← Model sum as rectangle area for means · Average from Total

Model sum as rectangle area for means · 12 practice problems

6.SP.A.36.SP.B.56.RP.A.36.NS.B.3

Generated variants — 12

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

Variant 1 easy answer: 12 girls

At one school, 3030 fifth-grade students took part in a math competition. The overall mean score was 5757 points. If the boys' mean score was 5555 points and the girls' mean score was 6060 points, find how many girls took part in the math competition.

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.
total score=students×mean\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.
30×57=171030 \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=2,60−57=357 - 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×3=(30−g)×2⇒g=12g \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.
18×55+12×60=171018 \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 answer: 20 girls

At one school, 5050 fifth-grade students took part in a math competition. The overall mean score was 61.861.8 points. If the boys' mean score was 6161 points and the girls' mean score was 6363 points, find how many girls took part in the math competition.

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.
total score=students×mean\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.
50×61.8=309050 \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=0.8,63−61.8=1.261.8 - 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×1.2=(50−g)×0.8⇒g=20g \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.
30×61+20×63=309030 \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 answer: 12 girls

At one school, 3636 fifth-grade students took part in a math competition. The overall mean score was 6161 points. If the boys' mean score was 5959 points and the girls' mean score was 6565 points, find how many girls took part in the math competition.

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.
total score=students×mean\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.
36×61=219636 \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=2,65−61=461 - 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×4=(36−g)×2⇒g=12g \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.
24×59+12×65=219624 \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 answer: 10 girls

At one school, 2525 fifth-grade students took part in a math competition. The overall mean score was 6666 points. If the boys' mean score was 6464 points and the girls' mean score was 6969 points, find how many girls took part in the math competition.

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.
total score=students×mean\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.
25×66=165025 \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=2,69−66=366 - 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×3=(25−g)×2⇒g=10g \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.
15×64+10×69=165015 \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 answer: 18 girls

At one school, 4545 fifth-grade students took part in a math competition. The overall mean score was 67.667.6 points. If the boys' mean score was 6666 points and the girls' mean score was 7070 points, find how many girls took part in the math competition.

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.
total score=students×mean\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.
45×67.6=304245 \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=1.6,70−67.6=2.467.6 - 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×2.4=(45−g)×1.6⇒g=18g \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.
27×66+18×70=304227 \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 answer: 30 girls

At one school, 7575 fifth-grade students took part in a math competition. The overall mean score was 71.671.6 points. If the boys' mean score was 7070 points and the girls' mean score was 7474 points, find how many girls took part in the math competition.

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.
total score=students×mean\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.
75×71.6=537075 \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=1.6,74−71.6=2.471.6 - 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×2.4=(75−g)×1.6⇒g=30g \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.
45×70+30×74=537045 \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 answer: 16 girls

At one school, 4040 fifth-grade students took part in a math competition. The overall mean score was 7272 points. If the boys' mean score was 7070 points and the girls' mean score was 7575 points, find how many girls took part in the math competition.

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.
total score=students×mean\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.
40×72=288040 \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=2,75−72=372 - 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×3=(40−g)×2⇒g=16g \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.
24×70+16×75=288024 \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 answer: 32 girls

At one school, 8080 fifth-grade students took part in a math competition. The overall mean score was 7474 points. If the boys' mean score was 7272 points and the girls' mean score was 7777 points, find how many girls took part in the math competition.

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.
total score=students×mean\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.
80×74=592080 \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=2,77−74=374 - 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×3=(80−g)×2⇒g=32g \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.
48×72+32×77=592048 \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 answer: 20 girls

At one school, 6060 fifth-grade students took part in a math competition. The overall mean score was 8383 points. If the boys' mean score was 8282 points and the girls' mean score was 8585 points, find how many girls took part in the math competition.

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.
total score=students×mean\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.
60×83=498060 \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=1,85−83=283 - 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×2=(60−g)×1⇒g=20g \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.
40×82+20×85=498040 \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 answer: 24 girls

At one school, 4848 fifth-grade students took part in a math competition. The overall mean score was 83.583.5 points. If the boys' mean score was 8181 points and the girls' mean score was 8686 points, find how many girls took part in the math competition.

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.
total score=students×mean\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.
48×83.5=400848 \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=2.5,86−83.5=2.583.5 - 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×2.5=(48−g)×2.5⇒g=24g \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.
24×81+24×86=400824 \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 answer: 30 girls

At one school, 9090 fifth-grade students took part in a math competition. The overall mean score was 6464 points. If the boys' mean score was 6262 points and the girls' mean score was 6868 points, find how many girls took part in the math competition.

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.
total score=students×mean\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.
90×64=576090 \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=2,68−64=464 - 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×4=(90−g)×2⇒g=30g \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.
60×62+30×68=576060 \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 answer: 40 girls

At one school, 100100 fifth-grade students took part in a math competition. The overall mean score was 79.279.2 points. If the boys' mean score was 7878 points and the girls' mean score was 8181 points, find how many girls took part in the math competition.

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.
total score=students×mean\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.
100×79.2=7920100 \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=1.2,81−79.2=1.879.2 - 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×1.8=(100−g)×1.2⇒g=40g \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.
60×78+40×81=792060 \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.