Nice Littler Earner
Smith and Jones were hired at the same time by Stainsbury’s Superdupermarket, with a starting salary of £10,000 per year. Every six months, Smith’s pay rose by £500 compared with that for the previous 6-month period. Every year, Jones’s pay rose by £1,600 compared with that for the previous 12-month period.
Three years later, who had earned more?
Iain
Jones? Or is this one of those ones where I miss something entirely elementary and go off in the wrong direction?
Smith = 13600
Jones = 14800
16 Nov 2009, 16:11
Steve Rumsby
I think the question is about overall earnings over the period, rather than just the final salary. Does that help?
16 Nov 2009, 16:39
Simon Whitehouse
Smith has earnt more.
Jones has had an average pay of £11,600 and so been paid a total of £34,800 over three years. Smith has earnt an average of £12,500 and so been paid a total of £37,500 over three years.
The question says that “Every six months, Smith’s pay rose by £500 compared with that for the previous 6-month period”.
An increase of £500 compared with the 6-month period equates to a rise of £1000 in his salary. And he gets that every 6 months. So, he’s getting a £2k rise every year and he’s seeing some of it before Jones does.
16 Nov 2009, 17:06
Eleanor Lovell
Sorry Iain – you missed something…Simon got it though. This is Ian Stewart’s explanation:
Surprisingly, Smith earned more – even though £1,600 per year is greater than Smith’s accumulated £500 + £1,000 over a year.
To see why, tabulate their earnings for each six-month period:
Note that Jones’s £1,600 splits into two amount of £800 for each half-year, so his half-yearly figures increase by £800 every year. Smith’s half-year figures increase by £500 every half-year. Despite that, Smith is ahead in every period after the first, and gets ever further ahead as time passes.
In fact, at the end of year n, Smith has earned a total of 10,000n + 5,00n(2n-1) pounds, while Jones has earned a total of 10,000n + 800n(n-1) pounds.
So Smith – Jones = 200n2 + 300n, which is positive and grows with n.
17 Nov 2009, 16:16
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