A city’s school board merged the city’s two rival high schools to form one new high school, which then had \(4500\) stud

[M7MBA]

A city’s school board merged the city’s two rival high schools to form one new high school, which then had \(4500\) students. \(10\%\) of the students at high school \(A\) and \(15\%\) of students at high school \(B\) were in the marching band. If all marching band members joined the new school’s marching band, which then had \(570\) members, how many students did school \(A\) have before the merge?

A. 2100
B. 2200
C. 2300
D. 2400
E. 2500

Answer: A

Source: Veritas Prep

[Scott@TargetTestPrep]

A city’s school board merged the city’s two rival high schools to form one new high school, which then had \(4500\) students. \(10\%\) of the students at high school \(A\) and \(15\%\) of students at high school \(B\) were in the marching band. If all marching band members joined the new school’s marching band, which then had \(570\) members, how many students did school \(A\) have before the merge?

A. 2100
B. 2200
C. 2300
D. 2400
E. 2500

Answer: A

Solution:

We can let A = the number of students in school A, and B = the number of students in school B. Since the combined schools had 4,500 students, we know that A + B = 4,500.

We are also given that 10% of the students at high school A and 15% of the students at high school B were in the marching band, which has 570 members. Thus:

0.1A + 0.15B = 570

10A + 15B = 57,000

We can isolate B in the first equation, and we have: B = 4,500 - A.

We now can substitute 4,500 - A for B in the equation 10A + 15B = 57,000:

10A + 15(4,500 - A) = 57,000

10A + 67,500 - 15A = 57,000

-5A = -10,500

A = 2,100

Answer: A