A certain number is first increased by 20%.\nTo return from this increased value back to the original number, by what percentage must the increased number be reduced?

Introduction / Context:
This question explores the idea that a percentage increase and a percentage decrease of the same numerical rate do not cancel each other out. When a quantity is increased by a certain percentage, the base value changes, so the percentage decrease required to return to the original value is different. This is a classic concept in percentage arithmetic, very common in aptitude exams, and it demonstrates why successive percentage changes must be handled carefully using multiplication rather than simple addition or subtraction.

Given Data / Assumptions:

• An original number is increased by 20%.
• We then want to reduce the increased number by some percentage so that we recover the original number.
• We assume the percentage decrease is calculated on the increased value, not the original value.
• The task is to find this required percentage decrease.

Concept / Approach:
If the original number is x, a 20% increase gives a new value of x * 1.20. To return from this value to x, we need to apply a reduction factor such that x = (1.20 * x) * (1 - r), where r is the required reduction rate expressed as a decimal. Solving this equation for r yields the percentage decrease needed. This method highlights that the increase and decrease are based on different bases, which is why the rates are not equal even though they seem symmetrical at first glance.

Step-by-Step Solution:
Step 1: Let the original number be x. Step 2: After a 20% increase, the new value is x * (1 + 20 / 100) = x * 1.20. Step 3: Let the required reduction rate be r (in decimal form) applied to the increased value. Step 4: To return to the original number, we need x = 1.20 * x * (1 - r). Step 5: Divide both sides by x (x is nonzero), giving 1 = 1.20 * (1 - r). Step 6: Solve for (1 - r): (1 - r) = 1 / 1.20. Step 7: Compute 1 / 1.20 = 1 / (6 / 5) = 5 / 6. Step 8: So 1 - r = 5 / 6, which means r = 1 - 5 / 6 = 1 / 6. Step 9: Convert r to a percentage: r = 1 / 6 which is 16 2/3%.

Verification / Alternative check:
Take a concrete example. Let the original number be 100. A 20% increase gives 120. Now we ask what percentage decrease on 120 will bring us back to 100. The required reduction is 120 - 100 = 20. The decrease as a percentage of 120 is 20 / 120 * 100 = 16.666..., which is 16 2/3%. This numeric check matches the algebraic derivation and confirms that the correct reduction rate is 16 2/3%, not 20%.

Why Other Options Are Wrong:
A reduction of 20% on 120 would give 96, not 100, so 20% is incorrect. A 21% reduction is even larger and would lead to 120 * 0.79 = 94.8, which is too low. A reduction of 14 1/3% would lead to 120 * (1 - 14 1/3 / 100) and also not reach 100. Only 16 2/3% produces the exact original number when applied to the increased value, so the other options do not satisfy the equality.

Common Pitfalls:
Many learners mistakenly assume that a 20% decrease will reverse a 20% increase, but this ignores the fact that the base for the decrease is larger than the base for the increase. This misunderstanding leads to incorrect answers in many percentage and profit and loss questions. The safe approach is always to define a variable for the original value, write out the multiplicative factors, and solve the resulting equation instead of relying on intuition.

Final Answer:
To return to the original number after a 20% increase, the new number must be reduced by 16 2/3%.