Jack's weight first increased by 20% and then his new weight decreased by 25%. His final weight is what percent of his beginning weight?

Options

A. `95%`
B. `92.5%`
C. `90%`
D. `88.5%`

Answer

To solve the problem step by step, we will follow the changes in Jack's weight based on the percentage increases and decreases mentioned in the question.

### Step 1: Define Jack's Initial Weight
Let Jack's initial weight be \( X \) kg.

### Step 2: Calculate the Weight After the First Increase
Jack's weight first increases by 20%. To find the new weight after this increase:
- Calculate 20% of \( X \):
\[ 20\% \text{ of } X = \frac{20}{100} \times X = \frac{1}{5}X \]
- Add this increase to the initial weight:
\[ \text{New Weight} = X + \frac{1}{5}X = \frac{5X}{5} + \frac{1X}{5} = \frac{6X}{5} \]

### Step 3: Calculate the Weight After the Decrease
Next, Jack's weight decreases by 25% from the new weight:
- Calculate 25% of the new weight \( \frac{6X}{5} \):
\[ 25\% \text{ of } \frac{6X}{5} = \frac{25}{100} \times \frac{6X}{5} = \frac{1}{4} \times \frac{6X}{5} = \frac{6X}{20} = \frac{3X}{10} \]
- Subtract this decrease from the new weight:
\[ \text{Final Weight} = \frac{6X}{5} - \frac{3X}{10} \]
To perform the subtraction, we need a common denominator:
- The common denominator of 5 and 10 is 10.
- Convert \( \frac{6X}{5} \) to have a denominator of 10:
\[ \frac{6X}{5} = \frac{12X}{10} \]
- Now perform the subtraction:
\[ \text{Final Weight} = \frac{12X}{10} - \frac{3X}{10} = \frac{12X - 3X}{10} = \frac{9X}{10} \]

### Step 4: Calculate the Final Weight as a Percentage of the Initial Weight
Now we need to find what percentage the final weight is of the initial weight:
- Use the formula for percentage:
\[ \text{Percentage} = \left( \frac{\text{Final Weight}}{\text{Initial Weight}} \right) \times 100\% \]
- Substitute the values:
\[ \text{Percentage} = \left( \frac{\frac{9X}{10}}{X} \right) \times 100\% = \left( \frac{9}{10} \right) \times 100\% = 90\% \]

### Final Answer
Jack's final weight is **90%** of his initial weight.

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