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# %-drop

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I've got a question that goes:

"Person drank 22 litres of alcohol in 1987. In 2001 it's down to 9.3 litres of alcohol per person."

Assignment: Find the %-drop.

The result list says that it's 5.97% - I'm having issues getting to that number.

Nov 7, 2015

#7
+10

Mmmmm.....OK.......I didn't see the words " yearly % percent drop" anywhere in the question......!!!

Here's how to solve this :

9.3 = 22 (1 - r)^14    where r is the percent decrease [ in decimal form] per year

Divide both sides by 22

9.3 / 22  = (1 - r)^14

Take the log of both sides :

log (9.3 / 22) =  log (1 - r)^14

And by a log property, we can write :

log (9.3 / 22)  = 14 * log (1 - r)

Divide both sides by 14

log (9.3 / 22) / 14  = log (1 - r)

And....in exponential form we have

10^ [ log (9.3 / 22) / 14 ]  = 1 - r

Rearrange

r = 1 - 10^ [ log (9.3 / 22) / 14 ]  = about 0.0596489...  or about  5.96489% per year   Nov 7, 2015
edited by CPhill  Nov 7, 2015

#1
+5

19.954%

Nov 7, 2015
#2
+5

The percent drop is given by :

( [22 - 9.3] / 22  * 100 ) %  = about 57.72 %   Nov 7, 2015
#3
+5

"Person drank 22 litres of alcohol in 1987. In 2001 it's down to 9.3 litres of alcohol per person."

Assignment: Find the %-drop.

The result list says that it's 5.97% - I'm having issues getting to that number.

1 - (9.3/22)=0.5773 X 100=57.73% decrease in alcohol consumption in 14 years from 1987 to 2001

So that 1 - .5773=.4227^1/14=.94035-1=-.05965 X 100=-5.97%decrease in alcohol consumption per year from 1987 to 2001.

Nov 7, 2015
#4
+5

I'll try to be a bit more thorough.

The things I'm told is:

"In 2001 each Greenlandish person drank 9.3 litres of pure alcohol, which is a lot less than what they did in the 1980's. Their alltime highest was in 1987, where they topped at 22 litres of alcohol per citizen per year."

Assignment:

"Read the thing above. Define, with two decimals, the average yearly percentage drop in alcohol consumption per person in the period 1987-2001."

It's Math where I have to explain what I do and such, so I'm given the results-list from the beginning.

The result according to this list is 5.97% (drop on average per year per person).

However, despite many attempts, different approaches I'm no where near this number.

$$r = 1+{\sqrt[-14]{9.3/22}}$$(add 9.3_14, and 22_0)

And I tried:

$$r = 1-{\sqrt[-15]{9.3/22}}$$(22_15/9.3_0)

Which  was 5.9081%

I've tried talking it with a math buddy of mine, none of us managed to get it to the right result of "a drop of 5.97%"

Nov 7, 2015
#5
+5

Oh, #3 you wrote that as I did mine.

You're way seems right, I'll take a look at it and try to see if I understand it correctly. Thank you!

Nov 7, 2015
#6
+5

Guest #5

It is actually very simple. If you don't understand it, just let me know. Guest #3.

Nov 7, 2015
#7
+10

Mmmmm.....OK.......I didn't see the words " yearly % percent drop" anywhere in the question......!!!

Here's how to solve this :

9.3 = 22 (1 - r)^14    where r is the percent decrease [ in decimal form] per year

Divide both sides by 22

9.3 / 22  = (1 - r)^14

Take the log of both sides :

log (9.3 / 22) =  log (1 - r)^14

And by a log property, we can write :

log (9.3 / 22)  = 14 * log (1 - r)

Divide both sides by 14

log (9.3 / 22) / 14  = log (1 - r)

And....in exponential form we have

10^ [ log (9.3 / 22) / 14 ]  = 1 - r

Rearrange

r = 1 - 10^ [ log (9.3 / 22) / 14 ]  = about 0.0596489...  or about  5.96489% per year   CPhill Nov 7, 2015
edited by CPhill  Nov 7, 2015
#8
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This is a primitive percentage calculation!!.

9.3/22=0.42272.....^(1/14)=0.94035... - 1=-0.05965 X 100=-5.965%

Nov 7, 2015
#9
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Thank you a ton for all of your help!

Much much appreciated.

Nov 9, 2015