4 below the girls'. Every boy is short of the overall mean by 2, and every girl is over it by 4.

61 - 59 = 2,\quad 65 - 61 = 4

The girls' surplus has to cover the boys' shortfall exactly.

4Balance the two pieces to split the 36 students

#6 Guess and Check 6.RP.A.3

If there are g girls then 36 minus g boys, and the surplus must equal the shortfall.

g \times 4 = (36 - g) \times 2 \Rightarrow g = 12

12 girls and 24 boys.

5Check the totals rebuild 2196

#7 Identify Subproblems 6.NS.B.3

Add the two group totals back up.

24 \times 59 + 12 \times 65 = 2196

It matches, so 12 girls is right.

Answer: 12 girls

4 · Reviewdoes it hold up?

The overall mean 61 is nearer the boys' mean, so that group should be the larger -- and it is, 24 against 12.

Another way: Work in totals instead: the boys alone would give 2124, which is 72 short, and each girl swapped in adds 6 -- again 12.

Standardsmin grade 6

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Reading each mean as a total shared over a count.
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Measuring how far the overall mean sits from each group's.
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems — Balancing surplus against shortfall to split the group.
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals — The decimal multiplications on both sides of the check.

💡Takeaway. An overall mean sits closer to the bigger group. How far it leans tells you how the group splits.

Variant 4 easy

At one school, $25$ fifth-grade students took part in a math competition. The overall mean score was $66$ points. If the boys' mean score was $64$ points and the girls' mean score was $69$ points, find how many girls took part in the math competition.

Your answer

Show solution

1 · Understandwhat's really being asked

25 students averaged 66 points. Split by sex, the boys averaged 64 and the girls 69. We need the number of girls.

Givens

25 students in all, mean 66.
Boys' mean: 64.
Girls' mean: 69.

Unknowns

How many girls took part.

Constraints

The overall mean lies between the two group means, since it is built from both.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

Draw each group as a rectangle: how many students wide, how many points tall, so its area is that group's total score. The overall mean is a rectangle of the same total area over the whole width, which turns the question into a balance.

3 · Execute5 carry out the plan

1Picture each mean as a rectangle's area

#1 Draw a Diagram 6.SP.A.3

A group of n students averaging m points has a total of n times m -- the area of an n-by-m rectangle.

\text{total score} = \text{students} \times \text{mean}

Means become heights and counts become widths.

2Find the area of the whole class rectangle

#7 Identify Subproblems 6.NS.B.3

All 25 students at a mean of 66.

25 \times 66 = 1650

1650 points between everyone.

3Compare the overall mean to each group's mean

#1 Draw a Diagram 6.SP.B.5

The overall mean sits 2 above the boys' and 3 below the girls'. Every boy is short of the overall mean by 2, and every girl is over it by 3.

66 - 64 = 2,\quad 69 - 66 = 3

The girls' surplus has to cover the boys' shortfall exactly.

4Balance the two pieces to split the 25 students

#6 Guess and Check 6.RP.A.3

If there are g girls then 25 minus g boys, and the surplus must equal the shortfall.

g \times 3 = (25 - g) \times 2 \Rightarrow g = 10

10 girls and 15 boys.

5Check the totals rebuild 1650

#7 Identify Subproblems 6.NS.B.3

Add the two group totals back up.

15 \times 64 + 10 \times 69 = 1650

It matches, so 10 girls is right.

Answer: 10 girls

4 · Reviewdoes it hold up?

The overall mean 66 is nearer the boys' mean, so that group should be the larger -- and it is, 15 against 10.

Another way: Work in totals instead: the boys alone would give 1600, which is 50 short, and each girl swapped in adds 5 -- again 10.

Standardsmin grade 6

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Reading each mean as a total shared over a count.
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Measuring how far the overall mean sits from each group's.
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems — Balancing surplus against shortfall to split the group.
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals — The decimal multiplications on both sides of the check.

💡Takeaway. An overall mean sits closer to the bigger group. How far it leans tells you how the group splits.

Variant 5 medium

At one school, $45$ fifth-grade students took part in a math competition. The overall mean score was $67.6$ points. If the boys' mean score was $66$ points and the girls' mean score was $70$ points, find how many girls took part in the math competition.

Your answer

Show solution

1 · Understandwhat's really being asked

45 students averaged 67.6 points. Split by sex, the boys averaged 66 and the girls 70. We need the number of girls.

Givens

45 students in all, mean 67.6.
Boys' mean: 66.
Girls' mean: 70.

Unknowns

How many girls took part.

Constraints

The overall mean lies between the two group means, since it is built from both.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

Draw each group as a rectangle: how many students wide, how many points tall, so its area is that group's total score. The overall mean is a rectangle of the same total area over the whole width, which turns the question into a balance.

3 · Execute5 carry out the plan

1Picture each mean as a rectangle's area

#1 Draw a Diagram 6.SP.A.3

A group of n students averaging m points has a total of n times m -- the area of an n-by-m rectangle.

\text{total score} = \text{students} \times \text{mean}

Means become heights and counts become widths.

2Find the area of the whole class rectangle

#7 Identify Subproblems 6.NS.B.3

All 45 students at a mean of 67.6.

45 \times 67.6 = 3042

3042 points between everyone.

3Compare the overall mean to each group's mean

#1 Draw a Diagram 6.SP.B.5

The overall mean sits 1.6 above the boys' and 2.4 below the girls'. Every boy is short of the overall mean by 1.6, and every girl is over it by 2.4.

67.6 - 66 = 1.6,\quad 70 - 67.6 = 2.4

The girls' surplus has to cover the boys' shortfall exactly.

4Balance the two pieces to split the 45 students

#6 Guess and Check 6.RP.A.3

If there are g girls then 45 minus g boys, and the surplus must equal the shortfall.

g \times 2.4 = (45 - g) \times 1.6 \Rightarrow g = 18

18 girls and 27 boys.

5Check the totals rebuild 3042

#7 Identify Subproblems 6.NS.B.3

Add the two group totals back up.

27 \times 66 + 18 \times 70 = 3042

It matches, so 18 girls is right.

Answer: 18 girls

4 · Reviewdoes it hold up?

The overall mean 67.6 is nearer the boys' mean, so that group should be the larger -- and it is, 27 against 18.

Another way: Work in totals instead: the boys alone would give 2970, which is 72 short, and each girl swapped in adds 4 -- again 18.

Standardsmin grade 6

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Reading each mean as a total shared over a count.
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Measuring how far the overall mean sits from each group's.
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems — Balancing surplus against shortfall to split the group.
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals — The decimal multiplications on both sides of the check.

💡Takeaway. An overall mean sits closer to the bigger group. How far it leans tells you how the group splits.

Variant 6 medium

At one school, $75$ fifth-grade students took part in a math competition. The overall mean score was $71.6$ points. If the boys' mean score was $70$ points and the girls' mean score was $74$ points, find how many girls took part in the math competition.

Your answer

Show solution

1 · Understandwhat's really being asked

75 students averaged 71.6 points. Split by sex, the boys averaged 70 and the girls 74. We need the number of girls.

Givens

75 students in all, mean 71.6.
Boys' mean: 70.
Girls' mean: 74.

Unknowns

How many girls took part.

Constraints

The overall mean lies between the two group means, since it is built from both.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

Draw each group as a rectangle: how many students wide, how many points tall, so its area is that group's total score. The overall mean is a rectangle of the same total area over the whole width, which turns the question into a balance.

3 · Execute5 carry out the plan

1Picture each mean as a rectangle's area

#1 Draw a Diagram 6.SP.A.3

A group of n students averaging m points has a total of n times m -- the area of an n-by-m rectangle.

\text{total score} = \text{students} \times \text{mean}

Means become heights and counts become widths.

2Find the area of the whole class rectangle

#7 Identify Subproblems 6.NS.B.3

All 75 students at a mean of 71.6.

75 \times 71.6 = 5370

5370 points between everyone.

3Compare the overall mean to each group's mean

#1 Draw a Diagram 6.SP.B.5

The overall mean sits 1.6 above the boys' and 2.4 below the girls'. Every boy is short of the overall mean by 1.6, and every girl is over it by 2.4.

71.6 - 70 = 1.6,\quad 74 - 71.6 = 2.4

The girls' surplus has to cover the boys' shortfall exactly.

4Balance the two pieces to split the 75 students

#6 Guess and Check 6.RP.A.3

If there are g girls then 75 minus g boys, and the surplus must equal the shortfall.

g \times 2.4 = (75 - g) \times 1.6 \Rightarrow g = 30

30 girls and 45 boys.

5Check the totals rebuild 5370

#7 Identify Subproblems 6.NS.B.3

Add the two group totals back up.

45 \times 70 + 30 \times 74 = 5370

It matches, so 30 girls is right.

Answer: 30 girls

4 · Reviewdoes it hold up?

The overall mean 71.6 is nearer the boys' mean, so that group should be the larger -- and it is, 45 against 30.

Another way: Work in totals instead: the boys alone would give 5250, which is 120 short, and each girl swapped in adds 4 -- again 30.

Standardsmin grade 6

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Reading each mean as a total shared over a count.
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Measuring how far the overall mean sits from each group's.
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems — Balancing surplus against shortfall to split the group.
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals — The decimal multiplications on both sides of the check.

💡Takeaway. An overall mean sits closer to the bigger group. How far it leans tells you how the group splits.

Variant 7 medium

At one school, $40$ fifth-grade students took part in a math competition. The overall mean score was $72$ points. If the boys' mean score was $70$ points and the girls' mean score was $75$ points, find how many girls took part in the math competition.

Your answer

Show solution

1 · Understandwhat's really being asked

40 students averaged 72 points. Split by sex, the boys averaged 70 and the girls 75. We need the number of girls.

Givens

40 students in all, mean 72.
Boys' mean: 70.
Girls' mean: 75.

Unknowns

How many girls took part.

Constraints

The overall mean lies between the two group means, since it is built from both.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

Draw each group as a rectangle: how many students wide, how many points tall, so its area is that group's total score. The overall mean is a rectangle of the same total area over the whole width, which turns the question into a balance.

3 · Execute5 carry out the plan

1Picture each mean as a rectangle's area

#1 Draw a Diagram 6.SP.A.3

A group of n students averaging m points has a total of n times m -- the area of an n-by-m rectangle.

\text{total score} = \text{students} \times \text{mean}

Means become heights and counts become widths.

2Find the area of the whole class rectangle

#7 Identify Subproblems 6.NS.B.3

All 40 students at a mean of 72.

40 \times 72 = 2880

2880 points between everyone.

3Compare the overall mean to each group's mean

#1 Draw a Diagram 6.SP.B.5

The overall mean sits 2 above the boys' and 3 below the girls'. Every boy is short of the overall mean by 2, and every girl is over it by 3.

72 - 70 = 2,\quad 75 - 72 = 3

The girls' surplus has to cover the boys' shortfall exactly.

4Balance the two pieces to split the 40 students

#6 Guess and Check 6.RP.A.3

If there are g girls then 40 minus g boys, and the surplus must equal the shortfall.

g \times 3 = (40 - g) \times 2 \Rightarrow g = 16

16 girls and 24 boys.

5Check the totals rebuild 2880

#7 Identify Subproblems 6.NS.B.3

Add the two group totals back up.

24 \times 70 + 16 \times 75 = 2880

It matches, so 16 girls is right.

Answer: 16 girls

4 · Reviewdoes it hold up?

The overall mean 72 is nearer the boys' mean, so that group should be the larger -- and it is, 24 against 16.

Another way: Work in totals instead: the boys alone would give 2800, which is 80 short, and each girl swapped in adds 5 -- again 16.

Standardsmin grade 6

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Reading each mean as a total shared over a count.
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Measuring how far the overall mean sits from each group's.
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems — Balancing surplus against shortfall to split the group.
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals — The decimal multiplications on both sides of the check.

💡Takeaway. An overall mean sits closer to the bigger group. How far it leans tells you how the group splits.

Variant 8 medium

At one school, $80$ fifth-grade students took part in a math competition. The overall mean score was $74$ points. If the boys' mean score was $72$ points and the girls' mean score was $77$ points, find how many girls took part in the math competition.

Your answer

Show solution

1 · Understandwhat's really being asked

80 students averaged 74 points. Split by sex, the boys averaged 72 and the girls 77. We need the number of girls.

Givens

80 students in all, mean 74.
Boys' mean: 72.
Girls' mean: 77.

Unknowns

How many girls took part.

Constraints

The overall mean lies between the two group means, since it is built from both.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

Draw each group as a rectangle: how many students wide, how many points tall, so its area is that group's total score. The overall mean is a rectangle of the same total area over the whole width, which turns the question into a balance.

3 · Execute5 carry out the plan

1Picture each mean as a rectangle's area

#1 Draw a Diagram 6.SP.A.3

A group of n students averaging m points has a total of n times m -- the area of an n-by-m rectangle.

\text{total score} = \text{students} \times \text{mean}

Means become heights and counts become widths.

2Find the area of the whole class rectangle

#7 Identify Subproblems 6.NS.B.3

All 80 students at a mean of 74.

80 \times 74 = 5920

5920 points between everyone.

3Compare the overall mean to each group's mean

#1 Draw a Diagram 6.SP.B.5

The overall mean sits 2 above the boys' and 3 below the girls'. Every boy is short of the overall mean by 2, and every girl is over it by 3.

74 - 72 = 2,\quad 77 - 74 = 3

The girls' surplus has to cover the boys' shortfall exactly.

4Balance the two pieces to split the 80 students

#6 Guess and Check 6.RP.A.3

If there are g girls then 80 minus g boys, and the surplus must equal the shortfall.

g \times 3 = (80 - g) \times 2 \Rightarrow g = 32

32 girls and 48 boys.

5Check the totals rebuild 5920

#7 Identify Subproblems 6.NS.B.3

Add the two group totals back up.

48 \times 72 + 32 \times 77 = 5920

It matches, so 32 girls is right.

Answer: 32 girls

4 · Reviewdoes it hold up?

The overall mean 74 is nearer the boys' mean, so that group should be the larger -- and it is, 48 against 32.

Another way: Work in totals instead: the boys alone would give 5760, which is 160 short, and each girl swapped in adds 5 -- again 32.

Standardsmin grade 6

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Reading each mean as a total shared over a count.
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Measuring how far the overall mean sits from each group's.
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems — Balancing surplus against shortfall to split the group.
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals — The decimal multiplications on both sides of the check.

💡Takeaway. An overall mean sits closer to the bigger group. How far it leans tells you how the group splits.

Variant 9 hard

At one school, $60$ fifth-grade students took part in a math competition. The overall mean score was $83$ points. If the boys' mean score was $82$ points and the girls' mean score was $85$ points, find how many girls took part in the math competition.

Your answer

Show solution

1 · Understandwhat's really being asked

60 students averaged 83 points. Split by sex, the boys averaged 82 and the girls 85. We need the number of girls.

Givens

60 students in all, mean 83.
Boys' mean: 82.
Girls' mean: 85.

Unknowns

How many girls took part.

Constraints

The overall mean lies between the two group means, since it is built from both.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

Draw each group as a rectangle: how many students wide, how many points tall, so its area is that group's total score. The overall mean is a rectangle of the same total area over the whole width, which turns the question into a balance.

3 · Execute5 carry out the plan

1Picture each mean as a rectangle's area

#1 Draw a Diagram 6.SP.A.3

A group of n students averaging m points has a total of n times m -- the area of an n-by-m rectangle.

\text{total score} = \text{students} \times \text{mean}

Means become heights and counts become widths.

2Find the area of the whole class rectangle

#7 Identify Subproblems 6.NS.B.3

All 60 students at a mean of 83.

60 \times 83 = 4980

4980 points between everyone.

3Compare the overall mean to each group's mean

#1 Draw a Diagram 6.SP.B.5

The overall mean sits 1 above the boys' and 2 below the girls'. Every boy is short of the overall mean by 1, and every girl is over it by 2.

83 - 82 = 1,\quad 85 - 83 = 2

The girls' surplus has to cover the boys' shortfall exactly.

4Balance the two pieces to split the 60 students

#6 Guess and Check 6.RP.A.3

If there are g girls then 60 minus g boys, and the surplus must equal the shortfall.

g \times 2 = (60 - g) \times 1 \Rightarrow g = 20

20 girls and 40 boys.

5Check the totals rebuild 4980

#7 Identify Subproblems 6.NS.B.3

Add the two group totals back up.

40 \times 82 + 20 \times 85 = 4980

It matches, so 20 girls is right.

Answer: 20 girls

4 · Reviewdoes it hold up?

The overall mean 83 is nearer the boys' mean, so that group should be the larger -- and it is, 40 against 20.

Another way: Work in totals instead: the boys alone would give 4920, which is 60 short, and each girl swapped in adds 3 -- again 20.

Standardsmin grade 6

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Reading each mean as a total shared over a count.
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Measuring how far the overall mean sits from each group's.
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems — Balancing surplus against shortfall to split the group.
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals — The decimal multiplications on both sides of the check.

💡Takeaway. An overall mean sits closer to the bigger group. How far it leans tells you how the group splits.

Variant 10 hard

At one school, $48$ fifth-grade students took part in a math competition. The overall mean score was $83.5$ points. If the boys' mean score was $81$ points and the girls' mean score was $86$ points, find how many girls took part in the math competition.

Your answer

Show solution

1 · Understandwhat's really being asked

48 students averaged 83.5 points. Split by sex, the boys averaged 81 and the girls 86. We need the number of girls.

Givens

48 students in all, mean 83.5.
Boys' mean: 81.
Girls' mean: 86.

Unknowns

How many girls took part.

Constraints

The overall mean lies between the two group means, since it is built from both.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

Draw each group as a rectangle: how many students wide, how many points tall, so its area is that group's total score. The overall mean is a rectangle of the same total area over the whole width, which turns the question into a balance.

3 · Execute5 carry out the plan

1Picture each mean as a rectangle's area

#1 Draw a Diagram 6.SP.A.3

A group of n students averaging m points has a total of n times m -- the area of an n-by-m rectangle.

\text{total score} = \text{students} \times \text{mean}

Means become heights and counts become widths.

2Find the area of the whole class rectangle

#7 Identify Subproblems 6.NS.B.3

All 48 students at a mean of 83.5.

48 \times 83.5 = 4008

4008 points between everyone.

3Compare the overall mean to each group's mean

#1 Draw a Diagram 6.SP.B.5

The overall mean sits 2.5 above the boys' and 2.5 below the girls'. Every boy is short of the overall mean by 2.5, and every girl is over it by 2.5.

83.5 - 81 = 2.5,\quad 86 - 83.5 = 2.5

The girls' surplus has to cover the boys' shortfall exactly.

4Balance the two pieces to split the 48 students

#6 Guess and Check 6.RP.A.3

If there are g girls then 48 minus g boys, and the surplus must equal the shortfall.

g \times 2.5 = (48 - g) \times 2.5 \Rightarrow g = 24

24 girls and 24 boys.

5Check the totals rebuild 4008

#7 Identify Subproblems 6.NS.B.3

Add the two group totals back up.

24 \times 81 + 24 \times 86 = 4008

It matches, so 24 girls is right.

Answer: 24 girls

4 · Reviewdoes it hold up?

The overall mean 83.5 is nearer the girls' mean, so that group should be the larger -- and it is, 24 against 24.

Another way: Work in totals instead: the boys alone would give 3888, which is 120 short, and each girl swapped in adds 5 -- again 24.

Standardsmin grade 6

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Reading each mean as a total shared over a count.
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Measuring how far the overall mean sits from each group's.
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems — Balancing surplus against shortfall to split the group.
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals — The decimal multiplications on both sides of the check.

💡Takeaway. An overall mean sits closer to the bigger group. How far it leans tells you how the group splits.

Variant 11 hard

At one school, $90$ fifth-grade students took part in a math competition. The overall mean score was $64$ points. If the boys' mean score was $62$ points and the girls' mean score was $68$ points, find how many girls took part in the math competition.

Your answer

Show solution

1 · Understandwhat's really being asked

90 students averaged 64 points. Split by sex, the boys averaged 62 and the girls 68. We need the number of girls.

Givens

90 students in all, mean 64.
Boys' mean: 62.
Girls' mean: 68.

Unknowns

How many girls took part.

Constraints

The overall mean lies between the two group means, since it is built from both.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

Draw each group as a rectangle: how many students wide, how many points tall, so its area is that group's total score. The overall mean is a rectangle of the same total area over the whole width, which turns the question into a balance.

3 · Execute5 carry out the plan

1Picture each mean as a rectangle's area

#1 Draw a Diagram 6.SP.A.3

A group of n students averaging m points has a total of n times m -- the area of an n-by-m rectangle.

\text{total score} = \text{students} \times \text{mean}

Means become heights and counts become widths.

2Find the area of the whole class rectangle

#7 Identify Subproblems 6.NS.B.3

All 90 students at a mean of 64.

90 \times 64 = 5760

5760 points between everyone.

3Compare the overall mean to each group's mean

#1 Draw a Diagram 6.SP.B.5

The overall mean sits 2 above the boys' and 4 below the girls'. Every boy is short of the overall mean by 2, and every girl is over it by 4.

64 - 62 = 2,\quad 68 - 64 = 4

The girls' surplus has to cover the boys' shortfall exactly.

4Balance the two pieces to split the 90 students

#6 Guess and Check 6.RP.A.3

If there are g girls then 90 minus g boys, and the surplus must equal the shortfall.

g \times 4 = (90 - g) \times 2 \Rightarrow g = 30

30 girls and 60 boys.

5Check the totals rebuild 5760

#7 Identify Subproblems 6.NS.B.3

Add the two group totals back up.

60 \times 62 + 30 \times 68 = 5760

It matches, so 30 girls is right.

Answer: 30 girls

4 · Reviewdoes it hold up?

The overall mean 64 is nearer the boys' mean, so that group should be the larger -- and it is, 60 against 30.

Another way: Work in totals instead: the boys alone would give 5580, which is 180 short, and each girl swapped in adds 6 -- again 30.

Standardsmin grade 6

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Reading each mean as a total shared over a count.
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Measuring how far the overall mean sits from each group's.
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems — Balancing surplus against shortfall to split the group.
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals — The decimal multiplications on both sides of the check.

💡Takeaway. An overall mean sits closer to the bigger group. How far it leans tells you how the group splits.

Variant 12 hard

At one school, $100$ fifth-grade students took part in a math competition. The overall mean score was $79.2$ points. If the boys' mean score was $78$ points and the girls' mean score was $81$ points, find how many girls took part in the math competition.

Your answer

Show solution

1 · Understandwhat's really being asked

100 students averaged 79.2 points. Split by sex, the boys averaged 78 and the girls 81. We need the number of girls.

Givens

100 students in all, mean 79.2.
Boys' mean: 78.
Girls' mean: 81.

Unknowns

How many girls took part.

Constraints

The overall mean lies between the two group means, since it is built from both.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

Draw each group as a rectangle: how many students wide, how many points tall, so its area is that group's total score. The overall mean is a rectangle of the same total area over the whole width, which turns the question into a balance.

3 · Execute5 carry out the plan

1Picture each mean as a rectangle's area

#1 Draw a Diagram 6.SP.A.3

A group of n students averaging m points has a total of n times m -- the area of an n-by-m rectangle.

\text{total score} = \text{students} \times \text{mean}

Means become heights and counts become widths.

2Find the area of the whole class rectangle

#7 Identify Subproblems 6.NS.B.3

All 100 students at a mean of 79.2.

100 \times 79.2 = 7920

7920 points between everyone.

3Compare the overall mean to each group's mean

#1 Draw a Diagram 6.SP.B.5

The overall mean sits 1.2 above the boys' and 1.8 below the girls'. Every boy is short of the overall mean by 1.2, and every girl is over it by 1.8.

79.2 - 78 = 1.2,\quad 81 - 79.2 = 1.8

The girls' surplus has to cover the boys' shortfall exactly.

4Balance the two pieces to split the 100 students

#6 Guess and Check 6.RP.A.3

If there are g girls then 100 minus g boys, and the surplus must equal the shortfall.

g \times 1.8 = (100 - g) \times 1.2 \Rightarrow g = 40

40 girls and 60 boys.

5Check the totals rebuild 7920

#7 Identify Subproblems 6.NS.B.3

Add the two group totals back up.

60 \times 78 + 40 \times 81 = 7920

It matches, so 40 girls is right.

Answer: 40 girls

4 · Reviewdoes it hold up?

The overall mean 79.2 is nearer the boys' mean, so that group should be the larger -- and it is, 60 against 40.

Another way: Work in totals instead: the boys alone would give 7800, which is 120 short, and each girl swapped in adds 3 -- again 40.

Standardsmin grade 6

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Reading each mean as a total shared over a count.
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Measuring how far the overall mean sits from each group's.
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems — Balancing surplus against shortfall to split the group.
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals — The decimal multiplications on both sides of the check.

💡Takeaway. An overall mean sits closer to the bigger group. How far it leans tells you how the group splits.

Model sum as rectangle area for means · 12 practice problems

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