Sharing Donuts: A Fun Math Problem Solved

Imagine you have 6 delicious donuts and you want to share 1/3 of them with a friend, while keeping the rest for yourself. How many donuts will you have left? Let’s break it down step by step to find out.

Step-by-Step Solution

Calculate 1/3 of 6 donuts:

The fraction 1/3 means that the donuts are divided into 3 equal parts, and you take 1 part. So, we divide 6 by 3:

6 ÷ 3 2

Subtract the number of donuts given to your friend from the total:

Now, subtract the 2 donuts you gave to your friend from the original 6 donuts:

6 - 2 4

Therefore, you will have 4 donuts left for yourself.

Common Misconceptions

Some people might think you will have 5.66 donuts, or that your friend gets two-thirds of a donut, or that there is a specific answer written in a book. However, the actual problem does not specify any fractional donuts, so according to the steps above, 4 whole donuts is the correct answer.

Alternative Scenarios

The problem is not limited to just the scenario described. You could also give away 2/3 of the donuts, which is the same as giving 4/6 of them.

Here's how you can solve this alternative:

Multiply the fraction 2/3 by 6 to find out how many donuts your friend will get:

2/3 of 6 4/6 of 6 (2/3) × 6 4

Your friend would get 4 donuts, and you would have 6 - 4 2 donuts left.

Multiple Possibilities

The problem allows for different interpretations. Here are some possible scenarios:

Give 2 donuts to your friend, leaving 4 for yourself. Give 1/3 of your 6 donuts (which is 2 donuts), leaving 4 for yourself. Give 1/3 of what you have left after giving 1 donut to your friend (which is 5 donuts), and you would have 5 2/3 donuts left. Eat 1 or more donuts and give 1/3 of the remainder to your friend. For example, if you eat 1 donut, you have 5 left. Giving 1/3 of 5 (which is 1.67) to your friend, you would have 3.33 donuts left.

These scenarios show the flexibility of the problem and highlight how different interpretations can lead to various results, making it a fun and educational math problem.