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My 9-year-old daughter is stuck on this question and normally I can help her, but I am also stuck on this! I have looked everywhere to find out how to do this but to no avail so any help/guidance is appreciated:

The possible answers are:

$1 \frac{29}{300}$

$\frac{269}{300}$

$9 \frac{29}{30}$

$\frac{299}{300}$

$1 \frac{299}{300}$

Whenever I try this I am doing $\frac{99.66} \times 100$ is $\frac{9966}{10000}$ and then simplify down, but I am not getting the answer. I get $\frac{4983}{5000}$? Then I can't simplify more — unless I am doing this totally wrong?

Hello All - just an update - I managed to teach this to her right now, and she cracked it pretty much the first time! Thanks sooooooo much for the answers and here is some of the working she did:

Enter image description here

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    $\begingroup$ $\frac{2}{3}$ does not equal $0.66$, although they are approximate. You've been simplifying the wrong number! You need to convert the "mixed fraction" $99\frac{2}{3}$ to the "improper fraction" $\frac{p}{q}$ for some whole numbers $p$ and $q$. $\endgroup$
    – anon
    Jun 9, 2015 at 15:15
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    $\begingroup$ side question: is there any reason why the lack of operation is interpreted as an addition? (especially since it is often used for multiplication) $\endgroup$
    – njzk2
    Jun 9, 2015 at 17:52
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    $\begingroup$ @njzk2 bad notation: "mixed fractions" are evil. $\endgroup$
    – Avi
    Jun 9, 2015 at 18:03
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    $\begingroup$ In general, $x \frac{p}{q}$ is equals to $\frac{(x * q) + p}{q}$. Therefore, $x \frac{p}{q}\%$ equals $\frac{(x * q) + p}{q*100}$. $\endgroup$
    – haccks
    Jun 10, 2015 at 15:27
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    $\begingroup$ Nine, huh? I'm impressed! $\endgroup$
    – Lightness Races in Orbit
    Jun 11, 2015 at 15:24

14 Answers 14

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You can almost eyeball this, which is a useful technique for dealing with multiple choice quizzes against the clock.

If 100% = 1, then you're looking for a fraction slightly less than one. That lets you rule out all but two of the answers. How much less? A third of a percent. One percent is 1/100. Divide that by three to get 1/300. That points you at 299/300 being the answer.

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    $\begingroup$ I'm not sure what techniques need to be invoked in order to teach this to a 9 yo, but the methods employed by this solution are incredibly useful and will help her later on. Especially, as noted, with multiple choice. $\endgroup$
    – Paddling Ghost
    Jun 9, 2015 at 16:43
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    $\begingroup$ The answer as written seems completely appropriate for a nine year old to me. $\endgroup$
    – ColinK
    Jun 9, 2015 at 23:47
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    $\begingroup$ I think anyone of any age benefits from seeing a problem being solved in more than one way. This way seems a useful alternative. $\endgroup$
    – trichoplax is on Codidact now
    Jun 10, 2015 at 0:33
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    $\begingroup$ This is the only answer that seems to take the age of the student into consideration. This is a very good way to approximate on the fly and probably what the teacher expects. $\endgroup$
    – user45623
    Jun 10, 2015 at 4:47
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    $\begingroup$ This is by far the best answer so far. The appropriate way to deal with multiple choice arithmetic answers is to use approximation, anything else is long-winded and painfully slow. $\endgroup$
    – Jack Aidley
    Jun 11, 2015 at 10:00
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So you want to find $$\frac{99+\frac{2}{3}}{100}.$$

The numerator is equal to $$\frac{3(99)}{3}+\frac{2}{3} = \frac{297}{3}+\frac{2}{3} = \frac{299}{3}.$$

Dividing by $100$ you get $$\frac{\frac{299}{3}}{100} = \frac{299}{3(100)}=\frac{299}{300}.$$

The reason you are getting the wrong answer is because you are rounding. The fraction $\frac{2}{3}$ is not equal to $0.66$ but rather is equal to $0.6666\cdots$ (often written $0.\overline{6}$).

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    $\begingroup$ This is great - I am getting this now thanks to this example (well, I hope I am understanding this as I need to teach it to my daughter!) $\endgroup$
    – lara400
    Jun 9, 2015 at 15:20
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    $\begingroup$ @lara400 Good luck and welcome to this site! $\endgroup$
    – Casteels
    Jun 9, 2015 at 15:21
  • $\begingroup$ thanks SOOO much - I have added a picture of her understanding it and doing some sums she has! thanks so much again! $\endgroup$
    – lara400
    Jun 9, 2015 at 18:06
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    $\begingroup$ @Casteels I was taught a dot over the number to show it's recurring (or over the first and last of a recurring sequence), not a line. $\endgroup$
    – Majenko
    Jun 12, 2015 at 15:45
  • $\begingroup$ @Majenko I was taught that both ways are used commonly iirc. Indeed the grand poobah of knowledge confirms this. $\endgroup$
    – Casteels
    Jun 12, 2015 at 18:18
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$99\frac{2}{3}\%$ is $\frac{1}{3}\%$ away from $100\%$ or $1$. $\frac{1}{3}\%$ is literally $\frac{1}{300}$. The answer is therefore $\frac{299}{300}$.

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  • $\begingroup$ This is what I would have done too. $\endgroup$
    – JK01
    Jun 11, 2015 at 4:42
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    $\begingroup$ The top answers don't do this because while this is a useful approach to solving this particular problem, it's not teaching an algorithm for solving any such problem. How does a 9y.o. know that 1/3% is "literally 1/300"? $\endgroup$
    – Michael Blackburn
    Jun 12, 2015 at 14:29
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    $\begingroup$ @jazmatician: simple. By teaching them that 1% = 1/100 and that 1/3% = 1/3 * 1/100 = 1/(3*100) = 1/300. I hope thet teach kids any of these steps. $\endgroup$
    – physicalattraction
    Jun 12, 2015 at 21:27
  • $\begingroup$ So your proposed algorithm for teaching kids to perform an additional multiplication and subtraction to get an answer, and just hope that they notice when "difficult" numbers (like 99 and 2/3%) are close to simpler numbers? While Nimrod's answer is correct, it depends on the facts of the particular problem (1/3% is 1/300, a mathematical fact he relied upon to derive his answer). When teaching children arithmetic, we don't force them to learn a large number of facts, rather teach them the concepts required to algorithmically solve the problem. Casteel's answer is correct, and pjc50's is useful. $\endgroup$
    – Michael Blackburn
    Jun 29, 2015 at 14:56
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    $\begingroup$ There is no multiplication involved. My answer got reformatted but there was a reason I wrote it the way I originally did, which read: "1/3% is literally 1/300." It is meant to be a visually suggestive mnemonic. If you know the history of the "per cento" sign -- and by extension, the permille and others -- you'll know that per-whatever just adds the number of o's as indicated to the denominator as zeros. It is also an important and transformative experience in one's education to discover the equivalence of fractions and division down to the way they are formatted on paper. $\endgroup$
    – Nimrod
    Jun 30, 2015 at 23:28
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$99 \frac{2}{3} \% = 99 \frac{2}{3} \cdot \frac{1}{100} = \frac{299}{3} \cdot \frac{1}{100} = \frac{299}{300}$.

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Separately, $$99\%=\frac{99}{100},$$ and $$\frac{2}{3}\%=\frac{\frac{2}{3}}{100}=\frac{2}{300}$$

Adding together we get $$99\frac{2}{3}\%=99\%+\frac{2}{3}\%=\frac{99}{100}+\frac{2}{300}.$$ Using algebra of fractions you then get $$\frac{99\times 300+2\times 100}{100\times 300}=\frac{99\times 3+2}{300}=\frac{299}{300}.$$

Remark As a teaching "aside", I like to think about what "percent" means. Break it down. "Percent" is a juxtaposition of the two words "per" and "cent". "Per" just means "for each" and "cent" literally means "100" (e.g. there are a 100 cents in a dollar!). So "percent" just means something per 100, i.e $1.2345\%$ is just $1.2345$ per $100$, or in other words $\frac{1.2345}{100}$. Similarly $10\%$ is just $10$ per $100$, or in other words $\frac{10}{100}$. And related to the above $\frac{2}{3}\%$ is just $\frac{2}{3}$ per $100$, or in other words $\frac{\frac{2}{3}}{100}$. And to finish $99\frac{2}{3}\%$ is just $99\frac{2}{3}$ per $100$, or in other words $\frac{99\frac{2}{3}}{100}$. Therefore, if you know your fractions, then percentages are a breeze !!

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  • $\begingroup$ This is how I would have explained it. Though similar to the answer by @pjc50 $\endgroup$
    – jiggunjer
    Jun 11, 2015 at 19:26
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Multiply both numerator and denominator by $3$ to get $$\frac{3 \times (99 \frac{2}{3})}{3 \times 100} = \frac{3 \times 99 + 3 \times \frac{2}{3}}{300}$$

Which is $$\frac{297 + 2}{300} = \frac{299}{300} $$

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Already it has been shown how to arrive at the correct answer using fractions. One may also consider using the process of elimination. $99 \frac{2}{3}\%$ is clearly less than $1$. So, by inspection, we can rule out choices 1,3, and 5. We are left with two possibilities, $\frac{269}{300}$ and $\frac{299}{300}$. Now, $\frac{269}{300} \approx \frac{270}{300}=.90$ or 90%. Since $269 <270$, $\frac{269}{300} < \frac{270}{300} = 90\%$. Thus, the answer cannot be $\frac{269}{300}$. This leaves one possibility, $\frac{299}{300}$.

While this method may not be immediately available to a 9 yo, it is easy to see that three possibilities are easily ruled out. This sort of thinking (process of elimination) can go a long way with multiple choice tests in the future and with math in general. (Think proof by contradiction.)

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A third of a percent is missing, i.e. one three-hundredth. Two hundred and ninety nine three-hundredths remain.

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Here's how I did it in my head:

$99\%$ is $\frac{99}{100}$. This leaves $\frac{2}{3}\%$ unaccounted for. So we need to go $\frac{2}{3}$rds of the $1\%$ distance from $\frac{99}{100}$ to $\frac{100}{100}$. Clearly, we need more than one step to do this; we need three. So, we need to go $\frac{2}{3}$rds of the distance from $\frac{297}{300}$ to $\frac{300}{300}$, which is $\frac{2}{300}$. So the final answer is $\frac{297}{300}+\frac{2}{300}=\frac{299}{300}$.

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Why are you even trying to do this as a math problem? Because it is a multiple choice question, figure out which answers are wrong.

  1. The First, third and fifth are all incorrect because they are over 1.00 and of course 99 2/3% is less than 1.00
  2. The second answer is 1.00 - 31/300 or 1/10 less than 1.00, or .90~something, and therefore wrong also
  3. The fourth answer is 1.00 - 1/300, or most likely 99 2/3% if you can't actually do the math

My wife is a fifth grade teacher, and what they are actually teaching is that if you are stuck on the mathematical equations, use inference and elimination to deduce the correct answer and work backward. That simple.

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    $\begingroup$ Hello Jim - I don't agree 100% with you on this - you are right about elimination and I try and teach her that; however, knowing the maths is equally, if not more, important. My daughter is not in fifth grade (I am in the UK and fifth grade is = Year 6 here (check Wikipedia)) but she is equivalent to third grade (Year 4 in UK) - there is no one in her class remotely close to her in maths and I doubt a lot of fifth graders can do this at this level thus understanding of maths is paramount to the success of elimination that will come further once she knows how to do sums like these. $\endgroup$
    – lara400
    Jun 9, 2015 at 19:57
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    $\begingroup$ -1 The question is not about strategies for multiple-choice questions, it's about how to do the math. "Why are you even trying to do this as a math problem?" Because it is a math problem. Understanding why percents and fractions behave the way they do and how to solve them is valuable. $\endgroup$
    – Reinstate Monica -- notmaynard
    Jun 10, 2015 at 15:59
  • $\begingroup$ Notmaynard, read the second-to-last sentence of my post - it says to work backward. I gave this to my wife last night and she said I was 100% correct. (B/T/W, she has MEd +30, Nat Board Certified, state director for Singl Gender Studies, and an adjunct professor at our primary state university.) My post was not about answering multiple choice questions, it is about how to use multiple choice to figure out how to do an equation. $\endgroup$
    – Jim S.
    Jun 10, 2015 at 18:33
  • $\begingroup$ Not only that, but the previous 15 people all gave the same answer that the OP could easily have looked up in any elementary school math textbook. I was offering an alternative way of understanding the method of equation solving since they obviously weren't understanding the traditional method. $\endgroup$
    – Jim S.
    Jun 10, 2015 at 18:40
  • $\begingroup$ Bang on. This is the way to solve it. It's sad this has got downvotes. $\endgroup$
    – Jack Aidley
    Jun 11, 2015 at 10:01
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99 and $2/3$ percent is $\frac{\displaystyle 99 + \frac{2}{3}}{100}=\frac{\displaystyle\frac{3\times 99 + 2}{3}}{100}$. Does this help?

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There are multiple roads that lead to Rome:

First, take out the $\%$. We find $\frac{299}{3}$

$99 \frac{2}{3} = 99 + \frac{2}{3} = \frac{1}{3} \cdot ( 297 + 2) = \frac{1}{3} \cdot 299 = \frac{299}{3}$

or

$99 \frac{2}{3} = 99 + \frac{2}{3} = 3 \cdot \frac{99}{3} + \frac{2}{3} = \frac{297}{3} + \frac{2}{3} = \frac{299}{3}$

Lastly, add the $\%$ back: $\frac{299}{3} \cdot \frac{1}{100} = \frac{299}{300}$

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$99 \frac{2}{3} = 99 + \frac{2}{3} = 99 + (1 - \frac{1}{3}) = 100 - \frac{1}{3} = \frac{300}{3} - \frac{1}{3} = \frac{299}{3}$

So $99 \frac{2}{3}\% = \frac{299}{3}\% = \frac{299}{300}$

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Is it too hard to mentally calculate the numerator?

$99*3 + 2 = 300 - 3 + 2 = 299$

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