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# The cost C, in dollars, to remove p percent of a certain pollutant fro

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The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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18 Oct 2015, 12:15
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35% (medium)

Question Stats:

73% (01:37) correct 27% (02:03) wrong based on 2009 sessions

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The cost C, in dollars, to remove p percent of a certain pollutant from a pond is estimated by using the formula C = 100,000p/(100 - p). According to this estimate, how much more would it cost to remove 90 percent of the pollutant from the pond than it would cost to remove 80 percent of the pollutant?

(A) \$500,000
(B) \$100,000
(C) \$50,000
(D) \$10,000
(E) \$5,000

Kudos for a correct solution.

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The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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Updated on: 25 Nov 2017, 14:03
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9
diegocml wrote:
After checking the solution for this exercise, it looks quite easy. I just don't understand one thing. Why can I just plug the 90 and the 80 for p in the formula and not 90% (90/100) or 80% (80/100)?

i.e.: 100,000(90/100) / 100 - (90/100)

Thanks a lot.

Hi diegocml, "per cent" means "per hundred." So when I say "90 percent," I am actually saying "90/100". Thus, if you plug the number .9 into this formula, which is already formatted for a percent input (in other words, the equation itself has already taken the division by 100 into account), then it will become .9/100 = .009, which is less than one percent!

Lesson learned: if the question or formula already includes the word "percent," then don't divide by 100, because you will be dividing by 100 twice.

Originally posted by mcelroytutoring on 06 Feb 2017, 08:23.
Last edited by mcelroytutoring on 25 Nov 2017, 14:03, edited 6 times in total.
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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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18 Oct 2015, 12:30
Bunuel wrote:
The cost C, in dollars, to remove p percent of a certain pollutant from a pond is estimated by using the formula C = 100,000p/(100 - p). According to this estimate, how much more would it cost to remove 90 percent of the pollutant from the pond than it would cost to remove 80 percent of the pollutant?

(A) \$500,000
(B) \$100,000
(C) \$50,000
(D) \$10,000
(E) \$5,000

Kudos for a correct solution.

Cost to remove 90% - Cost to remove 80% = 100,000*90/(100 - 90) - 100,000*80/(100 - 80) = 9*100,000 - 4*100,000 = 5*100,000 =500,000

Ans A
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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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18 Oct 2015, 21:19
1
Bunuel wrote:
The cost C, in dollars, to remove p percent of a certain pollutant from a pond is estimated by using the formula C = 100,000p/(100 - p). According to this estimate, how much more would it cost to remove 90 percent of the pollutant from the pond than it would cost to remove 80 percent of the pollutant?

(A) \$500,000
(B) \$100,000
(C) \$50,000
(D) \$10,000
(E) \$5,000

Kudos for a correct solution.

required cost = (100,000 * 90)/(100 - 90) - (100,000 * 80) /(100 - 80)
= 100,000 * (90/10 - 80/20)
= 100,000 * (9 - 4)
= 100,000 * (5)
= 500,000

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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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19 Oct 2015, 09:28
1
Hii,

Instead of putting 80 or 90 as P, will we not put 80% or 90% as P.

Kind Regards,
Richa
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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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19 Oct 2015, 18:14
1
2
It has to be A...

C = (100000*p)/100-p
Here we can check puuting the values for 90 and 80 i.e.

Let 100000 be x so
for 90 its => 90x/10 =>9x ----1

for 80 its => 80x/20 =>4x ------2

Now , subtracting 2 from 1 we get => 9x - 4x = 5x or 5*100000 => 500,000
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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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19 Oct 2015, 20:49
To remove 90% of pollutant
C= 100,000(90)/(100-90) = 900,000

To remove 80% of pollutant
C' = 100,000(80)/(100-80) = 400,000

Difference = C - C' = 500,000

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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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28 Oct 2015, 04:42
Richa07 wrote:
Hii,

Instead of putting 80 or 90 as P, will we not put 80% or 90% as P.

Kind Regards,
Richa

Would this approach be wrong?

I know it would make our calculations whole a lot more difficult, but one could put in as P 0.9 and 0.8 at the beginning and after throwing away 30 seconds only realise the flaw in the approach. But would this also be wrong mathematically?

Thanks,
Jay
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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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02 May 2016, 14:35
Trick here is to not think too hard about it. p doesn't need to be expressed as a fraction or a decimal

just plug and play. 90%-80% difference (first do 90, then do 80, and subtract)

90% : 100,000(90)/(10) = 900,000

80% : 100,000(80)/(20) = 400,000

9-4 = 5*100,000
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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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03 May 2016, 04:38
Bunuel wrote:
The cost C, in dollars, to remove p percent of a certain pollutant from a pond is estimated by using the formula C = 100,000p/(100 - p). According to this estimate, how much more would it cost to remove 90 percent of the pollutant from the pond than it would cost to remove 80 percent of the pollutant?

(A) \$500,000
(B) \$100,000
(C) \$50,000
(D) \$10,000
(E) \$5,000

Kudos for a correct solution.

The cost to remove 90 percent of the pollutant is determined by letting p = 90 in the cost formula:

100,000(90)/(100 – 90) = 9,000,000/10 = 900,000

Similarly, the cost to remove 80 percent of the pollutant is determined by letting p = 80 in the cost formula:

100,000(80)/(100 – 80) = 8,000,000/20 = 400,000

To determine how much more it costs to remove 90 percent of the pollutant than 80 percent of the pollutant, we calculate the difference of the two costs:

900,000 – 400,000 = 500,000

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The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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07 Dec 2016, 19:15
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Attached is a visual that should help. In order to solve this question correctly, you must realize that p is expressed as a percent and thus should not be converted to a decimal.
Attachments

Screen Shot 2016-12-07 at 7.14.39 PM.png [ 129.25 KiB | Viewed 27825 times ]

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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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21 Jan 2017, 08:06
1
After checking the solution for this exercise, it looks quite easy. I just don't understand one thing. Why can I just plug the 90 and the 80 for p in the formula and not 90% (90/100) or 80% (80/100)?

i.e.: 100,000(90/100) / 100 - (90/100)

Thanks a lot.
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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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05 Feb 2017, 21:07
Richa07 wrote:
Hii,

Instead of putting 80 or 90 as P, will we not put 80% or 90% as P.

Kind Regards,
Richa

Richa07 , do you know why we should approach this exercises as such?
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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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06 Feb 2017, 08:43
Cost to remove 90% = 900,000
Cost to remove 80% = 400,000

Excess cost = 900,000-400,000
=500,000
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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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12 Mar 2017, 07:28
done in a minute!:

(100 000*90)/10 - (100 000*80)/20=100 000*(9-4)=500 000\$
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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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16 Sep 2018, 04:21
Hi all,

Why is it not correct to take p= 10 as in the difference between 90 and 80 will be 10 percent?

Thanks!

Posted from my mobile device
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The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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17 Sep 2018, 08:38
Lionila wrote:
Hi all,

Why is it not correct to take p= 10 as in the difference between 90 and 80 will be 10 percent?

Thanks!

Posted from my mobile device

Because in this case cost is not directly proportional to the %, hence % change will not give you the same change in cost.
E.g. You can buy 1 phone for \$100 or 2 for \$150 per promo - you get 2 times the phones, but pay 1.5 time the price.
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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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16 Jun 2019, 04:14
Hi Bunuel ,

Why not just plug in the difference (10%) into the equation? That's what I did first but I got a wrong answer. I read Hero8888 's explanation but I don't understand.

How come getting the percent difference here will not work? For me it seems logically sound. Where does this logic go wrong here?

Thanks.
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Re: The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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22 Jun 2019, 00:21
Top Contributor
Diwabag wrote:
Hi Bunuel ,

Why not just plug in the difference (10%) into the equation? That's what I did first but I got a wrong answer. I read Hero8888 's explanation but I don't understand.

How come getting the percent difference here will not work? For me it seems logically sound. Where does this logic go wrong here?

Thanks.

I have the same question whey are you putting 90 or 80 instead of 0.90 and 0.80?
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The cost C, in dollars, to remove p percent of a certain pollutant fro  [#permalink]

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12 Nov 2019, 11:32
Diwabag wrote:
Hi Bunuel ,

Why not just plug in the difference (10%) into the equation? That's what I did first but I got a wrong answer. I read Hero8888 's explanation but I don't understand.

How come getting the percent difference here will not work? For me it seems logically sound. Where does this logic go wrong here?

Thanks.

Hi - the only way plugging in 10 % (difference between 90 and 80) will work is if the variable x is attached to ALL elements of the equation ...example below

A) 10 x - (4/5) x --> if you plug in 90 and 80 separately and subtract or just plug in 10 %, you will get the same answer
B) 18x + 5x ------> if you plug in 90 and 80 separately and subtract or just plug in 10 %, you will get the same answer
C) 10 x - 6x --------> if you plug in 90 and 80 separately and subtract or just plug in 10 %, you will get the same answer
D) 125 x / 6 --------> if you plug in 90 and 80 separately and subtract or just plug in 10 %, you will get the same answer

However If you have the following equation where X is not attached to the constant -- plugging in 10 % will fail
example : 100 + (x) --> if you plug in 90 and 80 separately and subtract or just plug in 10 , you will get DIFFERENT answers

Logic : when you are subtracting (90 - 80) to give you 10 % --> you are getting rid of the constant within the equation (in above case, 100 in blue color) ..when you plug in 10 %, the constant remains...

In this equation, the numerator has the x attached to it, but the 100 underline does not and thus will fail if you put in just a 10
I) 100,000 * X / (100-X)
The cost C, in dollars, to remove p percent of a certain pollutant fro   [#permalink] 12 Nov 2019, 11:32