What is the Base, Rate and Percentage? 225 is 25% of n
The Correct Answer and Explanation is:
To solve the problem “225 is 25% of n”, we must identify the Base, Rate, and Percentage.
Step-by-step Solution:
Given:
- Percentage = 225
- Rate = 25%
- Base = unknown (n)
The general formula to find the relationship between percentage, rate, and base is: Percentage=Rate×Base\text{Percentage} = \text{Rate} \times \text{Base}Percentage=Rate×Base
Substitute the known values: 225=25%×n225 = 25\% \times n225=25%×n
Convert 25% to a decimal: 25%=25100=0.2525\% = \frac{25}{100} = 0.2525%=10025=0.25
Now substitute and solve: 225=0.25×n225 = 0.25 \times n225=0.25×n
Divide both sides of the equation by 0.25: n=2250.25=900n = \frac{225}{0.25} = 900n=0.25225=900
Final Answer:
- Base (n) = 900
- Rate = 25%
- Percentage = 225
300-Word Explanation:
In percentage problems, the three key elements are the base, the rate, and the percentage. The base is the whole or the total amount that is being considered. The rate is the percent or part per hundred that is being applied. The percentage is the part of the base represented by the rate.
The basic formula used in percentage calculations is: Percentage=Rate×Base\text{Percentage} = \text{Rate} \times \text{Base}Percentage=Rate×Base
In the problem given, we are told that 225 is 25% of some number nnn. That means 225 is a part of a whole, and the rate at which we are taking the part is 25%.
First, convert the rate from a percent to a decimal, which makes calculations easier. Twenty-five percent is equal to 0.25 in decimal form. Using the formula: 225=0.25×n225 = 0.25 \times n225=0.25×n
To find the unknown base (n), divide both sides of the equation by 0.25: n=2250.25=900n = \frac{225}{0.25} = 900n=0.25225=900
This result tells us that 225 is one-fourth (25%) of 900. It checks out because if you take 25% of 900, you multiply 900 by 0.25 and get 225. This process demonstrates how percentages work and how understanding the relationship among the base, rate, and percentage allows you to solve many real-world math problems effectively.
