Question

Donations to an annual fundraiser are 15% greater this year than last year. Last year, donations were 10% greater than the year before. The amount raised this year is $10,120. How much was raised 2 years ago?

Answer 1

No

Answer 2

Let's denote the amount raised two years ago as \( x \).

According to the problem:
1. Last year, donations were 10% greater than the year before:
\[
\text{Last year's donations} = x + 0.1x = 1.1x
\]

2. This year, donations are 15% greater than last year:
\[
\text{This year's donations} = \text{Last year's donations} + 0.15 \times \text{Last year's donations} = 1.1x + 0.15 \times 1.1x = 1.1x(1 + 0.15) = 1.1x \times 1.15 = 1.265x
\]

3. We know that this year's donations amount to \( 10,120 \):
\[
1.265x = 10,120
\]

To find \( x \), we will divide both sides of the equation by 1.265:
\[
x = \frac{10,120}{1.265}
\]

Now we calculate \( x \):
\[
x \approx 7,993.70
\]

Thus, the amount raised two years ago was approximately \( \boxed{7,993.70} \).

Answer 3

I apologize for any confusion in my previous response. Let's carefully work through the problem step by step.

1. **Define the amounts raised:**
- Let \( A \) be the amount raised two years ago.
- Last year, donations were 10% greater than the amount raised two years ago:
\[
\text{Last year's amount} = A + 0.1A = 1.1A
\]

2. **Next, define this year's donations:**
- This year, donations are 15% greater than last year's:
\[
\text{This year's amount} = 1.1A + 0.15 \times 1.1A = 1.1A (1 + 0.15) = 1.1A \times 1.15
\]
- Performing the multiplication gives:
\[
1.1A \times 1.15 = 1.265A
\]

3. **Set the equation with this year's amount:**
- We know that this year's amount raised is $10,120:
\[
1.265A = 10,120
\]

4. **Solving for \( A \):**
\[
A = \frac{10,120}{1.265}
\]

5. **Calculate \( A \):**
- Now we perform the division:
\[
A \approx \frac{10,120}{1.265} \approx 7,993.70
\]

This indicates that the amount raised two years ago was approximately **$7,993.70**.

If this is different from what you expect, please clarify any details or specific values from the problem, and I will adjust accordingly. Thank you for your patience!