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DS from Knewton

Expert replies
by this_time_i_will » Mon Apr 05, 2010 5:30 pm
I found the following DS question in Knewton's diagonistic test:


In year Y, the high school dropout rate was 3.3 percent, compared with 4.1 percent in year X. If the same number of students enrolled in high school at the beginning of both years, how many students dropped out in year Y?

1. In year X the number of high school dropouts among those students enrolled at the beginning of the year was 615,000.
2. 120,000 more students enrolled at the beginning of the year dropped out in year X than in year Y.



My doubt: the stem says that same number of students enrolled in high school at the beginning of both years. However, statement B implies that the number of students enrolled for both the years, x & y, is different. Am i interpreting something wrong?
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Source: — Data Sufficiency |

by dxgamez » Mon Apr 05, 2010 7:00 pm
My interpretation is there were 120,000 more students who were enrolled at the beginning of the year dropped out in Year X than in Year Y.

That means the difference in drop out % would be equal to 120,000. Ans would be B?
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by mmon » Mon Apr 05, 2010 7:12 pm
Year Drop Out Rate
------------------------------
X 4.1%
Y 3.3%

1. In year X - > 615,000 dropped out
4.1% of n = 615,000
n = 615X10^4 / 41
= 15*10^4
easily can be found 3.3% of n
SUFFICIENT

2. drop out in X - drop out in Y = 120,000
.8% of n = 120,000
n easily can be found
SUFFICIENT

IMO D
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by eaakbari » Mon Apr 05, 2010 9:47 pm
IMO D

Stem:
We can figure we need to find the value of n being number of students enrolled or the number of drop-outs of either.

we get from stem 3.3n/100= drop out in x
4.1n/100= drop out in y

Statement one
gives us drop in x . hence suff


Statement two
tells us
Xdrop - Y drop = 120000

Hence Suff


D
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by this_time_i_will » Mon Apr 05, 2010 10:24 pm
dxgamez wrote:My interpretation is there were 120,000 more students who were enrolled at the beginning of the year dropped out in Year X than in Year Y.

That means the difference in drop out % would be equal to 120,000. Ans would be B?
AAh! got my mistake...seems i was sleeping while solving this one :D.
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