Need help calculating percentages

[MichiganGirl]

I'm really having a brain fart! I can't figure out how to get the percentage here. Can anyone help please?

I have a total of 537 votes. One person got 271, the other got 266.
I have this, but can't remember how to solve it!

x% of 537=271

Thanks!

[Ryda Wong, EBfCo.]

Originally Posted by MichiganGirl

I'm really having a brain fart! I can't figure out how to get the percentage here. Can anyone help please?

I have a total of 537 votes. One person got 271, the other got 266.
I have this, but can't remember how to solve it!

x% of 537=271

Thanks!

Divide the number for which you want the percentage by the total, so 271/537 = .504

So, 50%

[Auntie Witch]

Divide each number by 537.

Number of votes for x/total number of votes = % of votes for x.

[Pinecone]

The answer is 50.47%.

I did it by dividing 100 by 537 then multiplying the result by 271.

[wanderwoman]

Divide the smaller number by the larger number and multiply by 100. In this case it's pretty close to 50% (50.4655...)

Darn! Multiply spanked with numbers!

[Ryda Wong, EBfCo.]

Originally Posted by Pinecone

The answer is 50.47%.

I forget rounding. here, you'd have the .5. Do you round up or down with .5?

[Auntie Witch]

Originally Posted by Ryda Wong, EBfCo.

I forget rounding. here, you'd have the .5. Do you round up or down with .5?

Up. .4 and below, you round down.

[Minstrel]

In a percentage problem, "of" is the same as "times". So your x% of 537 = 271 is equivalent to x% * 537 = 271. Then you can solve algebraically from there by dividing both sides by 537.

x% = 271/537 = 50.47%

I learned the "of" = "times" I think in 7th or 8th grade, and I still have to think percent problems through that way!

[MichiganGirl]

Thanks so much!

I'm doing a paper on the Electoral College. I am putting a graph in of the 1888 and 2000 elections, where the person with the most popular vote did not get the most electoral votes.

[BringTheNoise]

Up. .4 and below, you round down.

I was taught to go with the second decimal place - .51 would round down, .56 would round up. If it was .55, I think it was supposed to round up.

[Pinecone]

Originally Posted by BringTheNoise

I was taught to go with the second decimal place - .51 would round down, .56 would round up. If it was .55, I think it was supposed to round up.

Interesting - we were taught to round up if it was .5 or above. Where did you go to school?

[Ieuan ab Arthur]

Hi All:

Originally Posted by Pinecone

Interesting - we were taught to round up if it was .5 or above.

That's what I was taught - if the last significant digit is 5 or more, round up. For example, 123.445 would be rounded up to two decimals as 123.45. By the same rule, 123.444 would be rounded down to two decimals as 123.44.

Ta ra 'wan,

Ieuan "Lies My Teacher Told Me" ab Arthur

[Floater]

Originally Posted by Minstrel

x% = 271/537 = 50.47%

Make it "x% = (271/537)*100 = 50.47%" instead.

[Mad Jay]

Originally Posted by Floater

Make it "x% = (271/537)*100 = 50.47%" instead.

Nah uh, % implies divide by 100. The "percent" really means "per cent", and cent is 100, so it really means "per 100" So it is perfectly valid to say

0.5047 = 50.47%

0.5047 and 50.47% are the exact same number , just expressed in differrent "units", so to speak

So

271/537 = 50.47%

is perfectly valid, because it means

271/537 = 50.47/100

[BringTheNoise]

Interesting - we were taught to round up if it was .5 or above. Where did you go to school?

Dumfriesshire - Hightae Primary and Lockerbie Academy to be exact.

[Silkenray]

And I was very recently taught that if a number ends in 5, and you're rounding it to the next digit above, to round whichever way will give you an even result.

For instance, .15 would round to .2, and .25 would also round to .2.

But apparently this rule is very new and not everyone adheres to it.

[Beachlife!]

Originally Posted by Silkenray

And I was very recently taught that if a number ends in 5, and you're rounding it to the next digit above, to round whichever way will give you an even result.

For instance, .15 would round to .2, and .25 would also round to .2.

But apparently this rule is very new and not everyone adheres to it.

Yep, that is the new rule though it does create it's own set of issues. The Business Intelligence software I work with uses that rule and I find myself constantly explaining it, especially to accontants.

[Floater]

Originally Posted by Mad Jay

Nah uh, % implies divide by 100. The "percent" really means "per cent", and cent is 100, so it really means "per 100" So it is perfectly valid to say

0.5047 = 50.47%

0.5047 and 50.47% are the exact same number , just expressed in differrent "units", so to speak

So

271/537 = 50.47%

is perfectly valid, because it means

271/537 = 50.47/100

I see your point, but I don't know if I altogether agree with your reasoning. In everyday life we just talk about so and so many percent of something and to point out that 5 percent really should be understood as 5 hundredths of the total seems a bit irrelevant if you just want to show someone how to calculate it. That understanding comes later.

To me the sign "%" is just a shorthand way of writing the word "percent". The important thing is to make the student understand (seeing from a teacher's point of view) how you do to reach that figure, in this case, 50.47, which I think is best done by setting up the equation as I do it.

[UEL]

Originally Posted by Silkenray

But apparently this rule is very new and not everyone adheres to it.

Originally Posted by Beachlife!

Yep, that is the new rule though it does create it's own set of issues. The Business Intelligence software I work with uses that rule and I find myself constantly explaining it, especially to accontants.

It was introduced to reduce something called an "upward bias".

I got to see an example where it was considerably visible. We used to get that all the time on some of our equipment. It was constantly performing more poorly than the condition of it would indicate. Our instrumentation for calibration would measure to the nearest half metre per second, but our computers for calculation would accept only whole numbers**. So, for every half m/s, it was increased by rounding. Get enough measurements like that and the velocities are significantly "increased" across the board, but the resultant performance is not.

When we switched to the even number rounding, the velocities became immediately consistent with the performance of our equipment.

**this event took place in about 1990.

[Jahungo]

Mad Jay, I would say what you are saying is correct, but it's essentially skipping the units conversion step (which is not wrong, since the two sides are still equal, but it doesn't show a step).

Saying:
271/537 = 50.5%

is like saying:
10.16 cm / 4 = 1.000 in

The equality is correct, but it skips a step.

To show all the steps, one would say:
271/537 = 0.505
0.505 x 100% = 50.5%
Or, combined as:
271/537 x 100% = 50.5%

This is like saying:
10.16 cm / 4 = 2.54 cm
2.54 cm x (1 in / 2.54 cm) = 1 in
Or combined as:
10.16 cm / 4 x (1 in / 2.54 cm) = 1 in



As for significant figures, the rule for division is that quotient should have as many significant figures as whichever out of the divisor and dividend has the least significant figures.

In this case, 271 and 537 both have three significant figures, so your quotient should also have three significant figures. So to express this in significant figures, you would say (as I did above):
271/537 x 100%= 50.5%

In 0.5046..., the three significant figures are 0.504. You go to the next place to decide whether to round the last digit up or down. Since the next place is a 6, it rounds up to 0.505 - or 50.5%