I must spend exactly $80 to buy exactly 100 budgies. Blue budgies cost three dollars each green budgies cost two dollars each and red budgies cost $.50 each. if

Question

A student must spend exactly $80 to buy exactly 100 budgies. Blue budgies cost three dollars each green veggies cost two dollars each and yellow budgies cost $.50 each. if the student wants to ensure that they have at least one budgie of each color how many of each color do they need to buy?

Answer 1

Ryleigh,

 

First, I want to say that this is a great question. These problems really make you think. Before you read my answer, I want to urge you to stop and try to figure out the next steps before reading my whole answer. Here we go.

 

Let's first write the equations for this problem. The total cost for all the budgies is $80. We know we want a number of blue (b), green (g), and yellow (y) budgies and that number must be at least 1 for each color. We also know the price for each color.

 

The total cost for blue budgies is 3b. The total cost for green budgies is 2g. Finally, the total cost for yellow budgies is 0.5y. We know that the total cost of all the budgies together is 80.

 

1.  3b + 2g + 0.5y = 80.

 

Another bit of information we are given is the total number of budgies. All the blue, green, and yellow birds add up to 100.

 

2.  b + g + y = 100.

 

There are multiple ways to approach this kind of problem. You can try brute force and just plug in whole numbers until you find the right combination. Another way is to use the two equations to write a simpler one that only involves two variables and try to reason what the choices should be. I like that approach better than trying all the possible combinations. If you want to try all the possible combinations, I suggest writing a short computer program.

 

Before you go on, try to write an equation in only two variables using equations 1 and 2 above.

 

Unfortunately, equation 1 has that pesky 0.5 in it. I'm going to multiply both sides of the equation by 2 to get rid of it. Now equation 1 is

 

3. 6b + 4g + y = 160.

 

This doesn't change the answer. Now solve equation 2 above for y.

 

4. y = 100 - b - g.

 

Substitute equation 4 into equation 3 and simplify to get

 

6b + 4g + 100 - b - g = 160

 

5. 5b + 3g = 60.

 

Now we can make some observations about equation 5. Before you read further, try to use equation 5 to figure out what values of b and g make sense. Here are my observations.

• The term 5b will only take on values 5, 10, 15, 20,...,55 for whole number values of b. Since that is the case, we know we have to add a multiple of 5 to any of those numbers get a sum of 60.
• Since 3g must be a multiple of 5, our only options are g=5, 10, 15 corresponding to 3g = 15, 30, 45. We don't use g=20 because that would be 3g = 60, so no blue budgies would be possible.
• We also can't use b = 12 because 5b = 60 and no green budgies would be possible.

With this limited range of values for 5b and 3g, we can only choose the following pairs of values: 

 

g = 5, b = 9

g = 10, b = 6

g = 15, b = 3

 

The number of yellow birds comes from equation 4. We have the following values for y given the three options of blue and green:

 

y = 100 - 5 - 9 = 86

y = 100 - 10 - 6 = 84

y = 100 - 15 - 3 = 82

 

You can check the answers by computing the total cost 3b + 2g + 0.5y for each triple (b,g,y).

 

(b,g,y) = (9,5,86): 27 + 10 + 43 = 80

(b,g,y) = (6,10,84): 18 + 20 + 42 = 80

(b,g,y) = (3,15,82): 9 + 30 + 41 = 80.

 

I hope this helps.

Answer 2

Let the number of Blue, Green, and Red items be represented by B, R, and G.

With a computer, you would consider:
   For Values of B from 1 to 100
     For Values of G from 1 to 100
       For Values of Y from 1 to 100
         If ((B+G+Y)=100) AND (3*B+2*B+0.5*Y)=80) THEN OUTPUT (B,G,Y)
      Next Y

    Next G

  Next B

Extremely quickly, this produces:
   3 15 82
   6 10 84
   9  5  86

Do these answers give us a clue for solving it logically using math?

Yes, notice whether the numbers are "even" or "odd". Also, multiples of 5 and multiples of 3 are listed. The number of Y items seems to be a “fill-in-the-rest” number.

First, you should realize that there MUST be an even number of Y (since exactly $80 must be spent). If there are an even number of Y items, the number of B items and the number of G items must both be odd [since the total is even (100)].

The important constraint is that the total number of items is even (100) and the total cost is even ($80). Let’s express that:

       B + G + Y = 100        [eq1]

     3B + 2G + 0.5Y = 80    [eq2]

    6B + 4G + Y = 160      [2*eq2]

      B +  G + Y = 100       [eq1]

   ----------------------    [eliminate Y; subtract equations]

    5B + 3G      =   60    [eq3]

    3G = 60 – 5B     [from eq3]

    G = (60-5B)/3       [eq4; and G must be an integer at least 1]

This limits G to 5, 10, and 15. Why?
      Possible values of 5B:      5 10 15 20 25 30 25 40 45 50 55
            Corresponding G:      *   *  15  *   *  10  *   *   5   *   *

       So, corresponding B:      *   *   3   *   *   6  *    *   9   *   *

       [the values of G are all odd and are multiples of 5]

        [the values of B are all odd and are multiples of 3]

How, “fill in” the value of Y to make both eq1 and eq2 true.
      The answers (B,G,Y) are (3,15,82), (6,10,84) and (9,5,86).