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The number of students enrolled at school X this year is 7 percent [#permalink]
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12 Nov 2014, 09:54
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Re: The number of students enrolled at school X this year is 7 percent [#permalink]
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12 Nov 2014, 11:26
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. 11272 = 40 Answer choice E!



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Re: The number of students enrolled at school X this year is 7 percent [#permalink]
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12 Nov 2014, 11:40
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) Equating & Solving we get, Y=2400 ANS is E



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Re: The number of students enrolled at school X this year is 7 percent [#permalink]
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12 Nov 2014, 11:58
Let us assume, number of student enrolled at school X last year is x number of student enrolled at school Y last year is y.
this year enrollment at school X increases by 7% i.e. 7x/100 this year enrollment at school Y increases by 3% i.e. 3y/100
As per question stem, school X grew by 40 more students than school Y did this means 7x/100 = 40+ 3y/100  equation(1)
last year total student at school was 4000 i.e. x+y = 4000  equation(2)
Equate Equation (1) & (2), y will be 2400
Answer: E
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Re: The number of students enrolled at school X this year is 7 percent [#permalink]
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12 Nov 2014, 20:26
......................... School X ................ School Y ................... Total Last Yr ................. 4000a ..................... a ............................ 4000 (Assumed "a" as number of students in last yr) Difference ......... \((4000a) * \frac{7}{100}\) ................. \(a * \frac{3}{100}\) .............. (Growth) Given that "Growth" of School X is 40 more than growth of School Y Setting up the equation \(\frac{3a}{100} + 40 = \frac{7}{100} * (4000a)\) \(10a = 7 * 4000  4000\) a = 2400 Answer = E
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Re: The number of students enrolled at school X this year is 7 percent [#permalink]
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13 Nov 2014, 09:20
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.
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Re: The number of students enrolled at school X this year is 7 percent [#permalink]
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25 Feb 2016, 05:26
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



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Re: The number of students enrolled at school X this year is 7 percent [#permalink]
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18 Feb 2017, 10:36
Since the problem states that: " If school X grew by 40 more students than school Y did " It talks about the individual growth and not the total number of students.
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Re: The number of students enrolled at school X this year is 7 percent [#permalink]
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18 Feb 2017, 22:17
Let x be the number of students in X and y in Y.
As per the question,
x +y = 4000 ... (1) 0.07x  0.03y = 40 => 7x  3y = 4000 ... (2)
Solving equation 1 and 2:
y = 2400



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The number of students enrolled at school X this year is 7 percent [#permalink]
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24 Feb 2017, 11:21
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?
These differences in language are so subtle..



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Re: The number of students enrolled at school X this year is 7 percent [#permalink]
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24 Feb 2017, 15:04
iliavkoSince 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. Hope this helps.



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Re: The number of students enrolled at school X this year is 7 percent [#permalink]
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24 Feb 2017, 15:38
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=4000y .07(4000y).03y=40 y=2400 students E



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Re: The number of students enrolled at school X this year is 7 percent [#permalink]
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18 Apr 2018, 09:21
Let the number of students last year be X and Y. This year the number of students will be 1.07X and 1.03Y. We have 2 equations  a) 0.07X=0.03Y+40 b) X+Y=4000 Solve the two and get the value of Y as 2400 or option E.
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