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# The number of students enrolled at school X this year is 7 percent

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Math Expert
Joined: 02 Sep 2009
Posts: 46307
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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Difficulty:

65% (hard)

Question Stats:

65% (02:19) correct 35% (02:30) wrong based on 197 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?

A. 480
B. 1600
C. 1920
D. 2080
E. 2400

Kudos for a correct solution.

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Posts: 98
Re: The number of students enrolled at school X this year is 7 percent [#permalink]

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12 Nov 2014, 11:26
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.

112-72 = 40

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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
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)

Equating & Solving we get, Y=2400

ANS is E
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Posts: 222
Re: The number of students enrolled at school X this year is 7 percent [#permalink]

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12 Nov 2014, 11:58
1
2
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

Regards,
Ammu
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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
1
1
......................... School X ................ School Y ................... Total

Last Yr ................. 4000-a ..................... a ............................ 4000 (Assumed "a" as number of students in last yr)

Difference ......... $$(4000-a) * \frac{7}{100}$$ ................. $$a * \frac{3}{100}$$ ..............
(Growth)

Given that "Growth" of School X is 40 more than growth of School Y

Setting up the equation

$$\frac{3a}{100} + 40 = \frac{7}{100} * (4000-a)$$

$$10a = 7 * 4000 - 4000$$

a = 2400

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Math Expert
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Posts: 46307
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: $$2400 = y$$.

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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: $$2400 = y$$.

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
1
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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Joined: 27 Aug 2014
Posts: 2
Re: The number of students enrolled at school X this year is 7 percent [#permalink]

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24 Feb 2017, 15:04
iliavko

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.

Hope this helps.
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Posts: 1020
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=4000-y
.07(4000-y)-.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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Re: The number of students enrolled at school X this year is 7 percent   [#permalink] 18 Apr 2018, 09:21
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