← Recompute mean after adding or removing data · Average from Total

Recompute mean after adding or removing data · 12 practice problems

6.SP.B.56.SP.A.3

Generated variants — 12

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

Variant 1 easy answer: 30 years

A swimming class has $24$ students, and their average age is $20$ years. Then $6$ new students join the class, and the average age becomes $22$ years. Find the average age, in years, of the $6$ new students.

Show solution

1 · Understandwhat's really being asked

24 students average 20 years old. 6 more join, and the average for all 30 becomes 22. We need the average age of just the 6 newcomers.

Givens

24 students, average age 20.
6 new students join.
The average across all 30 is then 22.

Unknowns

The average age of the 6 new students.

Constraints

The newcomers' average cannot be found by averaging the two class averages -- the groups are different sizes.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#8 Analyze the Units

Averages do not add, but totals do. Turn each average back into a total of years, and the difference between them belongs entirely to the newcomers.

3 · Execute4 carry out the plan

1Total age of the original 24 students

#11 Work Backwards 6.SP.B.5

An average of 20 over 24 students means their ages add to 24 lots of 20.

24 \times 20 = 480

480 years between them.

2Total age of all 30 students afterward

#11 Work Backwards 6.SP.B.5

The same move on the new average, now over 30 students.

30 \times 22 = 660

660 years in the bigger class.

3Combined age the 6 newcomers added

#7 Identify Subproblems 6.SP.A.3

The only difference between the two totals is the newcomers, so subtracting isolates them.

660 - 480 = 180

The 6 newcomers bring 180 years between them.

4Average age of the 6 new students

#8 Analyze the Units 6.SP.A.3

Share those 180 years evenly among the 6 of them.

180 \div 6 = 30

They average 30 years old.

Answer: 30 years

4 · Reviewdoes it hold up?

The average rose from 20 to 22, so the newcomers must be older than 20 -- and 30 is.

Another way: Think in shifts: each of the 24 originals gains 2 year of average, and the newcomers supply it.

Standardsmin grade 6

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Turning each stated average back into a total.
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Isolating the newcomers' combined age and averaging it.

💡Takeaway. You cannot average two averages of different-sized groups. Turn each one back into a total, and totals behave.

Variant 2 easy answer: 27 years

A swimming class has $30$ students, and their average age is $15$ years. Then $10$ new students join the class, and the average age becomes $18$ years. Find the average age, in years, of the $10$ new students.

Show solution

1 · Understandwhat's really being asked

30 students average 15 years old. 10 more join, and the average for all 40 becomes 18. We need the average age of just the 10 newcomers.

Givens

30 students, average age 15.
10 new students join.
The average across all 40 is then 18.

Unknowns

The average age of the 10 new students.

Constraints

The newcomers' average cannot be found by averaging the two class averages -- the groups are different sizes.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#8 Analyze the Units

Averages do not add, but totals do. Turn each average back into a total of years, and the difference between them belongs entirely to the newcomers.

3 · Execute4 carry out the plan

1Total age of the original 30 students

#11 Work Backwards 6.SP.B.5

An average of 15 over 30 students means their ages add to 30 lots of 15.

30 \times 15 = 450

450 years between them.

2Total age of all 40 students afterward

#11 Work Backwards 6.SP.B.5

The same move on the new average, now over 40 students.

40 \times 18 = 720

720 years in the bigger class.

3Combined age the 10 newcomers added

#7 Identify Subproblems 6.SP.A.3

The only difference between the two totals is the newcomers, so subtracting isolates them.

720 - 450 = 270

The 10 newcomers bring 270 years between them.

4Average age of the 10 new students

#8 Analyze the Units 6.SP.A.3

Share those 270 years evenly among the 10 of them.

270 \div 10 = 27

They average 27 years old.

Answer: 27 years

4 · Reviewdoes it hold up?

The average rose from 15 to 18, so the newcomers must be older than 15 -- and 27 is.

Another way: Think in shifts: each of the 30 originals gains 3 year of average, and the newcomers supply it.

Standardsmin grade 6

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Turning each stated average back into a total.
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Isolating the newcomers' combined age and averaging it.

💡Takeaway. You cannot average two averages of different-sized groups. Turn each one back into a total, and totals behave.

Variant 3 easy answer: 35 years

A swimming class has $36$ students, and their average age is $25$ years. Then $4$ new students join the class, and the average age becomes $26$ years. Find the average age, in years, of the $4$ new students.

Show solution

1 · Understandwhat's really being asked

36 students average 25 years old. 4 more join, and the average for all 40 becomes 26. We need the average age of just the 4 newcomers.

Givens

36 students, average age 25.
4 new students join.
The average across all 40 is then 26.

Unknowns

The average age of the 4 new students.

Constraints

The newcomers' average cannot be found by averaging the two class averages -- the groups are different sizes.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#8 Analyze the Units

Averages do not add, but totals do. Turn each average back into a total of years, and the difference between them belongs entirely to the newcomers.

3 · Execute4 carry out the plan

1Total age of the original 36 students

#11 Work Backwards 6.SP.B.5

An average of 25 over 36 students means their ages add to 36 lots of 25.

36 \times 25 = 900

900 years between them.

2Total age of all 40 students afterward

#11 Work Backwards 6.SP.B.5

The same move on the new average, now over 40 students.

40 \times 26 = 1040

1040 years in the bigger class.

3Combined age the 4 newcomers added

#7 Identify Subproblems 6.SP.A.3

The only difference between the two totals is the newcomers, so subtracting isolates them.

1040 - 900 = 140

The 4 newcomers bring 140 years between them.

4Average age of the 4 new students

#8 Analyze the Units 6.SP.A.3

Share those 140 years evenly among the 4 of them.

140 \div 4 = 35

They average 35 years old.

Answer: 35 years

4 · Reviewdoes it hold up?

The average rose from 25 to 26, so the newcomers must be older than 25 -- and 35 is.

Another way: Think in shifts: each of the 36 originals gains 1 year of average, and the newcomers supply it.

Standardsmin grade 6

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Turning each stated average back into a total.
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Isolating the newcomers' combined age and averaging it.

💡Takeaway. You cannot average two averages of different-sized groups. Turn each one back into a total, and totals behave.

Variant 4 easy answer: 20 years

A swimming class has $40$ students, and their average age is $30$ years. Then $10$ new students join the class, and the average age becomes $28$ years. Find the average age, in years, of the $10$ new students.

Show solution

1 · Understandwhat's really being asked

40 students average 30 years old. 10 more join, and the average for all 50 becomes 28. We need the average age of just the 10 newcomers.

Givens

40 students, average age 30.
10 new students join.
The average across all 50 is then 28.

Unknowns

The average age of the 10 new students.

Constraints

The newcomers' average cannot be found by averaging the two class averages -- the groups are different sizes.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#8 Analyze the Units

Averages do not add, but totals do. Turn each average back into a total of years, and the difference between them belongs entirely to the newcomers.

3 · Execute4 carry out the plan

1Total age of the original 40 students

#11 Work Backwards 6.SP.B.5

An average of 30 over 40 students means their ages add to 40 lots of 30.

40 \times 30 = 1200

1200 years between them.

2Total age of all 50 students afterward

#11 Work Backwards 6.SP.B.5

The same move on the new average, now over 50 students.

50 \times 28 = 1400

1400 years in the bigger class.

3Combined age the 10 newcomers added

#7 Identify Subproblems 6.SP.A.3

The only difference between the two totals is the newcomers, so subtracting isolates them.

1400 - 1200 = 200

The 10 newcomers bring 200 years between them.

4Average age of the 10 new students

#8 Analyze the Units 6.SP.A.3

Share those 200 years evenly among the 10 of them.

200 \div 10 = 20

They average 20 years old.

Answer: 20 years

4 · Reviewdoes it hold up?

The average fell from 30 to 28, so the newcomers must be younger than 30 -- and 20 is.

Another way: Think in shifts: each of the 40 originals gives up 2 year of average, which is 80 years in all, and that is what the newcomers fall short by.

Standardsmin grade 6

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Turning each stated average back into a total.
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Isolating the newcomers' combined age and averaging it.

💡Takeaway. You cannot average two averages of different-sized groups. Turn each one back into a total, and totals behave.

Variant 5 medium answer: 31 years

A swimming class has $48$ students, and their average age is $41$ years. Then $12$ new students join the class, and the average age becomes $39$ years. Find the average age, in years, of the $12$ new students.

Show solution

1 · Understandwhat's really being asked

48 students average 41 years old. 12 more join, and the average for all 60 becomes 39. We need the average age of just the 12 newcomers.

Givens

48 students, average age 41.
12 new students join.
The average across all 60 is then 39.

Unknowns

The average age of the 12 new students.

Constraints

The newcomers' average cannot be found by averaging the two class averages -- the groups are different sizes.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#8 Analyze the Units

Averages do not add, but totals do. Turn each average back into a total of years, and the difference between them belongs entirely to the newcomers.

3 · Execute4 carry out the plan

1Total age of the original 48 students

#11 Work Backwards 6.SP.B.5

An average of 41 over 48 students means their ages add to 48 lots of 41.

48 \times 41 = 1968

1968 years between them.

2Total age of all 60 students afterward

#11 Work Backwards 6.SP.B.5

The same move on the new average, now over 60 students.

60 \times 39 = 2340

2340 years in the bigger class.

3Combined age the 12 newcomers added

#7 Identify Subproblems 6.SP.A.3

The only difference between the two totals is the newcomers, so subtracting isolates them.

2340 - 1968 = 372

The 12 newcomers bring 372 years between them.

4Average age of the 12 new students

#8 Analyze the Units 6.SP.A.3

Share those 372 years evenly among the 12 of them.

372 \div 12 = 31

They average 31 years old.

Answer: 31 years

4 · Reviewdoes it hold up?

The average fell from 41 to 39, so the newcomers must be younger than 41 -- and 31 is.

Another way: Think in shifts: each of the 48 originals gives up 2 year of average, which is 96 years in all, and that is what the newcomers fall short by.

Standardsmin grade 6

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Turning each stated average back into a total.
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Isolating the newcomers' combined age and averaging it.

💡Takeaway. You cannot average two averages of different-sized groups. Turn each one back into a total, and totals behave.

Variant 6 medium answer: 30 years

A swimming class has $45$ students, and their average age is $50$ years. Then $5$ new students join the class, and the average age becomes $48$ years. Find the average age, in years, of the $5$ new students.

Show solution

1 · Understandwhat's really being asked

45 students average 50 years old. 5 more join, and the average for all 50 becomes 48. We need the average age of just the 5 newcomers.

Givens

45 students, average age 50.
5 new students join.
The average across all 50 is then 48.

Unknowns

The average age of the 5 new students.

Constraints

The newcomers' average cannot be found by averaging the two class averages -- the groups are different sizes.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#8 Analyze the Units

Averages do not add, but totals do. Turn each average back into a total of years, and the difference between them belongs entirely to the newcomers.

3 · Execute4 carry out the plan

1Total age of the original 45 students

#11 Work Backwards 6.SP.B.5

An average of 50 over 45 students means their ages add to 45 lots of 50.

45 \times 50 = 2250

2250 years between them.

2Total age of all 50 students afterward

#11 Work Backwards 6.SP.B.5

The same move on the new average, now over 50 students.

50 \times 48 = 2400

2400 years in the bigger class.

3Combined age the 5 newcomers added

#7 Identify Subproblems 6.SP.A.3

The only difference between the two totals is the newcomers, so subtracting isolates them.

2400 - 2250 = 150

The 5 newcomers bring 150 years between them.

4Average age of the 5 new students

#8 Analyze the Units 6.SP.A.3

Share those 150 years evenly among the 5 of them.

150 \div 5 = 30

They average 30 years old.

Answer: 30 years

4 · Reviewdoes it hold up?

The average fell from 50 to 48, so the newcomers must be younger than 50 -- and 30 is.

Another way: Think in shifts: each of the 45 originals gives up 2 year of average, which is 90 years in all, and that is what the newcomers fall short by.

Standardsmin grade 6

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Turning each stated average back into a total.
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Isolating the newcomers' combined age and averaging it.

💡Takeaway. You cannot average two averages of different-sized groups. Turn each one back into a total, and totals behave.

Variant 7 medium answer: 24 years

A swimming class has $54$ students, and their average age is $34$ years. Then $6$ new students join the class, and the average age becomes $33$ years. Find the average age, in years, of the $6$ new students.

Show solution

1 · Understandwhat's really being asked

54 students average 34 years old. 6 more join, and the average for all 60 becomes 33. We need the average age of just the 6 newcomers.

Givens

54 students, average age 34.
6 new students join.
The average across all 60 is then 33.

Unknowns

The average age of the 6 new students.

Constraints

The newcomers' average cannot be found by averaging the two class averages -- the groups are different sizes.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#8 Analyze the Units

Averages do not add, but totals do. Turn each average back into a total of years, and the difference between them belongs entirely to the newcomers.

3 · Execute4 carry out the plan

1Total age of the original 54 students

#11 Work Backwards 6.SP.B.5

An average of 34 over 54 students means their ages add to 54 lots of 34.

54 \times 34 = 1836

1836 years between them.

2Total age of all 60 students afterward

#11 Work Backwards 6.SP.B.5

The same move on the new average, now over 60 students.

60 \times 33 = 1980

1980 years in the bigger class.

3Combined age the 6 newcomers added

#7 Identify Subproblems 6.SP.A.3

The only difference between the two totals is the newcomers, so subtracting isolates them.

1980 - 1836 = 144

The 6 newcomers bring 144 years between them.

4Average age of the 6 new students

#8 Analyze the Units 6.SP.A.3

Share those 144 years evenly among the 6 of them.

144 \div 6 = 24

They average 24 years old.

Answer: 24 years

4 · Reviewdoes it hold up?

The average fell from 34 to 33, so the newcomers must be younger than 34 -- and 24 is.

Another way: Think in shifts: each of the 54 originals gives up 1 year of average, which is 54 years in all, and that is what the newcomers fall short by.

Standardsmin grade 6

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Turning each stated average back into a total.
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Isolating the newcomers' combined age and averaging it.

💡Takeaway. You cannot average two averages of different-sized groups. Turn each one back into a total, and totals behave.

Variant 8 medium answer: 23 years

A swimming class has $56$ students, and their average age is $33$ years. Then $14$ new students join the class, and the average age becomes $31$ years. Find the average age, in years, of the $14$ new students.

Show solution

1 · Understandwhat's really being asked

56 students average 33 years old. 14 more join, and the average for all 70 becomes 31. We need the average age of just the 14 newcomers.

Givens

56 students, average age 33.
14 new students join.
The average across all 70 is then 31.

Unknowns

The average age of the 14 new students.

Constraints

The newcomers' average cannot be found by averaging the two class averages -- the groups are different sizes.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#8 Analyze the Units

Averages do not add, but totals do. Turn each average back into a total of years, and the difference between them belongs entirely to the newcomers.

3 · Execute4 carry out the plan

1Total age of the original 56 students

#11 Work Backwards 6.SP.B.5

An average of 33 over 56 students means their ages add to 56 lots of 33.

56 \times 33 = 1848

1848 years between them.

2Total age of all 70 students afterward

#11 Work Backwards 6.SP.B.5

The same move on the new average, now over 70 students.

70 \times 31 = 2170

2170 years in the bigger class.

3Combined age the 14 newcomers added

#7 Identify Subproblems 6.SP.A.3

The only difference between the two totals is the newcomers, so subtracting isolates them.

2170 - 1848 = 322

The 14 newcomers bring 322 years between them.

4Average age of the 14 new students

#8 Analyze the Units 6.SP.A.3

Share those 322 years evenly among the 14 of them.

322 \div 14 = 23

They average 23 years old.

Answer: 23 years

4 · Reviewdoes it hold up?

The average fell from 33 to 31, so the newcomers must be younger than 33 -- and 23 is.

Another way: Think in shifts: each of the 56 originals gives up 2 year of average, which is 112 years in all, and that is what the newcomers fall short by.

Standardsmin grade 6

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Turning each stated average back into a total.
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Isolating the newcomers' combined age and averaging it.

💡Takeaway. You cannot average two averages of different-sized groups. Turn each one back into a total, and totals behave.

Variant 9 hard answer: 30 years

A swimming class has $60$ students, and their average age is $45$ years. Then $15$ new students join the class, and the average age becomes $42$ years. Find the average age, in years, of the $15$ new students.

Show solution

1 · Understandwhat's really being asked

60 students average 45 years old. 15 more join, and the average for all 75 becomes 42. We need the average age of just the 15 newcomers.

Givens

60 students, average age 45.
15 new students join.
The average across all 75 is then 42.

Unknowns

The average age of the 15 new students.

Constraints

The newcomers' average cannot be found by averaging the two class averages -- the groups are different sizes.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#8 Analyze the Units

Averages do not add, but totals do. Turn each average back into a total of years, and the difference between them belongs entirely to the newcomers.

3 · Execute4 carry out the plan

1Total age of the original 60 students

#11 Work Backwards 6.SP.B.5

An average of 45 over 60 students means their ages add to 60 lots of 45.

60 \times 45 = 2700

2700 years between them.

2Total age of all 75 students afterward

#11 Work Backwards 6.SP.B.5

The same move on the new average, now over 75 students.

75 \times 42 = 3150

3150 years in the bigger class.

3Combined age the 15 newcomers added

#7 Identify Subproblems 6.SP.A.3

The only difference between the two totals is the newcomers, so subtracting isolates them.

3150 - 2700 = 450

The 15 newcomers bring 450 years between them.

4Average age of the 15 new students

#8 Analyze the Units 6.SP.A.3

Share those 450 years evenly among the 15 of them.

450 \div 15 = 30

They average 30 years old.

Answer: 30 years

4 · Reviewdoes it hold up?

The average fell from 45 to 42, so the newcomers must be younger than 45 -- and 30 is.

Another way: Think in shifts: each of the 60 originals gives up 3 year of average, which is 180 years in all, and that is what the newcomers fall short by.

Standardsmin grade 6

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Turning each stated average back into a total.
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Isolating the newcomers' combined age and averaging it.

💡Takeaway. You cannot average two averages of different-sized groups. Turn each one back into a total, and totals behave.

Variant 10 hard answer: 40 years

A swimming class has $28$ students, and their average age is $60$ years. Then $7$ new students join the class, and the average age becomes $56$ years. Find the average age, in years, of the $7$ new students.

Show solution

1 · Understandwhat's really being asked

28 students average 60 years old. 7 more join, and the average for all 35 becomes 56. We need the average age of just the 7 newcomers.

Givens

28 students, average age 60.
7 new students join.
The average across all 35 is then 56.

Unknowns

The average age of the 7 new students.

Constraints

The newcomers' average cannot be found by averaging the two class averages -- the groups are different sizes.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#8 Analyze the Units

Averages do not add, but totals do. Turn each average back into a total of years, and the difference between them belongs entirely to the newcomers.

3 · Execute4 carry out the plan

1Total age of the original 28 students

#11 Work Backwards 6.SP.B.5

An average of 60 over 28 students means their ages add to 28 lots of 60.

28 \times 60 = 1680

1680 years between them.

2Total age of all 35 students afterward

#11 Work Backwards 6.SP.B.5

The same move on the new average, now over 35 students.

35 \times 56 = 1960

1960 years in the bigger class.

3Combined age the 7 newcomers added

#7 Identify Subproblems 6.SP.A.3

The only difference between the two totals is the newcomers, so subtracting isolates them.

1960 - 1680 = 280

The 7 newcomers bring 280 years between them.

4Average age of the 7 new students

#8 Analyze the Units 6.SP.A.3

Share those 280 years evenly among the 7 of them.

280 \div 7 = 40

They average 40 years old.

Answer: 40 years

4 · Reviewdoes it hold up?

The average fell from 60 to 56, so the newcomers must be younger than 60 -- and 40 is.

Another way: Think in shifts: each of the 28 originals gives up 4 year of average, which is 112 years in all, and that is what the newcomers fall short by.

Standardsmin grade 6

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Turning each stated average back into a total.
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Isolating the newcomers' combined age and averaging it.

💡Takeaway. You cannot average two averages of different-sized groups. Turn each one back into a total, and totals behave.

Variant 11 hard answer: 18 years

A swimming class has $72$ students, and their average age is $38$ years. Then $8$ new students join the class, and the average age becomes $36$ years. Find the average age, in years, of the $8$ new students.

Show solution

1 · Understandwhat's really being asked

72 students average 38 years old. 8 more join, and the average for all 80 becomes 36. We need the average age of just the 8 newcomers.

Givens

72 students, average age 38.
8 new students join.
The average across all 80 is then 36.

Unknowns

The average age of the 8 new students.

Constraints

The newcomers' average cannot be found by averaging the two class averages -- the groups are different sizes.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#8 Analyze the Units

Averages do not add, but totals do. Turn each average back into a total of years, and the difference between them belongs entirely to the newcomers.

3 · Execute4 carry out the plan

1Total age of the original 72 students

#11 Work Backwards 6.SP.B.5

An average of 38 over 72 students means their ages add to 72 lots of 38.

72 \times 38 = 2736

2736 years between them.

2Total age of all 80 students afterward

#11 Work Backwards 6.SP.B.5

The same move on the new average, now over 80 students.

80 \times 36 = 2880

2880 years in the bigger class.

3Combined age the 8 newcomers added

#7 Identify Subproblems 6.SP.A.3

The only difference between the two totals is the newcomers, so subtracting isolates them.

2880 - 2736 = 144

The 8 newcomers bring 144 years between them.

4Average age of the 8 new students

#8 Analyze the Units 6.SP.A.3

Share those 144 years evenly among the 8 of them.

144 \div 8 = 18

They average 18 years old.

Answer: 18 years

4 · Reviewdoes it hold up?

The average fell from 38 to 36, so the newcomers must be younger than 38 -- and 18 is.

Another way: Think in shifts: each of the 72 originals gives up 2 year of average, which is 144 years in all, and that is what the newcomers fall short by.

Standardsmin grade 6

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Turning each stated average back into a total.
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Isolating the newcomers' combined age and averaging it.

💡Takeaway. You cannot average two averages of different-sized groups. Turn each one back into a total, and totals behave.

Variant 12 hard answer: 42 years

A swimming class has $80$ students, and their average age is $27$ years. Then $20$ new students join the class, and the average age becomes $30$ years. Find the average age, in years, of the $20$ new students.

Show solution

1 · Understandwhat's really being asked

80 students average 27 years old. 20 more join, and the average for all 100 becomes 30. We need the average age of just the 20 newcomers.

Givens

80 students, average age 27.
20 new students join.
The average across all 100 is then 30.

Unknowns

The average age of the 20 new students.

Constraints

The newcomers' average cannot be found by averaging the two class averages -- the groups are different sizes.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#8 Analyze the Units

Averages do not add, but totals do. Turn each average back into a total of years, and the difference between them belongs entirely to the newcomers.

3 · Execute4 carry out the plan

1Total age of the original 80 students

#11 Work Backwards 6.SP.B.5

An average of 27 over 80 students means their ages add to 80 lots of 27.

80 \times 27 = 2160

2160 years between them.

2Total age of all 100 students afterward

#11 Work Backwards 6.SP.B.5

The same move on the new average, now over 100 students.

100 \times 30 = 3000

3000 years in the bigger class.

3Combined age the 20 newcomers added

#7 Identify Subproblems 6.SP.A.3

The only difference between the two totals is the newcomers, so subtracting isolates them.

3000 - 2160 = 840

The 20 newcomers bring 840 years between them.

4Average age of the 20 new students

#8 Analyze the Units 6.SP.A.3

Share those 840 years evenly among the 20 of them.

840 \div 20 = 42

They average 42 years old.

Answer: 42 years

4 · Reviewdoes it hold up?

The average rose from 27 to 30, so the newcomers must be older than 27 -- and 42 is.

Another way: Think in shifts: each of the 80 originals gains 3 year of average, and the newcomers supply it.

Standardsmin grade 6

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures — Turning each stated average back into a total.
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number — Isolating the newcomers' combined age and averaging it.

💡Takeaway. You cannot average two averages of different-sized groups. Turn each one back into a total, and totals behave.