Question

The ages of four sisters are consecutive odd integers with the sum of 200. How old are the sisters? N+ n+2 n+4 n+6=200

Answer 1

To find the ages of the four sisters, we can define their ages using consecutive odd integers. Let's denote the youngest sister's age as \( n \). The ages of the four sisters can then be expressed as:

- Youngest sister: \( n \)
- Second sister: \( n + 2 \)
- Third sister: \( n + 4 \)
- Oldest sister: \( n + 6 \)

According to the problem, the sum of their ages is 200:

\[
n + (n + 2) + (n + 4) + (n + 6) = 200
\]

Now, we can combine like terms:

\[
n + n + 2 + n + 4 + n + 6 = 200
\]

This simplifies to:

\[
4n + 12 = 200
\]

Next, we'll isolate \( n \) by subtracting 12 from both sides:

\[
4n = 200 - 12
\]
\[
4n = 188
\]

Now, divide both sides by 4:

\[
n = \frac{188}{4} = 47
\]

Now that we have \( n \), we can find the ages of each sister:

- Youngest sister: \( n = 47 \)
- Second sister: \( n + 2 = 49 \)
- Third sister: \( n + 4 = 51 \)
- Oldest sister: \( n + 6 = 53 \)

Thus, the ages of the sisters are:

- 47 years
- 49 years
- 51 years
- 53 years

To confirm, let's check the sum:

\[
47 + 49 + 51 + 53 = 200
\]

Therefore, the ages of the sisters are **47**, **49**, **51**, and **53**.