If ( x ) is increased by 30% and then decreased by 20%, the resulting value is 24% more than the original value. What is the original value of ( x )?


Let's solve this step-by-step by setting up the appropriate equations based on the information given.

### Step-by-step Solution:

1. **Increase \( x \) by 30%**:
   - Increasing \( x \) by 30% means the new value after the increase is \( x + 0.30x = 1.30x \).

2. **Decrease the new value by 20%**:
   - Decreasing this new value by 20% means the value becomes \( 1.30x - 0.20 \times 1.30x = 1.30x \times 0.80 = 1.04x \).

3. **Resulting value is 24% more than the original value**:
   - The problem states that this resulting value is 24% more than the original value \( x \).
   - This means \( 1.04x = x + 0.24x = 1.24x \).

4. **Set up the equation**:
   - We have two expressions for the resulting value after both percentage changes:
     \[
     1.04x = 1.24x
     \]

5. **Solving for \( x \)**:
   - To solve for \( x \), notice that we have an equation that shows a contradiction if it states \( 1.04x = 1.24x \). This suggests a misunderstanding in the setup of our understanding:
   
6. **Correcting the interpretation**:
   - Let's reconsider: After applying both percentage changes, \( 1.04x \) is the modified \( x \) value after operations, equating it to 1.24 times the initial value is already done as above. Which implies the initial value of \( x \) is self-resolving in that case:
   
   - By isolating \( x \), knowing it matches the manipulative outcome:
   
     \[
     x = x \times \frac{1.04}{1.24}
     \]

   - Here, let's re-isolate to solve for equivalence identity:
   
     \[
     \frac{1.04x}{1.24x} = 1 
     \]

    This confirms a setup implying equality:
     
     \[
     x = \frac{1.04}{1.24} x =  x
     \]

### Conclusion:

**Since we are not calculating \( x \) against an unknown value or variable driven difference: The original and resulting values represent a dynamic process reflecting the very variable itself as unchanged due to its own percentage operations effectively canceling one another in unique combinatory approach and outcome. Thus, effectively, any value of \( x \) remains, yielding constant original \( x \) irrespective of transformation operations.**

Hence, the solution demonstrates that \( x \) would not change but illustrates via transformations without adding external result deviations:
Thus: **Any valid real number x holds equivalence**.  

Would you like further clarification or explore more challenging examples or exercises around percentage changes, and equations?

Here are some follow-up questions for additional practice:
1. What happens if instead, \( x \) is decreased by 30% first and then increased by 40%?
2. If the original value of \( x \) is 100, what is the resulting value after the same changes?
3. How would the result change if the value was increased by 50% first and then decreased by 30%?
4. Can you derive a general formula for successive percentage increases and decreases?
5. What is the result if \( x \) is first increased by 20%, then increased again by 20%, instead of a decrease? 

**Tip:** When working with percentages in equations, it's crucial to translate percentage changes into decimal multipliers to manage the arithmetic easily.

Explore how to solve equations involving percentage changes, such as increases and decreases. This problem involves finding the original value of x after it is increased by 30% and then decreased by 20%, resulting in 24% more than the original value. Suitable for high school students (Grades 9-12) learning about percentage calculations and algebraic equations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation