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In very wealthy PSEland, the hourly rate is £1,000. A man works for 1,000 hours and earns £1,000,000. Nice going!

The next year, the government of PSEland decided to raise his wages by 10%, so to earn his million, he only worked for 909.090909 hours, which is 91% of 1000 hours (roughly!).

So a 10% increase in wages results in a 9% decrease in worked hours.

Where did the 1% go?

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closed as off-topic by Gareth McCaughan Apr 10 at 15:46

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  • $\begingroup$ The 1% of what? $\endgroup$ – noedne Mar 30 at 21:15
  • $\begingroup$ a 10% increase in wages should result in a 10% decrease in worked hours, @noedne $\endgroup$ – JonMark Perry Mar 30 at 21:18
  • $\begingroup$ I get that they look like the same number, but why would someone expect that to be true? $\endgroup$ – noedne Mar 30 at 21:20
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    $\begingroup$ Percentages confuse people a lot, I guess $\endgroup$ – PiIsNot3 Mar 30 at 21:22
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    $\begingroup$ This question has a distinct flavour to it that I don't very much like. Contrast with this question: "He got a 200% raise, why isn't he working negative 1000 hours?" $\endgroup$ – Bass Mar 31 at 6:55
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Short answer: that’s not how percentages work.

Long answer:

A $p\%$ increase in a value followed by a $p\%$ decrease does not give us back our original value. That’s because we’re taking the $p\%$ decrease of the increased value, so our discount is slightly more than if we took the discount of the original amount. Thus, if we wanted to get back to our original value, we need to take a percentage decrease that’s lower than $p\%.$

Mathematically speaking, if we let $P = \frac{p}{100},$ then we’re trying to find $d$ such that $(1 + P)(1 - d) = 1.$ The solution is $d = \frac{P}{1 + P},$ which is less than $P.$

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The question

doesn't make sense. You have a formula r x h = 1 M, where r = h = 1000. Therefore 1.1r x 1/1.1h = 1M and 1/1.1 = .9090909...

in general

whatever increase (1+r) of salary is accompanied by a decrease of 1/(1+r) of hours.

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