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21 Jul 2010, 02:43
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
82% (01:41) correct
18%
(02:09)
wrong
based on 282
sessions
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70 students are enrolled in Math, English, or German. 40 students are in Math, 35 are in English, and 30 are in German. 15 students are enrolled in all three of the courses. How many of the students are enrolled in exactly two of the courses: Math, English, and German?
A. 4
B. 5
C. 6
D. 7
E. 8
A. 4
B. 5
C. 6
D. 7
E. 8
please explain your answers.
what is the significance of 'exactly' (Exactly 40 are in Math,
30 in German, 35 in English ) in the stem?
what is the significance of 'exactly' (Exactly 40 are in Math,
30 in German, 35 in English ) in the stem?
21 Jul 2010, 08:18
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for venn diagrams in case of 3 cases:
Total = m(a) + m(b) + m(c) + m(a&b) + m(b&c) + m(c&a) - 2*m(a&b&c)
So, 70 = 40 + 30 + 35 - m(a&b) - m(b&c) - m(c&a) - 2*15
=> -35 = - [ m(a&b) + m(b&c) + m(c&a) ] - 30
Therefore, m(a&b) + m(b&c) + m(c&a) = 5
Total = m(a) + m(b) + m(c) + m(a&b) + m(b&c) + m(c&a) - 2*m(a&b&c)
So, 70 = 40 + 30 + 35 - m(a&b) - m(b&c) - m(c&a) - 2*15
=> -35 = - [ m(a&b) + m(b&c) + m(c&a) ] - 30
Therefore, m(a&b) + m(b&c) + m(c&a) = 5
whiplash2411
Joined: 09 Jun 2010
Last visit: 02 Mar 2015
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21 Jul 2010, 06:43
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I think the OA makes perfect sense. And you cannot just ignore the term exactly. Exactly two means that they are not enrolled in all three classes. The simplest way to extract info from what's given is to draw a Venn Diagram.
So, from this picture, we are asked to find out what \(x+y+z\) is.
Let's look at given information and form the constraints:
Total = 70
\(x+y+z+a+b+c+15 = 70\)
\((x+y+z) + (a+b+c) = 55\)
Total Math = 40
\(x+y+a+15 = 40\)
\(x+y = 25-a\)
Total German = 30
\(y+z+c+15 = 30\)
\(y+z = 15 -c\)
Total English = 35
\(x+z+b+15 = 35\)
\(x+z = 20-b\)
So now combining all the bolded equations regarding totals of each subject we get
\(2(x+y+z) = 15+25+20 - (a+b+c) = 60 - (a+b+c)\)
So \((a+b+c) = 60 - 2(x+y+z)\)
Now substituting this into the first equation regarding total students, we get
\((x+y+z) + 60 - 2(x+y+z) = 55\)
Hence \(x+y+z = 5\)
nravi4: The mistake you made in getting 50 is this. You counted the students enrolled in two of three subjects, but not strictly so. So your calculation includes the central space of 15 which is students enrolled in all three subjects for each subject you counted. So to get to the answer from your answer you need to do \(50 - (3*15) = 5\)
Hope this is clear.
Attachment:
File comment: Venn Diagram
c73058.jpg [ 19.92 KiB | Viewed 33167 times ]
c73058.jpg [ 19.92 KiB | Viewed 33167 times ]
So, from this picture, we are asked to find out what \(x+y+z\) is.
Let's look at given information and form the constraints:
Total = 70
\(x+y+z+a+b+c+15 = 70\)
\((x+y+z) + (a+b+c) = 55\)
Total Math = 40
\(x+y+a+15 = 40\)
\(x+y = 25-a\)
Total German = 30
\(y+z+c+15 = 30\)
\(y+z = 15 -c\)
Total English = 35
\(x+z+b+15 = 35\)
\(x+z = 20-b\)
So now combining all the bolded equations regarding totals of each subject we get
\(2(x+y+z) = 15+25+20 - (a+b+c) = 60 - (a+b+c)\)
So \((a+b+c) = 60 - 2(x+y+z)\)
Now substituting this into the first equation regarding total students, we get
\((x+y+z) + 60 - 2(x+y+z) = 55\)
Hence \(x+y+z = 5\)
nravi4: The mistake you made in getting 50 is this. You counted the students enrolled in two of three subjects, but not strictly so. So your calculation includes the central space of 15 which is students enrolled in all three subjects for each subject you counted. So to get to the answer from your answer you need to do \(50 - (3*15) = 5\)
Hope this is clear.
