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Came across this on the web

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by bingojohn » Fri Aug 17, 2007 12:07 pm
Anonymous wrote:In a group of 30 students, 8 are enrolled in an English class and 16 are enrolled in an Algebra class. How many students are enrolled in both an English and an Algebra class?

(1) 20 are enrolled in exactly one of these two classes.

(2) 3 are not enrolled in either of these classes.
Answer ?

Use formula:
total = first group + second group + neither group - both groups
=> 30 = 8 english + 16 algebra + neither - both

(1) gives us nothing, NS

(2) gives us the number of students enrolled in neither english nor algebra... which if plugged into the equation above, gives us 'both' as 3.

Hence Sufficient.
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Source: — Data Sufficiency |
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Re: Came across this on the web

by ratindasgupta » Sun Aug 19, 2007 3:46 am
Anonymous wrote:In a group of 30 students, 8 are enrolled in an English class and 16 are enrolled in an Algebra class. How many students are enrolled in both an English and an Algebra class?

(1) 20 are enrolled in exactly one of these two classes.

(2) 3 are not enrolled in either of these classes.
there seems to be some error in the data. cant put my finger on it. maybe i'm wrong.

And i dont think its B anyway. Coz goin by bingo's formula

30 = 8 + 16 + 3 - x. Which means x is -3.
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by beny » Sun Aug 19, 2007 9:27 am
Yes, there is an error in the data. Assuming that there are 30 students, and only 3 are not enrolled in either class, then 27 must be enrolled in at least 1 class.

Yet, the total number of students (8+16) is only 24.
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by Bharat » Mon Aug 27, 2007 1:57 am
Hi,
Please look at this:
Let E: student reading only English
A: Student reading only Algebra
X: student reading English & Algebra Both
E + X = 8
A + X = 16
From 1: E + A = 20

Using the above three eqns: X = 2. Hence 1 is sufficient.
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by Bharat » Mon Aug 27, 2007 1:58 am
Hi,
Please look at this:
Let E: student reading only English
A: Student reading only Algebra
X: student reading English & Algebra Both
E + X = 8
A + X = 16
From 1: E + A = 20

Using the above three eqns: X = 2. Hence 1 is sufficient.
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