A team of employees completed a large project. They were recognized with a monetary bonus to share among themselves. The ages of the team members are consecutive integers and no one on the team has the same age. The oldest team member is \(45\).
The bonus was paid out as follows:
\(\$1000\) to the oldest member of the project team plus \(\frac{1}{10}\) of what remains, then
\(\$2000\) to the second oldest member of the project team plus \(\frac{1}{10}\) of what then remains, then
\(\$3000\) to the third oldest member of the project team plus \(\frac{1}{10}\) of what then remains, and so on.
After all of the bonus money had been distributed, each member of the project team had received the same amount and no money was left over. What is the age of the youngest member of the project team?
Solution 1
Let \(x\) represent the amount that
each team member receives.
Let \(y\) represent the total amount of
the bonus.
Then \(y \div x\) is the number of team
members.
The first team member gets \(\$1000\) plus one-tenth of the remainder. Therefore, \[x = 1000 + \frac{1}{10}(y - 1000)\] Multiplying both sides by \(10\), \[10x = 10\,000 + y - 1000\] Simplifying and solving for \(y\), we have \[y = 10x - 9000\]
The second team member gets \(\$2000\) plus one-tenth of the remainder after the first team member’s share and \(\$2000\) is removed. Therefore, \[x = 2000+\frac{1}{10}(y - x - 2000)\] Multiplying both sides by \(10\), \[10x = 20\,000 + y - x - 2000\] Simplifying and solving for \(y\), we have \[y = 11x - 18\,000\]
Since \(y = 10x - 9000\) and \(y = 11x - 18\,000\), we have \[\begin{aligned} 10x - 9 000 &= 11x - 18\,000\\ x&=9000 \end{aligned}\] Therefore, \(y=10x-9000 = 10(9000) - 9000 =81\,000\).
Therefore, each team member receives \(\$9000\) and the total bonus is \(\$81\,000\). The number of team members is \(y \div x = 81\,000 \div 9000 = 9\). The oldest team member is \(45\) and the ages of the team members are consecutive integers, so the youngest team member is \(37\). The ages of the team members are \(37\), \(38\), \(39\), \(40\), \(41\), \(42\), \(43\), \(44\), \(45\).
Solution 2
Let \(n\) represent the number of team members.
Team member \(n\) (the youngest team member) receives \(n \times \$1 000\) or \(1000n\).
So \(1000n\) is \(\frac{9}{10}\) of what remains after team member \((n-1)\) is given \((n-1) \times 1000\).
Therefore, \(\frac{1}{10}\) of what remains is \(\frac{1000n}{9}.\)
Then team member \((n-1)\) receives \(1000(n-1) + \frac{1000n}{9}\).
But team member \((n-1)\) and team member \(n\) each receive the same amount.
Therefore, \[\begin{aligned} 1000(n-1) + \frac{1000n}{9}&=1000n\\[2mm] 1000n - 1000 + \frac{1000n}{9}&=1000n\\[2mm] \frac{1000n}{9}&=1000\\[2mm] 1000n &= 9000\\ n&=9 \end{aligned}\] Therefore, there are \(9\) team members and each receives \(\$1000n = \$1000(9) = \$9000\). Since the oldest team member is \(45\) and the ages of the team members are consecutive integers, the youngest team member is \(37\).