Hoozan wrote:
BUT please could you help me understand when do we include the % sign and when do we not? Since percent means per 100 can I say that if we divide by 100 we drop the % sign and if we multiply by 100 we include the % sign?
There are quite a few ways to think about this, and if others read my post, they might be thinking about percents differently (when I get to the algebra), and if they're getting right answers, there's no need to think about it the way I'm explaining things here. But the symbol "%" means "per 100", or algebraically, "/100", so "25%"
means "25/100" -- they're the same thing. And if you have a number like 0.4, then we can always multiply this by "100%" without changing its value, because 100% = 100/100 = 1. So
0.4 = 0.4 * 100% = (0.4*100) % = 40%
So that's why what you've said in the quote above is correct.
Hoozan wrote:
Also, could we also say let "p" be the percent greater than value so:
\(\frac{5+x}{60}\) = \(\frac{1 + p}{100}\) x \(\frac{x}{60}\)
Thus p = \(\frac{500}{x}\)%
I think there's an issue with the mathematical typesetting in what I quote above -- I'm sure you meant to write:
\(\\
\frac{5+x}{60} = \left( 1 + \frac{p}{100} \right) \left( \frac{x}{60} \right)\\
\)
which is perfectly correct. Because in this equation you put in "p/100" which is "p %", when you solve for p, you're done -- that's the percentage you're looking for, and you don't want to multiply or divide by 100. I often do these problems slightly differently, just to avoid fractions: instead I'll solve it like a ratio problem (so I'm finding the ratio of the increase to the smaller value), and then as the final step, I convert that ratio into a percentage so it becomes the answer to a percent increase problem. So you can also solve this way:
\(\\
\frac{5+x}{60} = \left( 1 + r \right) \left( \frac{x}{60} \right)\\
\)
then solve for r, knowing r is
not a percentage, then finally multiplying by "100%" at the end (imitating what I did with "0.4" above) to make it into a percentage.
Hoozan wrote:
Further, what if we were asked \(\frac{5+x}{60 }\) is \(\frac{500}{x}\) % greater than what value? How would we represent this question using the above two methods?
(1) \(\frac{5+x}{60 }\)= 1 + \(\frac{500}{100x }\) x (?) OR
(2) \(\frac{5+x}{60}\) - (?) / (?) x 100% = \(\frac{500}{x }\)%
While writing version (2) I often get confused when to add the % and when to not
There are a few ways, all perfectly good, to think about a question like that. If (5 + x)/60 is (500/x)% greater than some value V, then (5 + x)/60 is equal to that value V, plus (500/x)% of V. That's where your equation (1) comes from, though again there's an issue with the mathematical typesetting -- it should read:
\(\\
\frac{5 + x}{60} = V + \left( \frac{500}{100x} \right) V = \left( 1 + \frac{500}{100x} \right) V\\
\)
Whenever we're working with percents in an equation where you'll genuinely need to do some algebra (adding or subtracting on both sides, say), you'll want to avoid including "%" symbols anywhere, so you'd want to do what you did here -- replace "500/x %" with "500/100x" first, at which point we're completely done with the "%" sign, and then work with that new fraction. Otherwise you'd need to learn a whole set of unnecessary rules about what you're allowed and not allowed to do when you have a "%" sign in an equation.
It can be a bit faster to learn that when we increase something by z%, we are always multiplying it by 1 + (z/100), so for example, if we increase something by 60%, we are multiplying it by 1 + 60/100 = 1.6. You could use that principle here to produce the same equation above in one less step.
As for your equation (2), I think that's a potentially confusing way to think about situations like this. The danger in these abstract percent questions is that you can easily produce an answer that has too many zeros, or not enough, so you want a method where you can be completely certain if you're looking at a ratio or looking at a percentage. I think your equation (2) is incomplete, but in principle, you could use it: when we say one value is p% greater than another, what we mean is that the ratio of the difference in the two values to the smaller value is p%, or p to 100. So here, if (5+x)/60 is (500/x)% greater than V, that means the ratio of the difference in the two values, which is (5+x)/60 - V, to the smaller value, which is V, is equal to (500/x)%, which is 500/100x, or 5/x. So you get this equation:
\(\\
\begin{align}\\
\frac{\frac{5 + x}{60} - V}{V} &= \frac{5}{x} \\\\
x \left( \frac{5 + x}{60} - V \right) &= 5V \\\\
(x) \left( \frac{5 + x}{60} \right) &= 5V + xV \\\\
(x) \left( \frac{5 + x}{60} \right) &= V(5 + x) \\\\
V &= \frac{x}{60}\\
\end{align}\\
\)
I find that a confusing way to think through the problem though, and it also leads to messier algebra than the other method, so at least in an algebraic question like this one, I wouldn't use it (though in simpler situations I might). Notice though that I converted the percentage into a ratio immediately, so I had an ordinary number to work with, and I never needed to include any percent symbols in the equation anywhere -- I think it would be very easy to end up with the wrong number of zeros trying to write one or both sides of the equation as a percentage rather than as a simple ratio.
In simpler questions, where all we're doing is simplifying a fraction that will turn into the answer, I often leave the % symbol in the equation though. So if something increases from $300 to $360, and I want to find the percent increase, I'd often just write
\(\\
\left(\frac{360 - 300}{300} \right) \times 100\%\\
\)
then I'd leave the '%' symbol alone, and cancel the '100' with the denominator:
\(\\
\left(\frac{360 - 300}{300} \right) \times 100\% = \left( \frac{360 - 300}{3} \right) \% = 20\%\\
\)
and the '%' symbol here just reminds me that I've already taken care of the conversion to a percentage. But this situation is different from the question in this thread -- in this example, we're not manipulating quantities in an equation and moving things around, and we're just tacking the "%" symbol on at the end. In any more complicated situation, I'd convert percents to ordinary fractions first, and then you can use all the familiar rules of arithmetic and algebra, and then you'd just need to pay attention to whether you need to convert your answer to a percentage at the end.
Hope that clears things up!
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Updated on: 14 Oct 2019, 06:34
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