Problem
Average means the total shared out evenly, so I can rebuild the total: multiply the average age by how many students there were.
If everyone were exactly 34, all 54 of them together would be 54 thirty-fours — that combined pile of years is the same whether ages are equal or not.
6.SP.B.5Work BackwardsNow there are 60 students whose average is 33, so I rebuild their total age the same way.
The new average of 33 already accounts for everyone, so 60 thirty-threes is the whole class's age added up.
6.SP.B.5Work BackwardsThe only thing that changed the total was the 6 new students, so their combined age is the new total minus the old total.
Take away the years that were already there, and what's left is exactly the years the newcomers brought.
6.SP.A.3Identify SubproblemsThe six newcomers' ages together are the new total less the old total.
The only thing that changed the class total was the newcomers joining, so the gap between the two totals is exactly what they brought with them.
Each total can be rebuilt from its average, because an average is that total handed out in equal shares to the group.
Their 144 combined years are shared among 6 people, so divide to get the average for one newcomer.
An average is a fair share: split the 144 years equally among the 6 students and each 'gets' 24.
6.SP.A.3Analyze The UnitsTo redo an average after the group changes, turn each average back into a total (average × count), compare the totals, then share the difference out evenly.