Question

activity guide - candy store challenge
name(s)
period
date
candy: gummy bears, chocolate, licorice, mints
decimal price: $1.76/lb, $4.16/lb, $7.52/lb, $0.48/lb
binary price: (empty)
computer science principles

Answer:

To convert decimal prices to binary, we'll take each price (in dollars, considering the decimal part as cents for easier conversion, or directly converting the decimal fraction). Let's start with one of the candies, say Gummy Bears with a price of \$1.76/lb.

Step 1: Separate the integer and fractional parts

The price is \( 1.76 \) dollars. The integer part is \( 1 \), and the fractional part is \( 0.76 \).

Step 2: Convert the integer part to binary

The integer part \( 1 \) in binary is \( 1 \).

Step 3: Convert the fractional part to binary

To convert the fractional part \( 0.76 \) to binary, we multiply by 2 repeatedly and record the integer parts:

• \( 0.76 \times 2 = 1.52 \), integer part \( 1 \)
• \( 0.52 \times 2 = 1.04 \), integer part \( 1 \)
• \( 0.04 \times 2 = 0.08 \), integer part \( 0 \)
• \( 0.08 \times 2 = 0.16 \), integer part \( 0 \)
• \( 0.16 \times 2 = 0.32 \), integer part \( 0 \)
• \( 0.32 \times 2 = 0.64 \), integer part \( 0 \)
• \( 0.64 \times 2 = 1.28 \), integer part \( 1 \)
• \( 0.28 \times 2 = 0.56 \), integer part \( 0 \)
• \( 0.56 \times 2 = 1.12 \), integer part \( 1 \)
• \( 0.12 \times 2 = 0.24 \), integer part \( 0 \)

-... (we can stop here for a reasonable approximation)

Putting the fractional binary parts together, we have \( 0.1100001010... \)

Step 4: Combine integer and fractional parts

So, \( 1.76_{10} \approx 1.1100001010_2 \)

Let's do another one, Mints with a price of \$0.48/lb:

Step 1: Fractional part (integer part is 0)

\( 0.48 \)

Step 2: Convert fractional part to binary

• \( 0.48 \times 2 = 0.96 \), integer part \( 0 \)
• \( 0.96 \times 2 = 1.92 \), integer part \( 1 \)
• \( 0.92 \times 2 = 1.84 \), integer part \( 1 \)
• \( 0.84 \times 2 = 1.68 \), integer part \( 1 \)
• \( 0.68 \times 2 = 1.36 \), integer part \( 1 \)
• \( 0.36 \times 2 = 0.72 \), integer part \( 0 \)
• \( 0.72 \times 2 = 1.44 \), integer part \( 1 \)
• \( 0.44 \times 2 = 0.88 \), integer part \( 0 \)
• \( 0.88 \times 2 = 1.76 \), integer part \( 1 \)
• \( 0.76 \times 2 = 1.52 \), integer part \( 1 \)

-...

So, \( 0.48_{10} \approx 0.0111101011..._2 \)

For Licorice (\$7.52/lb):

Step 1: Integer part \( 7 \), fractional part \( 0.52 \)

• Integer part \( 7 \) in binary: \( 7 \div 2 = 3 \) remainder \( 1 \); \( 3 \div 2 = 1 \) remainder \( 1 \); \( 1 \div 2 = 0 \) remainder \( 1 \). So, \( 7_{10} = 111_2 \)
• Fractional part \( 0.52 \):
• \( 0.52 \times 2 = 1.04 \), integer part \( 1 \)
• \( 0.04 \times 2 = 0.08 \), integer part \( 0 \)
• \( 0.08 \times 2 = 0.16 \), integer part \( 0 \)
• \( 0.16 \times 2 = 0.32 \), integer part \( 0 \)
• \( 0.32 \times 2 = 0.64 \), integer part \( 0 \)
• \( 0.64 \times 2 = 1.28 \), integer part \( 1 \)

-...

So, \( 7.52_{10} \approx 111.100001..._2 \)

For Chocolate (\$4.16/lb):

Step 1: Integer part \( 4 \), fractional part \( 0.16 \)

• Integer part \( 4 \) in binary: \( 4 \div 2 = 2 \) remainder \( 0 \); \( 2 \div 2 = 1 \) remainder \( 0 \); \( 1 \div 2 = 0 \) remainder \( 1 \). So, \( 4_{10} = 100_2 \)
• Fractional part \( 0.16 \):
• \( 0.16 \times 2 = 0.32 \), integer part \( 0 \)
• \( 0.32 \times 2 = 0.64 \), integer part \( 0 \)
• \( 0.64 \times 2 = 1.28 \), integer part \( 1 \)
• \( 0.28 \times 2 = 0.56 \), integer part \( 0 \)
• \( 0.56 \times 2 = 1.12 \), integer part \( 1 \)

-...

So, \( 4.16_{10} \approx 100.00101..._2 \)

If we were to fill the table, we'd calculate the binary price for each candy using the above method (approximating the fractional part to a reasonable number of bits). For example, for Gummy Bears (\$1.76):

• Integer: \( 1_2 = 1 \)
• Fractional: \( 0.76 \approx 0.1100001010 \) (as above)
• So, binary price: \( 1.1100001010_2 \) (or rounded to a certain number of bits, e.g., 10 bits after the decimal: \( 1.1100001010_2 \))

Similarly, we can compute for the other candies. The key process is separating the integer and fractional parts, converting the integer via division by 2, and the fractional via multiplication by 2, then combining the results.