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.

Let $ x $ be the total number of coins Sasa originally had.

### Step 1: Determine the number of 20-cent and 50-cent coins.
- The number of 20-cent coins is $ \frac{7}{8}x $.
- The number of 50-cent coins is $ \frac{1}{8}x $.

### Step 2: Account for the coins spent.
- Sasa spent 145 of her 50-cent coins, leaving $ \frac{1}{8}x - 145 $ 50-cent coins.
- Sasa also spent $ \frac{5}{7} $ of her 20-cent coins, leaving $ \frac{2}{7} \times \frac{7}{8}x = \frac{1}{4}x $ 20-cent coins.

After spending these coins, the total number of coins left is given as $ \frac{2}{7}x $. So, the remaining coins consist of some 20-cent coins and some 50-cent coins:

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

### Step 3: Set up the equation and solve for $ x $.

Simplify the left-hand side:

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

Find a common denominator for $ \frac{1}{4}x $ and $ \frac{1}{8}x $:

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

Thus, the equation becomes:

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

### Step 4: Solve for $ x $.

Multiply both sides of the equation by 56 (the least common denominator of 8 and 7):

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

This simplifies to:

\[
21x - 8120 = 16x
\]

Now, subtract $ 16x $ from both sides:

\[
5x = 8120
\]

Finally, divide by 5:

\[
x = 1624
\]

### Step 5: Calculate the remaining money.
- The total number of 20-cent coins Sasa had originally is $ \frac{7}{8} \times 1624 = 1421 $ 20-cent coins.
- The total number of 50-cent coins Sasa had originally is $ \frac{1}{8} \times 1624 = 203 $ 50-cent coins.

After spending 145 50-cent coins, Sasa had $ 203 - 145 = 58 $ 50-cent coins left.

After spending $ \frac{5}{7} $ of her 20-cent coins, she had $ \frac{2}{7} \times 1421 = 406 $ 20-cent coins left.

The total amount of money left is:

\[
406 \times 0.20 + 58 \times 0.50 = 81.20 + 29 = 110.20
\]

### Final Answer:
Sasa had \$110.20 left.

Would you like a detailed explanation on any specific step? Here are some related questions:

1. How do you find fractional parts of a number in general?
2. Why is multiplying by the least common denominator helpful in solving equations with fractions?
3. What would happen if Sasa had more or fewer 50-cent coins?
4. How do you check that the solution satisfies all the conditions?
5. How do you solve similar problems with three or more types of coins?

**Tip:** When dealing with coins or percentages, always track how fractions of the total change after each transaction.

Learn how to solve a real-life algebra problem involving 20-cent and 50-cent coins. Use fractional arithmetic and linear equations to find the total amount of money left after spending. This solution is ideal for students in Grades 8-10.

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