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Question about percents

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Question about percents

by Stockmoose16 » Tue Oct 21, 2008 12:14 pm
Hello,

I'm getting confused about when you can use the whole number of a percent, and when you have to put it over 100.

Some questions say, "M is X percent of Z." If we say M=15, X=25%, how would you write M and X in terms of Z?

Would it be XM=Z or (x/100) * M?

When can you use the whole percent and when do you have to change it into a fraction?
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by cramya » Tue Oct 21, 2008 1:44 pm
M is X% of Z

Eg: 10 is 50% of 20

10 = 50/100*20
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by Stockmoose16 » Tue Oct 21, 2008 1:49 pm
cramya wrote:M is X% of Z

Eg: 10 is 50% of 20

10 = 50/100*20
That didn't answer my question. I wanted to know when you can use the full number (i.e. 25%) and when you must use the fraction (i.e. 25/100).

Anyone?
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by cramya » Tue Oct 21, 2008 2:37 pm
I thought I did but may be not.

When its a percent 100 is assumed to be the demoninator. I dont udnerstand what you mean when to use 25 versus 25/100

I am thinking if it says 25% its is always 25/100

I am sure there are other experts in this forum who can provide you the exact answer u r loking for if this does not help.

Good luck!
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by bekkilyn » Tue Oct 21, 2008 2:38 pm
When you see a question that asks, "What is 25% of X?" then you are going to want to either convert to the decimal number or use 25/100 in the equation. Either method would work, and you would decide which to use based on which calculation would be easiest for you for a particular problem.

For example, the question might be, "What is 25 percent of 200?"

Method a)
200 x 0.25 = 50.

Method b)
(200)(25/100) = 5000/100 = 50

You won't generally just use 25 in the calculation for 25% unless you can remember to move the decimal in the final answer, which can get confusing, so I'd recommend doing either of the above conversions first.
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by Stockmoose16 » Tue Oct 21, 2008 2:39 pm
cramya wrote:I thought I did but may be not.

When its a percent 100 is assumed to be the demoninator. I dont udnerstand what you mean when to use 25 versus 25/100

I am thinking if it says 25% its is always 25/100

I am sure there are other experts in this forum who can provide you the exact answer u r loking for if this does not help.

Good luck!
It's not always 25/100. Sometimes the questions give you a bunch of variables, and if you try to use 25/100 to fill in the VIC you'll get the wrong answer. For these problems, you MUST use 25 (the whole number), not 25/100. I just don't understand how you know when to use the whole number and when to use a fraction.
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by Stockmoose16 » Tue Oct 21, 2008 2:44 pm
bekkilyn wrote:When you see a question that asks, "What is 25% of X?" then you are going to want to either convert to the decimal number or use 25/100 in the equation. Either method would work, and you would decide which to use based on which calculation would be easiest for you for a particular problem.

For example, the question might be, "What is 25 percent of 200?"

Method a)
200 x 0.25 = 50.

Method b)
(200)(25/100) = 5000/100 = 50

You won't generally just use 25 in the calculation for 25% unless you can remember to move the decimal in the final answer, which can get confusing, so I'd recommend doing either of the above conversions first.
The following is an example where I HAD to use 25, not 25/100, to get the correct answer. It's a VIC question. Can someone explain why you have to use the whole number?


If an item that originally sold for z dollars was marked up by x percent and then discounted by y percent, which of the following expressions represents the final price of the item?


A)
10,000z + 100z(x – y) – xyz
----------------------------------
10,000

B)
10,000z + 100z(y – x) – xyz
-----------------------------------
10,000

C)
100z(x – y) – xyz
----------------------
10,000

D)
100z(y – x) – xyz
-------------------------
10,000


E)
10,000
------------------
100yz + xy
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by bekkilyn » Tue Oct 21, 2008 3:10 pm
You will use these values in the equation:

z = original amount
x percent = x/100
y percent = y/100

Then you will set up this equation:

z + z(x/100) - ([z + z(x/100)][y/100])

Basically what you are doing is first taking x percent of z and adding the result to the original amount, z, and then subtracting y percent of that same result.

Solve:

z + (zx/100) - [(zy/100) + (xyz/10000)]

z + zx/100 - zy/100 - xyz/10000

[10000z + 100zx - 100zy - xyz]/10000

[10000z + 100z(x-y) - xyz] / 10000


My result turns out to be ( A)
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by Stockmoose16 » Tue Oct 21, 2008 3:14 pm
bekkilyn wrote:You will use these values in the equation:

z = original amount
x percent = x/100
y percent = y/100

Then you will set up this equation:

z + z(x/100) - ([z + z(x/100)][y/100])

Basically what you are doing is first taking x percent of z and adding the result to the original amount, z, and then subtracting y percent from that result.

Solve:

z + (zx/100) - [(zy/100) + (xyz/10000)]

z + zx/100 + zy/100 = xyz/10000

[10000z + 100zx - 100zy - xyz]/10000

[10000z + 100z(x-y) - xyz] / 10000


My result turns out to be ( A)
The answer is A, but you're answering the question using algebra. Try answering it using VIC, and enter 25/100 instead of 25 (for X) and 10 instead 10/100 for Y, and you'll get the incorrect answer. I don't get it.
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by bekkilyn » Tue Oct 21, 2008 3:27 pm
You wouldn't want to plug in 25/100 into the answer equation because the answer equation already accounts for the 1/100 from the algebra derivations, so in this case you would just plug in the 25 for the x:


[10000z + 100z(25-y) - 25yz] / 10000

If you were to work backwards in place of x, you would end up getting 25/100 instead of x/100 in the original equation:

z + z(25/100) - ([z + z(25/100)][y/100])

By the way, what is VIC?
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by Stockmoose16 » Tue Oct 21, 2008 3:47 pm
bekkilyn wrote:You wouldn't want to plug in 25/100 into the answer equation because the answer equation already accounts for the 1/100 from the algebra derivations, so in this case you would just plug in the 25 for the x:


[10000z + 100z(25-y) - 25yz] / 10000

If you were to work backwards in place of x, you would end up getting 25/100 instead of x/100 in the original equation:

z + z(25/100) - ([z + z(25/100)][y/100])

By the way, what is VIC?
VIC = Variable in answer choices -- they're questions that usually allow you to plug in.

Can you please explain how, if you were to work backwards in place of x, you would end up getting 25/100? It looks to me like you'd get 25 (100Z) for the middle term. I don't get it.
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by bekkilyn » Tue Oct 21, 2008 4:15 pm
Working backwards from original equation:

[10000z + 100z(25-y) - 25yz] / 10000

Distribute the 10000 as the denominator for each term and distribute 100z with each term in (25-y):

10000z/10000 + 100(25)z/10000 - 100zy/10000 - 25xy/10000

Now reduce the fractions for each term since originally 10000 was used as the greatest common denominator for adding the terms:

z + z(25)/100 - zy/100 - 25yz/10000

Which becomes:

z + z(25)/100 - [zy/100 + 25yz/10000]

Factor out y/100 from the term in [ ]'s.

z + z(25)/100 - [z + z(25)/100](y/100)

The terms with (25)/100 or y/100 are the original x and y percents.
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by mpaudena » Tue Oct 21, 2008 4:17 pm
I have a similar problem but I think I can answer stockmoose's questions as it relates to VIC. First my question so it doesn't get lost. Stockmoose, read below to see if that helps you out.

A certain movie star's salary for each film she makes consists of a fixed amount, along with a percentage of the gross revenue the film generates. In her last two roles, the star made $32 million on a film that grossed $100 million, and $24 million on a film that grossed $60 million. If the star wants to make at least $40 million on her next film, what is the minimum amount of gross revenue the film must generate?

Okay, I set up the equation like this: [$32M = f + (p/100)($100M)] - [$24M = f + (p/100)($60M)

My answer was wrong. Turns out I don't need to put in the p/100 and only need to put the p and then solve for p. So question: When do you put the variable over 100 cause I've seen instances where you do?

Stockmoose's question:
If you are doing VIC then you are picking numbers. When the question says was discounted by "X percent and then discounted by Y percent" you literally have to put in the numbers in place of X and Y. That is to say if you used 25 for X then you will be thinking 25%. When you are doing your calculation to find the answer that you will then match with the answer from the answer choices you will take into account the fact that the question says percent and you will use either .25 or 25/100 in your calculation. HOWEVER, when you are plugging in the numbers YOU DO NOT take into account the fact that it says percent. Consider it like this - the answer choices say nothing of percentage, only the question does. And because the question does you use that for your calculation but when matching the answer from your calculation with the answer choices you only put in the actual value of X that you chose and not the extra "percent" that is only found in the question.
Last edited by mpaudena on Tue Oct 21, 2008 4:19 pm, edited 1 time in total.
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by Stockmoose16 » Tue Oct 21, 2008 4:18 pm
bekkilyn wrote:Working backwards from original equation:

[10000z + 100z(25-y) - 25yz] / 10000

Distribute the 10000 as the denominator for each term and distribute 100z with each term in (25-y):

10000z/10000 + 100(25)z/10000 - 100zy/10000 - 25xy/10000

Now reduce the fractions for each term since originally 10000 was used as the greatest common denominator for adding the terms:

z + z(25)/100 - zy/100 - 25yz/10000

Which becomes:

z + z(25)/100 - [zy/100 + 25yz/10000]

Factor out y/100 from the term in [ ]'s.

z + z(25)/100 - [z + z(25)/100](y/100)

The terms with (25)/100 or y/100 are the original x and y percents.
Ok, but how do you know that 100 was put in the equation to make it a percentage. Maybe you were still supposed to use 25/100, and then put it over 100. I just don't understand how you can assume that 100 was put there to make a percent.
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by Stockmoose16 » Tue Oct 21, 2008 4:22 pm
Stockmoose16 wrote:
bekkilyn wrote:Working backwards from original equation:

[10000z + 100z(25-y) - 25yz] / 10000

Distribute the 10000 as the denominator for each term and distribute 100z with each term in (25-y):

10000z/10000 + 100(25)z/10000 - 100zy/10000 - 25xy/10000

Now reduce the fractions for each term since originally 10000 was used as the greatest common denominator for adding the terms:

z + z(25)/100 - zy/100 - 25yz/10000

Which becomes:

z + z(25)/100 - [zy/100 + 25yz/10000]

Factor out y/100 from the term in [ ]'s.

z + z(25)/100 - [z + z(25)/100](y/100)

The terms with (25)/100 or y/100 are the original x and y percents.
Ok, but how do you know that 100 was put in the equation to make it a percentage. Maybe you were still supposed to use 25/100, and then put it over 100. I just don't understand how you can assume that 100 was put there to make a percent.
Mpaudena,

But both the questions above involve plugging in numbers. Taking M percent of X means you could plug 25 in for X and 10 in for X. Why is this question different from the more involved VIC posted above, where you DO NOT put the numbers over 100? I really can't grasp how you can tell, from the question, whether to put the numbers you choose over 100 or not.
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