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S95-09

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:49
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Difficulty:

75% (hard)

Question Stats:

56% (02:48) correct 44% (03:50) wrong based on 78 sessions

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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
[Reveal] Spoiler: OA

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Joined: 02 Sep 2009
Posts: 41601

Kudos [?]: 124053 [0], given: 12070

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16 Sep 2014, 01:49
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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13 Feb 2015, 03:41
Bunuel wrote:
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: $$2400 = y$$.

X + Y =4000
Difference in no. of stds joined in current year = 40
Let us assume Y as 2400 and X 1600

X increased by 7% and Y by 3%.

X = 1712, Y - 2472
Diff in no. of students joined in current year is 40.

Hi Bunel ,

I was not clear on this equation. .07X - .03 Y = 40

It should be 1.07X -1.03Y = 40. The questions states that 7% more than last year.

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03 Apr 2015, 10:33
It states that X grew by.......
therefore,check "Grew" word...which means the difference between final state and original state.
Thus,
0.07X....
Similarly for Y.....

Hope this helps!!

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07 Jul 2015, 03:34
So, I have a question. The stem says that for school X the number of students this year was 7% more than that of ast year. So, if x is the number of students last year, shouldn't the number of students this year be x + 0.07x = 1.07x (the total students this year for school X).

Similarly for Y, this year there would be 1.03Y.

The stem also says that this year X grew by 40 more students than Y did. Does not this mean that the number of students X has this year (1.07X) is 40 more than the number of students school Y has this year (1.03Y)? Because this translates into 1.07X = 1.03Y+40.

Then, using the same relationship X+Y = 4000, from where we result in X = 4000 - Y, and substituting into the relationship above (1.07X = 1.03Y+40) we result in a different number for Z.

So, what I do not understand is why 40 relates only to the % of the increase (0.07X or 0.03Y), and not to the increased student number itself (1.07X and 1.03Y)?

Thanks!

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01 Oct 2016, 03:08
I have the same doubt as pacifist85.On increasing from x by 7% it becomes 1.07x-1.03y=40

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01 Oct 2016, 03:10
bhamini1 wrote:
I have the same doubt as pacifist85.On increasing from x by 7% it becomes 1.07x-1.03y=40

Check the discussion here: the-number-of-students-enrolled-at-school-x-this-year-is-7-percent-188462.html
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23 Aug 2017, 11:16
For what it's worth, (A) (C) and (D) can all be eliminated right away because they cannot be increased by 3% without creating partial students (gross). This leaves (B) and (E). From here it doesn't take long to figure which is correct by plugging each into the stem.

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Re: S95-09   [#permalink] 23 Aug 2017, 11:16
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