Extra money is always a good thing. But sometimes you need a helping hand from maths to make sure everyone gets an equal share of the extra money.
A music school has managed to come in under budget this year. It’s great news, because it means the surplus can be distributed in the form of vouchers to six music groups in the school.
There are four trios and two quartets in total. The only problem is that the school board didn’t really think it through, and spent the money on different amounts in vouchers. The vouchers are in amounts of €10, €20, €30, €40, €50, €60, €70, €80 and €90.
The principal doesn’t think any group should get more than another; they all worked so hard all year and deserve a fair budgetary treat. He considered the musicians that made up the groups. Anna, Ben, Carl, Daisy, Edward, Felicity, Gwen, Harry and Irene were all so committed to their music. Just nine people made up all six groups, and they all played in at least two groups each. That was commitment.
He thought about the nine students, and looked at the nine vouchers of differing amounts. “If only there was a way I could divide the money up evenly,” he said to himself.
Then he hatched a plan to give the vouchers out to the musicians in such a way that meant each group had an equal share at the end. But first he needed to figure out which group had who, so he drew a chart.
“That’s it,” he said, after poring over the chart for the evening. “That’s how I’ll give the vouchers out.”
How did the music principal give out the vouchers to ensure every group got an equal share of money?
Scroll down for the solution to this week’s puzzle.
The principal decided to create an equation to fairly distribute the vouchers. First he thought about the trios. He knew that Anna was in all four trios, so he decided to take her out of the equation to see what he was left with, letting s represent what each club would get:
s – a = b + f = c + g = d + h = e + i
Then he added the funds of the two quartets together:
2s = (b + c + d + e) + (f + g + h + i) = (b + f) + (c + g) + (d + h) + (e + i)
He knew that meant:
2s = 4(s – a) which, when solved, gave s = 2a.
He knew the nine vouchers added up to €450, which was the total sum of the two quartets plus whatever funding he would need to give Anna.
s + s + a = 2a + 2a + a = 5a
And 5a = 450
“So I’ll give Anna the €90 voucher,” he said triumphantly. “Now I just have to divide up the rest. If the total sum of funds for each club is twice Anna’s amount, then the amount each club gets must be €180.”
He knew, with Anna holding the €90 voucher, the other two members of each trio would have to make up the other €90, which would be:
10 + 80 = 20 + 70 = 30 + 60 = 40 + 50 = 90
He also made sure the two quartets total funding worked out:
10 + 40 + 60 + 70 = 180
80 + 50 + 30 + 20 = 180
He leaned back in his chair and smiled, thinking of all the new instruments the music groups could get.
This question was inspired by a problem from the 1st Round of the Austrian Mathematics Olympiad. The International Mathematical Olympiad 2017 will take place in Rio de Janeiro on 12-23 July 2017. Secondary school students in Ireland who wish to participate in Olympiads are encouraged to attend free Mathematics Enrichment Programmes organised as part of the Irish Mathematical Olympiad.
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