Sasa had some 20-cent and 50-cent coins. Let x be the total number of coins. Then, 7/8 x were 20-cent coins, and the rest, 1 - 7/8 = 1/8 x, were 50-cent coins. After Sasa spent $72.50 worth of 50-cent coins, which is equivalent to 72.50 / 0.50 = 145 50-cent coins, and 5/7 of the 20-cent coins, she had 2/7 x coins left. Find the total amount of money Sasa left.

We are given the following information:

- $ x $ is the total number of coins.
- $ \frac{7}{8}x $ are 20-cent coins, and $ \frac{1}{8}x $ are 50-cent coins.
- Sasa spent 145 of her 50-cent coins and $ \frac{5}{7} $ of her 20-cent coins.
- After spending the coins, Sasa had $ \frac{2}{7}x $ coins left.

### Step 1: Express the total number of coins spent

We know Sasa had $ \frac{1}{8}x $ 50-cent coins initially. After spending 145 50-cent coins, the number of remaining 50-cent coins is:

\[
\text{Remaining 50-cent coins} = \frac{1}{8}x - 145
\]

Next, Sasa had $ \frac{7}{8}x $ 20-cent coins. She spent $ \frac{5}{7} $ of these, so the remaining 20-cent coins are:

\[
\text{Remaining 20-cent coins} = \frac{7}{8}x \times \left(1 - \frac{5}{7}\right) = \frac{7}{8}x \times \frac{2}{7} = \frac{1}{4}x
\]

Thus, the total number of coins left after spending both types of coins is:

\[
\text{Total remaining coins} = \left( \frac{1}{8}x - 145 \right) + \frac{1}{4}x
\]

### Step 2: Set up the equation

We are told that after spending the coins, Sasa had $ \frac{2}{7}x $ coins left. So we equate this to the expression for the total remaining coins:

\[
\frac{2}{7}x = \left( \frac{1}{8}x - 145 \right) + \frac{1}{4}x
\]

### Step 3: Solve the equation

First, let's combine like terms on the right-hand side. To do this, we need a common denominator between $ \frac{1}{8}x $ and $ \frac{1}{4}x $. The least common denominator is 8, so:

\[
\frac{1}{4}x = \frac{2}{8}x
\]

Thus, the equation becomes:

\[
\frac{2}{7}x = \frac{1}{8}x + \frac{2}{8}x - 145 = \frac{3}{8}x - 145
\]

Now, solve for $ x $ by eliminating the fractions. Multiply through by 56 (the least common multiple of 7 and 8):

\[
56 \times \frac{2}{7}x = 56 \times \left( \frac{3}{8}x - 145 \right)
\]

This simplifies to:

\[
16x = 21x - 8120
\]

Now, isolate $ x $:

\[
16x - 21x = -8120
\]

\[
-5x = -8120
\]

\[
x = \frac{-8120}{-5} = 1624
\]

So, the total number of coins $ x $ is 1624.

### Step 4: Calculate the total amount of money Sasa had left

We now know that Sasa had 1624 coins in total. Let's calculate how much money she had left.

- The remaining 50-cent coins are $ \frac{1}{8} \times 1624 - 145 = 203 - 145 = 58 $.
- The remaining 20-cent coins are $ \frac{1}{4} \times 1624 = 406 $.

The total amount of money Sasa had left is:

\[
\text{Total money left} = 58 \times 0.50 + 406 \times 0.20 = 29 + 81.2 = 110.2
\]

Thus, Sasa had **$110.20** left.

Would you like further clarifications or details on any step?

Here are 5 related questions:
1. How do you calculate total money from coin counts?
2. What is the relationship between fractions of a whole and percentages?
3. How can you set up equations based on word problems?
4. What are some strategies for solving linear equations with fractions?
5. How do you combine terms with different denominators in an equation?

**Tip:** When dealing with fractions in equations, multiplying through by the least common denominator simplifies the calculations.

Learn how to solve a coin problem involving fractions and linear equations, where Sasa spends 50-cent and 20-cent coins. The problem explores concepts in algebra, such as solving equations with fractions, and is suitable for students in Grades 9-12.

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