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There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German.

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.

Attachment:

File comment: Venn Diagram

c73058.jpg [ 19.92 KiB | Viewed 1535 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 \(50 - (3*15) = 5\)

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.

Attachment:

c73058.jpg

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.

Hi

cant we take 'a' as 40 here as it is mentioned exactly 40 on math?

It says that exactly 40 are in math, not that 40 are in ONLY math. The people who take Math and English or Math and German or even all three are also in math, aren't they not? So you can't take a to be 40 since 40 is the sum of a,x, y and 15, i.e. people who take only Math, people who take Math and English, people who take Math and German and people who take all three. Hope this is clear.

It says that exactly 40 are in math, not that 40 are in ONLY math. The people who take Math and English or Math and German or even all three are also in math, aren't they not? So you can't take a to be 40 since 40 is the sum of a,x, y and 15, i.e. people who take only Math, people who take Math and English, people who take Math and German and people who take all three. Hope this is clear.

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