70 students are enrolled in Math, English, or German. 40 students are

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.

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

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.

Hi, there! I'm happy to help with this question.

The Barron's answer is correct, but they shot through some complicated stuff awfully quickly.

Fact #1 = There are 70 students total.

Fact #2 = There are 40 in maths, 35 in English, 30 in German, for a total of 105.

This 105 is, of course, more than the total number of students ---- this is because the folks taking exactly two courses (the "doublers") have been counted twice, and the folks taking three courses (the "triplers") have been counted three times.

Fact #3 = 15 students are taking all three, i.e. there are 15 triplers.

So, if we want to make the 105 number jive with the actual total of 70, we have do two things
a) subtract the doublers once -- they are counted twice, and we only want them counted once
b) subtract the triplers twice --- they are counted three times, and we only want them counted once, so we have to subtract twice that total. That's where the mysterious factor of 2 arises ------ it's what you have to subtract from how many times they've been counted (3x) so that you are left with only counting them once. (If we subtracted the number of triplers times 3, the result would be that they wouldn't be counted at all!)

If we call the number of doublers N, this logic leads us to the Barron's equation

70 = 105 - N - 2*15

which leads to the Barron's answer of N = 5.

Does that make sense? Please let me know if you have any further question.

Mike
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Bunuel did excellent discussion on the topic covering whole concept on the following link:

formulae-for-3-overlapping-sets-69014.html

Hi Mike
Thank you very much.

I worked through this as follows and took quiet some time. I am not very good at understanding and applying formulas...

Fact 1 :- 70 students are taking either 3 or 2 or 1 subject
Fact 2 :- 15 are definitely taking all 3

so now 70-15= 55 remaining are taking 2 or 1 subject

When you draw a Venn diagram ..it is seen..
Students taking Maths remaining are 40 - 15 = 25
Students taking English are 35-15=20
Students taking German are 30 - 15 = 15

So sum of these are total students taking only 1 subject = 60

Therefore 60-55=5 are students who take only 2 subjects.

Cheers

Letting M = math, E = English, and G = German, we can use the formula for a 3-category scenario:

Total = n(M) + n(E) + n(G) - n(M and E) - n(M and G) - n(E and G) + n(all 3) - n(none)

70 = 40 + 35 + 30 - n(M and E) - n(M and G) - n(E and G) + 15 - 0

70 = 120 - n(M and E) - n(M and G) - n(E and G)

50 = n(M and E) + n(M and G) + n(E and G)

Note that the term n(M and E) includes those taking M and E, but it also includes the 15 who are taking all three. Similarly, the term n(M and G) includes those taking M and G, but it also includes the 15 who are taking all three. And, finally, the term n(E and G) includes those taking E and G, but it also includes the 15 who are taking all three.

Thus, we have added an extra 15 individuals three times. The total taking exactly 2 courses is not 50. Rather, it is

50 - 15 x 3 = 50 - 45 = 5.

Answer: A

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70 is the total number of students
X is students taking only math
Y students taking two courses
15 students are taking 3 courses

Thus, X+Y+15 = 70
X= 55-Y

40+30+35 =105 which is the total amount of courses being taken (not students)

105= X + 2Y + 3(15)
X=55-Y

Solve for Y gives you 5.
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Exactly Two = 40+30+35 - 70 - 2*15 = 5 (Ans)

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