Question
Beth and Joanne have $2880 in all. Beth gave 1/4 of her money to Joanne. Then Joanne gave 1/3 of the total amount that she had to Beth. Now both girls have the same amount of money. How much did Beth have at first?
Answers
GPT-4o mini
Let \( B \) be the initial amount of money Beth had, and \( J \) be the initial amount of money Joanne had. According to the problem, we have the following equation for the total amount:
\[
B + J = 2880
\]
Next, Beth gave \( \frac{1}{4}B \) to Joanne. After this transaction:
- Beth's new amount:
\[
B - \frac{1}{4}B = \frac{3}{4}B
\]
- Joanne's new amount:
\[
J + \frac{1}{4}B
\]
Then, Joanne gave \( \frac{1}{3} \) of her new total amount to Beth. Calculating what Joanne had after receiving from Beth:
\[
\text{Joanne's new amount} = J + \frac{1}{4}B
\]
Now, Joanne gives \( \frac{1}{3} \) of this total amount to Beth:
\[
\frac{1}{3} \left( J + \frac{1}{4}B \right)
\]
Beth receives this amount, so her total after Joanne's gift becomes:
\[
\frac{3}{4}B + \frac{1}{3} \left( J + \frac{1}{4}B \right)
\]
We simplify the expression for Beth's new amount:
\[
\text{Beth's new amount} = \frac{3}{4}B + \frac{1}{3}(J + \frac{1}{4}B)
\]
Distributing \( \frac{1}{3} \):
\[
\frac{1}{3}J + \frac{1}{12}B
\]
Now adding this to Beth's amount:
\[
\text{Beth's new amount} = \frac{3}{4}B + \frac{1}{3}J + \frac{1}{12}B
\]
We need to convert fractions to a common base to simplify. The least common multiple of 4, 3, and 12 is 12:
\[
\frac{3}{4}B = \frac{9}{12}B, \quad \frac{1}{3}J = \frac{4}{12}J, \quad \frac{1}{12}B = \frac{1}{12}B
\]
Now, combine these:
\[
\text{Beth's new amount} = \frac{9}{12}B + \frac{4}{12}J + \frac{1}{12}B = \frac{10}{12}B + \frac{4}{12}J = \frac{5}{6}B + \frac{1}{3}J
\]
Now we calculate the total amount Joanne has after her own transaction. After giving some to Beth, her amount is:
\[
\text{Joanne's new amount} = J + \frac{1}{4}B - \frac{1}{3}(J + \frac{1}{4}B)
\]
Using a similar process to distribute \( -\frac{1}{3} \):
\[
-\frac{1}{3}J - \frac{1}{12}B
\]
Then we have:
\[
\text{Joanne's new amount} = J + \frac{1}{4}B - \frac{1}{3}J - \frac{1}{12}B
\]
Converting these terms to a common base of 12:
- \( J = \frac{12}{12}J \)
- \( \frac{1}{4}B = \frac{3}{12}B \)
So Joanne's amount simplifies to:
\[
\text{Joanne's new amount} = \frac{12}{12}J - \frac{4}{12}J + \frac{3}{12}B - \frac{1}{12}B = \frac{8}{12}J + \frac{2}{12}B = \frac{2}{3}J + \frac{1}{6}B
\]
Now, according to the problem, now both girls have the same amount, so we set their amounts equal:
\[
\frac{5}{6}B + \frac{1}{3}J = \frac{2}{3}J + \frac{1}{6}B
\]
Clearing fractions by multiplying by 6:
\[
5B + 2J = 4J + B
\]
Now we isolate terms:
\[
5B - B = 4J - 2J
\]
\[
4B = 2J
\]
\[
2B = J
\]
Now substitute \( J = 2B \) into the total equation \( B + J = 2880 \):
\[
B + 2B = 2880
\]
\[
3B = 2880
\]
\[
B = 960
\]
Now, substituting back for \( J \):
\[
J = 2B = 2 \times 960 = 1920
\]
Thus, Beth initially had:
\[
\boxed{960}
\]
\[
B + J = 2880
\]
Next, Beth gave \( \frac{1}{4}B \) to Joanne. After this transaction:
- Beth's new amount:
\[
B - \frac{1}{4}B = \frac{3}{4}B
\]
- Joanne's new amount:
\[
J + \frac{1}{4}B
\]
Then, Joanne gave \( \frac{1}{3} \) of her new total amount to Beth. Calculating what Joanne had after receiving from Beth:
\[
\text{Joanne's new amount} = J + \frac{1}{4}B
\]
Now, Joanne gives \( \frac{1}{3} \) of this total amount to Beth:
\[
\frac{1}{3} \left( J + \frac{1}{4}B \right)
\]
Beth receives this amount, so her total after Joanne's gift becomes:
\[
\frac{3}{4}B + \frac{1}{3} \left( J + \frac{1}{4}B \right)
\]
We simplify the expression for Beth's new amount:
\[
\text{Beth's new amount} = \frac{3}{4}B + \frac{1}{3}(J + \frac{1}{4}B)
\]
Distributing \( \frac{1}{3} \):
\[
\frac{1}{3}J + \frac{1}{12}B
\]
Now adding this to Beth's amount:
\[
\text{Beth's new amount} = \frac{3}{4}B + \frac{1}{3}J + \frac{1}{12}B
\]
We need to convert fractions to a common base to simplify. The least common multiple of 4, 3, and 12 is 12:
\[
\frac{3}{4}B = \frac{9}{12}B, \quad \frac{1}{3}J = \frac{4}{12}J, \quad \frac{1}{12}B = \frac{1}{12}B
\]
Now, combine these:
\[
\text{Beth's new amount} = \frac{9}{12}B + \frac{4}{12}J + \frac{1}{12}B = \frac{10}{12}B + \frac{4}{12}J = \frac{5}{6}B + \frac{1}{3}J
\]
Now we calculate the total amount Joanne has after her own transaction. After giving some to Beth, her amount is:
\[
\text{Joanne's new amount} = J + \frac{1}{4}B - \frac{1}{3}(J + \frac{1}{4}B)
\]
Using a similar process to distribute \( -\frac{1}{3} \):
\[
-\frac{1}{3}J - \frac{1}{12}B
\]
Then we have:
\[
\text{Joanne's new amount} = J + \frac{1}{4}B - \frac{1}{3}J - \frac{1}{12}B
\]
Converting these terms to a common base of 12:
- \( J = \frac{12}{12}J \)
- \( \frac{1}{4}B = \frac{3}{12}B \)
So Joanne's amount simplifies to:
\[
\text{Joanne's new amount} = \frac{12}{12}J - \frac{4}{12}J + \frac{3}{12}B - \frac{1}{12}B = \frac{8}{12}J + \frac{2}{12}B = \frac{2}{3}J + \frac{1}{6}B
\]
Now, according to the problem, now both girls have the same amount, so we set their amounts equal:
\[
\frac{5}{6}B + \frac{1}{3}J = \frac{2}{3}J + \frac{1}{6}B
\]
Clearing fractions by multiplying by 6:
\[
5B + 2J = 4J + B
\]
Now we isolate terms:
\[
5B - B = 4J - 2J
\]
\[
4B = 2J
\]
\[
2B = J
\]
Now substitute \( J = 2B \) into the total equation \( B + J = 2880 \):
\[
B + 2B = 2880
\]
\[
3B = 2880
\]
\[
B = 960
\]
Now, substituting back for \( J \):
\[
J = 2B = 2 \times 960 = 1920
\]
Thus, Beth initially had:
\[
\boxed{960}
\]