+0

# %-drop

+5
432
9

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.

Guest Nov 7, 2015

#7
+81029
+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
Sort:

#1
+104
+5

19.954%

cars_456  Nov 7, 2015
#2
+81029
+5

The percent drop is given by :

( [22 - 9.3] / 22  * 100 ) %  = about 57.72 %

CPhill  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.

Guest 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%"

Guest 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!

Guest Nov 7, 2015
#6
+5

Guest #5

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

Guest Nov 7, 2015
#7
+81029
+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
0

This is a primitive percentage calculation!!.

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

Guest Nov 7, 2015
#9
0

Thank you a ton for all of your help!

Much much appreciated.

Guest Nov 9, 2015

### 4 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details