School A has graduated 1,700 students and graduates 50 students every year. School B has graduated 600 students, but graduates 150 every year. How many years (Y) will it be before School B has as many graduates as School A?
[tex]
\begin{array}{l}
g=50 Y+1,700 \\
g=150 Y+600
\end{array}
[/tex]
[?] years
Answer (1)
Define the number of graduates for School A as g A = 50 Y + 1700 .
Define the number of graduates for School B as g B = 150 Y + 600 .
Set g A = g B and solve for Y : 50 Y + 1700 = 150 Y + 600 .
Solve for Y to find the number of years: Y = 11 .
Explanation
Problem Analysis Let's analyze the problem. We have two schools, A and B, with different initial numbers of graduates and different graduation rates per year. We need to find out after how many years the total number of graduates from School B will be equal to the total number of graduates from School A.
Define Variables Let g A be the number of graduates from School A after Y years, and g B be the number of graduates from School B after Y years. We can express the number of graduates for each school as a function of the number of years, Y .
Graduates from School A For School A, the number of graduates after Y years is given by: g A = 50 Y + 1700 This equation represents the initial number of graduates (1700) plus the number of graduates added each year (50) times the number of years ( Y ).
Graduates from School B Similarly, for School B, the number of graduates after Y years is given by: g B = 150 Y + 600 This equation represents the initial number of graduates (600) plus the number of graduates added each year (150) times the number of years ( Y ).
Equating the Graduates We want to find the number of years ( Y ) when the number of graduates from both schools is equal. So, we set g A = g B :
50 Y + 1700 = 150 Y + 600
Rearranging the Equation Now, let's solve for Y . First, we rearrange the equation to isolate the terms with Y on one side and the constant terms on the other side: 1700 − 600 = 150 Y − 50 Y
Simplifying Simplify the equation: 1100 = 100 Y
Solving for Y Now, divide both sides by 100 to solve for Y :
Y = 100 1100 Y = 11
Final Answer Therefore, it will take 11 years for School B to have as many graduates as School A.
Examples
This type of problem is useful in business when comparing the growth of two companies. For example, Company A starts with 1700 customers and gains 50 customers per year, while Company B starts with 600 customers but gains 150 customers per year. Determining when Company B will have as many customers as Company A helps in strategic planning and forecasting.
