Justine is visiting Switzerland, a country known for its world-famous 
chocolate. Justine would like to purchase some chocolate to give to her 
friends as souvenirs from her trip. The Petit Chocolatier shop offers pieces 
of chocolate for $0.10 each, bars of chocolate for $2.00, and boxes of 
chocolate for $5.00. If Justine spent exactly $100.00 and purchased 100 
total items, how many of each type of chocolate did she purchase?

This problem can be solved using a system of equations. Let's define variables for the number of each type of chocolate Justine buys:

- Let $ x $ be the number of pieces of chocolate.
- Let $ y $ be the number of bars of chocolate.
- Let $ z $ be the number of boxes of chocolate.

We have two key conditions:
1. The total number of items purchased is 100:
\[
x + y + z = 100
\]
2. The total amount of money spent is $100. Given the prices:
   - $0.10 per piece of chocolate,
   - $2.00 per bar of chocolate,
   - $5.00 per box of chocolate,
   
The total cost equation is:
\[
0.10x + 2y + 5z = 100
\]

Now, we can work on solving these equations. To simplify the second equation, multiply it by 10 to remove the decimal:
\[
x + 20y + 50z = 1000
\]

Now we have the system of equations:
\[
1. \, x + y + z = 100
\]
\[
2. \, x + 20y + 50z = 1000
\]

We can solve this system by substitution or elimination. Let's solve for $ x $ in terms of $ y $ and $ z $ using the first equation:
\[
x = 100 - y - z
\]

Substitute this expression for $ x $ into the second equation:
\[
(100 - y - z) + 20y + 50z = 1000
\]
Simplify:
\[
100 - y - z + 20y + 50z = 1000
\]
\[
100 + 19y + 49z = 1000
\]
Subtract 100 from both sides:
\[
19y + 49z = 900
\]

Now, we can solve this equation using trial and error or algebraic methods, looking for integer solutions for $ y $ and $ z $. Let's start with trial and error:

1. If $ z = 10 $:
   \[
   19y + 49(10) = 900
   \]
   \[
   19y + 490 = 900
   \]
   \[
   19y = 410
   \]
   \[
   y = \frac{410}{19} = 21.58 \, (\text{not an integer, discard this})
   \]

2. If $ z = 12 $:
   \[
   19y + 49(12) = 900
   \]
   \[
   19y + 588 = 900
   \]
   \[
   19y = 312
   \]
   \[
   y = \frac{312}{19} = 16.42 \, (\text{not an integer, discard this})
   \]

3. If $ z = 18 $:
   \[
   19y + 49(18) = 900
   \]
   \[
   19y + 882 = 900
   \]
   \[
   19y = 18
   \]
   \[
   y = \frac{18}{19} = 0.95
   \

Justine is visiting Switzerland, a country known for its world-famous chocolate. Justine would like to purchase some chocolate to give to her friends as souvenirs from her trip. The Petit Chocolatier shop offers pieces of chocolate for $0.10 each, bars of chocolate for $2.00, and boxes of chocolate for $5.00. If Justine spent exactly $100.00 and purchased 100 total items, how many of each type of chocolate did she purchase?

Discover how to solve a real-life problem using algebra by determining the number of pieces, bars, and boxes of chocolate Justine purchased in Switzerland, given a total of 100 items and a $100 budget. This problem involves solving a system of linear equations, suitable for students in Grades 9-11.

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