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A city’s school board merged the city’s two rival high schools to form

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A city’s school board merged the city’s two rival high schools to form  [#permalink]

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New post 16 Feb 2017, 03:33
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Question Stats:

70% (01:57) correct 30% (01:59) wrong based on 63 sessions

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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

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Re: A city’s school board merged the city’s two rival high schools to form  [#permalink]

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New post 16 Feb 2017, 04:27
lets say total number of school a is x and total number of school b is y

x + y = 4500
0.1x + 0.15y = 570

to equations 2 variables, if we solve

x =2100 and y = 2400

answer is A
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A city’s school board merged the city’s two rival high schools to form  [#permalink]

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New post 16 Feb 2017, 04:32
Bunuel wrote:
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


A + B = 4500 (total number of students of both schools)
0.1A + 0.15B =570 (10% of the students at high school A and 15% of students at high school B were in the marching band which consisted of 570 students)

On solving the above 2 equations, we have A = 2100 and B = 2400

Hence Option A is correct
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Re: A city’s school board merged the city’s two rival high schools to form  [#permalink]

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New post 16 Feb 2017, 05:59
A+B=4500
A=x
B=4500-X
0.1x + 0.15(4500-X)=570
Solving this equation gives x=2100
So correct Ans is A.

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Re: A city’s school board merged the city’s two rival high schools to form  [#permalink]

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New post 16 Feb 2017, 09:35
Bunuel wrote:
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


One of those sums in which solving via options is quicker. Thanks to the answer being in the first option.

A+B=4500,

Option A) A=2100, so B=2400. 10%A=210, 15%B=360. Total members of band => 210+360=570. Hence the answer.

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Re: A city’s school board merged the city’s two rival high schools to form  [#permalink]

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New post 21 Feb 2017, 10:22
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Bunuel wrote:
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


We can let A = the number of students from school A, and B = the number of students from 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
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Re: A city’s school board merged the city’s two rival high schools to form  [#permalink]

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New post 22 Sep 2018, 11:43
With this weighted average problem, there are two core methods you could use to solve it: algebraically, or with the weighted average mapping strategy. For this problem, the algebra requires a bit less creativity to set up. You're looking at:

a+b=4500
and
0.1a+0.15b=570
If you multiply the second equation by −10, that allows you to use the Elimination Method to get rid of the variable a
−10(0.1a+0.15b=570)
means that:
−a+−1.5b=−5700
Stack that up with the other equation and add:

a+b=4500
−a+−1.5b=−5700
And you have one equation, all with the variable b
−0.5b=−1200, so multiply by −2 to arrive at:
b=2400
That means that a, the variable in question, must be 2100
2100, making answer choice A correct.
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Re: A city’s school board merged the city’s two rival high schools to form &nbs [#permalink] 22 Sep 2018, 11:43
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A city’s school board merged the city’s two rival high schools to form

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