Swiss Chocolates, Lollipops, Gumdrops

Date: 02/10/2003 at 23:53:47
From: Della DA
Subject: Math

You have $100 and must buy 100 pieces of candy.

   Imported Swiss chocolates cost $15 each.
   Lollipops are $1 each.
   Gumdrops are $.25 each.

You must purchase at least one of each item of candy.
Pieces of candy must be whole.

How many of each candy do you buy to spend $100 and to get exactly 100 
pieces?

Date: 02/11/2003 at 01:14:24
From: Doctor Jeremiah
Subject: Re: Math

Hi Della,

You should have tried figuring it out for 3 Swiss chocolates. Do that 
now, and if you still can't get an answer you like then read what 
follows.

So that I don't have to type so much I am going to use C for the 
number of Swiss chocolates, L for the number of lollipops, and G for 
the number of gumdrops.

If there is a total of 100 pieces, then the number of chocolates (C)
plus the number of lollipops (L) plus the number of gumdrops (G)
must equal 100, because the sum of the number of all three types
must be 100. So:   C + L + G = 100

   If 1 chocolate costs $15 and C is the number of Swiss chocolates, 
   then the cost of all the chocolates must be $15 times C.

   If 1 lollipop costs $1 and L is the number of lollipops, then the 
   cost of all the lollipops must be $1 times L.
 
   If 1 gumdrop costs $0.25 and "G" is the number of gumdrops, then 
   the cost of all the gumdrops must be $0.25 times G.

The total cost of all three types is $100, so the cost of the Swiss 
chocolates ($15C) plus the cost of the lollipops ($1L) plus the cost 
of the gumdrops ($0.25G) must be $100. The equation for that is:   
$15C + $1L + $0.25G = $100

Now, we have two equations:

     C + L + G = 100
     15C + L + 0.25G = 100

But we have three unknown values (C, L, and G), so we can't yet solve 
this. We either need one more piece of information or we need to 
assume a value for the number of one of the types. Swiss chocolate 
costs $15, so we need at least one but not more than six. This is true 
because $15 times 7 = $105. So let's start by assuming that the number 
of Swiss chocolates is one.

Now we have three equations:

     C + L + G = 100
     15C + L + 0.25G = 100
     C = 1

Since the third equation says that C=1 we can change all the C in the 
other two equations to 1 and get rid of the dependence on C. That 
leaves us with two equations with two unknowns:

     1 + L + G = 100
     15 + L + 0.25G = 100

To solve this for L and G we must get one of the unknowns by itself. I 
will pick L purely at random. After a bit of rearrangement of the 
first equation we get:

     L = 99 - G

Since L is equal to 99-G we can change all the L in the other equation 
to 99-G and get this equation:

     15 + 99-G + 0.25G = 100

Which can be solved for G like this:

     15 + L + 0.25G = 100
     15 + 99-G + 0.25G = 100
     15 + 99 - 100 = G - 0.25G
     14 = 0.75G
     14/0.75 = G
     56/3 = G

Notice that this isn't a whole number. Since you can't buy part of a 
gumdrop, the number of Swiss chocolates cannot be one. So let's start 
over and assume that the number of Swiss chocolates is two.  

Now we have these three equations:

     C + L + G = 100
     15C + L + 0.25G = 100
     C = 2

Since the third equation says that C=2 we can change all the C in the 
other two equations to 2 and get rid of the dependence on C. That 
leaves us with two equations with two unknowns:

     2 + L + G = 100
     30 + L + 0.25G = 100

To solve this for L and G we must get one of the unknowns by itself. I 
will pick L purely at random. After a bit of rearrangement of the 
first equation we get:

     L = 98 - G

Since L is equal to 98-G we can change all the L in the other equation 
to 98-G and get this equation:

     30 + L + 0.25G = 100
     15 + 98-G + 0.25G = 100
     15 + 98 - 100 = G - 0.25G
     13 = 0.75G
     13/0.75 = G
     52/3 = G

Notice that this isn't a whole number. Since you can't buy part of a 
gumdrop, the number of Swiss chocolates cannot be two. So let's go 
back and assume that the number of Swiss chocolates is three.  

Now we have three equations:

     C + L + G = 100
     15C + L + 0.25G = 100
     C = 3

Since the third equation says that C=3 we can change all the C in the 
other two equations to 3 and get rid of the dependence on C. That 
leaves us with two equations with two unknowns:

     3 + L + G = 100
     45 + L + 0.25G = 100

To solve this for L and G we must get one of the unknowns by itself. I 
will pick L purely at random. After a bit of rearrangement of the 
first equation we get:

     L = 97 - G

Since L is equal to 97-G we can change all the L in the other equation 
to 97-G and get this equation:

     45 + L + 0.25G = 100
     45 + 97-G + 0.25G = 100
     45 + 97 - 100 = G - 0.25G
     42 = 0.75G
     42/0.75 = G
     56 = G

This is a whole number but it might still not be the right answer. So 
we need to solve for L and then check the results. 

Remember that:

     L = 97 - G

And since G=56 we can solve for L and get:  L = 41

The question is, will the original two equations work with L = 41, 
G = 56, and C = 3?

     C + L + G = 100
     15C + L + 0.25G = 100

Becomes:

     3 + 41 + 56 = 100
     45 + 41 + 14 = 100

And those two equations are definitely true.

    3 + 41 + 56 does equal 100
   45 + 41 + 14 does equal 100

This means that the number of each type is: L = 41, G = 56, and C = 3.

Which means that there are 3 Swiss chocolates, 41 lollipops, and 56 
gumdrops.

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/