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

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 2

No

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!