Question

Difficulty: Very hardPercentage Calculations

A military supply depot receives an initial stock of fuel. Over the first month, the stock is increased by 25%25\%. During the second month, the new stock is decreased by x%x\%. If the final amount of fuel at the end of the second month is exactly 10%10\% less than the initial stock, what is the value of xx?

  1. 28Answer
  2. B
    30
  3. C
    35
  4. D
    72

Answer

The value of x is 28.
The correct answer is derived by setting up an algebraic equation for successive percentage changes. An initial increase of 25% results in 1.25 times the original amount. To reach a final amount that is 10% less than the original (0.90 times the original), the intermediate amount must be multiplied by 0.72 (since 0.90 / 1.25 = 0.72). A multiplier of 0.72 means that 72% of the amount remains, representing a 28% decrease.

Step-by-Step Solution

1
Define a variable for the initial stock.
Let the initial stock be S.
Establishing a variable allows for clear algebraic translation of the successive changes.
2
Express the amount of fuel after the first month.
S + 0.25S = 1.25S
A 25% increase means multiplying the initial base by 1.25.
3
Express the amount of fuel after the second month.
1.25S×(1−x100)1.25S \times (1 - \frac{x}{100})
The x% decrease must be applied to the new intermediate base of 1.25S, not the original base.
4
Set up an equation based on the final given condition.
1.25S \times (1 - \frac{x}{100}) = 0.90S
The problem states the final amount is exactly 10% less than the initial stock (which is 0.90S).
5
Solve the equation for the remaining percentage multiplier.
1 - \frac{x}{100} = \frac{0.90}{1.25} = \frac{90}{125} = \frac{18}{25} = 0.72
Dividing both sides by 1.25S isolates the multiplier associated with the decrease.
6
Calculate the value of x.
\frac{x}{100} = 1 - 0.72 = 0.28, so x = 28
Subtracting the remaining percentage (72%) from 100% yields the percentage decrease.

Key Concept

Successive Percentage Changes and Base Identification
Rate this question