enrollment in the ski club increased by 30% this year. there are now 182 students in the club. how many students were there last year

Answer 1

Let \( x \) be the number of students in the ski club last year. According to the problem, the enrollment increased by 30%, so this year's enrollment can be represented as:

\[ x + 0.30x = 1.30x \]

We know that this year there are 182 students:

\[ 1.30x = 182 \]

To find \( x \), we can solve for it by dividing both sides by 1.30:

\[ x = \frac{182}{1.30} \]

Calculating that gives:

\[ x = \frac{182}{1.30} \approx 140 \]

So, there were approximately 140 students in the ski club last year.

Answer 2

can you do it as a multiplication problem

Answer 3

Certainly! You can set it up as a multiplication problem.

The enrollment this year is 30% more than last year. If we represent last year’s enrollment as \( x \), then the increase can be represented as:

\[ x \times 1.30 = 182 \]

To find last year’s enrollment \( x \), you can rearrange this to solve for \( x \):

\[ x = \frac{182}{1.30} \]

Now, you can perform the multiplication first:

1. Set up the equation based on the percentage increase:

   \[ x \times 1.30 = 182 \]

2. To solve for \( x \), divide both sides by 1.30:

   \[ x = 182 \div 1.30 \]

Now, let's do the calculation:

\[ x = 182 \div 1.30 \approx 140 \]

Therefore, last year there were approximately 140 students in the ski club.

Answer 4

that is not mutiplecation

Answer 5

I apologize for misunderstanding your request. Let’s express it purely as a multiplication problem without division.

Since the ski club increased by 30%, you can express the number of students this year, 182, in terms of last year's enrollment:

1. Let’s call last year’s enrollment \( x \).

2. The relationship can be set up as:

   \[ x \times 1.30 = 182 \]

To solve it as a multiplication problem, think of last year's enrollment as being multiplied by 1.30 to give 182.

Now, if we rewrite the equation as:

\[ 1.30 \times x = 182 \]

We can still find \( x \) as follows:

To express this with multiplication:

\[ x = 182 \div 1.30 \]

But rather than divide directly, you could think of the calculation as finding the number that when multiplied by 1.30 equals 182. Hence, recognizing the relationship:

\[ x = 182 \times \left(\frac{1}{1.30}\right) \]

And if you'd like to directly see the math without division or rephrasing it:

1. We can consider that for multiplying it out:

   • This means that last year's students \( x \) multiplied by 1.30 equals this year’s students.

   • If you were solving purely within the function of your requirement for multiplication, you could simply note that if \( 182 = 1.30 \times x \):

If we just note it like this and rearrange intuitively:

\[ x = 182 \times \left(\text{something that derives from } 1.30\right) \]

However, the essential solution comes to the concept of knowing that:

Last year's count is represented as:

\[ x = 182 \times 0.769230769 , (which is \frac{1}{1.30} ) \]

Approximately, you multiply to confirm:

\[ 140 , \text{(the number last year matches to confirm slightly less)} \]

Therefore, the core essence was \( 1.30 \cdot x \) relates directly as a multiplication resulting in 182.

To summarize:

• Original \( x \) multiplied by 1.30 confirms the new total without needing direct division if you are defining a multiplier situation.

If you want a simpler multiplication example (as in student's growth):

• Start with the original \( x \) = 140 and see \( 140 \) times \( 1.30 \) gives 182 directly.

Thank you for your patience, and I hope this clarifies the approach!