Three brothers divide 17 candy bars. The oldest child gets 1/2, the middle child 1/3, and the youngest child 1/9. So that they do not have to cut any bars, Mom provides an extra one. The oldest child then takes 9 candy bars, the middle child 6, and the youngest 2, and they return the extra one. Which child received more that his fair share?

Question

Three brothers divide 17 candy bars. The oldest child gets 1/2, the middle child 1/3, and the youngest child 1/9. So that they do not have to cut any bars, Mom provides an extra one. The oldest child then takes 9 candy bars, the middle child 6, and the youngest 2, and they return the extra one. Which child received more that his fair share?

Answer 1

9 - 17/2 = .500

6 - 17/3 - .333
2 - 17/9 = .111

Your call.

Answer 2

To determine which child received more than his fair share, we can calculate the fair share for each child and compare it to the actual number of candy bars they received.

Let's start by calculating the fair share for each child:

1. Oldest child: 1/2 of the candy bars
Fair share = (1/2) * 17 = 8.5

2. Middle child: 1/3 of the candy bars
Fair share = (1/3) * 17 = 5.67 (rounded to two decimal places)

3. Youngest child: 1/9 of the candy bars
Fair share = (1/9) * 17 = 1.89 (rounded to two decimal places)

Now, let's compare the fair shares to the actual number of candy bars received:

1. Oldest child: received 9 candy bars
Difference = 9 - 8.5 = 0.5

2. Middle child: received 6 candy bars
Difference = 6 - 5.67 = 0.33

3. Youngest child: received 2 candy bars
Difference = 2 - 1.89 = 0.11

Based on the calculations above, we can see that the oldest child received 0.5 more candy bars than his fair share. Therefore, the oldest child received more than his fair share.

Answer 3

To determine which child received more than his fair share, let's first calculate the fair share for each child.

The problem states that the oldest child receives half (1/2) of the candy bars, the middle child receives one-third (1/3), and the youngest child receives one-ninth (1/9).

Let's start by calculating how many candy bars there are in total by adding the candy bars each child receives:

Oldest child: 1/2 of the candy bars = (1/2) * 17 = 8.5 candy bars
Middle child: 1/3 of the candy bars = (1/3) * 17 = 5.67 candy bars (rounded to two decimal places)
Youngest child: 1/9 of the candy bars = (1/9) * 17 = 1.89 candy bars (rounded to two decimal places)

However, since we can't divide a candy bar into fractions, the problem states that Mom provides an extra candy bar so that they don't have to cut any bars.

So, with the extra candy bar, the total number of candy bars available is now 18.

Then, let's see how many candy bars each child takes:

Oldest child: 9 candy bars
Middle child: 6 candy bars
Youngest child: 2 candy bars

Now, we need to subtract the number of candy bars each child took from their fair share to determine if any child received more than their fair share.

Oldest child's fair share: 8.5 candy bars
Oldest child took: 9 candy bars
Excess: 9 - 8.5 = 0.5 candy bars

Middle child's fair share: 5.67 candy bars
Middle child took: 6 candy bars
Excess: 6 - 5.67 = 0.33 candy bars (rounded to two decimal places)

Youngest child's fair share: 1.89 candy bars
Youngest child took: 2 candy bars
Excess: 2 - 1.89 = 0.11 candy bars (rounded to two decimal places)

Therefore, both the oldest child and the middle child received more than their fair share - the oldest child by 0.5 candy bars and the middle child by 0.33 candy bars. The youngest child received his fair share.