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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Mike McGarry
Magoosh Test PrepEducation is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)