Problem
Model sum as rectangle area for means
Fifty fifth-graders took a test. Everyone together averaged 61.8 points. The boys averaged 61 and the girls averaged 63. I need to find how many of the 50 students were girls.
Data & probability
Your answer
How to solve
Strategy Draw a Diagram — A mean is the total of all scores divided by how many people there are, so (total) = (count) x (mean). That product is exactly the area of a rectangle whose width is the count and whose height is the mean. Drawing the boys' rectangle and the girls' rectangle side by side turns the whole question into 'fit two rectangles into one big rectangle of area 50 x 61.8.' I break that into subproblems (find the big total, then split it), and I can confirm the split with a quick guess-and-check.
1STEP 1
Picture each mean as a rectangle's area
Draw each group as a rectangle: width the head count, height the mean, area the group's total.
total = count × mean
A sixth grader knows the mean is the 'fair share' each person would get, so giving every person that share back rebuilds the total. Area = width x height is the same multiplication, just drawn.
6.SP.A.3Draw A Diagram2STEP 2
Find the area of the whole class rectangle
The whole class is a rectangle 50 wide and 61.8 tall, so its area is 3090 points.
50 × 61.8 = 3090
Knowing the average and the number of people, I can run the average backwards to recover the whole pile of points the class earned.
6.NS.B.3Identify Subproblems3STEP 3
Compare the overall mean to each group's mean
The girls stand 1.2 above the class mean while the boys sit 0.8 below it.
61.8 - 61 = 0.8, 63 - 61.8 = 1.2
The average is a balance point: how high the girls float above it, times how many girls, must equal how far the boys sit below it, times how many boys.
6.SP.B.5Draw A Diagram4STEP 4
Balance the two pieces to split the 50 students
Balancing 1.2 times the girls against 0.8 times the boys splits the 50 in the ratio 2 : 3.
1.2 × g = 0.8 × b → g : b = 2 : 3 → g = 2/5 × 50 = 20
The group that is farther from the average needs fewer people to pull the balance; the closer group needs more. So the smaller distance (boys, 0.8) goes with the bigger count (boys, 30).
6.RP.A.3Guess And CheckThe amount the girls sit above the class mean exactly cancels the amount the boys sit below it.
Why?
The total of everyone's scores comes out the same whether it is counted group by group or all in one go.
🧱Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.
Why?
Measuring every score from the class mean instead of from zero, what is left over has to come to nothing.
Why?
The class total is the mean taken once for every student, so removing one mean per student leaves zero behind.
🧱Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.
5STEP 5
Check the totals rebuild 3090
Rebuilding the totals gives 1260 + 1830 = 3090, the same class total as before.
30 × 61 + 20 × 63 = 1830 + 1260 = 3090
If the two smaller rectangles really fill the big one, their areas must add back to the class total.
6.NS.B.3Identify SubproblemsAnswer
20 girls
The overall mean 61.8 is much closer to the boys' 61 than to the girls' 63, so there should be more boys than girls. Getting 30 boys and 20 girls fits that, and the rebuilt total 3090 divided by 50 gives exactly 61.8, so the answer is consistent.
Takeaway
A mean is just 'count times mean = total,' which is the area of a rectangle — so mixing two groups is really fitting two rectangles into one.
- Picture each mean as a rectangle's area
- Find the area of the whole class rectangle
- Compare the overall mean to each group's mean
- Balance the two pieces to split the 50 students
- Check the totals rebuild 3090
Where next?
Another one like thissuggested
Step back and solidifysuggested