Converting Package Prices from Francs to Euros

  • Thread starter Thread starter chawki
  • Start date Start date
AI Thread Summary
In 2001, a product was sold in 450 g packages for 35 francs each. By 2002, the package size was reduced by 24%, resulting in a new weight of 342 g. The price per kilogram increased by 26%, leading to a new price of 98 francs per kilogram. The total price for the new package size is calculated to be 44.1 francs, which converts to approximately 7.42 euros using the exchange rate of 1 € = 5.94573 francs. The final price for the new package is confirmed as 7.42 euros.
chawki
Messages
504
Reaction score
0

Homework Statement


In the year 2001, a product was sold in 450 g packages for 35 francs per package. In the
beginning of the year 2002 the size of the package was reduced by 24 percent and the
price per kilogram was increased by 26 percent.

Homework Equations


Determine the price of the new package in euros (€) and cents rounded to the nearest cent when 1 € equals 5.94573 francs.

The Attempt at a Solution


the new size = 450*0.76 = 342g
the new price/kilogram = 35*1.26/0.45 = 98 francs

it means that the new price of the new package is 98*0.45 = 44.1 francs
1€ = 5.94573 francs
x = 44.1 francs
x = 7.41708 € and that's the new price per package in euro.
 
Physics news on Phys.org


The new price for the new package is: 98*0.342. It's not 98*0.45.
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Marketing Research (Cost, Revenue, Etc.)
Replies
11
Views
5K

Hot Threads

Recent Insights

Back
Top