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The number of students enrolled at school X this year is 7 percent
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12 Nov 2014, 08:54
8
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A
B
C
D
E
Difficulty:
65% (hard)
Question Stats:
65% (02:48) correct 35% (03:12) wrong based on 205 sessions
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Tough and Tricky questions: Percents.
The number of students enrolled at school \(X\) this year is 7 percent more than it was last year. The number of students enrolled at school \(Y\) this year is 3 percent more than it was last year. If school \(X\) grew by 40 more students than school \(Y\) did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school \(Y\) last year?
Re: The number of students enrolled at school X this year is 7 percent
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12 Nov 2014, 10:26
3
I started with answer choice C and worked backwards and used some ballparking. If school Y had 1920 students last year, 3% increase would be slightly more than 57. X would have had 2080 students and a 7% increase would be around 145 students. The difference in increases is much too large so we should aim for a larger Y population in the answer choices.
I went to answer E next, knowing that if this was correct, I am good to go, but if not, D must be the answer.
With answer E, school Y had 2400 students and a 3% increase is 72. X would have had 1600 students. 7% of 1600 is 112.
Re: The number of students enrolled at school X this year is 7 percent
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12 Nov 2014, 10:40
2
1
Bunuel wrote:
Tough and Tricky questions: Percents.
The number of students enrolled at school \(X\) this year is 7 percent more than it was last year. The number of students enrolled at school \(Y\) this year is 3 percent more than it was last year. If school \(X\) grew by 40 more students than school \(Y\) did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school \(Y\) last year?
A. 480 B. 1600 C. 1920 D. 2080 E. 2400
Kudos for a correct solution.
Given Info: No. of students enrolled in school X is= 7% more than previous year's strength= 7X/100 No. of students enrolled in school Y is= 3% more than previous year's strength= 3Y/100
Given that "THIS YEAR" school X have 40 students more than school Y. So the increase in strength can be written as
7X/100= 40 + 3Y/100
7X= 4000 + 3Y....................(1)
And total students on previous year are X + Y=4000......................(2)
Re: The number of students enrolled at school X this year is 7 percent
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13 Nov 2014, 08:20
2
Bunuel wrote:
Tough and Tricky questions: Percents.
The number of students enrolled at school \(X\) this year is 7 percent more than it was last year. The number of students enrolled at school \(Y\) this year is 3 percent more than it was last year. If school \(X\) grew by 40 more students than school \(Y\) did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school \(Y\) last year?
A. 480 B. 1600 C. 1920 D. 2080 E. 2400
Kudos for a correct solution.
Official Solution:
The number of students enrolled at school \(X\) this year is 7 percent more than it was last year. The number of students enrolled at school \(Y\) this year is 3 percent more than it was last year. If school \(X\) grew by 40 more students than school \(Y\) did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school \(Y\) last year?
A. 480 B. 1600 C. 1920 D. 2080 E. 2400
We must determine the number of students enrolled at school \(Y\) last year.
We can set up two equations to describe this situation.
First: last year, the number of students enrolled at school \(X\) (which we will call \(x\)) and the number of students enrolled at school \(Y\) (which we will call \(y\)) together were equal to 4000. So \(x + y = 4000\).
Second: the number of additional students at school \(X\) this year (equal to 7 percent of last year's student body of \(x\), or \(0.07x\)) is 40 more than the number of additional students at school \(Y\) this year (equal to 3 percent of last year's student body of \(y\), or \(0.03y\)). Therefore, \(0.07x - 0.03y = 40\).
Now we use substitution to solve for \(y\). Rearrange the first equation to express \(x\) in terms of \(y\): \(x = 4000 - y\). Then, substitute for \(x\) in the second equation: \(0.07(4000 - y) - 0.03y = 40\).
Distribute the 0.07 to get: \(0.07(4000) - 0.07y - 0.03y = 40\)
Simplify and combine like terms: \(280 - 40 = (0.07 + 0.03)y\), or \(240 = 0.1y\).
Re: The number of students enrolled at school X this year is 7 percent
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25 Feb 2016, 04:26
Bunuel wrote:
Bunuel wrote:
Tough and Tricky questions: Percents.
The number of students enrolled at school \(X\) this year is 7 percent more than it was last year. The number of students enrolled at school \(Y\) this year is 3 percent more than it was last year. If school \(X\) grew by 40 more students than school \(Y\) did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school \(Y\) last year?
A. 480 B. 1600 C. 1920 D. 2080 E. 2400
Kudos for a correct solution.
Official Solution:
The number of students enrolled at school \(X\) this year is 7 percent more than it was last year. The number of students enrolled at school \(Y\) this year is 3 percent more than it was last year. If school \(X\) grew by 40 more students than school \(Y\) did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school \(Y\) last year?
A. 480 B. 1600 C. 1920 D. 2080 E. 2400
We must determine the number of students enrolled at school \(Y\) last year.
We can set up two equations to describe this situation.
First: last year, the number of students enrolled at school \(X\) (which we will call \(x\)) and the number of students enrolled at school \(Y\) (which we will call \(y\)) together were equal to 4000. So \(x + y = 4000\).
Second: the number of additional students at school \(X\) this year (equal to 7 percent of last year's student body of \(x\), or \(0.07x\)) is 40 more than the number of additional students at school \(Y\) this year (equal to 3 percent of last year's student body of \(y\), or \(0.03y\)). Therefore, \(0.07x - 0.03y = 40\).
Now we use substitution to solve for \(y\). Rearrange the first equation to express \(x\) in terms of \(y\): \(x = 4000 - y\). Then, substitute for \(x\) in the second equation: \(0.07(4000 - y) - 0.03y = 40\).
Distribute the 0.07 to get: \(0.07(4000) - 0.07y - 0.03y = 40\)
Simplify and combine like terms: \(280 - 40 = (0.07 + 0.03)y\), or \(240 = 0.1y\).
Multiply both sides by 10: \(2400 = y\).
Answer: E.
Hey - I'm just wondering why it isnt 1.07X and 1.03Y as the question states that its "7% more than last year"... same with Y
The number of students enrolled at school X this year is 7 percent
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24 Feb 2017, 10:21
why cant we say that 1.07X=1.03Y+40 ?
can you elaborate more?
is it because it talks about the growth of students?
How would the question have to be if we were to write 1.07X=1.03Y+40 ? "at the end of the year the school X had 40 students more than Y" ? or this would be the same as the stem?
Since the question stem says - 'If school X grew by 40 more students than school Y did' i.e. the number of additional students at school X this year (equal to 7 percent of last year's student body of x, or 0.07x) is 40 more than the number of additional students at school Y this year (equal to 3 percent of last year's student body of y, or 0.03y), hence .07x=.03y+40.
Re: The number of students enrolled at school X this year is 7 percent
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24 Feb 2017, 14:38
Bunuel wrote:
Tough and Tricky questions: Percents.
The number of students enrolled at school \(X\) this year is 7 percent more than it was last year. The number of students enrolled at school \(Y\) this year is 3 percent more than it was last year. If school \(X\) grew by 40 more students than school \(Y\) did, and if there were 4000 total enrolled students last year at both schools, how many students were enrolled at school \(Y\) last year?
A. 480 B. 1600 C. 1920 D. 2080 E. 2400
Kudos for a correct solution.
let x and y=number of students at X and Y respectively last year x=4000-y .07(4000-y)-.03y=40 y=2400 students E
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65% (02:48) correct 
