Problem Solving

In a group of 68 students, each student is registered for at least one of three classes - History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?
(A)13
(B)10
(C) 9
(D) 8
(E) 7

Correct Answer:B

I've tried to find an easier approach for this kind of problem, but I've got to one point and I can not finish it.

25+25+34= 84
84- 68= 16 (duplicate registrations). We know that 3 of these duplicates come from those students who registered for all three classes-> who registered for exactly 2 classes 16-3=13. At this point I don't know how to get 10.

Thanks for your help.

There are 3 students who are present in all the cases...so subtract 3 from all

No : 65

Hist : 22

Math : 22

English : 31

hence no of students in 2 classes only = (22+22+31) - 65 = 10

karanrulz4ever wrote:There are 3 students who are present in all the cases...so subtract 3 from all

No : 65

Hist : 22

Math : 22

English : 31

hence no of students in 2 classes only = (22+22+31) - 65 = 10

Thanks karanrulz4ever.


What if you have 3 students registered for exactly two classes and we have to find the number of students registered for all three classes? Are we using the same approach?

lavinia wrote:In a group of 68 students, each student is registered for at least one of three classes - History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?
(A)13
(B)10
(C) 9
(D) 8
(E) 7

Correct Answer:B

I've tried to find an easier approach for this kind of problem, but I've got to one point and I can not finish it.

25+25+34= 84
84- 68= 16 (duplicate registrations). We know that 3 of these duplicates come from those students who registered for all three classes-> who registered for exactly 2 classes 16-3=13. At this point I don't know how to get 10.

Thanks for your help.

This problem is testing your knowledge of overlapping groups. Here is the formula for 3 overlapping groups in which sometimes 2 groups overlap and sometimes all 3 groups overlap:

T = G1 + G2 + G3 - (those in 2 of the groups) - 2*(those in all 3 groups)

The big idea with overlapping groups is to subtract the overlaps. When we count everyone in the 3 groups, those in 2 of the groups will be counted twice, so they need to subtracted from the total once. Those in all 3 groups will be counted 3 times, so they need to be subtracted from the total twice.

In the problem above:
T = 68
G1+G2+G3 = history + math + english = 25+25+34 = 84
Those registered for exactly 2 subjects = x
Those registered for all 3 subjects = 3

Plugging into the formula, we get:

68 = 84 - x - 2*3
68 = 78 - x
x = 10.

The correct answer is B.

T = G1 + G2 + G3 - (those in 2 of the groups) - 2*(those in all 3 groups)

In this formulma, those in 2 of the groups are NOT those with exactly 2 of the groups , that's why the formula includes "2*(those in all 3 groups)" .

Am I wrong in my reasoning ? please...

GMATMadeEasy wrote:

T = G1 + G2 + G3 - (those in 2 of the groups) - 2*(those in all 3 groups)

In this formulma, those in 2 of the groups are NOT those with exactly 2 of the groups , that's why the formula includes "2*(those in all 3 groups)" .

Am I wrong in my reasoning ? please...

hmhm lets me rewrite the formular for you

true # of objects=(total#in group1)+(total # in group 2)+total#in group3)-(# in exactly 2 groups)- 2( # in all 3 groups)
thus for this question
68= 25+25+34-2*3-x ( x is number of people in exactly 2 group)
so x=10

@GMATGuruNY:This is one of THE bet formulas I have seen for overlapping 3 sets!!.

Can we usually use this formula in most 3 set cases or is it limited in its application?.

Many thx!

Aslan